将char *数组中的字符串分配给另一个char *数组

Question

I am trying to convert the hex values in array a to binary values and assign converted values to array b then print the array b. But all values in array b are same. The output is:

 111100001011000100010111101010001101 111100001011000100010111101010001101 111100001011000100010111101010001101 

If I use b[i] = strdup(hexToBin(a[i])); instead of b[i] = hexToBin(a[i]); the output will be:

 111100001011 111100001011000100010111 111100001011000100010111101010001101 

Is this something about pointers? Char * is a pointer which points first character of the string and are all characters after the pointer printed? What is right way of doing this?

#include <stdio.h>
#include <string.h>

char bin[100] = "";

char * hexToBin(char  hex[50]);

int main(void) {

    char * a[] = {
        "f0b",
        "117",
        "a8d",
    };

    char * b[3];

    for(int i = 0; i < 3; i++) {
        b[i] = hexToBin(a[i]);
    }

    for(int i = 0; i < 3; i++) {
        printf("%s\n", b[i]);
    }

}

char * hexToBin(char  hex[50]) {    

    for(int i=0; hex[i]!='\0'; i++)
    {
        switch(hex[i])
        {
            case '0':
                strcat(bin, "0000");
                break;
            case '1':
                strcat(bin, "0001");
                break;
            case '2':
                strcat(bin, "0010");
                break;
            case '3':
                strcat(bin, "0011");
                break;
            case '4':
                strcat(bin, "0100");
                break;
            case '5':
                strcat(bin, "0101");
                break;
            case '6':
                strcat(bin, "0110");
                break;
            case '7':
                strcat(bin, "0111");
                break;
            case '8':
                strcat(bin, "1000");
                break;
            case '9':
                strcat(bin, "1001");
                break;
            case 'a':
            case 'A':
                strcat(bin, "1010");
                break;
            case 'b':
            case 'B':
                strcat(bin, "1011");
                break;
            case 'c':
            case 'C':
                strcat(bin, "1100");
                break;
            case 'd':
            case 'D':
                strcat(bin, "1101");
                break;
            case 'e':
            case 'E':
                strcat(bin, "1110");
                break;
            case 'f':
            case 'F':
                strcat(bin, "1111");
                break;
            default:
                printf("Invalid hexadecimal input.");
        }
    }
    return bin;
}

Answer 1

The hexToBin function returns a pointer to the first element of the global bin array. Everytime! .

That means all pointers in b will be the very same pointer to the very same first element of the bin array.

If you know the maximum length of the strings, I recommend that you make b an array of arrays of char . For example

char b[3][500];  // 3 arrays of 499-character strings (+1 for the null-terminator)

Then instead of hexToBin returning a pointer to a single global array, pass a pointer to the string to be filled as argument to hexToBin :

void hexToBin(char *hex, char *bin);

and call it as

hexToBin(a[i], b[i]);

Answer 2

You only have one bin . What your hexToBin does is appending to that one bin and then returning that bin . In other words, when you call it multiple times, the result is always the same pointer, because you always return bin; .

So if you do this:

b[i] = hexToBin(a[i]);

Then in the end, all elements of b are pointing to bin , that's why you get the same output when you print them. If you do this instead:

b[i] = strdup(hexToBin(a[i]));

Then the result is not the same, because they all don't get bin assigned, but a copy of what bin has been at that time. That's why the results are different. So b[0] points to one copy, then bin is appended to again, but that doesn't change the b[0] copy.

If you use strdup , don't forget to free the memory it allocated:

for(int i = 0; i < 3; i++) {
    free(b[i]);
}

Answer 3

[My answer is mistaken. I leave it posted here for reference but the other answers are preferable.]

Your code looks pretty good for beginner's code and I like your style. I especially like this line:

    char * b[3];

Unfortunately, for this particular application, you must replace this with a less elegant line like

    char b[3][5];

The former line reserves no storage for your output. The latter line reserves five bytes per hex digit. You need storage, somehow.