在C ++中返回局部变量（有效C ++中的规则21，第3版） [英] Returning local variables in C++ (Rule 21 in Effective C++, 3rd edition)

问题描述

众所周知，从C ++中的函数返回局部变量是不安全的，由于范围。
在Effective C ++ Third Edition中，Scott Meyers在第21页第101页讲述了这个问题。但是，他总结说，正确的决定是写：

  inline const Rational运算符*（const Rational& lhs，const Rational& rhs）{
 return Rational（lhs.n * rhs.h，lhs.d * rhs .d）; 
} 
  

这不是一个不好的做法， / p>

UPD：感谢大家解释。

解决方案

实际返回一个局部变量。您可以返回局部变量的值，或指向它的指针（即其地址）或对其的引用。

返回局部变量的值是完全安全的：

  int good_func（）{
 int local = 42; 
 return local; //返回值local的副本
} 
  

返回 >指针或引用到局部变量是不安全的，因为变量在函数终止时不再存在：

  int * bad_func（）{
 int local = 42; 
 return& local; //返回指向local的指针
} 
  

除了C没有C ++风格的引用。）

As known, returning local variable from function in C++, is unsafe, due to scoping. In Effective C++ Third Edition, Scott Meyers tells about this problem in item 21, at page 101. However, in conclusion he said, that right decision will be to write:

inline const Rational operator*(const Rational& lhs, const Rational& rhs) {
    return Rational(lhs.n * rhs.h, lhs.d * rhs.d);
}

Isn't this also a bad practice, and this function is unsafe?

UPD: Thanks everybody for explanation.

解决方案

You can't actually return a local variable. You can return the value of a local variable, or a pointer to it (i.e., its address), or a reference to it.

Returning the value of a local variable is perfectly safe:

int good_func() {
    int local = 42;
    return local; // returns a copy of the value of "local"
}

Returning a pointer or reference to a local variable is unsafe, because the variable ceases to exist when the function terminates:

int* bad_func() {
    int local = 42;
    return &local; // returns a pointer to "local"
}

(The same applies in C, except that C doesn't have C++-style references.)

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