Is it possible to completely avoid C-style casts in C++?

Question

I do not believe that it is possible to completely avoid C-style casts when writing C++. I was surprised to find out that I needed to use a C-style cast to avoid a compiler truncation warning:

short value_a = 0xF00D;                     // Truncation warning in VS2008
short value_b = static_cast<short>(0xF00D); // Truncation warning in VS2008
short value_c = (short)0xF00D;              // No warning!

Are there other scenarios where there is no C++-style substitute for a C-style cast?

Answer 1

In C++, the C-style cast is defined (§5.4) in terms of C++-style casts. So for every cast you can do C-style, there's a matching C++-style cast (almost).

The "almost" is that C-style casts ignore base class accessibility. That is, there is no equivalent C++-style cast for the following:

struct foo {};
struct bar : private foo {};

bar b;
foo* f = (foo*)&b; // only way this can be done in a well-defined manner

So, no it's not strictly-speaking possible to completely ditch C-style casts. But the number of areas where a (combination of) C++-style casts doesn't suffice is few in count.

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The above is the "language answer". What you're experiencing has nothing to do with C-style casts versus C++ casts, but just compiler implementation. Warnings are absolutely implementation-specific, and have nothing to do with C++.

So don't make the mistake of using your findings on this particular compiler in this particular situation for concluding things about C++ in general.