Cout' Does Not Name a Type

‘cout’ does not name a type

The problem is that the code you have that does the printing is outside of any function. Statements that aren't declarations 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;
}

Error, cout does not name a type

Is it possible you forgot to include iostream?

#include <iostream>

This has to go to the top of your code.

Also in order for cout to work without std:: in front of it you have to use the namespace:

using namespace std;

You cant write code outside a function. You need at least an entry point:

//includes and namespaced go here

int main()
{
    //enter your code here
}

I strongly suggest you work through basic tutorials.

Your updated code has some errors:

system("CLS")

You forgot the ";" at the end

if (button == n)

You dont have a variable named "n". You probably meant to write:

    if (button == 'n')

Also in line 41 and 42, you have "\"s in your string. Those characters have a special meaning. Write another "\" before them to fix those errors.

The error is here:

int main(button){

It should be

int main(){

Ok i fixed your code:

#include <iostream> 
#include <stdlib.h>

using namespace std;
char button = 'a';

int main() {
    cout << "                                                                               \n";
    cout << "   *********           ********         *          ****     ********           \n";
    cout << "       *              *                * *        *     *   *                  \n";
    cout << "       *             *                *   *       *      *  *                  \n";
    cout << "       *             *               *     *      *     *   *****              \n";
    cout << "       *             *              *********     * ****    *                  \n";
    cout << "       *             *             *         *    * *       *                  \n";
    cout << "       *              *           *           *   *   *     *                  \n";
    cout << "   *********           ********  *             *  *     *   ********           \n";
    cout << "                                                                               \n";
    cin >> button;

    if (button == 'n')
    {
        system("CLS");
        cout << "    ***                                                \n";
        cout << "   * ..*     ***                                       \n";
        cout << "   * u *    * ..*     ***        .                     \n";
        cout << "    ***     * u *    * ..*      ***                    \n";
        cout << "     *       ***     * u *     * ..*                   \n";
        cout << "    ***       *       ***      * u *                   \n";
        cout << "   * * *     ***      ***       ***                    \n";
        cout << "   * * * *   * * *   * * *     *****   O               \n";
        cout << "   *   * * *   *  * * * * ** ** *** * /|\\              \n";
        cout << "   *    *    *   *   *   *     *   *   /\\              \n";
    }
    else
    {
        system("CLS");
        cout << "Invalid key enter n.";
    }
}

Why do I get error 'cout' in namespace 'std' does not name a type when I use using cout = std::cout;?

using cout = std::cout; refers to type alias declaration syntax. It's similar to typedef; so you're trying to declare a type named cout that refers to a previously defined type std::cout. But std::cout is not a type name, it's an object with type of std::ostream.

As the error message said, it's just trying to tell you that std::cout doesn't refer to a type name.

Multilevel Inheritance Error cout does not name a type

cout<<"\nEnter Patient ID\n"; is a statement that needs to be inside a function.

This, and your other statements, are floating somewhere in the class definition. This is not syntactically valid.

This is confusing the compiler and is issuing a somewhat cryptic error.

cout/cin does not name a type error

The body of the class declaration can only contain members, which can either be data or function declarations, and optionally access specifiers.

Wrap your code inside a function and then call that in main by an object

class distanceFormula {
public:
int speed;
int time;
int distance;
void init()
{
    cout << "What is the speed?" << endl;
    cin >> speed;

    cout << "How long did the action last?" << endl;
    cin >> time;
    distance = speed * time;
    cout << "The distance traveled was " << distance << endl;
}
};

int main()
{
    distanceFormula ao;
    ao.init();
    return 0;
};