在 C++ 中将 64 位值左移 64 位给出奇怪的结果 [英] Shifting 64 bit value left by 64 bits in C++ giving weird results

问题描述

可能的重复:
64 位移位问题

我在 Windows 8 64 位上使用 Visual Studio 2012，在调试模式下针对 x64，使用 AMD Phenom II.
所以基本上...

I'm using Visual Studio 2012 on Windows 8 64-bit, targeting x64 in debug mode, using an AMD Phenom II.
So Basically...

uint64_t Foo = 0xFFFFFFFFFFFFFFFF << 64;//Foo is now 0x0000000000000000
uint64_t Derp = 64;
uint64_t Bar = 0xFFFFFFFFFFFFFFFF << Derp;//Foo is now 0xFFFFFFFFFFFFFFFF

使用较低的值(例如 63)可恢复正常行为.
为什么会发生这种情况，我该如何解决?

Using a lower value such as 63 restores normal behavior.
Why is this happening and how can I get around it?

更新:我切换到发布模式.瞧，问题消失了，都返回了 0.但问题仍然处于调试模式，我需要在调试模式下调试我的代码.

Update: I switched to release mode. Lo and behold, the issue vanished and both returned 0. But the issue remains in debug mode which is where I need to be in order to debug my code.

推荐答案

如果移位大于或等于位宽的值，则移位操作具有未定义的行为.

Shift operation has undefined behavior if you shift by values greater than or equal to the bit width.

来自 C++11 草案中的第 5.8 p1 节:

From Section 5.8 p1 in the C++11 draft:

操作数应为整数或无作用域枚举类型，并执行整数提升.结果的类型是提升的左操作数的类型.如果正确的操作数，则行为未定义为负，或大于或等于提升的左操作数的位长度.

The operands shall be of integral or unscoped enumeration type and integral promotions are performed. The type of the result is that of the promoted left operand. The behavior is undefined if the right operand is negative, or greater than or equal to the length in bits of the promoted left operand.

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