In this example we will be learning very typical example of C Program where we will be using pre increment and post increment operators in same expression.

Re-commanded Reading : Increment Operator in C | Decrements Operator in C

Postfix and Prefix Expression in Same Statement

#include<stdio.h>
#include<conio.h>
void main() {
   int i = 0, j = 0;
   j = i++ + ++i;
   printf("%d\n", i);
   printf("%d\n", j);
}

Output :

2
2

Explanation of Program

A picture worth 100 lines of text. Consider the below program picture which explains the above program.

j = i++ + ++i;

Above line of code have total four operators -

1. Assignment Operator
2. Post Increment Operator
3. Pre Increment Operator
4. Arithmetic Operator

Now we need to arrange all the operators in the decreasing order of the Precedence -

Now we have arranged all the operators in the decreasing order of the precedence. (Operator Precedence and Associativity Document)

Execution flow

1. In the first step pre-increment operator gets an opportunity for execution. It will increment the value of variable
2. In the second step post increment operator will be executed. It does not increment the actual value of variable but mark it as pending execution.
3. Here pre-increment operation uses new Value of ‘i’ so [ i = 1 ].
4. This ‘i = 1’ is used for Post Increment Operation which uses Old Value i.e [i = 1] .
5. Both ‘1’s are added and value 2 is assigned to ‘j’.

Summary :

Step 1 : j = i++ + 1;
Step 2 : j = 1 + 1; (post-increment pending on i)
Step 3 : j = 2; (Addition)
Step 4 : j = 2; (Assignment)
Step 5 : Pending increment on i will be executed

We have already learnt different ways of using increment operators in c programming. In this tutorial we will study the impact of using multiple increment / decrement operators in printf.

Re-commanded Article : Use of Pre/Post increment operator in same expression

Multiple increment operators inside printf

#include<stdio.h>
void main() {
   int i = 1;
   printf("%d %d %d", i, ++i, i++);
}

Output :

3 3 1

Pictorial representation

Explanation of program

I am sure you will get confused after viewing the above image and output of program.

1. Whenever more than one format specifiers (i.e %d) are directly or indirectly related with same variable (i,i++,++i) then we need to evaluate each individual expression from right to left.
2. As shown in the above image evaluation sequence of expressions written inside printf will be - i++,++i,i
3. After execution we need to replace the output of expression at appropriate place

Below is programmatic representation of intermediate steps -

After Step 1 : printf("%d %d %d",i,++i,1);
After Step 2 : printf("%d %d %d",i,3,1);
After Step 3 : printf("%d %d %d",3,3,1);
After Step 4 : Output displayed - 3,3,1

After solving expressions printing sequence will : i, ++i, i++

Examples of typical expresssions

We have already studied the pre and post increment operators in c programming. In this tutorial we will be learning different examples of post increment operator.

Examples of post increment operator in expression

Before going through the different examples we can consider below values as initial values for each program explained below -

int i=0,j=0;
int k = 0;
int temp=3;

Depending upon these initial values try to grasp the below programs -

1.Post Increment Verity No 1

int main() {
   int num = 0, result = 0, i = 0;
   for (i = 0; i < 5; i++) {
      result = num++;
      printf("%d", result);
   }
   return 0;
}

Output : [ 0 1 2 3 4 ]

Consider the above program where we have used following statement -

result = num++;

num++ is a post increment which means we are firstly assigning the old value of variable ‘num’ to ‘result’. After assigning the older value of ‘result’ we will be increment value of ‘num’

2.Post Increment Verity No 2

int main() {
   int num = 0, i = 0;
   for (i = 0; i < 5; i++) {
      printf("%d", num++);
   }
   return 0;
}

Output : [ 0 1 2 3 4 ]

Inside printf function we have used post increment operator. It will print the value of variable first and then it will increment the value of variable after execution of printf statement.

3.Post Increment Verity No 3

for(k=0;k<5;k++)
  {
  j = i++ + temp;
  printf("%d",j);
  }

Firstly old value of ‘i’ is added with temp and result of addition is assigned to ‘j’. After addition value of ‘i’ will be incremented

4.Post Increment Verity No 4

for(k=0;k<5;k++)
  {
  j = i++ < temp;
  printf("%d\n",j);
  }

Re-commanded Article : Conditional Operator in Expression

Firstly old value of ‘i’ is compared with ‘temp’ . Then result is assigned to j, and new value is used in second iteration.

5.Post Increment Verity No 5

for(k=0;k<5;k++)
  {
  j = i++;
  printf("%d",i);
  }

• Old value of ‘i gets reflected only within same expression
• In the above snippet Old value is assigned to ‘j’ . but in the next statement always ‘new value’ is used.
• So while printing i Output will be = [1 2 3 4 5].

6.Post Increment Verity No 6

for(k=0;k<5;k++)
  {
  i = i++;
  printf("%d",i);
  }

• Old value of ‘i is assigned to ‘i’ itself.
• And again value is incremented after assigning.
• So while printing i Output will be = [1 2 3 4 5].
• But this Type is not supported by many compiles .

Note :

1. Same things happens while using Post-Decrement.
2. Post Increment is Unary Operator.
3. Post-Increment Operator has Highest Priority