Difference Between i++ and ++i in C

• Notes
• C

A Brief Explanation

Let’s say we have int i = 0;, and imagine there’s a temporary object (rvalue?) to store the result of i + 1 for the ++ operator.

Object is an area of memory that is used by our program, and temporary object in here means that the object has a temporary duration and will be deleted when the containing full expression ends.

Full expression here means:

• The complete expression that forms an expression statement (with terminating semicolon (;) at the end).
• One of the controlling expression if, switch, while, for, or do-while statement.
• The expression of an initializer (like int i = 0;).
• return statement.

This is what probably happen if we use i++ without assigned the result to another variable:

// i++;
tmp = i;
i = i + 1;

This is what probably happen if we use ++i without assigned the result to another variable:

// ++i;
i = i + 1;
tmp = i;

This is what probably happen if we use i++ when assigned the result to variable j:

// int j = i++;
tmp = i;
j = tmp;
i = i + 1;

This is what probably happen if we use ++i when assigned the result to variable j:

// int j = ++i;
i = i + 1;
tmp = i;
j = tmp;

The only difference between i++ and ++i is when we use the value of the operation in the same statement, like int j = ++i.

P.S:
At the time of writing this post, i am still learning about C, so this illustration might be wrong. Please let me know if there are a better illustration about this.

i++ or ++i in For Loop

When using either i++ or ++i in for-loop like this:

int i;

for (i = 0; i < 5; i++)
    printf("%d\n", i);

for (i = 0; i < 5; i++)
    printf("%d\n", i);

Both of them will operate identically because the increment of i and the print statement is in different line. It’s like we are using i++ or ++i without any assignment like this:

int i = 0;
i++;
++i;
printf("%d\n", i);

Side Note

Please keep in mind that operation i++ and ++i is also prone to integer overflow.

Here an overflow example:

#include <stdio.h>
#include <limits.h>

int main(void)
{
    unsigned int i = UINT_MAX;
    printf("before increment i: %d\n", i);

    i++;
    printf("after increment i: %d\n", i);

    return 0;
}

In the example above, as we store the maximum value of unsigned integer in i and when we increase the value of i, the result would be 0 because the value is more than the maximum value of unsigned integer, so it result in overflow.

References

• Stackoverflow about difference between i++ and ++i .
• About temporary object .
• About integer overflow .
• Unsigned integers overflow .
• Cppreference object lifetime .
• Cppreference statements .