Question about function with variable number of parameters

xailer · 01-13-2004, 08:36 AM

hi

I saw an example of variadic function with optional parameter list and its parameter list ended with (char *)0

here is function's prototype

int execl ( const char *path , const char *arg , ... , ( char *)0 ) ;

In practice

execl( "ps" , "ps" , "-ax" , 0 ) ;

Zero gets cast to char * and pointer to char points to nowhere .

But why do we need to give to pointer NULL value to terminate parameter list ?

And why does this have to be pointer to char ? Why not (int *)0 ?

thank yo

kev82 · 01-13-2004, 08:46 AM

by the man page for exec
The const char *arg and subsequent ellipses in the execl,
execlp, and execle functions can be thought of as arg0,
arg1, ..., argn. Together they describe a list of one or
more pointers to null-terminated strings that represent
the argument list available to the executed program. The
first argument, by convention, should point to the file
name associated with the file being executed. The list of
arguments must be terminated by a NULL pointer.

i think that makes it reasonably clear, its taking arguments of type char * and the last one must be null, you should technically use the constant NULL rather than 0 because its value isnt necessarily 0.

xailer · 01-13-2004, 09:36 AM

hi

I imagine yo have to end with NULL cos argv[] is an array of pointers ending with NULL

If I was to make my own variadic function I could end parameter list however I choosed to ?

thank you

kev82 · 01-13-2004, 09:57 AM

yes, you can end it how you like as long as you can identify the last argument, look at the man page for stdarg to see an example.

xailer · 01-13-2004, 11:37 AM

hi

thank you for your help