Thread: How to convert char array of 64 bytes to long long int

Here's an extension of the original program I gave. Choose an "endianness" and copy the key.ints into the rd_key of your struct.

Code:

#include <stdio.h>

//#define METHOD_ONE

int main() {
    char dek[] = "016211c7f60824a8ec5df54737a08ad0"
                 "a11b2e451160601a94ccd53555a335ab";
    union {
        unsigned char bytes[32];
        unsigned int  ints[8];
    } key;
    unsigned x;
    int i, j;

#ifdef METHOD_ONE
    for (i = 0; i < 32; i++) {
        sscanf(dek+i*2, "%2x", &x);
        key.bytes[i] = x;
    }
#else
    for (i = 0; i < 8; i++)
        for (j = 0; j < 4; j++) {
            sscanf(dek+(i*4+j)*2, "%2x", &x);
            key.bytes[i*4+3-j] = x;
        }
#endif

    // Print string
    printf("%s\n", dek);

    // Print bytes
    for (i = 0; i < 32; i++) printf("%02x", key.bytes[i]);
    printf("\n");

    // Print ints
    for (i = 0; i < 8; i++) printf("%08x", key.ints[i]);
    printf("\n");

    return 0;
}

i think i am getting it... will try and get back to the forum.. thanks a lot oogabooga and also to other users who have replied...

Actually, I've made the above code too complicated. You want something more like this:

Code:

void hexStringToUIntArray(char *str, unsigned *a) {
    int i;
    for (i = 0; i < 8; i++)
        sscanf(str + i * 8, "%8x", &a[i]);
}

int main() {
    char dek[] = "016211c7f60824a8ec5df54737a08ad0"
                 "a11b2e451160601a94ccd53555a335ab";
    unsigned a[8];
    int i;

    hexStringToUIntArray(dek, a);

    // Print string
    printf("%s\n", dek);

    // Print ints
    for (i = 0; i < 8; i++) printf("%08x", a[i]);
    printf("\n");

    return 0;
}