问题
Learning the hard way, I tried to left shift a long long and uint64_t to more than 32 bits on an x86 machine resulted 0. I vaguely remember to have read somewhere than on a 32 bit machine shift operators only work on the first 32 bits but cannot recollect the source.
I would like to know is if Shifting more than 32 bits of a uint64_t integer on an x86 machine is an Undefined Behavior?
回答1:
The standard says (6.5.7 in n1570):
3 The integer promotions are performed on each of the operands. The type of the result is that of the promoted left operand. 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.
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 × 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 × 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.
5 The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 / 2E2 . If E1 has a signed type and a negative value, the resulting value is implementation-defined.
Shifting a uint64_t a distance of less than 64 bits is completely defined by the standard.
Since long long must be at least 64 bits, shifting long long values less than 64 bits is defined by the standard for nonnegative values, if the result doesn't overflow.
Note, however, that if you write a literal that fits into 32 bits, e.g. uint64_t s = 1 << 32 as surmised by @drhirsch, you don't actually shift a 64-bit value but a 32-bit one. That is undefined behaviour.
The most common results are a shift by shift_distance % 32 or 0, depending on what the hardware does (and assuming the compiler's compile-time evaluation emulates the hardware semantics, instead of nasal demons.)
Use 1ULL < 63 to make the shift operand unsigned long long before the shift.
回答2:
The C standard requires the shift to work correctly. A particular buggy compiler might have the defect you describe, but that is buggy behaviour.
This is a test program:
#include <stdio.h>
#include <inttypes.h>
int main(void)
{
uint64_t x = 1;
for (int i = 0; i < 64; i++)
printf("%2d: 0x%.16" PRIX64 "\n", i, (x << i));
return 0;
}
This is the output on an i686 machine running RHEL 5 with GCC 4.1.2, and also on x86/64 machine (also running RHEL 5 and GCC 4.1.2), and on a x86/64 Mac (running Mac OS X 10.7.3 with GCC 4.7.0). Since that's the expected result, I conclude that there is no necessary problem on the 32-bit machine, and that GCC at least has not exhibited any such bug since GCC 4.1.2 (and probably never has exhibited such a bug).
0: 0x0000000000000001
1: 0x0000000000000002
2: 0x0000000000000004
3: 0x0000000000000008
4: 0x0000000000000010
5: 0x0000000000000020
6: 0x0000000000000040
7: 0x0000000000000080
8: 0x0000000000000100
9: 0x0000000000000200
10: 0x0000000000000400
11: 0x0000000000000800
12: 0x0000000000001000
13: 0x0000000000002000
14: 0x0000000000004000
15: 0x0000000000008000
16: 0x0000000000010000
17: 0x0000000000020000
18: 0x0000000000040000
19: 0x0000000000080000
20: 0x0000000000100000
21: 0x0000000000200000
22: 0x0000000000400000
23: 0x0000000000800000
24: 0x0000000001000000
25: 0x0000000002000000
26: 0x0000000004000000
27: 0x0000000008000000
28: 0x0000000010000000
29: 0x0000000020000000
30: 0x0000000040000000
31: 0x0000000080000000
32: 0x0000000100000000
33: 0x0000000200000000
34: 0x0000000400000000
35: 0x0000000800000000
36: 0x0000001000000000
37: 0x0000002000000000
38: 0x0000004000000000
39: 0x0000008000000000
40: 0x0000010000000000
41: 0x0000020000000000
42: 0x0000040000000000
43: 0x0000080000000000
44: 0x0000100000000000
45: 0x0000200000000000
46: 0x0000400000000000
47: 0x0000800000000000
48: 0x0001000000000000
49: 0x0002000000000000
50: 0x0004000000000000
51: 0x0008000000000000
52: 0x0010000000000000
53: 0x0020000000000000
54: 0x0040000000000000
55: 0x0080000000000000
56: 0x0100000000000000
57: 0x0200000000000000
58: 0x0400000000000000
59: 0x0800000000000000
60: 0x1000000000000000
61: 0x2000000000000000
62: 0x4000000000000000
63: 0x8000000000000000
回答3:
Daniel Fischer's answer answers the question about the C language specification. As for what actually happens on an x86 machine when you issue a shift by a variable amount, refer to the Intel Software Developer Manual Volume 2B, p. 4-506:
The count is masked to 5 bits (or 6 bits if in 64-bit mode and REX.W is used). The count range is limited to 0 to 31 (or 63 if 64-bit mode and REX.W is used).
So if you attempt to shift by an amount larger than 31 or 63 (for 32- and 64-bit values respectively), the hardware will only use the bottom 5 or 6 bits of the shift amount. So this code:
uint32_t RightShift(uint32_t value, uint32_t count)
{
return value >> count;
}
Will result in RightShift(2, 33) == 1 on x86 and x86-64. It's still undefined behavior according to the C standard, but on x86, if the compiler compiles it down to a sar instruction, it will have defined behavior on that architecture. But you should still avoid writing this sort of code that depends on architecture-specific quirks.
回答4:
No it is ok.
ISO 9899:2011 6.5.7 Bitwise shift operators
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.
That isn't the case here, so it is all fine and well-defined.
回答5:
Shifting by a number comprised between 0 and the predecessor of the width of the type does not cause undefined behavior, but left-shifting a negative number does. Would you be doing that?
On the other hand, right-shifting a negative number is implementation-defined, and most compilers, when right-shifting signed types, propagate the sign bit.
来源：https://stackoverflow.com/questions/10499104/is-shifting-more-than-32-bits-of-a-uint64-t-integer-on-an-x86-machine-undefined