'cout'不是一个类型 [英] ‘cout’ does not name a type

问题描述

我正在学习Adam Drozdek的书C ++中的数据结构和算法，我在vim中输入了第15页的代码，并将其编译到Ubuntu 11.10的终端。

I was learning Adam Drozdek's book "Data Structures and Algorithms in C++", well, I typed the code in page 15 in my vim and compiled it in terminal of my Ubuntu 11.10.

#include <iostream>
#include <cstring>
using namespace std;

struct Node{
    char *name;
    int age;
    Node(char *n = "", int a = 0){
        name = new char[strlen(n) + 1];
        strcpy(name, n);
        age = a;
    }
};

Node node1("Roger", 20), node2(node1);
cout << node1.name << ' ' << node1.age << ' ' << node2.name << ' ' << node2.age;
strcpy(node2.name, "Wendy");
node2.name = 30;
cout << node1.name << ' ' << node1.age << ' ' << node2.name << ' ' << node2.age;

但有一些错误：

oo@oo:~$ g++ unproper.cpp -o unproper
unproper.cpp:15:23: warning: deprecated conversion from string constant to ‘char*’ [-Wwrite-strings]
unproper.cpp:16:1: error: ‘cout’ does not name a type
unproper.cpp:17:7: error: expected constructor, destructor, or type conversion before ‘(’ token
unproper.cpp:18:1: error: ‘node2’ does not name a type
unproper.cpp:19:1: error: ‘cout’ does not name a type

我搜索过此，此， this 和这，但我找不到答案。

I have searched this,this,this and this, but I can't find the answer.

）

推荐答案

我认为问题是，你有打印的代码是在任何函数之外。 C ++中的语句需要在函数内部。例如：

I think the problem is that the code you have that does the printing is outside of any function. Statements in C++ need to be inside a function. For example:

#include <iostream>
#include <cstring>
using namespace std;

struct Node{
    char *name;
    int age;
    Node(char *n = "", int a = 0){
        name = new char[strlen(n) + 1];
        strcpy(name, n);
        age = a;
    }
};


int main() {
    Node node1("Roger", 20), node2(node1);
    cout << node1.name << ' ' << node1.age << ' ' << node2.name << ' ' << node2.age;
    strcpy(node2.name, "Wendy");
    node2.name = 30;
    cout << node1.name << ' ' << node1.age << ' ' << node2.name << ' ' << node2.age;
}

希望这有助！

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