CODEWITHSUNDEEP
cmalloc
Sae
Sae

Reputation: 79

the function of malloc(using malloc correctly)

so I'm quite new in this, sorry if it sound like a dumb question

I'm trying to understand malloc, and create a very simple program which will print "ABC" using ASCII code

here is my code (what our professor taught us) so far 

char *i;
i = malloc(sizeof(char)*4); 
*i = 65;
*(i+1) = 66;
*(i+2) = 67;
*(i+3) = '\0';

what I don't understand is, why do I have to put malloc there? the professor told us the program won't run without the malloc, but when I tried and run it without the malloc, the program run just fine. so what's the function of malloc there? am I even using it right?

any help and or explanation would be really appreciated

Upvotes: 0

Views: 528

Answers (3)

LLL
LLL

Reputation: 117

You have to allocate space for memory. In the example below, I did not allocate for memory for i, which resulted in a segmentation fault (you are trying to access memory that you don't have access to)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
  char *i;
  strcpy(i, "hello");
  printf("%s\n", i);
  return (0);
}

Output: Segmentation fault (core dumped)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
  char *i;
  /*Allocated 6 spots with sizeof char +1 for \0 character*/
  i = malloc(sizeof(char) * 6);
  strcpy(i, "hello");
  printf("%s\n", i);
  return (0);
}

Result: hello

Malloc allows you to create space, so you can write to a spot in memory. In the first example, "It won't work without malloc" because i is pointing to a spot in memory that doesn't have space allocated yet.

Upvotes: 1

user2371524
user2371524

Reputation:

the professor told us the program won't run without the malloc

This is not quite true, the correct wording would be: "The program's behavior is undefined without malloc()".

The reason for this is that

char *i;

just declares a pointer to a char, but there's no initialization -- this pointer points to some indeterminate location. You could be just lucky in that writing values to this "random" location works and won't result in a crash. I'd personally call it unlucky because this hides a bug in your program. undefined behavior just means anything can happen, including a "correct" program execution.

malloc() will dynamically request some usable memory and return a pointer to that memory, so after the malloc(), you know i points to 4 bytes of memory you can use. If malloc() fails for some reason (no more memory available), it returns NULL -- your program should test for it before writing to *i.

All that said, of course the program CAN work without malloc(). You could just write

char i[4];

and i would be a local variable with room for 4 characters.

Final side note: sizeof(char) is defined to be 1, so you can just write i = malloc(4);.

Upvotes: 4

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726479

Unfortunately, "runs fine" criterion proves nothing about a C program. Great deal of C programs that run to completion have undefined behavior, which does not happen to manifest itself on your particular platform.

You need special tools to see this error. For example, you can run your code through valgrind, and see it access uninitialized pointer.

As for the malloc, you do not have to use dynamic buffer in your code. It would be perfectly fine to allocate the buffer in automatic memory, like this:

char buf[4], *i = buf;

Upvotes: 3

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