A simple question about type coercion in C++

问题

Given a function prototype, and a type definition:

int my_function(unsigned short x);
typedef unsigned short blatherskite;

Is the following situation defined by standard:

int main(int argc, char** argv) {
  int result;
  blatherskite b;

  b=3;
  result = my_function(b);
}

Do I get type coercion predictably via the function prototype?

回答1:

If your question is really about whether the types of the argument and the parameter match, then the answer is yes. typedef does not introduce a new type, it only creates alias for an existing one. Variable b has type unsigned int, just like the parameter, even though b is declared using typedef-name blatherskite.

Your example is not very good for demonstrating that though. All integral types are convertible to each other in C++, so (ignoring range issues) the code would have defined behavior even if blatherskite designated a different type (a new type). But it doesn't. So this is also perfectly valid

void foo(unsigned int* p);
...
blatherskite *pb = 0;
foo(pb); // <- still valid

回答2:

No type coercion is needed. The typedef is just an alias for the same type, so you're passing an unsigned short to a function that takes an unsigned short.

来源：https://stackoverflow.com/questions/2907473/a-simple-question-about-type-coercion-in-c