Question

What is wrong with the following code fragment?

char *src = "hello";

char *dst = (char *) malloc(strlen(src)); //too small!

strcpy (dst, src); //work properly

Answer 1

char *src = "hello";
char *dst = (char *) malloc(strlen(src)); //too small!      <-  this line is the issue.
strcpy (dst, src); //work properly

strlen(src) gives a value of 5. because there are 5 characters in string "hello"
but there is also a hidden character of '\0', which indicates the end of a string.
so, to copy the string src into dst properly, we need to allocate one extra character to hold the end of the string character.
so, line 2 should be changed to the code below.
char *dst = (char *) malloc(1+strlen(src));

Modified code:
---------------
char *src = "hello";
char *dst = (char *) malloc(1 + strlen(src));
strcpy (dst, src);