shifting an unsigned int by 32bits to the left gives bad result

[samup]

When I try to shift a 32bit unsigned integer value to the left by 32 bits, I should have a result value of 0. But instead of that, I get back the original content of the shifted variable.

Example code:

unsigned int uiVal = 0x01;
unsigned int uiResult;

    uiResult = uiVal << 32;      //uiResult should have the value of 0, but I get a value of 1.

How it can be?
   

[samup]

I have found the answer in ANSI C specification, so sorry for bothering you.
So the answer is:

The result of the shift operation is undefined if the right operand is negative or if its value is greater than the number of bits in an int .

[dilution]

1) a << b is equivalent to a*(2^b). I don't see why the result shoud be zero.
2) to me, the result is undefined because as far as I remember, one can not shift by a number of bit greater or equal to the number of bit of the left expression. Unsigned int is 32 bit no?
BR