Why should C++ permit a variable name and a class name to be the same? [closed]

Question

#include <iostream>
#include <vector>   

using namespace std; 

typedef int UINT4;

class Hack
{};

Hack& operator <(Hack& a , Hack& b) 
{
    std::cerr << "1";  
    return a; 
}

Hack& operator >(Hack& a, Hack& b)
{
    std::cerr <<  "2";    
    return a; 
}

int main()
{
    Hack vector; // It SHOULD be OK!?
    Hack UINT4; // It SHOULD be OK!?
    Hack v;
    Hack Hack; // It SHOULD be OK!?
    Hack = v; // It SHOULD be OK!?

    // Please stop to think for a moment: What will the following output?
    vector<UINT4> v; 

    // Oh, my god! The whole world goes mad!

    return 0; 
} 

Answer 1

Ever heard of the following? Please look close at the parameter types :)

int stat(const char *path, struct stat *buf);

What you have shown in your question really isn't too dramatic. You declare a variable name in a local scope, but the type names in your program are in global scope. I will show you some real pervert thing now:

int nooo = 0;
int main() {
   int nooo = 0;
}

W00t? Why does C++ allow us to create two variables of the same name!?!

Alright, I was kidding. I will now introduce you into the real dark sides of messing with duplicate names. Consider what C++ allows us!

struct A {
  int A; // noes!
};

A A; // another A!

In essence, this one is all for C compatibility :)

Answer 2

You're asking the question from the point of view or someone who very well knows that he is using a class name as a variable identifier : obviously, this isn't a best practice.

Now, let's take it the other way around : suppose you have no idea that there is a weird typedef int i; in some obscure header file you are including. Would you expect it to break the following innocent code ?

int i = 0;