Unsigned Bitwise Shift Operators [C11]

Question

Edit: As pointed out below I missed the first part of the ANSI C standard: "If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand, the behavior is undefined." The errors (or rather lack of errors / difference in errors) are due to the particular compiler I was using.

I've come across something a bit strange, and I hope that someone can shed some light on my ignorance here. The necessary sample code is as follows:

#include <stdio.h>
int main(void)
{
    unsigned a, b;
    int w, x, y;

    a = 0x00000001;
    b = 0x00000020;
    w = 31;
    x = 32;
    y = 33;

    a << w; /*No error*/
    a << x; /*No error*/
    a << y; /*No error*/

    a << 31; /*No error*/
    a << 32; /*Error*/
    a << 33; /*Error*/

    a << 31U; /*No error*/
    a << 32U; /*Error*/
    a << 33U; /*Error*/

    a << w + 1; /*No error*/
    a << b; /*No error*/

    return 0;
}

My question is this: why is it that an error is returned for a raw number, but not for any of the variables? They, I think, should be treated the same. According to the C11 standard

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.

The right side, since the left is unsigned type, should be 2^E2 reduced modulo one more than the maximum value representable in the result type.... That sentence isn't entirely clear to me, but in practice it seems that it is E1 << (E2%32) - despite that 32 is not the maximum representable in the result type. Regardless, it is not undefined for the C11 standard, yet the error

left shift count >= width of type [enabled by default]

shows up when trying to compile. I cannot deduce why it is that some values of >31 work (e.g. x = 33; a <

I am using the GCC compiler on 64-Bit Fedora. Thanks in advance. -Will

I imagine the reason the variable ones aren't giving any errors is because the compiler can't know for sure what the value will be at runtime — Drew McGowen
The C spec only defines minimums, it's possible int is a different size. — lrobb
Shifting a 32 bit integer, signed or unsigned, by a negative count or by a count >= 32, is undefined behaviour. Anything can happen. — gnasher729
As far as the "maximum value representable" spiel goes, it's referring to the result of the bit shift; in other words, it's (E1 x 2^E2)%(2^32), since the integers are only 32 bits, hence the warnings. — Drew McGowen

Pascal Cuoq Pascal Cuoq · Accepted Answer · 2014-04-27T12:32:16

My question is this: why is it that an error is returned for a raw number, but not for any of the variables?

Because absence of compiler warnings is not a guarantee of good program behavior. The compiler would be right to emit a warning for a << x, but it does not have to.

They, I think, should be treated the same

The compiler is doing you a favor when it warns for a << 33. It is not doing you any favor when it doesn't warn for a << y, but the compiler does not have to do you any favor.

If you want to be certain that your program does not contain undefined behavior, you cannot rely on the absence of compiler warnings, but you can use a sound static analyzer. If a sound static analyzer for undefined behavior does not detect any in your program, then you can conclude that it does not produce any (modulo the conditions of use that would be documented for the analyzer in question). For instance:

$ frama-c -val t.c ... t.c:13:[kernel] warning: invalid RHS operand for shift. assert 0 ≤ x < 32;

in practice it seems that it is E1 << (E2%32)

The reason you are seeing this is that this is the behavior implemented by the shift instructions in x86_64's instruction set. However, shifting by a negative number or by a number larger than the width of the type is undefined behavior. It works differently on other architectures, and even some compiler for your architecture may compute it at compile-time (as part of the constant propagation phase) with rules that differ from the one you have noticed. Do not rely on the result being E1 << (E2%32) any more than you would rely on memory still containing the correct results after being free()d.

Unsigned Bitwise Shift Operators [C11]

4 Answers