Why does C++ say mynumber does not name a type

Question

1.00/5 (1 vote)

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My Code:

#include <iostream>

using namespace std;

int main();
myNumber == '2';
{
   if (myNumber == '200')
 {
    return 0
 }
else
{
    myNumber + 2;
    cout << myNumber << endl;
}
}

What I have tried:

I've tried changing the variable

Posted 23-Apr-20 5:42am

Bahooka

Updated 24-Jun-22 20:05pm

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raddevus · Accepted Answer · 2020-04-23T05:52:00

C++

mynumber == '2'

is an evaluation meaning is mynumber equal to the character '2'.

You have to declare the type of a variable before using it:
char mynumber = '2';

And keep in mind that you are using numbers as a char -- that is what the single-quotes are defining in your code.

Also, mynumber == '200' will not work. It's incorrect syntax.
You really want to declare the myNumber as an integer type, like:
int myNumber = 2; // no single-quotes.
You probably want something like:

C++

int main()
{
   int myNumber = 2;
   if (myNumber == 200)
   {
     return 0;
   }
  else
  {
    myNumber += 2;
    cout << myNumber << endl;
  }
} // end of main

Note: this code only runs one time, because there is no looping mechanism (while or for loop).

OriginalGriff · Accepted Answer · 2020-04-23T05:55:00

Couple of reasons for this.
0) You don't need a semicolon here:

C++

int main();

1) Your curly bracket is in the wrong place:

C++

int main();
myNumber == '2';
{

It should be immediately below the function declaration:

C++

int main()
   {
   myNumber == '2';

2) myNumber is not defined, so it doesn't have any type that C++ can know. Try

C++

int myNumber == '2';

3) "==" is a comparison operator, not an assignment, so even if you had declared it myNumber would be have no "real" value. Use "=" instead - that's the assignment operator.
4) '2' is a character, not an integer - you probably want to use numbers here.
5) '200' is not a character, it's three characters - so that won't compile either, and you can't compare that with a single character even if you wanted to.
6)

C++

myNumber + 2;

Does nothing: it does not change any values in the program at all. Probably, you meant

C++

myNumber += 2;

Or

C++

myNumber = myNumber + 2;

7) What did you expect '2' plus 2 to give you? Would it still work if you had assigned '9' instead?
8) Indent your code! pick a style, and stuck to it - poor indentation makes code harder to read, and that means it's harder to get right.

So little code, so many errors! :laugh:

Why does C++ say mynumber does not name a type

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