Why is the code working for the second case?

Question

#include <iostream> 
int *parray(int a[100][100],int m , int n) 
{   int b[200],i, t , j;
    i = 1;
    j = t = 0;
    for(int i = 0; i < m ; i++)
    for(int j = 0; j < n ; j++)
    b[t++] = a[i][j];

    //      for(int i = 0; i < m*n ; i++)      
    //      std::cout<<b[i]<<' ';
    return b;
}

int main()
{
  int a[100][100],m,n,*b;
  std::cin>>m>>n;

  for(int i = 0; i < m ; i++)
  for(int j = 0; j < n ; j++)
   std::cin>>a[i][j];

  b = parray(a,m,n);

  for(int i = 0; i < m*n ; i++)
   std::cout<<b[i]<<' ';

  return 0;
}

when input is

     2 2 1 2 3 4

output is

     1 2 612675848 32767

but when I remove the two "//" just before the return of parray() function the output becomes:

     1 2 3 4 1 2 3 4

Can someone please tell me why code works when I print the value before passing?

`int b[200]` 200 ain't gonna be big enough bub. It should be `int b[10000]` — scohe001, Sep 02 '14 at 17:43

score 0 · Accepted Answer · answered Sep 02 '14 at 17:41

0

int b[200] in parray() is local to parray(). It is removed from the stack from the function returns.(when the scope is over).

If you access it later in main() ,you get undefined behavior.

answered Sep 02 '14 at 17:41

Sagar D

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Thank You. But Why does it work the 2nd time? – jymspaul Sep 02 '14 at 17:45
@jymspaul By chance. – Overv Sep 02 '14 at 17:47
Because when you remove the comments,i.e. `//` , you print the local `b[]` in `parray()`, which will show you what b contains,because it is still in memory. – Sagar D Sep 02 '14 at 17:48

Why is the code working for the second case?

1 Answers1