Re: assignment of bigger type to smaller type

From:
"Maxim Yegorushkin" <maxim.yegorushkin@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
7 Jan 2007 18:34:23 -0500
Message-ID:
<1168187145.349600.130930@42g2000cwt.googlegroups.com>
Ivan Novick wrote:

#include <iostream>

int main()
{
    int x = LONG_MAX;
    std::cout << x << std::endl;
    short y = x;
    std::cout << y << std::endl;
}

Why would the C++ standard allow this code without requiring an error
message? Surely assigning a larger type to a smaller type could be
caught at compile time and allowing assignments like this is inherintly
dangerous?


We hardly can change the built-in assignment operator, but you might
like a little helper function, something like this:

int main()
{
    int x = LONG_MAX;
    short y = safe_cast(x); // compile time error
}

Please note, that safe_cast deduces its only template argument, so no
additional code maintenance required should type of x or y change.

The internals:

template<class T>
struct safe_cast_proxy
{
    T t;

    safe_cast_proxy(T t) : t(t) {}

    template<class U> operator U() const
    {
        typedef int static_assert[sizeof(U) >= sizeof(T) ? 1 : -1];
        return t;
    }
};

template<class T>
inline safe_cast_proxy<T> safe_cast(T t)
{
    return safe_cast_proxy<T>(t);
}

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