Static array, variable size allowed?

Question

I have a function like this

void foo( int i)
{
   ...
   uint8_t buf[ i];
   ...
}

and I don't understand why the compiler is not complaining... I'm using

g++ -c -g -Wall

to compile

Any ideas??

It's a compiler extension. I'm sure you can find a duplicate on here. — chris, Mar 01 '13 at 23:10
[Variable Length Array](http://en.wikipedia.org/wiki/Variable-length_array) — Mahesh, Mar 01 '13 at 23:10
I think you'll get a warning if you specify e.g. `-std=c++11` and have a `-pedantic` option. — Daniel Fischer, Mar 01 '13 at 23:11
@StephaneRolland: One might expect it to complain since this isn't valid C++. — Mike Seymour, Mar 02 '13 at 00:56
@MikeSeymour oh sorry, it's been so long I no longer use array I didn't notice the dynamic i in buf... thanx for reminding me. — Stephane Rolland, Mar 02 '13 at 10:31

score 1 · Accepted Answer · answered Mar 01 '13 at 23:11

1

It is a GCC compiler extension. It is allowed by the standard for a conforming implementation because it does not break any well-formed code (as long as it issues a diagnostic). It is, of course, not portable and therefore not recommended.

With the -pedantic option, you'll get the following warning:

warning: ISO C++ forbids variable length array ‘buf’ [-Wvla]

answered Mar 01 '13 at 23:11

Joseph Mansfield

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It's allowed **if** the compiler issues a diagnostic. – Pete Becker Mar 01 '13 at 23:12
oh! Is that new or something? I mean the extension – santi Mar 01 '13 at 23:13
@santi It's been there for a good while now. – Joseph Mansfield Mar 01 '13 at 23:14

Static array, variable size allowed?

1 Answers1