explain this nested printf statement

Question

int a = 5;

#include <stdio.h>

int main() {
    printf("%d", printf("hi!") * printf("bye"));
    return 0;
}

Output:

hi!bye9

I would like to know how the order in which the output has occurred. Does this mean printf function returns a value?

What is the reason behind inner printf statements being executed first?

Answer 1

This program exhibits unspecified behavior since the order of evaluation of sub expressions is unspecified except where it is specifically defined:

printf("%d",printf("hi!")*printf("bye"));
            ^             ^
            1             2

So either 1 or 2 could be evaluated first and you can not determine which. We can see this from the C99 draft standard section 6.5 Expressions paragraph 3 which says (emphasis mine going forward):

The grouping of operators and operands is indicated by the syntax.74) Except as specified later (for the function-call (), &&, ||, ?:, and comma operators), the order of evaluation of subexpressions and the order in which side effects take place are both unspecified.

Yes, printf does have a return value, which is the number of character printed or -1 if there is an error, in this case assuming no error the return value will be 3 for both the inner printfs

The arguments to a function are evaluated before the function is called which is why the inner printfs are executed first, this is covered in section 6.5.2.2 Function calls paragraph 4:

An argument may be an expression of any object type. In preparing for the call to a function, the arguments are evaluated, and each parameter is assigned the value of the corresponding argument.81)

Answer 2

The order of the internal printfs is unspecified. Another implementation, the same implementation with different compiler settings, or the same implementation with the same code in a different place might produce byehi!9.

The 9 comes because printf returns the number of characters printed, so the two internal printfs return 3 and the * is the familiar multiplication operator, giving 9.