Simple C malloc question that prints garbage on multiple invocations [closed]

I'm refreshing my C skills (been decades) and ran across something I don't quite understand. I'm working on some code that involves a lot of bit shifting, masking, etc. I have one function that creates a string with the binary representation of a number. I've found it's a lot more complicated than it needs to be so I've adjusted it, but here is the original code below.

Note: The caller is responsible for freeing memory, so I've taken care of that. This is running on Linux x64 in Eclipse C/C++ IDE.

char* get_bit_str(int num, int num_bits) {

    // XXX: This no worky worky...
    // When having multiple malloc'd string it eventually
    // get garbage.  Not too sure why...

    num_bits = num_bits > 0 ? num_bits : sizeof(int);

    char* buf = (char*)malloc(num_bits + 1);

    if (!buf) {
        printf("\nMemory allocation failed !\n");
        return "";
    }

    // Start at least significant bit.
    for (int i = num_bits - 1; i >= 0; i--) {
        char* bit = (num & (1 << i)) ? "1" : "0";

//      strncat(buf, bit, 1);
        strcat(buf, bit);
    }
//  strncat(buf, "\0", 1);
//  strcat(buf, "\n");

    return buf;
}

This works fine most of the time, but for one specific example, I get garbage. I've tried multiple approaches like overwriting the same char* or assigning a new one, but the result is always same. I also tried:

char* buf = (char*)malloc((num_bits + 1) * sizeof(char));

I've used strcpy and strncpy as well, but they produce the same issue. Both strcpy and strncpy append a null terminator, which might be an issue. Normally, the output is ok when I:

1. Call the function

2. Print the result

3. Free the char* pointer

However, if I:

1. Call the function with two different values (e.g., 1 and then 2)

2. Print out both char* pointers

3. Free both char* pointers

I get garbage for the second one using this test code (Like I say, I've tried to assign new variables but here I'm reusing the same "c" and "d"):

void test_get_bit_str() {
    char* c = get_bit_str(0, 4);
    printf("c (0) %s\n", c);

    char* d = get_bit_str(1, 4);
    printf("d (1) %s\n", d);

    free(c);
    free(d);

    c = get_bit_str(1, 4);
    printf("c (1) %s\n", c);

    d = get_bit_str(2, 4);
    printf("d (2) %s\n", d);

    free(c);
    free(d);
}

The output is:

c (0) 0000
d (1) 0001
c (1) ���D�U0001
d (2) 0010

You can see on the second c is garbage. I printed the pointer addresses, and it is reusing the same memory addresses after I free everything, which is fine, I think.

I really don't understand why this works when malloc and free is done for individual invocations but when I have two pointers defined (malloc twice, print out results, free both pointers), it's doing this.

Now, having stated all of that, I've made it a lot less complex with the following, which does work:

char* get_bit_str(int num, int num_bits) {
    num_bits = num_bits > 0 ? num_bits : sizeof(int);

    // Allocate enough memory for the number of bits including room for NULL.
    char* buf = (char*)malloc(num_bits + 1);

    if (!buf) {
        printf("\nMemory allocation failed !\n");
        return NULL;
    }

    *buf = 0;

    for (int i = 0; i < num_bits; i++) {
        char bit = (num & (1 << i)) ? '1' : '0';;

        // Here we want to but the least significant bit at the other end of
        // the array.  This is effectively swapping the array positions.
        buf[num_bits - 1 - i] = bit;
    }

    // Remember to append the NULL char.
    buf[num_bits] = '\0';

    return buf;
}

Any help understanding what's going on would be greatly appreciated.