Shifting 64 bit value left by 64 bits in C++ giving weird results [duplicate]

Question

Possible Duplicate:
64bit shift problem

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

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

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.

Answer 1

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

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

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.