Re: assignment of bigger type to smaller type
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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