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‘cout’ does not name a type

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Born Creative answered on June 17, 2020 Popularity 6/10 Helpfulness 3/10

answer ‘cout’ does not name a type

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‘cout’ does not name a type

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#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;

Popularity 6/10 Helpfulness 3/10 Language whatever

Source: stackoverflow.com

Tags: whatever

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Contributed on Jul 23 2022

Born Creative

0 Answers Avg Quality 2/10

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cout does not name a type

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//Statements in C++ need to be inside of a function
int main(){
	std::cout << "Hello World" << std::endl;
    //Works because we are inside of a function
}
std::cout << "Hello World" << std::endl;
//Doesn't work because we are not inside of a function

Popularity 9/10 Helpfulness 9/10 Language cpp

Source: Grepper

Tags: cout c++ cou

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Contributed on Jun 17 2020

Upset Unicorn

0 Answers Avg Quality 2/10

‘cout’ does not name a type

Comment

#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;

Popularity 6/10 Helpfulness 1/10 Language whatever

Source: stackoverflow.com

Tags: whatever

Link to this answer
Share Copy Link

Contributed on Jul 23 2022

Born Creative

0 Answers Avg Quality 2/10

‘cout’ does not name a type

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‘cout’ does not name a type

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cout does not name a type

‘cout’ does not name a type

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