Not declared in this scope error, anyone can help?

[Gregzksi]

It says int i is not declared in this scope (do while loop) and I can't figure out how it can't be as I'm declaring it in the while condition, can anyone help me? Thanks!
( please keep in mind I can't use any other statements or anything unless it's in this code as I haven't learned it and I'm trying to build this with what if learned so far, this is a test for chapter 5: loops in JumpingintoC++ (I also tryd with bool but seems to make no difference) )

This is the full code, the part that's giving the "i was not declared in this scope" is in the do while loop.

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

int main()
{
    int a;
    cout << "Press enter to start the 99 bottles song.\nPress enter everytime to continue the song. (or hold enter)\n";
    cin.ignore();
    for (a = 99; a > 0; -a) // Counts down until 0
    {
    cout << "" << a << " bottle(s) of beer on the wall," " " << a << " bottle(s) of beer.\n";
    cout << "Take one down, pass it around," " " << --a << " bottle(s) of beer on the wall.\n";
    cin.ignore ();
    if (a < 1)
    {
        cin.ignore();
        cout << "I hope you sang along!\n";
        cin.ignore ();
        break;
    }
    }
    do
    {
    int i;
    string stop;
    stop = "quit";
    cout << "What would you like to do? Type the option number to iniate it. (ex. 1)\nType quit to stop the program\n";
    cout << "Option 1: Login\nOption 2: Math\nOption3: Bottlesong ";
    cin >> i;
    }while (i != 1 || != 2 || != 3);
    if (i == 1)
    {

    }
    else if (i == 2)
    {

    }
    else if (i == 3)
    {

    }
}

[AbstractionAnon]

You're getting the errors on the following statement:

 
    } while (i != 1 || != 2 || != 3);

i is only defined within the braces of your do-while loop. Your conditionals are outside the braces and therefore out of scope

edit: Also your conditionals are defective. Your loop should look like this:

 
    int i;
    do
    {   string stop;
        
        stop = "quit";
        cout << "What would you like to do? Type the option number to iniate it. (ex. 1)\nType quit to stop the program\n";
        cout << "Option 1: Login\nOption 2: Math\nOption3: Bottlesong ";
        cin >> i;
    } while (i < 1 || i > 3);

Note that i has been moved outside the loop and that i is included in each term of the while condition.

PLEASE YOU CODE TAGS (the <> formatting button) when posting code.

[Gregzksi]

It seems I've fixed it so far by putting int i next to my a ontop, and changing the while code to:
}while (!(i=1) || !(i=2) || !(i=3));

I'm going to test if it works now, atleast it compiles now, is this the correct way?

UPDATE: It compiles now, but I know ! equals not, so how could I make it work so it loops aslong as int i is not 1,2 or 3?

[AbstractionAnon]

Your while loop isn't going to work as you expect it to.

(i=1)

is an assignment statement, not a comparison.
The result of that will always be true.

[Gregzksi]

Thanks for the replies, I've fixed it myself by doing this:
while ((i!=1) && (i!=2) && (i!=3));
but you're right Abstraction, you're statement makes more sense and is logical and short, thanks for the advice!

[Gregzksi]

It seems to be working now, but the problem I'm having now is that because I'm using integers and using while (i < 1 || i > 3) the moment I type characters it goes into rage-loop and it won't stop unless I stop it trough my Windows task manager, how could I fix this?

[AbstractionAnon]

A classic issue with cin >> intval;
Two alternatives:
1) Use getline with a string and test the string is numeric, then convert it to an integer.
2) Use a char value for your input, then test against the char value.

char choice;
  cin >> choice;
  switch (choice)
  { case '1': // do choice 1
                 break;
     //  other cases
     default: cout << Not a valid input' << endl;
   } 

Note the use of the character value in the cases.

[Gregzksi]

    int a;
    string i;
    do
    {
    cout << "What would you like to do? Type the option number to iniate it. (ex. 1)\n";
    cout << "Option 1: Login\nOption 2: Math\nOption 3: 99 Bottles song\nStart option: ";
    getline (cin, i);
    }while (i !="1" || i!="2" || i!="3");        // ((i!=1) && (i!=2) && (i!=3)) same thing, longer
    if (i == "1")
    {

    }
    else if (i == "2")
    {

    }
    else if (i == "3")

it works this way, and it doesnt crash/rageloop when pressing chars, but now it won't accept any numeric inputs, even tryed "3"

[AbstractionAnon]

The problem is your while condition again.
Your loop is going to repeat while any of the conditions is true.
if i is "1", then (i!="2") and (i!="3") will both be true, therefore your loop will repeat.

Use && instead of ||

while (i !="1" && i!="2" && i!="3");        

or the while condition I showed you earlier (using quoted values).

while (i < "1" || i > "3");

[Gregzksi]

Thanks, that helped! I dind't use and because I thought it checks && before anything else, or atleast I read it that way. Thanks!