How to avoid this typedef?

[Martin Gieseking]

Hello

I've the following piece of code that compiles fine:

typedef int (*(*T)[3])[4];
T *t = new T[k];

Now I would like to avoid the typedef but don't really know how to do this.
Is it possible at all?

Any help is really appreciated.

Alex

 

[Karl Heinz Buchegger]

Hello

I've the following piece of code that compiles fine:

typedef int (*(*T)[3])[4];
T *t = new T[k];

Now I would like to avoid the typedef but don't really know how to do this.
Is it possible at all?

It is possible. But why would you want to do it?
The code doesn't get clearer or simpler by doing so.

 

[Martin Gieseking]

I've the following piece of code that compiles fine:

typedef int (*(*T)[3])[4];
T *t = new T[k];

Now I would like to avoid the typedef but don't really know how to do
this. Is it possible at all?

It is possible. But why would you want to do it?
The code doesn't get clearer or simpler by doing so.

Because I have to create lots of dynamic arrays with various compound types
like those above. I don't want to add typedefs for all of them. Ok, maybe
the code becomes less readable but in these special cases I can live with
that.

 

[tom_usenet]

Hello

I've the following piece of code that compiles fine:

typedef int (*(*T)[3])[4];
T *t = new T[k];

Now I would like to avoid the typedef but don't really know how to do this.
Is it possible at all?

Don't do it!
int (*(*t)[3])[4] = new (int (*[10][3])[4]);
You have the choice of 1 nasty line and 1 simple line, or the above
completely hideous line.

I have to ask why you have such strange types anyway - containers
might provide a better approach, depending on your requirements.

Tom

C++ FAQ: http://www.parashift.com/c++-faq-lite/
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html

 

[Karl Heinz Buchegger]

I've the following piece of code that compiles fine:

typedef int (*(*T)[3])[4];
T *t = new T[k];

Now I would like to avoid the typedef but don't really know how to do
this. Is it possible at all?

It is possible. But why would you want to do it?
The code doesn't get clearer or simpler by doing so.

Because I have to create lots of dynamic arrays with various compound types
like those above. I don't want to add typedefs for all of them. Ok, maybe
the code becomes less readable but in these special cases I can live with
that.

If you think this is better:

void foo( int k )
{
int (*(* *t )[3])[4];

t = new ( int (*(*[k])[3])[4] ) ;
}

int main()
{
foo( 5);
}

But your description suggests, that your design needs some polish.

 

[Nick Hounsome]

I've the following piece of code that compiles fine:

typedef int (*(*T)[3])[4];
T *t = new T[k];

Now I would like to avoid the typedef but don't really know how to do
this. Is it possible at all?

It is possible. But why would you want to do it?
The code doesn't get clearer or simpler by doing so.

Because I have to create lots of dynamic arrays with various compound types
like those above. I don't want to add typedefs for all of them. Ok, maybe
the code becomes less readable but in these special cases I can live with
that.

I'm not sure what sort of variation you are intending but templates might
make it simpler
(i'm not sure you wont need a typename in here somewhere):

template <typename T,int D1,int D2>
struct Confusion
{
typedef T* (*(*TYPE)[D!])[D2];
static TYPE* make(int n) { return new TYPE[n]; }
};

Confusion<T,3,4>::TYPE* x = Confusion<T,3,4>::make(42);

This at least gets rid of all the brackets.

 

[Martijn Lievaart]

template <typename T,int D1,int D2>
struct Confusion
{
typedef T* (*(*TYPE)[D!])[D2];
static TYPE* make(int n) { return new TYPE[n]; }
};

Confusion<T,3,4>::TYPE* x = Confusion<T,3,4>::make(42);

This at least gets rid of all the brackets.

Good thinking. I would do:

template <typename T,int D1,int D2>
struct Confusion
{
typedef T* (*(*type)[D1])[D2];
};

Confusion<T,3,4>::type *x = new Confusion<T,3,4>::type(42);

HTH,
M4

 

[Nick Hounsome]

template <typename T,int D1,int D2>
struct Confusion
{
typedef T* (*(*TYPE)[D!])[D2];
static TYPE* make(int n) { return new TYPE[n]; }
};

Confusion<T,3,4>::TYPE* x = Confusion<T,3,4>::make(42);

This at least gets rid of all the brackets.

Good thinking. I would do:

template <typename T,int D1,int D2>
struct Confusion
{
typedef T* (*(*type)[D1])[D2];
};

Confusion<T,3,4>::type *x = new Confusion<T,3,4>::type(42);

I think you meant:

 

[Martijn Lievaart]

I think you meant:
Confusion<T,3,4>::type *x = new Confusion<T,3,4>::type[42];

Duh! Yes, thanks.

M4