Cannot understand shift operator behavior in C code -

Cannot understand shift operator behavior in C code -

look @ sample c code (extracted test case example):

main() { unsigned long a, b; int c; c = 32; = 0xffffffff << 32; b = 0xffffffff << c; printf ("a=%x, b=%x\n", a, b); }

prints: a=0, b=ffffffff

i cannot understand why b not zero, a. tested on microsoft c , gcc.

update: fixed stupid typo (should have been << c , not << b of course). question still stands, e.g. result still same.

in c undefined behavior left shift more width of promoted operand.

(c99, 6.5.7p3) "if value of right operand negative or greater or equal width of promoted left operand, behavior undefined."

the c rules types of integer constant different between c89 , c99.

i assume in examples below int , unsigned int types 32-bit.

in c99

the type of unsuffixed hexadecimal integer constant first in corresponding list in value can represented: int, unsigned int, long, unsigned long, long long, unsigned long long.

so here 0xffffffff of type unsigned int , 0xffffffff << 32 undefined behavior.

in c89

the type of unsuffixed hexadecimal integer constant first in corresponding list in value can represented: int, long, unsigned long

0xffffffff of type unsigned long, if unsigned long 32-bit , of type long if long 64-bit (or larger).

and 0xffffffff << 32 undefined behavior, if unsigned long , long 32-bit.

c bit-shift