Each time a method executes, any parameter variables listed in the method header are ____.

Getting Started - Write your First Hello-world C Program

Let's begin by writing our first C program that prints the message "Hello, world!" on the display console:

Hello, world!

Step 1: Write the Source Code: Enter the following source codes using a programming text editor (such as NotePad++ for Windows or gEdit for UNIX/Linux/Mac) or an Interactive Development Environment (IDE) (such as CodeBlocks, Eclipse, NetBeans or MS Visual Studio - Read the respective "How-To" article on how to install and get started with these IDEs).

Do not enter the line numbers (on the left panel), which were added to help in the explanation. Save the source file as "Hello.c". A C source file should be saved with a file extension of ".c". You should choose a filename which reflects the purpose of the program.

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#include <stdio.h>  
 
int main() {                    
    printf("Hello, world!\n");  
    return 0;                   
}                               

Step 2: Build the Executable Code: Compile and Link (aka Build) the source code "Hello.c" into executable code ("Hello.exe" in Windows or "Hello" in UNIX/Linux/Mac).

• On IDE (such as CodeBlocks), push the "Build" button.
• On Text editor with the GNU GCC compiler, start a CMD Shell (Windows) or Terminal (Mac, Linux) and issue these commands:

  > gcc -o Hello.exe Hello.c
   
  
  $ gcc -o Hello Hello.c

  where gcc is the name of GCC C compiler; -o option specifies the output filename ("Hello.exe" for Windows or "Hello" for UNIX/Linux/Mac); "Hello.c" is the input source file.

Step 3: Run the Executable Code: Execute (Run) the program.

• On IDE (such as CodeBlocks), push the "Run" button.
• On Text Editor with GNU GCC compiler, issue these command from CMD Shell (Windows) or Terminal (UNIX/Linux/Mac):

  > Hello
  Hello, world!
   
  
  $ ./Hello
  Hello, world!

Brief Explanation of the Program

/* ...... */
// ... until the end of the line
These are called comments. Comments are NOT executable and are ignored by the compiler. But they provide useful explanation and documentation to your readers (and to yourself three days later). There are two kinds of comments:

1. Multi-line Comment: begins with /* and ends with */. It may span more than one lines (as in Lines 1-3).
2. End-of-line Comment: begins with // and lasts until the end of the current line (as in Lines 4, 6, 7, 8, and 9).

#include <stdio.h>
The "#include" is called a preprocessor directive. A preprocessor directive begins with a # sign, and is processed before compilation. The directive "#include <stdio.h>" tells the preprocessor to include the "stdio.h" header file to support input/output operations. This line shall be present in all our programs. I will explain its meaning later.

int main() { ...... }
defines the so-called main() function. The main() function is the entry point of program execution. main() is required to return an int (integer).

printf("Hello, world!\n");
We invoke the function printf() to print the string "Hello, world!" followed by a newline (\n) to the console. The newline (\n) brings the cursor to the beginning of the next line.

return 0;
terminates the main() function and returns a value of 0 to the operating system. Typically, return value of 0 signals normal termination; whereas value of non-zero (usually 1) signals abnormal termination. This line is optional. C compiler will implicitly insert a "return 0;" to the end of the main() function.

C Terminology and Syntax

Statement: A programming statement performs a piece of programming action. It must be terminated by a semi-colon (;) (just like an English sentence is ended with a period) as in Lines 7 and 8.

Preprocessor Directive: The #include (Line 4) is a preprocessor directive and NOT a programming statement. A preprocessor directive begins with hash sign (#). It is processed before compiling the program. A preprocessor directive is NOT terminated by a semicolon - Take note of this rule.

Block: A block is a group of programming statements enclosed by braces { }. This group of statements is treated as one single unit. There is one block in this program, which contains the body of the main() function. There is no need to put a semi-colon after the closing brace.

Comments: A multi-line comment begins with /* and ends with */. An end-of-line comment begins with // and lasts till the end of the line. Comments are NOT executable statements and are ignored by the compiler. But they provide useful explanation and documentation. Use comments liberally.

