How do I convert the contents of an unsigned char * to a const char *?

Question

我可以跨越重新解释的类型转换，并且在大多数情况下都会发出警告，因此我想知道是否还有其他选择（或者当然是重新解释的类型的干净实现）。

Answer 1

You don't say what warning was given or what the problem was, but casting to char* with reinterpret_cast should work without warnings:

unsigned char *a;
const char *b = reinterpret_cast<char*>(a);

Answer 2

It depends on what you're trying to do.

If you just want to access the contents as char , then a simple static_cast or using the value in a context where a char is expected will do the trick.

If you need to pass the buffer to a function expecting a char const* , a reinterpret_cast is about the only solution.

If you want a string, using the pointers into the buffer will be fine:

std::string
bufferToString( unsigned char const* buffer, size_t length )
{
    return std::string( buffer, buffer + length );
}

or you can copy into an existing string:

myString.assign( buffer, buffer + length );
myString.append( buffer, buffer + length );
//  etc.

Any string function (or algorithm, like std::copy ) which takes two iterators can be used. All that is required is that dereferencing the iterator result in a type which converts implicitly to char , which is the case of unsigned char .

(You cannot use the string functions which take a buffer address and a length, as these are not templates, and require the buffer address to have type char const* . And while unsigned char converts implicitly to char , unsigned char* requires a reinterpret_cast to convert it to char* .)