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?
You are just trying to obfuscate your code, it is as simple as that. And the compiler is completely correct in telling you so.
If you have a precise idea what the assigned value should be, use that. My guess is that you have some unfounded presumption of short being 16 bit wide and that the sign representation of the target machine is two's complement. If that is so, assign -4083 to your variable. If you just need your variable as a bit vector, use an unsigned type.
As far as C is concerned the standard simply says about conversion from one integer type to another:
Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.
I imagine the the point of view of C++ with this respect is not much different. Other answers mention border cases where in C++ you would need a `C'-style cast to overrule all typechecks that C++ gives you. Feeling the need for them is an indication of bad design.
The case that you give as an example is certainly not one for which I would find any valid circumstances.