Whitespaces: Blank, tab, and newline are collectively called whitespaces. Extra whitespaces are ignored, i.e., only one whitespace is needed to separate the tokens. But they could help you and your readers better understand your program. Use extra whitespaces liberally.

Case Sensitivity: C is case sensitive - a ROSE is NOT a Rose, and is NOT a rose.

The Process of Writing a C Program

Step 1: Write the source codes (.c) and header files (.h).

Step 2: Pre-process the source codes according to the preprocessor directives. The preprocessor directives begin with a hash sign (#), such as #include and #define. They indicate that certain manipulations (such as including another file or replacement of symbols) are to be performed BEFORE compilation.

Step 3: Compile the pre-processed source codes into object codes (.obj, .o).

Step 4: Link the compiled object codes with other object codes and the library object codes (.lib, .a)to produce the executable code (.exe).

Step 5: Load the executable code into computer memory.

Step 6: Run the executable code.

C Program Template

You can use the following template to write your C programs. Choose a meaningful filename for you source file that reflects the purpose of your program with file extension of ".c". Write your programming statements inside the body of the main() function. Don't worry about the other terms for the time being. I will explain them later.

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#include <stdio.h>
 
int main() {
   // Your Programming statements HERE!
 
   return 0;
}

Let's Write a Program to Add a Few Numbers

Example: Adding Two Integers

Let's write a C program called "Add2Integers.c" to add two integers as follows:

Add2Integers.c

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#include <stdio.h>
 
int main() {
   int integer1;
   int integer2;
   int sum;     
 
   integer1 = 55;  
   integer2 = 66;  
   sum = integer1 + integer2;  
 
  
   printf("The sum of %d and %d is %d.\n", integer1, integer2, sum);
 
   return 0;
}

The sum of 55 and 66 is 121.

Dissecting the Program

int integer1;
int integer2;
int sum;
We first declare three int (integer) variables: integer1, integer2, and sum. A variable is a named storage location that can store a value of a particular data type, in this case, int (integer). You can declare one variable in one statement. You could also declare many variables in one statement, separating with commas, e.g.,

int integer1, integer2, sum;

integer1 = 55;
integer2 = 66;
sum = integer1 + integer2;
We assign values to variables integer1 and integer2; compute their sum and store in variable sum.

printf("The sum of %d and %d is %d.\n", integer1, integer2, sum);
We use the printf() function to print the result. The first argument in printf() is known as the formatting string, which consists of normal texts and so-called conversion specifiers. Normal texts will be printed as they are. A conversion specifier begins with a percent sign (%), followed by a code to indicate the data type, such as d for decimal integer. You can treat the %d as placeholders, which will be replaced by the value of variables given after the formatting string in sequential order. That is, the first %d will be replaced by the value of integer1, second %d by integer2, and third %d by sum. The \n denotes a newline character. Printing a \n bring the cursor to the beginning of the next line.

Example: Prompting User for Inputs

In the previous example, we assigned fixed values into variables integer1 and integer2. Instead of using fixed values, we shall prompt the user to enter two integers.

PromptAdd2Integers.c

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#include <stdio.h>
 
int main() {
   int integer1, integer2, sum;
 
   printf("Enter first integer: ");  
   scanf("%d", &integer1);           
   printf("Enter second integer: "); 
   scanf("%d", &integer2);           
 
   sum = integer1 + integer2;        
 
  
   printf("The sum of %d and %d is %d.\n", integer1, integer2, sum);
 
   return 0;
}

Enter first integer: 55
Enter second integer: 66
The sum of 55 and 66 is 121.

Disecting the Program

int integer1, integer2, sum;
We first declare three int (integer) variables: integer1, integer2, and sum in one statement.

printf("Enter first integer: ");
We use the printf() function to put out a prompting message.

scanf("%d", &integer1);
We then use the scanf() function to read the user input from the keyboard and store the value into variable integer1. The first argument of scanf() is the formatting string (similar to printf()). The %d conversion specifier provides a placeholder for an integer, which will be substituted by variable integer1. Take note that we have to place an ampersand sign (&), which stands for address-of operator, before the variable, I shall explain its significance later. It is important to stress that missing ampersand (&) in scanf() is a common error, which leads to abnormal termination of the program.

