Bitwise left shifting unexpected result

Question

I found some wired problems when I tried to do left shift for 32 times. The code in the test function should print the same results of 0x0, but I got "ffffffff, 0" instead. Anybody could give a hint of what's wrong with the code? Thank you!

int test(int n) {
    int mask = ~0 << (32 + ~n + 1);
    int mask1 = ~0 << (32 + ~0 + 1);
    printf("%x, %x\n", mask, mask1);
    return mask;
}

int main(){
    test(0);
}

same with http://stackoverflow.com/questions/3784996/why-does-left-shift-operation-invoke-undefined-behaviour-when-the-left-side-oper — MOHAMED, May 07 '13 at 16:30
Can you try making the int declaration an unsigned int and tell us if it will make any difference? — Manolis Ragkousis, Jan 07 '14 at 19:54

score 4 · Answer 1 · edited May 23 '17 at 10:32

4

In C, shifting with size greater than the type size (int in your case) is an undefined behavior

From this topic: Relevant quote from ISO C99 (6.5.7/4)

The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are ﬁlled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2^E2, reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undeﬁned.

edited May 23 '17 at 10:32

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answered May 07 '13 at 16:23

MOHAMED

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You mean it is still undefined if the shifting distance is the same a s the type size? – zhengbli May 07 '13 at 16:25
yes. if your type size is 32 then you are allowed to shift with number <= 31 – MOHAMED May 07 '13 at 16:28

score 2 · Answer 2 · answered May 07 '13 at 16:25

Assuming you have 32-bit integers, the results of left shifting more than 31 times are undefined

From C11 §6.5.7 Bitwise shift operators

The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 x 2E2 , reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 x 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

Bitwise left shifting unexpected result

2 Answers2