CODEWITHSUNDEEP
cpointerscharmalloc
VMi
VMi

Reputation: 366

Why char pointer saving data more than allocated memory in C?

While working on dynamic memory allocation in C, I am getting confused when allocating size of memory to a char pointer. While I am only giving 1 byte as limit, the char pointer successfully takes input as long as possible, given that each letter corresponds to 1 byte.

Also I have tried to find sizes of pointer before and after input. How can I understand what is happening here? The output is confusing me.

Look at this code:

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

int main()
{
    int limit;

    printf("Please enter the limit of your string - ");
    gets(&limit);

    char *text = (char*) malloc(limit*4);

    printf("\n\nThe size of text before input is %d bytes",sizeof(text));

    printf("\n\nPlease input your string - ");
    scanf("%[^\n]s",text);

    printf("\n\nYour string is %s",text);

    printf("\n\nThe size of char pointer text after input is %d bytes",sizeof(text));

    printf("\n\nThe size of text value after input is %d bytes",sizeof(*text));

    printf("\n\nThe size of ++text value after input is %d bytes",sizeof(++text));

    free(text);



    return 0;


}

Check this output:

enter image description here

Upvotes: 0

Views: 1113

Answers (2)

Mostafa
Mostafa

Reputation: 468

It works because malloc usually doesn't allocate the same number of bytes you pass to it. It reserves memory multiple of "blocks". It usually reserve more memory to "cache" it for next malloc calls as an optimization. (it is an implementation specific) check glibc malloc internals for example.

Using more memory than allocated by malloc is an undefined behavior. you may overwrite metadata of malloc saved on heap or corrupt other data.

Also I have tried to find sizes of pointer before and after input. How can I understand what is happening here? The output is confusing me.

The size of pointer is fixed for all pointer types in a machine, it is usually 4/8 bytes depending on the address size. It doesn't have anything to do with data size.

Upvotes: 5

Serge Ballesta
Serge Ballesta

Reputation: 149145

Welcome to the world of Undefined Behaviour!

char *text = malloc(limit*4); (don't cast malloc in C) will make text point the the first element of an array of size limit*4.

C will not prevent you to write past the end of any array, simply the behaviour is undefined by the standard. It may work fine, or it may crash immediately, or you may experience abnormal behaviour later in the program.

Here, the underlying system call has probably allocated a full memory page (often 4k), and as you have not used another malloc you have just used a memory belonging to the process but still officially unused. But do not rely on that and never use it in production code.

And sizeof does not make sense with pointers. sizeof(text) is sizeof(char *) (same for sizeof(++text) for same reason) and is the size of a pointer (generaly 2, 4 or 8 bytes) and sizeof(*text) is sizeof(char) which by definition is 1.

C is confident that you as the programmer know how much memory you have asked, and will not try to use more. Anything can happen if you do (including expected result) but do not blame the language or the compiler if it breaks: only you will be guilty.

Upvotes: 3

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