string - Difference between using strcpy() and copying the address of a the char* in C

Question

I have two dynamically allocated arrays. c

char **a = (char**)malloc(sizeof(char*) * 5));
char **b = (char**)malloc(sizeof(char*) * 5));

for (int i = 0; i < 7, i++) {
  a[i] = (char*)malloc(sizeof(char)*7);
  b[i] = (char*)malloc(sizeof(char)*7);
}

If a[0] was "hello" and I wanted to copy a[0] to b[0], would strcpy and pointer assignment be the same thing? For example:

1. strcpy(b[0], a[0])
2. b[0] = a[0]

Would these both do the same exact thing?

Answer 1

NO. Both are not same. In this case, strcpy(b[0], a[0]) is correct way to copy the string pointed by a[0] to b[0].

In case of b[0] = a[0], memory allocated to b[0] will lost and it will cause memory leak. Now freeing both of a[0] and b[0] will invoke undefined behavior. This is because both of them are pointing to same memory location and you are freeing same allocated memory twice.

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NOTE: It should be noted that, as Matt McNabb pointed in his comment, memory leak does not invokes undefined behavior.