Re: printing array in function

Steffen Dettmer
2010-09-03T10:24:36Z

On Fri, Sep 3, 2010, Nicolas Sabouret <Nicolas.Sabouret@lip6.fr> wrote:
> The problem we have is that arrays passed to functions are seen as
> pointers by gdb. Here is a simple example :
>
> 1:  void f(int tab[]) {
> 5:    int t[] = {-1,-1};
>
> (gdb) p tab     -> (int *) 0xbffff440

I think this is absolutely correct.

tab is an int pointer, the storage size of a referenced "array" isn't known.

> The only solution we found to display tab as an array is to use "p
> *tab@2", but this requires knowing the exact size of the array (2 in
> this example).

Yes, absolutely, same as in C:
------------------------------------------------------------------->8=======
#include <stdio.h>

void f(int tab[]) {
tab[0] = 1;
/* (%z is not known to all compilers) */
printf("sizeof(tab) = %lu\n", (unsigned long)sizeof(tab));
}
int main() {
int t[] = {-1,-1};
printf("sizeof(t) = %lu\n", (unsigned long)sizeof(t));
f(t);
return 0;
}
=======8<-------------------------------------------------------------------

produces

sizeof(t) = 8
sizeof(tab) = 4

so function f() also needs to know the size or number of
elements, even when a "size" is given:

void f(int tab[2]) {
assert(sizeof(tab) == sizeof(int*));
}

to illustrate this, development rules may forbid array notation
for function notation instead of pointer, leading to:

void f(int *tab) {
tab[0] = 1;
}

which, as far as I know, the same as above and should be portable.

> Our problem is that the gdb calls are integrated in a front-end for
> students (they do not type gdb commands directly) and that our frontend
> has no way of "guessing" what is the correct size for the array.

maybe by passing a size parameter (which is usually needed anyway
to be able to know the size of the array):

void f(int *tab, size_t tab_size) {
tab[0] = 1;
}
{
int t[] = {-1,-1};
f(t, sizeof(t)/sizeof(t[0]));
}

and using tab_size in the frontend?

oki,

Steffen