More C Questions ....

[Roger Oberholtzer]

On Tue, 2005-09-13 at 20:18, Ben Duncan wrote:
> Ok, in the following code snippet:
> ----------------------------------------------------------------------
> #ifdef SEMLOCK
> #if 1
>    if ( ( sem_id = semget ( key, 1, 0 ) ) == -1 )
>      if ( ( sem_id = semget ( key, 1, IPC_CREAT | 0660 ) ) == -1 )
>        {
>          shmdt ( ( char * ) db->shm );
>          if ( created )
>            shmctl ( db->shm_id, IPC_RMID, NULL );
>          return -1;
>        }
> #else
>    if ( ( sem_id = sem_open ( key ) ) == -1 )
>      if ( ( sem_id = sem_create ( key, 1 ) ) == -1 )
>        {
>          shmdt ( ( char * ) db->shm );
>          if ( created )
>            shmctl ( db->shm_id, IPC_RMID, NULL );
>          return -1;
>        }
> #endif
> #endif
> ------------------------------------------------------------------------
> 
> I know that if there is a #define SEMLOCK somewhere in the code/header
> then the #ifdef SEMLOCK is going to get taken. BUT .. What is it with
> that #if 1 ? What is it doing, how does it get set ?

It is a programming trick. It does not get set. Change the '1' to a '0'
to select which method gets used when SEMLOCK is defined. Sometimes old
code is kept around for a bit this way as a sort of documentation. So,
you edit the file to change which bit of code gets used. It cannot be
set any other way. I am guilty of such behavior. But a comment telling
what is going on is never a bad thing to have here as well.

> 
> Thanks ...
> 
> I
> 
> 
> 
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