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cstrtol

C warning: passing argument 2 of ‘strcpy’ makes pointer from integer without a cast

EXPLANATION:

simply trying to convert a a char to hexadecimal but i keep getting this error and I'm not sure how to get around this

PROBLEM:

warning: passing argument 1 of ‘strtol’ makes pointer from integer without a cast[cs214111@cs lab3]$ vi lab3.c

CODE:

void print_group(char array[])
{
    int num,a;
    char ch[10];

    printf("here ");    
    for (a = 0 ; a < 8 ; a++)
    {
        strcpy(ch,array[3]);  
        num = strtol(ch,0,16);//------------------THIS IS IT//
        printf("%i",num);
    }   
}
Solution

  • You are passing char where char * is expected, maybe

    strcpy(ch, &array[3]);

    But from your code it would seem like this, is what you actually need

    num = strtol(&array[3], 0, 16);

    if strcpy() works in this case, then this will work.