Using sizeof with a dynamically allocated array

gcc 4.4.1 c89

I have the following code snippet:

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

 char *buffer = malloc(10240);
 /* Check for memory error */
 if(!buffer)
 {
    fprintf(stderr, "Memory error\n");
    return 1;
 }
 printf("sizeof(buffer) [ %d ]\n", sizeof(buffer));

However, the sizeof(buffer) always prints 4. I know that a char* is only 4 bytes. However, I have allocated the memory for 10kb. So shouldn't the size be 10240? I am wondering am I thinking right here?

Many thanks for any suggestions,

You are asking for the size of a char* which is 4 indeed, not the size of the buffer. The sizeof operator can not return the size of a dynamically allocated buffer, only the size of static types and structs known at compile time.

sizeof doesn't work on dynamic allocations (with some exceptions in C99). Your use of sizeof here is just giving you the size of the pointer. This code will give you the result you want:

char buffer[10240];
printf("sizeof(buffer) [ %d ]\n", sizeof(buffer));

If malloc() succeeds, the memory pointed to is at least as big as you asked for, so there's no reason to care about the actual size it allocated.

Also, you've allocated 10 kB, not 1 kB.

It is up to you to track the size of the memory if you need it. The memory returned by malloc is only a pointer to "uninitialized" data. The sizeof operator is only working on the buffer variable.