Reading Multiple Integers

You can read multiple items in one scanf() statement as follows:

printf("Enter two integers: ");      
scanf("%d%d", &integer1, &integer2); 

In the scanf(), the first %d puts the first integer entered into variable integer1, and the second %d puts into integer2. Again, remember to place an ampersand (&) before the variables in scanf().

Exercises

1. Print each of the following patterns. Use one printf() statement for each line of outputs. End each line by printing a newline (\n).

   * * * * *      * * * * *    * * * * *
    * * * * *     *       *     *     *
   * * * * *      *       *      *   * 
    * * * * *     *       *       * *
   * * * * *      * * * * *        *
      (a)            (b)          (c)

2. Print the above patterns using ONE printf() statement.
3. Write a program to prompt user for 5 integers and print their sum. Use five int variables integer1 to integer5 to store the five integers.
4. Write a program to prompt user for 5 integers and print their product. Use an int variable product to store the product and operator * for multiplication.

What is a Program?

A program is a sequence of instructions (called programming statements), executing one after another - usually in a sequential manner, as illustrated in the previous example and the following flow chart.

Example (Sequential): The following program (CircleComputation.c) prompts user for the radius of a circle, and prints its area and circumference. Take note that the programming statements are executed sequentially - one after another in the order that they are written.

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#include <stdio.h>
 
int main() {
   double radius, circumference, area;
   double pi = 3.14159265;             
 
   printf("Enter the radius: "); 
   scanf("%lf", &radius);        
 
  
   area = radius * radius * pi;
   circumference = 2.0 * radius * pi;
 
  
   printf("The radius is %lf.\n", radius);
   printf("The area is %lf.\n", area);
   printf("The circumference is %lf.\n", circumference);
 
   return 0;
}

Enter the radius: 1.2
The radius is 1.200000.
The area is 4.523893.
The circumference is 7.539822.

Dissecting the Program

double radius, circumference, area;
double pi = 3.14159265;
We declare three double variables called radius, circumference and area. A double variable, unlike int, can hold real number (or floating-point number) such as 1.23 or 4.5e6. We also declare a double variable called pi and initialize its value to 3.1416.

printf("Enter the radius: ");
scanf("%lf", &radius);
We use print() to put up a prompt message, and scanf() to read the user input into variable radius. Take note that the %lf conversion specifier for double (lf stands for long float). Also remember to place an ampersand (&) before radius.

area = radius * radius * pi;
circumference = 2.0 * radius * pi;
perform the computation.

printf("The radius is %lf.\n", radius);
printf("The area is %lf.\n", area);
printf("The circumference is %lf.\n", circumference);
Again, we use %lf conversion specifier to print a double.

Take note that the programming statements inside the main() are executed one after another, sequentially.

Exercises

1. Follow the above example, write a program to print the area and perimeter of a rectangle. Your program shall prompt the user for the length and width of the rectangle, in doubles.
2. Follow the above example, write a program to print the surface area and volume of a cylinder. Your program shall prompt the user for the radius and height of the cylinder, in doubles.

What is a Variable?

Computer programs manipulate (or process) data. A variable is used to store a piece of data for processing. It is called variable because you can change the value stored.

More precisely, a variable is a named storage location, that stores a value of a particular data type. In other words, a variable has a name, a type and stores a value of that type.

• A variable has a name (or identifier), e.g., radius, area, age, height. The name is needed to uniquely identify and reference a variable, so as to assign a value to the variable (e.g., radius=1.2), and retrieve the value stored (e.g., area = radius*radius*pi).
• A variable has a type. Examples of type are:
  • int: for integers (whole numbers) such as 123 and -456;
  • double: for floating-point or real numbers, such as 3.1416, -55.66, 7.8e9, 1.2e3, -4.5e-6 having a decimal point and fractional part, in fixed or scientific notations.
• A variable can store a value of the declared type. It is important to take note that a variable is associated with a type, and can only store value of that particular type. For example, a int variable can store an integer value such as 123, but NOT real number such as 12.34, nor texts such as "Hello". The concept of type was introduced into the early programming languages to simplify interpretation of data.

