scoping issue

scoping issue

[Jason_Haynberg]

This strikes me as a bug.  What do you think?  It seems the first program should get the same error as the second?

$ cat t.cpp
#include <iostream>
using namespace std;

main()
{
   for (int i = 0; i < 5; i++) {
      cout << i << endl;
      double i = 10.0;
      cout << i << endl;
   }
}

$ g++ t.cpp

$ a.out
0
10
1
10
2
10
3
10
4
10

$ cat t.cpp
#include <iostream>
using namespace std;

main()
{
      int i = 0;
      cout << i << endl;
      double i = 10.0;
      cout << i << endl;
}

$ g++ t.cpp
t.cpp: In function `int main()':
t.cpp:9: conflicting types for `double i'
t.cpp:7: previous declaration as `int i'

$ g++ -v
Reading specs from /usr/local/gcc-3.2.3/bin/../lib/gcc-lib/i686-pc-linux-gnu/3.2.3/specs
Configured with: ./configure --prefix=/usr/local/gcc/gcc-3.2.3
Thread model: posix
gcc version 3.2.3

$ uname -a
Linux zolataylor 2.6.5-7.282-bigsmp #1 SMP Tue Aug 29 10:40:40 UTC 2006 i686 athlon i386 GNU/Linux

$ cat /etc/issue

Welcome to SUSE LINUX Enterprise Server 9 (i586) - Kernel \r (\l).



Jay Haynberg

Re: scoping issue

[Andrew Haley]

Jason_Haynberg wrote:
> This strikes me as a bug.  What do you think?  
> It seems the first program should get the same error as the second?

Your first program is equivalent to

main()
{
  {
    int i;
    for (i = 0; i < 5; i++) {
      cout << i << endl;
      double i = 10.0;
      cout << i << endl;
    }
  }
}

Andrew.

Re: scoping issue

[me22]

On Thu, Sep 4, 2008 at 09:33, Jason_Haynberg <haynberg@sig.com> wrote:
> This strikes me as a bug.  What do you think?  It seems the first program should get the same error as the second?
>
>   for (int i = 0; i < 5; i++) {
>      cout << i << endl;
>      double i = 10.0;
>      cout << i << endl;
>   }
>

This is just intuition, but my understanding is that the double is
defined inside the for loop body block, and the int is defined outside
of it (it has to be, to persist its value), so there are 2 different
scopes, so it's not an error.

The analogous non-loop code would therefore be this, rather than what
you posted:

     int i = 0;
     {
          cout << i << endl;
          double i = 10.0;
          cout << i << endl;
     }

HTH,
~ Scott