h101 h101 - 18 days ago 4
C Question

How to set/flip bitfields in C

I am trying to have a

uint64_t
bitfield be all set to 0. Then when I call the function within the given string, and it matches with the static global array that I have set it will flip the bit to 1. Currently I have the following code but for some reason when giving it different strings it is following the same behavior. So for example when I input the following string of "ABC" it should be printing out
111000
. How would I get the following behavior.

const size_t SETSIZE = sizeof(uint64_t) << 3;
char key[5] = { 'A', 'B', 'C', 'D', 'E', 'F' }

uint64_t set_encode(char *st) {
int i, j;
uint64_t set = 0;
int length = strlen(st);
for (i = 0; i < length; i++) {
for (j = 0; j < 5; j++) {
if (st[i] == key[j]) {
printf("%c", st[0]);
set = set | 1 << (SETSIZE - 1 - i);
}
}
}
printf("%lu\n", set);
return set;
}

Answer

Multiple problems in your code:

const size_t SETSIZE = sizeof(uint64_t) << 3;

Bytes might not be 8 bits, you should use const size_t SETSIZE = 64; since the type uint64_t, if present, is defined to be exactly 64-bit wide with 2's complement representation.

char key[5] = { 'A', 'B', 'C', 'D', 'E', 'F' }

The initializer has 6 characters but the explicit size is set to 5. Use char key[] = { 'A', 'B', 'C', 'D', 'E', 'F' };. Note that key is not a C string as not '\0' is present in the initializer. Note also that you have a missing ; at the end of the initializer.

uint64_t set_encode(char *st) {

You do not modify the string pointed to by st, use const char *st

    int i, j;

i and j should be defined as size_t for consistency with length.

    uint64_t set = 0;
    int length = strlen(st);

The return type is size_t because the length of the string might be larger than the range of int. In your particular case it is not fundamental as the function is only useful for strings with at most 64 characters, and you should also test for that.

    for (i = 0; i < length; i++) {
        for (j = 0; j < 5; j++) {

j should be compared to sizeof(key).

            if (st[i] == key[j]) {
                printf("%c", st[0]);

You probably want to print st[i] instead of st[0].

                set = set | 1 << (SETSIZE - 1 - i);

It would probably be more consistent to use the bits from the lowest value to the highest value, and 1 must be cast to (uint64_t) to avoid arithmetic overflow on int for strings longer than 31 characters (if int is 32-bit wide): set = set | (uint64_t)1 << i;. Note however that even with the cast, the shift operation is still undefined for shift amounts larger than 63 or negative.

            }
        }
    }
    printf("%lu\n", set);

set is not necessarily a long. You can print it as an unsigned long long that is at least 64-bit wide: printf("%llu\n", (unsigned long long)set); Or you can use the format specifiers from <inttypes.h>: printf("%"PRIu64"\n", set);

    return set;
}

Here is corrected version:

const size_t SETSIZE = 64;
char key[] = { 'A', 'B', 'C', 'D', 'E', 'F' };

uint64_t set_encode(const char *st) {
    uint64_t set = 0;
    size_t length = strlen(st);

    if (length > SETSIZE) {
        printf("string too long: %zd bytes\n", length);
        length = SETSIZE;
    }

    for (size_t i = 0; i < length; i++) {
        for (size_t j = 0; j < sizeof(key); j++) {
            if (st[i] == key[j]) {
                printf("%c", st[i]);
                set |= (uint64_t)1 << i;
            }
        }
    }
    printf("%llu\n", (unsigned long long)set);
    return set;
}