Re: Passing address of multidimentional array

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Doug Harrison [MVP] wrote:
On Tue, 03 Jun 2008 21:25:14 +0200, "Cezary H. Noweta"
<chncc@xxxxxxxxxxxxxxx> wrote:

Doug Harrison [MVP] wrote:

All the following are equivalent:

void f(int a[20][30]);
void f(int a[2][30]);
void f(int a[][30]);
void f(int (*a)[30]);

This is the *****only***** context in which array and pointer
declaration syntax is sort of interchangeable, and it's the reason
the magnitude of the first dimension doesn't matter - all four
declare "a" to be a pointer to an array of 30 ints.
No - they are still different with all consequences.

It is as I said.

For example array
declarators (the first three) are not modifiable lvalues, while the last
declarator is.

You are mistaken.

So it is impossible to use ,,a = sth;''

I don't know what that means.


I think Cezary meant to imply that the assignment in 'foo' below is supposed to fail for some reason:

void foo(int a[20][30])
{
int b[5][30];
a = b; // line 4
}

void bar(int (*a)[30])
{
int b[5][30];
a = b;
}

int main()
{
int arr[100][30];
foo(arr);
bar(arr);
}

Of course, since the two declarations are equivalent ('foo' and 'bar' have the same type, essentially), there is no error on line 4.

V
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