Why doesn't left bit-shift, "<<", for 32-bit integers work as expected when used more than 32 times?

Question

When I write the following program and use the GNU C++ compiler, the output is 1 which I think is due to the rotation operation performed by the compiler.

#include <iostream>

int main()
{
    int a = 1;
    std::cout << (a << 32) << std::endl;

    return 0;
}

But logically, as it's said that the bits are lost if they overflow the bit width, the output should be 0. What is happening?

The code is on ideone, http://ideone.com/VPTwj.

Answer 1

In C++, shift is only well-defined if you shift a value less steps than the size of the type. If int is 32 bits, then only 0 to, and including, 31 steps is well-defined.

So, why is this?

If you take a look at the underlying hardware that performs the shift, if it only has to look at the lower five bits of a value (in the 32 bit case), it can be implemented using less logical gates than if it has to inspect every bit of the value.

Answer to question in comment

C and C++ are designed to run as fast as possible, on any available hardware. Today, the generated code is simply a ''shift'' instruction, regardless how the underlying hardware handles values outside the specified range. If the languages would have specified how shift should behave, the generated could would have to check that the shift count is in range before performing the shift. Typically, this would yield three instructions (compare, branch, shift). (Admittedly, in this case it would not be necessary as the shift count is known.)