free dynamically allocated memory

Question

int i;
char *s = (char*)malloc(sizeof(char)*10);
for(i = 0; i <= 4; ++i)
    s[i] = 'a';
s[5] = '\0';
printf("%s\n", s);
free(s).

Will the above code have memory leak problem? How does the function "free" know how many memory need to be freed?

Answer 1

There is no memory leak in your code. As to your second question, when you call malloc, more happens than just returning memory to you. The C library reserves some space to put its own headers and bookkeeping information so that when you call free it can do the right thing.

Some editorial notes about your code:

• you don't need to cast the return value of malloc() in a C program. Conversions between void * (which malloc() returns) and other pointer types are implicit in C.

• sizeof(char) is 1, why bother writing it out?

• your loop writes three characters into s, and then your program skips a character (s[4]) before adding the \0. That's a bit weird. Did you mean to use i <= 4 in your loop, or maybe s[4] = '\0' after it?

• you have a . rather than a ; after your free() call. I guess that's just a typo here, rather than in your program, since it won't compile that way, though.