如何在不取消引用的情况下调用函数指针？ [英] How come pointer to a function be called without dereferencing?

问题描述

我在Py ++生成的C ++程序中有一个奇怪的typedef语句。

I have a weird typedef statement in a C++ program, generated by Py++.

double radius(int);  // function to be wrapped
typedef double (*radius_function_type)(int);    
bp::def("radius", radius_function_type(&radius));   // bp::def is a function for wrapping

的函数typedef statemnt不是这种类型，我们大多数人都很熟悉。

What I figured out so far is that the above typedef statemnt is not of the type, most of us are familiar with,

typedef complex_type simple_alias;

这是一种声明指向函数的指针的方法，该函数将int作为参数并返回double（相同作为原型）。因此，我现在的问题是，如何以函数的地址作为参数（不进行取消引用）？这也与原型不符。有人请解释！

Rather it is a way to declare pointer to a function which takes int as argument and returns double (same as the prototype). So my question now is that, how come pointer to a function (without dereferencing) be called with address of a function as an argument? This also doesn't match with the prototype. Somebody please explain!

推荐答案

您的问题令人困惑。您是否在问：

Your question is confusing. Are you asking what this does:

radius_function_type(&radius)"

这只是C ++类型转换，有点像：

This is just a C++ typecast, a bit like:

radius (int (42));

但由于半径已经是radius_function_type类型，因此您可以一样容易做到：

but since radius is already of type radius_function_type then you can just as easily do:

bp::def("radius", radius);

但是由于这是Py ++生成的代码，因此输出时可能要格外小心。

but as this is code generated by Py++, it's probably being extra careful with the output.

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