Execution of printf With ++ Operators in C

Consider the following statement in C and predict its output.

printf("%d %d %d", i, ++i, i++);

This statement invokes undefined behavior by referencing both ‘i’ and ‘i++’ in the argument list.

In C, the evaluation order of function arguments is not specified. It means the compiler is free to evaluate arguments in any order. That is why, Similar thing happens in statement printf("%d %d %d", i, ++i, i++); the order of evaluating i, ++i, and i++ is undefined, that leads to inconsistent and unpredictable results as different compilers or even the same compiler may choose different evaluation orders at different times.

For example, below three printf() statements may also cause undefined behaviour:

// C Program to demonstrate the three printf() statements
// that cause undefined behavior
#include <stdio.h>

// Driver Code
int main() {
    volatile int a = 10;
    printf("%d %d\n", a, a++);

    a = 10;
    printf("%d %d\n", a++, a);

    a = 10;
    printf("%d %d %d\n", a, a++, ++a);
    return 0;
}

11 10
10 10
12 11 11

Explanation: Usually, the compilers read parameters of printf() from right to left. So, 'a++' will be executed first as it is the last parameter of the first printf() statement. It will print 10. Although now the value has been increased by 1, so the second last argument, i.e., will print 11. Similarly, the other statements will also get executed.

Therefore, it is not recommended to do two or more than two pre or post-increment operators in the same statement. This means that there's absolutely no temporal ordering in this process. The arguments can be evaluated in any order, and the process of their evaluation can be intertwined in any way.

Note: In pre-increment, i.e., ++a, it will increase the value by 1 before printing, and in post-increment, i.e., a++, it prints the value at first, and then the value is incremented by 1.

How to avoid the undefined behaviour in this case?

To ensure predictable behaviour and avoid undefined behaviour, it is recommended to separate operations into distinct statements. For example, instead of writing printf("%d %d %d", i, ++i, i++);, break it down into multiple steps like illustrated in the below example:

// C Program to demonstrate the three printf() statements
// without causing undefined behavior
#include <stdio.h>

int main(){
    volatile int i = 10;
  
    int temp1 = i;
    int temp2 = ++i;
    int temp3 = i++;
  
    printf("%d %d %d", temp1, temp2, temp3);
  	return 0;
}

14 14 12