Re: Array Size : Error : Why ??

Am 09.11.2011 21:20, schrieb Vipin:


To make the problem statement more clear, I deliberately simplified your
example to

int Data[] = {0};

int main (){
   sizeof Data/sizeof Data[0];
   sizeof (int [])(&(Data[0]))/ sizeof Data[0];
}

The question is: Why is the second statement in main ill-formed?

The reason is some curiosity of the C++ language related to incomplete
types. To determine the size of some object or type, sizeof needs a
complete type. E.g. if you declare (but do not define=

class X;

the expression sizeof(X) is ill-formed, because the type X has not been
completely defined yet, so quiring sizeof cannot succeed.

We have a similar situation for a declaration of an array type without
bounds, like 'int[]'. Therefore testing the expression

sizeof(int[])

is also ill-formed, because sizeof has no information about the array
length.

The special situation is, when we have an array definition of the form

int Data[] = {0};

Here, the type of Data is incomplete until the closing brace, but after
the closing brace it is complete and has type int[1], because there is a
single element initializer.

Now writing a casting expression of the form

(int [])(&(Data[0])

is ill-formed for several reasons:

a) You cannot cast to an incomplete type
b) Even if we would write

(int[1])(&(Data[0]))

instead, this won't work, because arrays immediately undergo
array-to-pointer conversion unless directly bound to a reference to
array. The question anyway is, *why* you want to perform this code
transformation?

HTH & Greetings from Bremen,

Daniel Kr?gler

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