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。
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.
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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