C left shift on 64 bits fail

Question

I have this code in C (it\'s for study only):

    char x;
    uint64_t total = 0;

    for(x = 20; x < 30; x++){
        total = (((((1 << x) * x) /         


        

Answer 1

Because 1 is an int, 32 bits, so (1 << 27)*27 overflows. Use 1ull.

Regarding your comment, if x is a uint64_t, then 1 << x is still an int, but for the multiplication it would be cast to uint64_t, so there'd be no overflow. However, if x >= 31, 1 << x would be undefined behaviour (as the resulting value cannot be represented by a signed 32 bit integer type).

Answer 2

I guess your problem is, you calculate with 32bit and assign it later to a 64 bit value

division by 64 is the same as not shift 6 bit

char x;
uint64_t one = 1;
uint64_t total = 0;

for(x = 20; x < 30; x++){
    total = ((((one << (x - 6)) * x) + 1) * sizeof(uint64_t));
    printf("%d - %llu\n", x, total);        
}

not compiled yet