[英]Global and local variables in C++
我是C ++的新手,在打印局部变量和全局变量时遇到了一些问题。 考虑这段简单的代码:
#include <cstdlib>
#include <iostream>
using namespace std;
/*
*
*/
int x = 10; // This x is global
int main() {
int n;
{ // The pair of braces introduces a scope
int m = 10; // The variable m is only accessible within the scope
cout << m << "\n";
int x = 25; // This local variable x hides the global x
int y = ::x; // y = 10, ::x refers to the global variable x
cout << "Global: " << y << "\n" << "Local: " << x << "\n";
{
int z = x; // z = 25; x is taken from the previous scope
int x = 28; // it hides local variable x = 25 above
int t = ::x; // t = 10, ::x refers to the global variable x
cout << "(in scope, before assignment) t = " << t << "\n";
t = x; // t = 38, treated as a local variableout was not declared in this scope
cout << "This is another hidden scope! \n";
cout << "z = " << z << "\n";
cout << "x = " << x << "\n";
cout << "(in scope, after re assignment) t = " << t << "\n";
}
int z = x; // z = 25, has the same scope as y
cout << "Same scope of y. Look at code! z = " << z;
}
//i = m; // Gives an error, since m is only accessible WITHIN the scope
int m = 20; // This is OK, since it defines a NEW VARIABLE m
cout << m;
return 0;
}
我的目标是在各种范围内实现变量的可访问性,然后打印它们。 但是,我无法弄清楚为什么当我尝试打印最后一个变量z
,NetBeans会返回输出2025
。 这是我的示例输出:
10
Global: 10
Local: 25
(in scope, before assignment) t = 10
This is another hidden scope!
z = 25
x = 28
(in scope, after re assignment) t = 28
Same scope of y. Look at code! z = 2520
RUN FINISHED; exit value 0; real time: 0ms; user: 0ms; system: 0ms
希望有人能帮我理解发生的事情! :)
是不是z保持值2520是你在打印z和打印m之间省略添加新的行操作符的事实...
你在做:
cout << "Same scope of y. Look at code! z = " << z;
}
int m = 20;
cout << m;
但你应该这样做:
std::cout << "Same scope of y. Look at code! z = " << z << std::endl;
}
int m = 20;
std::cout << m << std::endl;
如果你只是遵循相同的标记输出和做类似的标准
std::cout << "M is: "<<m << std::endl;
你会通过观察输出更快地发现问题:
25M is: 20
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.