Return the contiguous block in c - c

Question

I create an array (char *charheap;) of length 32 bytes in the heap, and initialize all the elements to be \0. Here is my main function:
int main(void) {
char *str1 = alloc_and_print(5, "hello");
char *str2 = alloc_and_print(5, "brian");
}
char *alloc_and_print(int s, const char *cpy) {
char *ncb = char_alloc(s);// allocate the next contiguous block
if (ret == NULL) {
printf("Failed\n");
} else {
strcpy(ncb, cpy);
arr_print();// print the array
}
return ncb;
}
Here is what I implement:
/char_alloc(s): find the FIRST contiguous block of s+1 NULL ('\0')
characters in charheap that does not contain the NULL terminator
of some previously allocated string./
char *char_alloc(int s) {
int len = strlen(charheap);
for (int i = 0; i < len; i++) {
if (charheap[0] == '\0') {
char a = charheap[0];
return &a;
} else if (charheap[i] == '\0') {
char b = charheap[i+1];
return &b;
}
}
return NULL;
}
Expected Output: (\ means \0)
hello\\\\\\\\\\\\\\\\\\\\\\\\\\\
hello\brian\\\\\\\\\\\\\\\\\\\\\
This solution is completely wrong and I just print out two failed. :(
Actually, the char_alloc should return a pointer to the start of contiguous block but I don't know how to implement it properly. Can someone give me a hint or clue ?

Answer 1

Your function is returning a pointer to a local variable, therefore the caller receives a pointer to invalid memory. Just return the pointer into the charheap, which is what you want.
return &charheap[0]; /* was return &a; which is wrong */
return &charheap[i+1]; /* was return &b; which is wrong */
Your for loop uses i < len for the terminating condition, but, since charheap is \0 filled, strlen() will return a size of 0. You want to iterate through the whole charheap, so just use the size of that array (32 in this case).
int len = 32; /* or sizeof(charheap) if it is declared as an array */
The above two fixes should be enough to get your program to behave as you expect (see demonstration).
However, you do not place a check to make sure there is enough room in your heap to accept the allocation check. Your allocation should fail if the distance between the start of the available memory and the end of the charheap is less than or equal to the desired size. You can enforce this easily enough by setting the len to be the last point you are willing to check before you know there will not be enough space.
int len = 32 - s;
Finally, when you try to allocate a third string, your loop will skip over the first allocated string, but will overwrite the second allocated string. Your loop logic needs to change to skip over each allocated string. You first check if the current location in your charheap is free or not. If it is not, you advance your position by the length of the string, plus one more to skip over the '\0' terminator for the string. If the current location is free, you return it. If you are not able to find a free location, you return NULL.
char *char_alloc(int s) {
int i = 0;
int len = 32 - s;
while (i < len) {
if (charheap[i] == '\0') return &charheap[i];
i += strlen(charheap+i) + 1;
}
return NULL;
}