The above diagram illustrates 2 types of variables: int and double. An int variable stores an integer (whole number). A double variable stores a real number.

To use a variable, you need to first declare its name and type, in one of the following syntaxes:

var-type var-name;                    
var-type var-name-1, var-name-2,...;  
var-type var-name = initial-value;    
var-type var-name-1 = initial-value-1, var-name-2 = initial-value-2,... ;  

Take note that:

• Each declaration statement is terminated with a semi-colon (;).
• In multiple-variable declaration, the names are separated by commas (,).
• The symbol =, known as the assignment operator, can be used to assign an initial value (of the declared type) to the variable.

For example,

int sum;              
int number1, number2; 
double average;       
int height = 20;       

Once a variable is declared, you can assign and re-assign a value to a variable, via the assignment operator "=". For example,

int number;   
number = 99;  
number = 88;  
number = number + 1;  
int sum = 0;  
sum = sum + number;  
int num1 = 5, num2 = 6;  
double radius = 1.5;  
int number;           
sum = 55.66;          
sum = "Hello";        

Take note that:

• Each variable can only be declared once.
• You can declare a variable anywhere inside the program, as long as it is declared before it is being used.
• Once the type of a variable is declared, it can only store a value belonging to this particular type. For example, an int variable can hold only integer such as 123, and NOT floating-point number such as -2.17 or text string such as "Hello".
• The type of a variable cannot be changed inside the program.

x=x+1?

Assignment (=) in programming is different from equality in Mathematics. e.g., "x=x+1" is invalid in Mathematics. However, in programming, it means compute the value of x plus 1, and assign the result back to variable x.

"x+y=1" is valid in Mathematics, but is invalid in programming. In programming, the RHS of "=" has to be evaluated to a value; while the LHS shall be a variable. That is, evaluate the RHS first, then assign to LHS.

Some languages uses := as the assignment operator to avoid confusion with equality.

Basic Arithmetic Operations

The basic arithmetic operators are:

Addition, subtraction, multiplication, division and remainder are binary operators that take two operands (e.g., x + y); while negation (e.g., -x), increment and decrement (e.g., x++, --x) are unary operators that take only one operand.

Example

The following program (TestArithmetics.c) illustrates these arithmetic operations.

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#include <stdio.h>
 
int main() {
 
   int number1, number2; 
   int sum, difference, product, quotient, remainder; 
 
  
   printf("Enter two integers (separated by space): ");
   scanf("%d%d", &number1, &number2);  
 
  
   sum = number1 + number2;
   difference = number1 - number2;
   product = number1 * number2;
   quotient = number1 / number2;
   remainder = number1 % number2;
 
   printf("The sum, difference, product, quotient and remainder of %d and %d are %d, %d, %d, %d, %d.\n",
        number1, number2, sum, difference, product, quotient, remainder);
 
  
   ++number1;   
                
   --number2;   
                
   printf("number1 after increment is %d.\n", number1);
   printf("number2 after decrement is %d.\n", number2);
 
   quotient = number1 / number2;
   printf("The new quotient of %d and %d is %d.\n", number1, number2, quotient);
 
   return 0;
}

Enter two integers (separated by space): 98 5
The sum, difference, product, quotient and remainder of 98 and 5 are 103, 93, 49
0, 19, 3.
number1 after increment is 99.
number2 after decrement is 4.
The new quotient of 99 and 4 is 24.

Dissecting the Program

int number1, number2;
int sum, difference, product, quotient, remainder;
declare all the int (integer) variables number1, number2, sum, difference, product, quotient, and remainder, needed in this program.

printf("Enter two integers (separated by space): ");
scanf("%d%d", &number1, &number2);
prompt user for two integers and store into number1 and number2, respectively.

sum = number1 + number2;
difference = number1 - number2;
product = number1 * number2;
quotient = number1 / number2;
remainder = number1 % number2;
carry out the arithmetic operations on number1 and number2. Take note that division of two integers produces a truncated integer, e.g., 98/5 → 19, 99/4 → 24, and 1/2 → 0.

printf("The sum, difference, product, quotient and remainder of %d and %d are %d, %d, %d, %d, %d.\n",
   number1, number2, sum, difference, product, quotient, remainder);
prints the results of the arithmetic operations, with the appropriate string descriptions in between.

