Array parring in C

Question

Playing with array passing in c. Defining array in function Get_Present_Location() and passing it as a pointer into main where i simply print it. It seems to work except for the last element. Instead of -7 i get 0. Probably there is a stupid mistake on my part but I cannot understand why.

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

double *Get_Present_Location();

int main(){

    double *Present_Point;
    Present_Point=Get_Present_Location();

    for (int i=0;i<6;i++)
    {
        printf("Joint[%d] = %f\n",i+1,Present_Point[i]);
    }

    return 0;
}

double *Get_Present_Location()  
{

    double point[6]={4,1,5,-3,5,-7};   //Temporary

    return &point;
}

Ed Heal · Accepted Answer · 2014-11-01 10:47:32Z

3

You have answered your own question.

i.e.

  double point[6]={4,1,5,-3,5,-7};   //Temporary

It is temporary (also the code does not compile with the warnings on)

answered Nov 1, 2014 at 10:47

Ed Heal

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Comments

djna · Accepted Answer · 2014-11-01 10:57:54Z

0

In C once you have invalid pointers all manner of mysterious things may happen. You have two problems here. First

return &point;

this is returning a double**, a pointer to a pointer, you should instead

return point;

The array itself is effectively a pointer to double.

More important, as you have noted you are returning a pointer to a temporary, or local, piece of storage. On return from the function this memory, which is probably on the stack, is open to reuse, probably in the next function call. You are returning a pointer to some stack location that is subject to arbitrary change, in effect garbage.

You should instead allocate some memory and return a pointer to that.

answered Nov 1, 2014 at 10:57

djna

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