C Branching Statements

In the term software or computer programming, it has a set of instruction (in simple or complex form) called program. These instructions are also called statements, which occurs sequentially or in either conditional way or in the iterative way. To handle such types of statements some flow controls required. These flow controls are called Control Statements

In other words, the control statements are used to control the cursor in a program according to the condition or according to the requirement in a loop. There are mainly three types of control statements or flow controls. These are illustrated as below:

Branching

if statement

The if statement is a powerful decision making statement which can handle a single condition or group of statements. These have either true or false action....
When only one condition occurs in a statement, then simple if statement is used having one block.

/*Any Number is input through the keyboard. write a If program.*/

#include< stdio.h >
#include< conio.h >
void main()
{
    int n;
    n=1;
    clrscr(); //clrscr() is the function of #include< conio.h > header file which will clear previous output of program
    printf("Enter the Number");
    scanf("%d",&n);
    if(n>0)
    {
        printf("It is If Statement");
    }
    getch();
}

if-else statement

This statement also has a single condition with two different blocks. One is true block and other is false block...

/*Any Number is input through the keyboard. write a program to find out whether It is and Odd Number or Even Number.*/

#include< stdio.h >
#include< conio.h >
void main()
{
    int n;
    n=1;
    clrscr();
    printf("Enter the Number");
    scanf("%d",&n);
    if(n%2==0)
    {
        printf("This is Even Number");
    }
    else
    {
        printf("This is Odd Number");
    }
    getch();
}

Output is as :
Enter the Number
4
This is Even Number

nested if statement

When an if statement occurs within another if statement, then such type of is called nested if statement.

/*If the ages of Ram, sham and Ajay are input through the keyboard, write a program to determine the youngest of the three*/

#include< stdio.h >
#include< conio.h >
void main()
{
  int ram,sham,ajay;
  clrscr();
  printf("Enter the Three Ages of Ram,Sham and Ajay\n");
  scanf("%d%d%d",&ram,&sham,&ajay);
  if(ram < sham)
  {
    if(ram < ajay)
    {
      printf("Ram is Youngest");
    }
    else
    {
      printf("Ajay is Youngest");
    }
  }
  else
  {
    if(sham < ajay)
    {
      printf("Sham is Youngest");
    }
    else
    {
      printf("Ajay is Youngest");
    }
  }
  getch();
}

Output is as :
Enter the three Ages of Ram,Sham and Ajay
14
17
19
Ram is Youngest

Ladder if or else if statement

When in a complex problem number of condition arise in a sequence, then we can use Ladder-if or else if statement to solve the problem in a simple manner.

The marks obtained by a student in 5 different subjects are input through the keyboard. The student gets a division as per the following rules:

=> Percentage above or equal to 60 - First Division
=> Percentage between 50 and 59 - Second Division
=> Percentage between 40 and 49 - Third Division
=> Percentage less than 40 - Fail

Method-1

/*Write a program to calculate the division obtained by the student. There are two ways in which we can write a program for this example. These methods are given below: */

//Method-1
#include< stdio.h >
#include< conio.h >
void main()
{
  int eng,math,com,sci,ss,total;
  float per;
  clrscr();
  printf("Enter Five Subjects Marks\n");
  scanf("%d%d%d%d%d",&eng,&math,&com,&sci,&ss);

  total = eng + math + com + sci + ss ;
  printf("Toal Marks : %d",total);
  per = total * 100.00 / 500;
  printf("\n Percentage : %f", per);

  if(per >= 60)
  {
    printf("\n 1st Division");
  }
  else if(per >= 50)
  {
    printf("\n 2nd Division");
  }
  else if(per >= 40)
  {
    printf("\n 3rd Division");
  }
  else
  {
    printf("\n Sorry Fail");
  }
  getch();
}

Method-2

//Method-2
#include< stdio.h >
#include< conio.h >
void main()
{
  int eng,math,com,sci,ss,total;
  float per;
  clrscr();
  printf("Enter Five Subjects Marks\n");
  scanf("%d%d%d%d%d",&eng,&math,&com,&sci,&ss);

  total = eng + math + com + sci + ss ;
  printf("Toal Marks : %d",total);
  per = total * 100.00 / 500;
  printf("\n Percentage : %f", per);

  if(per >= 60)
  {
    printf("\n 1st Division");
  }
  if(per >= 50 && per < 60)
  {
    printf("\n 2nd Division");
  }
  if(per >= 40 && per < 50)
  {
    printf("\n 3rd Division");
  }
  if(per < 40)
  {
    printf("\n Sorry Fail");
  }
  getch();
}

switch statement

When number of conditions (multiple conditions) occurs in a problem and it is very difficult to solve such type of complex problem with the help of ladder if statement, then there is need of such type of statement which should have different alternatives or different cases to solve the problem in simple and easy way. For this purpose switch statement is used. .

/*WAP to print the four-days of week from monday to thrusday which works upon the choice as S,M,T,H using switch case*/

#include< stdio.h >
#include< conio.h >
void main()
{
  char n;
  clrscr();
  printf("Enter the Choice from Four Days...\n")
  printf("S = Sunday \n")
  printf("M = Monday \n")
  printf("T = Tuesday \n")
  printf("H = Thursday \n\n")

  scanf("%c",&n);

  switch(n)
  {
     case 'S':
     printf("Sunday");
     break;

     case 'M':
     printf("Monday");
     break;

     case 'T':
     printf("Tuesday");
     break;

     case 'H':
     printf("Thursday");
     break;

     default:
     printf("Out of Choice");
     break;

   }
  getch();
}

Output is as :
Enter the Choice from Four Days...
S = Sunday
M = Monday
T = Tuesday
H = Thursday

S
Sunday

Conditional Control Statement

This statement is based on conditional operator. This statement solves the problem's condition in a single line and is a fast executable operation. For this purpose we can take combination of ? and :

/*The easiest way to use conditional control statement*/

#include< stdio.h >
#include< conio.h >
void main()
{
  int n;
  clrscr();

  n%4==0 ? printf("Leap Year") : printf("Not Leap Year");

            //OR

  //printf(n%4==0 ? "Leap Year" : "Not Leap Year");

  getch();
}