Assignment of malloc'ed array to stack'ed array

Question

I was trying to return an array from child_prog() to main(). The code looks like this:

#include<stdio.h>
#include<stdlib.h>
int* child_prog(int some_input);

void main(void){
  int C[10];
  int some_input;
  C = child_prog(some_input);
}

int* child_prog(int some_input){
  static int out[10];
  // ...
  // ... some wizardry
  return out;
}

Now the compiler generates an error saying that it can not assign to C (which is an int[] type) the value returned from child_prog (which is an int* type). Indeed, the program works fine when I make C an int* and malloc it 10 ints of memory. But I don't understand why the compiler can't assign to C (an array defined as C[10] and hence a pointer) the value returned from child_prog (an array defined as static int out[10] and hence again a pointer).

MByD · Accepted Answer · 2012-05-19 06:16:08Z

4

You cannot assign to an array. You need to memcpy it.
int* != int[] while the first is a pointer that might point to an int of a sequence of ints, the second is a sequence of ints
You can use int *C; and pass the length of the array (if unknown at compile time) as an out parameter.
static int out[10]; is not a malloced, but static.

edited May 19, 2012 at 6:16

answered May 19, 2012 at 6:09

MByD

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3 Comments

Abhinav Over a year ago

Ok, understood point number-1 that int*!=int[]. But, in the case when I define C as int* C instead of int C[], and malloc it the known amount of memory, and child_prog returns to C the pointer to the static int array, shouldn't I be able to bypass the need to do memcpy?

MByD Over a year ago

If you are not assigning to an array, but to a pointer, you don't need memcpy.

Abhinav Over a year ago

Thanks for slaying a lot of my dragons ... rather quickly

nav_jan · Accepted Answer · 2012-05-19 06:24:22Z

1

One solution can be to declare C as:

int *C;

as said by Binyamin You cannot change the address of an array as it is statically allocated, which precisely you are trying to do with:

C = child_prog(some_input);

answered May 19, 2012 at 6:24

nav_jan

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Comments

vvnraman · Accepted Answer · 2012-05-19 06:26:38Z

0

The type of int [] is int * const which means the memory it points to is a constant and trying to change it will give a compiler error. Obviously it isn't equal to int *.
The variable out in child_prog() is statically allocated which means its not on the stack but somewhere in the global data section. Hence no matter how many times you call child_prog(), you'll always be returned the same memory location.
Hence to copy the array, do a memcpy(dest,src,bytes) if you want to save the data returned from child_prog().

answered May 19, 2012 at 6:26

vvnraman

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Comments

Jay D · Accepted Answer · 2012-05-19 06:34:14Z

0

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

#define SUCCESS    0 
#define FAILURE    1 

int child_prog(int some_input, int *output, int output_size);

void main(void){
  int C[10];
  int some_input;
  C = child_prog(some_input, c, 10);
}

int  child_prog(int some_input, int *output, int output_size)
{
  static int source[10];
      ...
      .../*some wizardry*/

  memcpy(output, source, output_size);


     ... if (erronous Condition)
             return FAILURE;
         else

  return SUCCESS;
}

answered May 19, 2012 at 6:34

Jay D

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