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

They are different. In strcpy() case you are copying the characters, in the assignment case you adjust the pointer to point to the same array.

 a[0] -> allocated memory containing "hello"
 b[0] -> allocated memory not yet initialised

After b[0] = a[0]

both now point to same memory

 a[0] -> allocated memeory containing "hello"
 b[0] ---^ 

and worse nothing now points to b[0]'s

 allocated memory not yet initialised

so you can never free it; memory leak.