Execution of printf with ++ operators

Consider below C++ program and predict its output.

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

The above invokes undefined behaviour by referencing both ‘i’ and ‘i++’ in the argument list. It is not defined which order the arguments are evaluated. Different compilers may choose different orders. A single compiler can also choose different orders at different times.

For example below three printf statements also cause undefined behavior.

// All three pritf() statements in this cause undefined behavior
#include<stdio.h>

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

Therefore, it is not recommended to do two or more than two pre or post increment operators in the same statement.

This article is contributed by Spoorthi Aman. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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