Question

# include <stdio.h>
# include <stdbool.h>
# include <string.h>
# include <stdlib.h>

int main ()
{
  char * buffer;
  buffer = malloc (2);

  if (buffer == NULL){
    printf("big errors");
  }
  strcpy(buffer, "hello");
  printf("buffer is %s\n", buffer);
  free(buffer);
  return 0;
}

I allocated 2 bytes of memory to the pointer/char buffer yet if I assign the C-style string hello to it, it still prints the entire string, without giving me any errors. Why doesn’t the compiler give me an error telling me there isn’t enough memory allocated? I read a couple of questions that ask how to check how much memory malloc actually allocates but I didn’t find a concrete answer. Shouldn’t the free function have to know exactly how much memory is allocated to buffer?

Answer 1

The compiler doesn’t know. This is the joy and terror of C. malloc belongs to the runtime. All the compilers knows is that you have told it that it returns a void*, it has no idea how much, or how much strcpy is going to copy.

Tools like valgrind detect some of these errors. Other programming languages make it harder to shoot yourself in the foot. Not C.