C++核心準(zhǔn)則Con.2:默認(rèn)情況下，將成員函數(shù)定義為const類型
秋英
Con.2: By default, make member functions?const
Con.2:默認(rèn)情況下，將成員函數(shù)定義為const類型
Reason（原因）
A member function should be marked?const?unless it changes the object's observable state. This gives a more precise statement of design intent, better readability, more errors caught by the compiler, and sometimes more optimization opportunities.
只要沒有修改對(duì)象的可觀察狀態(tài)，就應(yīng)該將成員函數(shù)定義為const類型。這是設(shè)計(jì)意圖的更清晰表達(dá)，可以帶來更好的可讀性，方便編譯器捕捉更多的錯(cuò)誤，而且有時(shí)還會(huì)帶來更多的優(yōu)化機(jī)會(huì)。
Example, bad（反面示例）
class Point {
int x, y;
public:
int getx() { return x; } // BAD, should be const as it doesn't modify the object's state
// ...
};
void f(const Point& pt)
{
int x = pt.getx(); // ERROR, doesn't compile because getx was not marked const
}Note（注意）
It is not inherently bad to pass a pointer or reference to non-const, but that should be done only when the called function is supposed to modify the object. A reader of code must assume that a function that takes a "plain"?T*?or?T&?will modify the object referred to. If it doesn't now, it might do so later without forcing recompilation.
傳遞指針或者引用給非常量也不是說一定不好，但是它只應(yīng)該發(fā)生在調(diào)用一個(gè)假定會(huì)修改對(duì)象值的情況下。代碼的讀者必須假設(shè)接受原始T*或T&參數(shù)的函數(shù)會(huì)修改（指針或引用，譯者注）參照的對(duì)象。如果（修改，譯者注）沒有發(fā)生在現(xiàn)在，那么可能是以后會(huì)發(fā)生而且不需要重新編譯。
Note（注意）
There are code/libraries that offer functions that declare a T*?even though those function do not modify that?T. This is a problem for people modernizing code. You can
有些代碼/庫(kù)提供的函數(shù)定義了T*參數(shù)卻沒有修改T。這對(duì)于更新代碼使其適用現(xiàn)在C/C++的人來說是一個(gè)問題，你可以
update the library to be?const-correct; preferred long-term solution
將庫(kù)更新到正確定義const屬性的版本；優(yōu)先使用長(zhǎng)期的解決方案。
"cast away?const";?best avoided
使用常數(shù)類型轉(zhuǎn)換；最好避免這種做法。
provide a wrapper function
提供一個(gè)包裝函數(shù)
Example（示例）:
void f(int* p); // old code: f() does not modify `*p`
void f(const int* p) { f(const_cast(p)); } // wrapper
Note that this wrapper solution is a patch that should be used only when the declaration of?f()?cannot be modified, e.g. because it is in a library that you cannot modify.
注意包裝方式是一種只能在f()的聲明無法改變的情況下使用的“補(bǔ)丁”方案。例如當(dāng)函數(shù)使庫(kù)的一部分而無法修改時(shí)。
Note（注意）
A?const?member function can modify the value of an object that is?mutable?or accessed through a pointer member. A common use is to maintain a cache rather than repeatedly do a complicated computation. For example, here is a?Date?that caches (memoizes) its string representation to simplify repeated uses:
const類型的成員函數(shù)可以通過mutable對(duì)象或者借助指針成員修改對(duì)象的值。一個(gè)通常的用法是維護(hù)一個(gè)緩存以回避重復(fù)的計(jì)算。例如，下面的代碼中的Data類就為簡(jiǎn)單的重復(fù)使用緩存（記憶）了一個(gè)字符串表現(xiàn)。
class Date {
public:
// ...
const string& string_ref() const
{
if (string_val == "") compute_string_rep();
return string_val;
}
// ...
private:
void compute_string_rep() const; // compute string representation and place it in string_val
mutable string string_val;
// ...
};
Another way of saying this is that?constness is not transitive. It is possible for a?const?member function to change the value of?mutable?members and the value of objects accessed through non-const?pointers. It is the job of the class to ensure such mutation is done only when it makes sense according to the semantics (invariants) it offers to its users.
說明這件事的另一個(gè)方式是常量屬性是不可傳遞的。一個(gè)const成員函數(shù)修改mutable成員的值，或者通過一個(gè)非const指針訪問對(duì)象值都是可能的。保證這種修改只有在符合其向用戶提供的語(yǔ)義（不變量）是這個(gè)類的工作。
See also:?Pimpl
參見：I.27 考慮使用指向?qū)崿F(xiàn)的指針技術(shù)獲得穩(wěn)定的ABI
Enforcement（實(shí)施建議）
Flag a member function that is not marked?const, but that does not perform a non-const?operation on any member variable.
如果一個(gè)函數(shù)沒有定義為const類型，有沒有執(zhí)行針對(duì)任何成員變量的非常量操作，標(biāo)記它。
原文鏈接
https://github.com/isocpp/CppCoreGuidelines/blob/master/CppCoreGuidelines.md#con2-by-default-make-member-functions-const
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