++number1;
--number2;
illustrate the increment and decrement operations. Unlike '+', '-', '*', '/' and '%', which work on two operands (binary operators), '++' and '--' operate on only one operand (unary operators). ++x is equivalent to x = x + 1, i.e., increment x by 1. You may place the increment operator before or after the operand, i.e., ++x (pre-increment) or x++ (post-increment). In this example, the effects of pre-increment and post-increment are the same. I shall point out the differences in later section.

Exercises

1. Introduce one more int variable called number3, and prompt user for its value. Print the sum and product of all the three integers.
2. In Mathematics, we could omit the multiplication sign in an arithmetic expression, e.g., x = 5a + 4b. In programming, you need to explicitly provide all the operators, i.e., x = 5*a + 4*b. Try printing the sum of 31 times of number1 and 17 times of number2 and 87 time of number3.

What If Your Need To Add a Thousand Numbers? Use a Loop!

Suppose that you want to add all the integers from 1 to 1000. If you follow the previous examples, you would require a thousand-line program! Instead, you could use a loop in your program to perform a repetitive task, that is what the dumb computers are good at.

Example

Try the following program SumNumbers.c, which sums all the integers from 1 to an upperbound provided by the user, using a so-called while-loop.

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#include <stdio.h>
 
int main() {
   int sum = 0;   
                  
   int upperbound;
 
  
   printf("Enter the upperbound: ");
   scanf("%d", &upperbound); 
 
  
   int number = 1;
   while (number <= upperbound) {
      sum = sum + number;  
      ++number;            
   }
  
   printf("The sum from 1 to %d is %d.\n", upperbound, sum);
 
   return 0;
}

Enter the upperbound: 1000
The sum from 1 to 1000 is 500500.

Dissecting the Program

int sum = 0;
declares an int variable named sum and initializes it to 0. This variable will be used to accumulate numbers over the steps in the repetitive loop.

printf("Enter the upperbound: ");
scanf("%d", &upperbound);
prompt user for an upperbound to sum.

int number = 1;
while (number <= upperbound) {
   sum = sum + number;
   ++number;
}
This is the so-called while-loop. A while-loop takes the following syntax:

initialization-statement;
while (test) {
   loop-body;
}
next-statement;

As illustrated in the flow chart, the initialization statement is first executed. The test is then checked. If the test is true, the body is executed. The test is checked again and the process repeats until the test is false. When the test is false, the loop completes and program execution continues to the next statement after the loop.

In our program, the initialization statement declares an int variable named number and initializes it to 1. The test checks if number is equal to or less than the upperbound. If it is true, the current value of number is added into the sum, and the statement ++number increases the value of number by 1. The test is then checked again and the process repeats until the test is false (i.e., number increases to upperbound+1), which causes the loop to terminate. Execution then continues to the next statement (in Line 22).

In this example, the loop repeats upperbound times. After the loop is completed, Line 22 prints the result with a proper description.

Exercises

1. Modify the above program to sum all the number between a lowerbound and an upperbound provided by the user.
2. Modify the above program to sum all the odd numbers between 1 to an upperbound. (Hint: Use "number = number + 2".)
3. Modify the above program to sum all the numbers between 1 to an upperbound that are divisible by 7. (Hint: Use "number = number + 7")
4. Modify the above program to find the sum of the square of all the numbers from 1 to an upperbound, i.e. 1*1 + 2*2 + 3*3 +...
5. Modify the above program to compute the product of all the numbers from 1 to 10. (Hint: Use a variable called product instead of sum and initialize product to 1. Ans: 3628800.) Based on this code, write a program to display the factorial of n, where n is an integer between 1 to 12.

