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         


        

Answer 1

Understand that a[0] and b[0] in this case are themselves arrays (or pointers to characters arrays, to be more precise)

strcpy() would traverse through each individual element of the character array which a[0] points to. While the assignment b[0] = a[0] will just make b[0] point where a[0] is pointing, resulting in memory leak (the memory to which b[0] was pointing cannot be free'd).

Visualize the state of affairs using a simple figure like this -

     +---+ -->  +------+          +---+---+---+---+---+----+
     | a |      | a[0] | ------>  |'H'|'E'|'L'|'L'|'O'|'\0'| 
     +---+      +------+          +---+---+---+---+---+----+
                | a[1] |
                +------+
                | ...  |
                +------+    
                | a[n] |
                +------+