I have a program with a parent and a child process. Before the fork(), the parent process called malloc() and filled in an array with some data. After the fork(), the child needs that data. I know that I could use a pipe, but the following code appears to work:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
int main( int argc, char *argv[] ) {
char *array;
array = malloc( 20 );
strcpy( array, "Hello" );
switch( fork() ) {
case 0:
printf( "Child array: %s\n", array );
strcpy( array, "Goodbye" );
printf( "Child array: %s\n", array );
free( array );
break;
case -1:
printf( "Error with fork()\n" );
break;
default:
printf( "Parent array: %s\n", array );
sleep(1);
printf( "Parent array: %s\n", array );
free( array );
}
return 0;
}
The output is:
Parent array: Hello
Child array: Hello
Child array: Goodbye
Parent array: Hello
I know that data allocated on the stack is available in the child, but it appears that data allocated on the heap is also available to the child. And similarly, the child cannot modify the parent's data on the stack, the child cannot modify the parent's data on the heap. So I assume the child gets its own copy of both stack and heap data.
Is this always the case in Linux? If so, where the is the documentation that supports this? I checked the fork() man page, but it didn't specifically mention dynamically allocated memory on the heap.
Thank you
Each page that is allocated for the process (be it a virtual memory page that has the stack on it or the heap) is copied for the forked process to be able to access it.
Actually, it is not copied right at the start, it is set to Copy-on-Write, meaning once one of the processes (parent or child) try to modify a page it is copied so that they will not harm one-another, and still have all the data from the point of fork() accessible to them.
For example, the code pages, those the actual executable was mapped to in memory, are usually read-only and thus are reused among all the forked processes - they will not be copied again, since no one writes there, only read, and so copy-on-write will never be needed.