Conditional (or Decision)

What if you want to sum all the odd numbers and also all the even numbers between 1 and 1000? There are many way to do this. You could declare two variables: sumOdd and sumEven. You can then use a conditional statement to check whether the number is odd or even, and accumulate the number into the respective sum. The program SumOddEven.c is as follows:

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#include <stdio.h>
 
int main() {
   int sumOdd  = 0;
   int sumEven = 0;
   int upperbound; 
 
  
   printf("Enter the upperbound: ");
   scanf("%d", &upperbound);  
 
  
   int number = 1;
   while (number <= upperbound) {
      if (number % 2 == 0) { 
         sumEven = sumEven + number;
      } else {               
         sumOdd = sumOdd + number;
      }
      ++number;
   }
 
  
   printf("The sum of odd numbers is %d.\n", sumOdd);
   printf("The sum of even numbers is %d.\n", sumEven);
   printf("The difference is %d.\n", (sumOdd - sumEven));
 
   return 0;
}

Enter the upperbound: 10000
The sum of odd numbers is 25000000.
The sum of even numbers is 25005000.
The difference is -5000.

Dissecting the Program

int sumOdd = 0;
int sumEven = 0;
declare two int variables named sumOdd and sumEven and initialize them to 0, for accumulating the odd and even numbers, respectively.

if (number % 2 == 0) {
   sumEven = sumEven + number;
} else {
   sumOdd = sumOdd + number;
}
This is a conditional statement. The conditional statement can take one these forms: if-then or if-then-else.

if ( test ) {
   true-body;
}

if ( test ) {
   true-body;
} else {
   false-body;
}

For a if-then statement, the true-body is executed if the test is true. Otherwise, nothing is done and the execution continues to the next statement. For a if-then-else statement, the true-body is executed if the test is true; otherwise, the false-body is executed. Execution is then continued to the next statement.

In our program, we use the remainder operator (%) to compute the remainder of number divides by 2. We then compare the remainder with 0 to test for even number.

Comparison Operators

There are six comparison (or relational) operators:

Take note that the comparison operator for equality is a double-equal sign (==); whereas a single-equal sign (=) is the assignment operator.

Combining Simple Conditions

Suppose that you want to check whether a number x is between 1 and 100 (inclusive), i.e., 1 <= x <= 100. There are two simple conditions here, (x >= 1) AND (x <= 100). In programming, you cannot write 1 <= x <= 100, but need to write (x >= 1) && (x <= 100), where "&&" denotes the "AND" operator. Similarly, suppose that you want to check whether a number x is divisible by 2 OR by 3, you have to write (x % 2 == 0) || (x % 3 == 0) where "||" denotes the "OR" operator.

There are three so-called logical operators that operate on the boolean conditions:

For examples:

(x >= 0) && (x <= 100)  

  

(x < 0) || (x > 100)       
!((x >= 0) && (x <= 100))  
 


((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)

Exercises

1. Write a program to sum all the integers between 1 and 1000, that are divisible by 13, 15 or 17, but not by 30.
2. Write a program to print all the leap years between AD1 and AD2010, and also print the number of leap years. (Hints: use a variable called count, which is initialized to zero. Increment the count whenever a leap year is found.)

Type double & Floating-Point Numbers

Recall that a variable in C has a name and a type, and can hold a value of only that particular type. We have so far used a type called int. A int variable holds only integers (whole numbers), such as 123 and -456.

In programming, real numbers such as 3.1416 and -55.66 are called floating-point numbers, and belong to a type called double. You can express floating-point numbers in fixed notation (e.g., 1.23, -4.5) or scientific notation (e.g., 1.2e3, -4E5.6) where e or E denote the exponent of base 10.

Example

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#include <stdio.h>
 
int main() {
   double celsius, fahrenheit;
 
   printf("Enter the temperature in celsius: ");
   scanf("%lf", &celsius); 
   fahrenheit = celsius * 9.0 / 5.0 + 32.0;
   printf("%.2lf degree C is %.2lf degree F.\n\n", celsius, fahrenheit);
     
 
   printf("Enter the temperature in fahrenheit: ");
   scanf("%lf", &fahrenheit);
   celsius =  (fahrenheit - 32.0) * 5.0 / 9.0;
   printf("%.2lf degree F is %.2lf degree C.\n\n", fahrenheit, celsius);
 
   return 0;
}

Enter the temperature in celsius: 37.2
37.20 degree C is 98.96 degree F.

