C++:编写自己的函数时要keep in mind

2014.2

函数的返回类型可以是reference to an object，例如：

// return a reference to the shorter of two strings(C++ primer 5th edition)

const string &shorterString( const string &s1, const string &s2)

{

  return s1.size() <= s2.size() ? s1:s2;

}

但是不能返回reference或者pointer to a “local” object

原因是“After a function terminates, references to local objects refer to memory that is no longer valid.”

以上是两部分原理。

思考：最后看到一面那段程序，parameter list中两个parameter都是pass by reference。

函数返回类型是reference to an object，如果将两个parameter的“&”去掉，就使得paramter变为pass by value，那么以此方式传入的argument就是只在函数里面存在的一个argument，函数结束时会被destroyed，所以改动后的程序应该是错的，并且其错误原因应该就是返回了local object的引用。

经过验证，以下这段程序确实如预料的。

//  intend to return a reference to the shorter of two strings, but may be erroneous

const string &shorterString( const string s1, const string s2)

{

  return s1.size() <= s2.size() ? s1:s2;

}

2014.4

C++ primer 5th edition

15.2 "Using Members of the Base Class from the Derived Class"

重要观点“Interactions with an object of a class-type should use the interface of that class, even if that object is the base-class part of a derived object.”

"Declarations of Derived Classes"

重要观点"The purpose  of a declaration is to make known that a name exists and what kind of entity it  denotes, for example, a class, function, or variable."

So, declaration with inheritance list like:"class Bulk_quote: public Quote;" is wrong, "class Bulk_quote" is right.