how to use uint64_t in C

Question

#include <stdio.h>
#include <stdint.h>

int main(){
    uint64_t a = 1 << 63;
    /* do some thing */
    return 0;
}

$ gcc -Wall -Wextra -std=c99 test.c -o test  
warning: left shift count >= width of type [-Wshift-count-overflow]

Q: uint64_t should have 64 bits width, why the left shift operation overflows?

@LưuVĩnhPhúc C specifies `1` as an _integer constant_ of type `int`. The only _literals_ specified in C are _string literals_ and _compound literals_. `1` is neither of those 2 literals. — chux - Reinstate Monica, Mar 28 '17 at 04:17
"why the left shift operation overflows?" --> which happens first: `1 << 63` or its assignment to `uint64_t a`? Since `1 << 63` occurs first, the type it is assigned to is irrelevant to `1 << 63` evaluation. — chux - Reinstate Monica, Mar 28 '17 at 04:24

user253751 · Answer 1 · 2017-03-28T04:21:21.523

11

1 is an int which is either only 32 bits on your platform, or it could be 64 bits but signed.

Use (uint64_t)1 << 63 to cast 1 to 64-bit unsigned integer first. (Or ((uint64_t)1) << 63 if you prefer)

edited Mar 28 '17 at 04:21

answered Mar 28 '17 at 03:58

user253751

1

no need to cast, `1ULL` looks better. And `(uint64_t)1 << 64` invokes undefined behavior because you shifted more than the value's bit length – phuclv Mar 28 '17 at 04:16
@LưuVĩnhPhúc Fixed, 64 should've been 63. – user253751 Mar 28 '17 at 04:22

1 Answers1