Question

What does the rule about sequence points say about the following code?

int main(void) {
    int i = 5;
    printf("%d", ++i, i); /* Statement 1 */
}

There is just one %d. I am confused because I am getting 6 as output in compilers GCC, Turbo C++ and Visual C++. Is the behavior well defined or what?

This is related to my last question.

Answer 1

It’s undefined because of 2 reasons:

1. The value of i is twice used without an intervening sequence point (the comma in argument lists is not the comma operator and does not introduce a sequence point).

2. You’re calling a variadic function without a prototype in scope.

3. The number of arguments passed to printf() are not compatible with the format string.

4. the default output stream is usually line buffered. Without a '\n' there is no guarantee the output will be effectively output.