Enter the temperature in fahrenheit: 100
100.00 degree F is 37.78 degree C.

Mixing int and double, and Type Casting

Although you can use a double to keep an integer value (e.g., double count = 5), you should use an int for integer. This is because int is far more efficient than double, in terms of running times and memory requirement.

At times, you may need both int and double in your program. For example, keeping the sum from 1 to 100 (=5050) as an int, and their average 50.5 as a double. You need to be extremely careful when different types are mixed.

It is important to note that:

• Arithmetic operations ('+', '-', '*', '/') of two int's produce an int; while arithmetic operations of two double's produce a double. Hence, 1/2 → 0 (take note!) and 1.0/2.0 → 0.5.
• Arithmetic operations of an int and a double produce a double. Hence, 1.0/2 → 0.5 and 1/2.0 → 0.5.

You can assign an integer value to a double variable. The integer value will be converted to a double value automatically, e.g., 3 → 3.0. For example,

int i = 3;
double d;
d = i;              
d = 88;             
double nought = 0;  

However, if you assign a double value to an int variable, the fractional part will be lost. For example,

double d = 55.66;
int i;
i = d;   

Some C compilers signal a warning for truncation, while others do not. You should study the "warning messages" (if any) carefully - which signals a potential problem in your program, and rewrite the program if necessary. C allows you to ignore the warning and run the program. But, the fractional part will be lost during the execution.

Type Casting Operators

If you are certain that you wish to carry out the type conversion, you could use the so-called type cast operator. The type cast operation returns an equivalent value in the new-type specified.

(new-type)expression;

For example,

double d = 5.5;
int i;
i = (int)d;        i = (int)3.1416;   

Similarly, you can explicitly convert an int value to double by invoking type-casting operation too.

Example

Try the following program and explain the outputs produced:

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#include <stdio.h>
 
int main() {
   int sum = 0;     
   double average;  
 
  
   int number = 1;
   while (number <= 100) {
      sum = sum + number;
      ++number;
   }
   printf("The sum is %d.\n", sum);
 
   
   average = sum / 100;
   printf("Average 1 is %lf.\n", average);
   average = (double)sum / 100;
   printf("Average 2 is %lf.\n", average);
   average = sum / 100.0;
   printf("Average 3 is %lf.\n", average);
   average = (double)(sum / 100);
   printf("Average 4 is %lf.\n", average);
 
   return 0;
}

The sum is 5050.
Average 1 is 50.000000.  
Average 2 is 50.500000.
Average 3 is 50.500000.
Average 4 is 50.000000.  

The first average is incorrect, as int/int produces an int (of 50).

For the second average, the value of sum (of int) is first converted to double. Subsequently, double/int produces double.

For the third average, int/double produces double.

For the fourth average, int/int produces an int (of 50), which is then casted to double (of 50.0) and assigned to average (of double).

Exercises

1. Write a program called HarmonicSeriesSum to compute the sum of a harmonic series 1 + 1/2 + 1/3 + 1/4 + .... + 1/n, where n = 1000. Your program shall prompt user for the value of n. Keep the sum in a double variable, and take note that 1/2 gives 0 but 1.0/2 gives 0.5.
   Try computing the sum for n=1000, 5000, 10000, 50000, 100000.
   Hints:

   #include <stdio.h>
    
   int main() {
      int maxDenominator;   
      double sum = 0.0;  
   
         ......
    
      int denominator = 1;
      while (denominator <= maxDenominator) {
         
         ......
         ++denominator;  
      }
      
      ......
   }

2. Write a program called GeometricSeriesSum to compute the sum of a geometric series 1 + 1/2 + 1/4 + 1/8 + .... + 1/n. You program shall prompt for the value of n. (Hints: Use post-processing statement of denominator = denominator * 2.)

Summary

I have presented the basics for you to get start in programming. To learn programming, you need to understand the syntaxes and features involved in the programming language that you chosen, and you have to practice, practice and practice, on as many problems as you could.

Link to "C Language References & Resources"

What happens when you call a method and the method ends?

When you a method you write multiple methods with a shared name but different parameter lists?

When methods must share data you can pass?

What are methods with identical names that have identical parameter lists but different return types?