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From:	Hans-Bernhard Broeker <broeker AT physik DOT rwth-aachen DOT de>
Newsgroups:	comp.os.msdos.djgpp
Subject:	Re: array size inside of a function
Date:	4 Mar 2004 13:57:20 GMT
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Andrey <huli_ebtyt AT list DOT ru> wrote:
> > sizeof(char **) == address size = 4 on 32bit machines.
> > 

> so in main(), array name is not a pointer?

Right.  It will "decay" into a pointer if you use it as one, but it
"is" not a pointer.  This is a very common misunderstanding among new
C programmers, and unfortunately one that is being stated as fact by
many careless C book authors.

Passing an array as an argument to some function is one of the usages
that counts as "using it as a pointer".  The only cases that don't
count as such, actually, are use of the 'sizeof' or '&' operators on
the name of the array.

> and there no way free from passing size?

Essentially: no.  It's thus usually a good plan to join the pointer
and the length into a struct:

	typedef	struct array_of_foo {
	   foo *pointer;
	   size_t length;
	} array_of_foo;

-- 
Hans-Bernhard Broeker (broeker AT physik DOT rwth-aachen DOT de)
Even if all the snow were burnt, ashes would remain.

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