Bitwise shift for unsigned long long type

Question

In a c program. I am trying to use the left shift operator on uint64_t variable.

E.g.

 // with shift of 24 bits uint64_t x = 0; x = (((uint64_t)76) << (24)); Output is: x = 1275068416 --------------------------------------------- // with shift of 32 bits uint64_t x = 0; x = (((uint64_t)76) << (32)); Output is: x = 0

If I perform left shift till 24 bits then it works fine, but at 32 bits it outputs 0. Whereas what I think is as the size of uint64_t i.e. unsigned long long is 64 bits. So shouldn't it work till the 64 bit shift ?

I am just printing out the value of x using printf as printf("....%d\n", x); — TechJ
– TechJ, Commented Feb 25, 2016 at 18:07
Ah, that's just that. To print a value of type uint64_t you need printf("%"PRIu64"\n", x);. — ach
– ach, Commented Feb 25, 2016 at 18:09
@EugeneSh. and Andrey Chernyakhovskiy ... thanks it works as suggested by both of u :-) — TechJ
– TechJ, Commented Feb 25, 2016 at 18:11

dbush · Accepted Answer · 2016-02-25 18:11:10Z

You're using the wrong format specifier to print the output. The %d format specifier expects an int, which apparently is 32-bit on your system. So passing a 64-bit value (and an unsigned one at that) leads to undefined behavior.

You should use the PRIu64 macro to get the correct format specifier for an unsigned 64-bit value.

printf("%"PRIu64"\n", x);

%lu is solely reserved for unsigned long. On a Windows machine that's typically 32 bits.

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