Question

I am having hard time understanding typedef pattern for arrays.

typedef char Char10[10];
void fun (Char10 a)  // not passing reference (interested in pass by value)
{
  if(typeid(Char10) == typeid(char*))
    throw 0;  // <--- never happens
}

int main ()
{
  char a[10];  fun(a);  // ok
  char b[11];  fun(b);  // why works ?
}

Why the different sizes of array by value are accepted by fun() ? Are char[10] and char[11] not different types ?

Edit: For those who says it decays to pointer, see my edited code. char[10] and char* doesn’t seem to match.

Answer 1

They are different types, you’re right.

It’s a misleading quirk of C++ that you can be seen to have a function

void fun(char a[10])

Since you cannot pass arrays by value, and C++ is silly, this is actually the function

void fun(char* a)

And, of course, both inputs degrade happily to char*.

It would be nice if C++ did not let you even pretend to accept an array by value, but it is silly in this way.. inherited from C.