Shift operators

[vrakas]

Hello i am reading an article in topcoder and it says
"When executing shift instructions for a << b, the x86 architecture uses only the bottom 5 bits of b (6 for 64-bit integers). This means that shifting left (or right) by 32 does nothing, rather than clearing all the bits. "
He means that you can shift at most with 31?
Thanks

[Cubbi]

The C++ a << b is not defined as "execute the CPU instruction SHL", what that article says is irrelevant.

It is defined as "calculate a*2^b" and, for unsigned types, if the result does not fit the result type, "reduce the result modulo maximum value of the return type plus 1"

For 32-bit unsigned types, shifting left by 32 results in zero: http://ideone.com/pM8xtr

Oh, right, Peter87 is correct: modulo reduction only applies to overflows when right operand is less than 32 in this case

[Peter87]

The standard says this about the shift operators.
§5.8/1

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.