Where C stores dynamically a string inside a struct-CodePudding

I wonder where the literal string that str points to is allocated, given that (I assume) malloc only makes room for the pointer into the Heap.

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

typedef struct
{
    int a;
    char* str;
} Bar;

int main(int argc, char *argv[])
{
    Bar* bar_ptr = (Bar*)malloc(sizeof(*bar_ptr));

    bar_ptr->a = 51;
    bar_ptr->str = "hello world!";

    printf("%d\n", bar_ptr->a);
    printf("%s\n", bar_ptr->str);

    return 0;
}

CodePudding user response：

As already pointed out the string is stored in a read-only data segment. You can print the address where the string is stored (in hexadecimal format) like that:

printf("0x%p\n", bar_ptr->str);

CodePudding user response：

Correct - all your struct type stores is the address of the first character in the string. The string contents are stored "somewhere else" - in a string literal, in another dynamically-allocate block, in a static or auto array, etc.

You could declare everything auto:

void foo( void )
{
  char aStr[] = "this is not a test";
  Bar barInstance;

  barInstance.a = 51;
  barInstance.str = aStr;

  ...
}

You could allocate everything dynamically:

Bar *barInstance = malloc( sizeof *barInstance );
if ( barInstance )
{
  size_t size = strlen( "This is not a test" );
  barInstance->str = malloc( size   1 );
  if ( barInstance->str )
    strcpy ( barInstance->str, "This is not a test" );
  ...
  /**
   * You must free barInstance->str before freeing
   * barInstance - just freeing barInstance won't
   * free the memory barInstance->str points to,
   * since that was a separate allocation.
   */
  free( barInstance->str );
  free( barInstance );
}