c++ - 编译器警告 re : references to local variables

Question

所以，我有这两个函数ref()和 pointy()分别创建本地引用和指向本地定义的 int 的指针:

#include <iostream>

int& ref() {
  int knuckles = 6;
  int &chuckles = knuckles;
  return chuckles;
};

int* pointy() {
    int buckles = 8;
    return &buckles;
};

int main(int argc, char **argv) {
    int a = ref(), *b = pointy();
    int c = 14, d = 20;
    std::cout << a << ' ' << *b << ' ' << c+d;
};

代码编译正常，并给出关于返回局部变量地址的警告 buckles ，但令我担心的是它没有说明任何关于 ref() 的内容返回对 knuckles 的引用.我的编译器(g++ 通过 MinGW，如果它有所作为)只是在 sleep 吗？是否有关于引用的内容可以防止引用对象超出范围？还是我的引用语法不好？

非常感谢!

Answer 1

您的两个测试用例不是平行的。这个程序:

#include <iostream>

int& ref() {
  int knuckles = 6;
  return knuckles;
};

int* pointy() {
    int buckles = 8;
    return &buckles;
};

int main(int argc, char **argv) {
    int a = ref(), *b = pointy();
    int c = 14, d = 20;
    std::cout << a << ' ' << *b << ' ' << c+d;
};

产生这些警告:

$ g++ -O4 -std=c++0x    x.cc   -o x
x.cc: In function ‘int& ref()’:
x.cc:4:7: warning: reference to local variable ‘knuckles’ returned [enabled by default]
x.cc: In function ‘int* pointy()’:
x.cc:9:9: warning: address of local variable ‘buckles’ returned [enabled by default]