3

所以,我有这两个函数ref()pointy()它们分别创建一个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;
};

代码编译得很好,它给出了一个关于返回局部变量地址的警告buckles,但我担心的是它没有说任何关于ref()返回引用的内容knuckles。我的编译器(通过 MinGW 的 g++,如果有影响的话)只是在工作中睡觉吗?引用是否有一些东西可以防止所指对象超出范围?还是我的参考语法不好?

非常感谢!

4

1 回答 1

4

您的两个测试用例不平行。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]
于 2012-04-11T16:20:19.663 回答