why the code can compile -- about function pointer

[George2]

Hello everyone,


I am learning code from others, but I do not know why the following
code section can compile?

Why assignment p = new (void (*[3])()) is ok? Are the left side and
right side of assignment having compatible type?

void (**p)();

int main()
{
	p = new (void (*[3])());
	return 0;
}

thanks in advance,
George

 

[Rolf Magnus]

Hello everyone,


I am learning code from others, but I do not know why the following
code section can compile?

Why assignment p = new (void (*[3])()) is ok? Are the left side and
right side of assignment having compatible type?

Not just compatible. They have the very same type.

void (**p)();[/QUOTE]

So it's a pointer to pointer to function taking no parameters and returning
void.
[QUOTE]
int main()
{
p = new (void (*[3])());[/QUOTE]

Here you create an array of 3 pointers to functions taking no parameters and
returning void. The operator new returns a pointer to the first element of
that array, which happens to have the same type as p.
[QUOTE]
return 0;
}

A simpler example that is similar in that regard would be one that creates
an array of int instead of an array of pointers to functions:

int *p;

int main()
{
p = new int[3];
return 0;
}

 

[Bo Persson]

Hello everyone,


I am learning code from others, but I do not know why the following
code section can compile?

Why assignment p = new (void (*[3])()) is ok? Are the left side and
right side of assignment having compatible type?

void (**p)();

int main()
{
p = new (void (*[3])());
return 0;
}

Why bother?

In real code you don't want to do this, ever!


Bo Persson