Printf outputs is in reverse order

Question

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Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)

i tried this simple C program in GCC

#include<stdio.h>
int main(){
int x=5;
printf("%d,%d,%d,%d,%d",++x,x++,x,x++,++x);
return 0;
}

and the output was 9,7,7,6,6 i traced it and assumed that it will print 6,6,7,7,9 but i found my assumption in reverse order, how come!

This is undefined behavior. You basically can increment only once each incremented variable or location inside a statement. — Basile Starynkevitch, Nov 28 '12 at 01:44

score 4 · Accepted Answer · answered Nov 28 '12 at 01:44

4

Because your program has undefined behaviour. There is no sequence point between the evaluations of function arguments, and it is undefined behaviour to mutate the same object more than once without intervening sequence point.

The program is simply ill-formed. It is not a valid C program.

answered Nov 28 '12 at 01:44

Kerrek SB

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And if compiling with `gcc -Wall` you get several warnings e.g. `eslam.c:4:39: warning: operation on 'x' may be undefined` – Basile Starynkevitch Nov 28 '12 at 01:45

Printf outputs is in reverse order

1 Answers1