CODEWITHSUNDEEP
cmallocrealloc
user1807181
user1807181

Reputation: 3

Why won't realloc work in this example?

My professor gave us an "assignment" to find why realloc() won't work in this specific example. I tried searching this site and I think that it won't work because there is no real way to determine the size of a memory block allocated with malloc() so realloc() doesn't know the new size of the memory block that it needs to reallocate.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <windows.h>
int main ()
{
    MEMORYSTATUS memInfo;
    memInfo.dwLength = sizeof(MEMORYSTATUS);
    GlobalMemoryStatus(&memInfo);
    double slobodno = memInfo.dwAvailVirtual/1024./1024.;

    printf("%g MB\n",slobodno);
    int br=0,i,j;

    char **imena,*ime,*temp,*bbb=NULL;
    imena=(char**) malloc(sizeof(char*)*(br+1));
    while(1)
    {
        printf("Unesite ime: ");
        ime=(char*) malloc(sizeof(char)*4000000);
        gets(ime);
        printf("%u\n", strlen(ime));
        ime=(char*) realloc(ime,strlen(ime)+1);

        GlobalMemoryStatus(&memInfo);
        slobodno = memInfo.dwAvailVirtual/1024./1024.;
        printf("%g MB\n",slobodno);

        if (strcmp(ime,".")==0)
           {free(ime);free(imena[br]);break;}
        imena[br++]=ime;
        imena=(char**) realloc(imena,sizeof(char*)*(br+1));
    }

    for (i=0;i<br-1;i++)
        for (j=i+1;j<br;j++)
            if (strcmp(imena[i],imena[j])>0)
            {
                temp=imena[i];
                imena[i]=imena[j];
                imena[j]=temp;
            }


    //ovde ide sortiranje
    for (i=0;i<br;i++)
        printf("%s\n",imena[i]);

    for(i=0;i<br;i++)
       free(imena[i]);

    free(imena);

    return 0;
}

Note: Professor added the lines for printing out the available memory so we can see that realloc() doesn't work. Every new string we enter just takes up sizeof(char)+4000000 bytes and can't be reallocated. I'm trying to find out why. Thanks in advance

Upvotes: 0

Views: 917

Answers (4)

Bart van Ingen Schenau
Bart van Ingen Schenau

Reputation: 15758

The problem is not that realloc doesn't know the size of the original block. Even though that information is not available for us programmers, it is required to be available to realloc (even if the block was allocated with malloc or calloc).

The line

ime=(char*) realloc(ime,strlen(ime)+1);

looks like it is shrinking the previously allocated block to fit the contents exactly, but there is no requirement that is actually shrinks the block of memory and makes the remainder available again for a new allocation.

Edit

Another thing I just thought of: The shrinking with realloc might work OK, but the memory is not returned by the runtime library to the OS because the library keeps it around for a next allocation. Only, the next allocation is for such a large block that it does not fit the memory freed up with realloc.

Upvotes: 0

prprcupofcoffee
prprcupofcoffee

Reputation: 2970

realloc doesn't free memory. These functions work with a big block of memory (called a "heap") and carve chunks out when you call realloc/malloc/calloc. If you need more memory than is in the heap at the moment, then the heap is expanded by asking the operating system for more memory.

When you call realloc to make a memory block smaller, all that happens is that the memory you don't need any more is made available for *alloc to hand out again on a different request. Neither realloc nor free ever shrink the heap to return memory back to the operating system. (If you need that to happen, you need to call the operating system's native memory allocation procedures, such as VirtualAlloc on Windows.)

Upvotes: 0

Omri Barel
Omri Barel

Reputation: 9480

I have a feeling that it has something to do with the page sizes on Windows. For example, if you change 4000000 to 400000, you can see that the memory can be re-used.

I think that allocating 4000000 forces Windows to use "huge" page sizes (of 4MB) and for some (unknown to me) reason, realloc doesn't work on them in the way that you would expect (i.e. making unused memory available for other allocations).

This seems to be related to Realloc() does not correctly free memory in Windows, which mentions VirutalAlloc, but I'm not sure it clarifies the exact reason that realloc doesn't work.

Upvotes: 3

prprcupofcoffee
prprcupofcoffee

Reputation: 2970

From MSDN:

The memblock argument points to the beginning of the memory block. If memblock is NULL, realloc behaves the same way as malloc and allocates a new block of size bytes.

So the line ime=(char*) realloc(NULL,sizeof(char)*4000000); just malloc's new memory each time.

Upvotes: 0

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