While loop with OR conditions, beginner

This is a discussion on While loop with OR conditions, beginner within the C++ Programming forums, part of the General Programming Boards category; I am just beginning and going through Jumping Into C++ . I am working #2 in chapter 5 which states... ...

  1. #1
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    While loop with OR conditions, beginner

    I am just beginning and going through Jumping Into C++. I am working #2 in chapter 5 which states... "Write a menu program that lets the user select from a list of options, and if the input is not one of the options, reprint the list"
    Code:
    #include <iostream>
    #include <string>
    using namespace std;
    int main()
    {
        string input = "unknown";
        while ( input != "1" || input != "2" || input != "3" || input != "4" || input != "5")
        {
            cout << "file (1)\t edit (2)\t view (3)\t search (4)\t project (5)\n";
            cout << "Select a function: ";
            cin >> input;
        }
    }
    No matter what I enter the loop never breaks. I am pretty sure my "while" line is not right but I don't know. I know this is elementary stuff but I want to understand the easy stuff before it gets really hairy.

  2. #2
    C++ Witch laserlight's Avatar
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    Suppose input == "1". Then input != "1" is false, so we check if input != "2". Well, it is true that input != "2", so we continue looping. Problem is, when input == "1", you want to stop looping. So, under what conditions does (input != "1" || input != "2" || input != "3" || input != "4" || input != "5") ever evaluate to false? Well, when (input == "1" && input == "2" && input == "3" && input == "4" && input == "5"). Unfortunately, input can only hold one of these values at any one time.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

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    Let me put my logic hat on and ponder a bit.

    Thanks for the reply. I will re-tool a little.

  4. #4
    Registered User manasij7479's Avatar
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    You could simplify it a little:
    Code:
    while( input[0]>'0' && input[0] <'6')
    Manasij Mukherjee | gcc-4.9.2 @Arch Linux
    Slow and Steady wins the race... if and only if :
    1.None of the other participants are fast and steady.
    2.The fast and unsteady suddenly falls asleep while running !



  5. #5
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by manasij7479 View Post
    You could simplify it a little:
    Code:
    while( input[0]>'0' && input[0] <'6')
    That would be wrong though. It should be:
    Code:
    while (input.length() != 1 || input[0] < '1' || input[0] > '5')
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

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    If you want a bit more structure, you can do something like:

    Code:
    #include <iostream> 
    #include <string> 
    
    using namespace std; 
    
    int main() 
    { 
       int choice = 0;
       do
       {
          cout << "file (1)\t edit (2)\t view (3)\t search (4)\t project (5)\t quit (6)\n"; 
          cout << "Select a function: "; 
          cin >> choice; 
          switch( choice )
          {
             case 1: cout << "file" << endl;
             case 2: cout << "edit" << endl;
             case 3: cout << "view" << endl;
             case 4: cout << "search" << endl;
             case 5: cout << "project" << endl;
          }
       } 
       while( choice != 6 ) //loop stops on 6
    }


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