c - Unexpected output when executing left-shift by 32 bits

Question

When I do a left shift of a hex I get -1 as output with the following code:

unsigned int i,j=0;
i= (0xffffffff  << (32-j));
printf("%d",i);

Similarly when I changed the shift value to 32, the output is 0, but I get warnings from compiler as (left shift count >= width of type)

unsigned int i,j=32;
i= (0xffffffff  << (32));
printf("%d",i);

I was expecting the same results in both the cases (ie, 0), but got confused why is displaying wrong output in case #1, and in case #2 the result is correct but the compiler warns!

The result is same in 32 and 64 bit x86 machines.

Can someone explain the results above?

Answer 1

It's undefined behavior to left-shit 32 or greater on a 32-bit integer. That's what the error is about.

C11 6.5.7 Bitwise shift operators

The integer promotions are performed on each of the operands. The type of the result is that of the promoted left operand. If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand, the behavior is undefined.