const 和非常量函数的重载是如何工作的?
STL 充满了这样的定义:
iterator begin ();
const_iterator begin () const;
由于返回值不参与重载决议，这里唯一的区别是函数是const.这是重载机制的一部分吗?编译器的解析算法是什么:
As return value does not participate in overloading resolution, the only difference here is the function being const. Is this part of the overloading mechanism? What is the compiler's algorithm for resolving a line like:
vector<int>::const_iterator it = myvector.begin();
推荐答案
在你给出的例子中:
vector<int>::const_iterator it = myvector.begin();
如果 myvector 不是 const，则将调用 begin() 的非 const 版本，您将依赖从迭代器到 const_iterator 的隐式转换.
if myvector isn't const the non-const version of begin() will be called and you will be relying on an implicit conversion from iterator to const_iterator.
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