Re: typedef float float4[4]; std::vector<float4>; does not compile, why?

From:

Jeff Schwab <jeff@schwabcenter.com>

Newsgroups:

comp.lang.c++

Date:

Tue, 19 May 2009 12:34:22 -0400

Message-ID:

<_omdnTLXw7ASQo_XnZ2dnUVZ_hOdnZ2d@giganews.com>

Brian Cole wrote:

On May 19, 2:01 am, SG <s.gesem...@gmail.com> wrote:

On 19 Mai, 00:12, Brian Cole <col...@gmail.com> wrote:

The following code will not compile in gcc 4.3.2 on Ubuntu 8.10
#include <vector>
typedef float float4[4];
int main()
{
std::vector<float4> vals;
}

As Alf said, float4 doesn't satisfy the vector templates' requirements
as a value type.

If your C++ implementation ships with the TR1 library extension or you
can install some TR1 implementation and/or Boost you could solve this
problem via:

typedef std::tr1::array<float,4> float4;

Apparently GCC considers this a bug (http://gcc.gnu.org/bugzilla/
show_bug.cgi?id=40192). So now I'm doubly confused.

It wasn't a bug. They should not have "fixed" it.

I'm not a language lawyer, is "float4" a pointer to an array of 4
floats? Or is it an array of 4 floats?

It's an array (in your original post). C++ raw arrays decay to
pointers, but that's not something you should need to worry about unless
you're manually laying out memory. You almost never should need a raw
array. Forget about raw arrays.

Basically if we define float4_t
as "struct float4_t { float x,y,z,w; };" is std::vector<float4>
equivalent to std::vector<float4_t> or std::vector<float4_t *>.

Neither. std::vector<float[4]> shouldn't even be legal.