If command problems.

This is a discussion on If command problems. within the C Programming forums, part of the General Programming Boards category; I feel so bad making another topic so soon after the other one. I was just wondering if there was ...

  1. #1
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    If command problems.

    I feel so bad making another topic so soon after the other one. I was just wondering if there was something wrong with my if statements here. When I choose an option other than option 1,2, or 3 it works fine the program looks back and asks for another option to be inputted.

    However when I input 1, 2, or 3 as the option, it's not working out.
    The positive if commands seem to be working. When I select option 1 it prints option 1, however it then prints option 3 also. When I select option 2, it prints option 2 however it then prints option 3 also. If I select option 3, it continues in the while loop even though it should pass it by.

    Hopefully this is just a small thing I've done wrong and quick to solve.
    Alex

    Code:
    //A program that calculates the reduced mass of diatomic molecules.
    //The following libraries will be used.
    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    
    Manual_Input()
    {
    
    }
    
    Textfile_Input()
    {
    
    }
    
    int main()
    {
    int selection;//Defines the variable used to decide on a keyboard or text file input.
    while (selection!=1||selection!=2)
    {
    
        printf("A program to calculate the reduced mass of diatomic molecules.\n");//Intro Message.
        Data_input://Allows the program to loop if the user selects a value other than 1,2, or 3.
        printf("Input data from the keyboard, from a text\nfile, or close the program?\n");
        printf("1. From the keyboard.\n");printf("2. From a text file.\n");printf("3. Close program.\n");printf(":");
        scanf("%d",&selection);//Allows the user to choose whether or not to input from a text file or the keyboard.
    
        if (selection==1)
            {
                printf("Hello");
            }
        if (selection==2)
            {
                printf("how are you?");
            }
        if (selection!=1 || selection!=2 || selection!=3)
            {
                printf("Invalid Option, please choose again. ");
            }
    
    }
    return 0;
    }

  2. #2
    SAMARAS std10093's Avatar
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    Code:
    int selection;
    while (selection!=1||selection!=2)
    Tell me what this code does
    Code - functions and small libraries I use


    It’s 2014 and I still use printf() for debugging.


    "Programs must be written for people to read, and only incidentally for machines to execute. " —Harold Abelson

  3. #3
    and the hat of wrongness Salem's Avatar
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    > if (selection!=1 || selection!=2 || selection!=3)
    You want && here, not ||

    At the moment, you have say
    selection = 1
    which means you have
    if ( false || true || true )

    But then if you have
    selection = 2
    you have
    if ( true || false || true )

    Your if statement as written is always true.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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    Quote Originally Posted by std10093 View Post
    Code:
    int selection;
    while (selection!=1||selection!=2)
    Tell me what this code does
    Well that code was providing the loop, it seemed to be working because it kept looping back.

  5. #5
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    Quote Originally Posted by Salem View Post
    > if (selection!=1 || selection!=2 || selection!=3)
    You want && here, not ||

    At the moment, you have say
    selection = 1
    which means you have
    if ( false || true || true )

    But then if you have
    selection = 2
    you have
    if ( true || false || true )

    Your if statement as written is always true.
    But if it wasn't 1 it doesn't necessarily mean False || True || True, because someone could have put in the wrong number like 4. No? Then they would all be false.

  6. #6
    SAMARAS std10093's Avatar
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    When I reach for first time the line of while (line 2 of the quoted code) want is the value of selection?
    Code - functions and small libraries I use


    It’s 2014 and I still use printf() for debugging.


    "Programs must be written for people to read, and only incidentally for machines to execute. " —Harold Abelson

  7. #7
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    Ermm it wouldn't have a value?

  8. #8
    SAMARAS std10093's Avatar
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    Exactly. So what's the behavior of the code?

    Better initialize your variable
    Code - functions and small libraries I use


    It’s 2014 and I still use printf() for debugging.


    "Programs must be written for people to read, and only incidentally for machines to execute. " —Harold Abelson

  9. #9
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    Well this is the thing, because I have a chemistry teacher and not a programming teacher who taught us, we didn't go into that depth of detail. I assumed before you assigned the value, it still existed just the value had an arbritray value of 0 or something like that.

  10. #10
    SAMARAS std10093's Avatar
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    What happens to a declared, unitialized variable in C?

    You may use the do while loop, if you don't want to initialize it
    Code - functions and small libraries I use


    It’s 2014 and I still use printf() for debugging.


    "Programs must be written for people to read, and only incidentally for machines to execute. " —Harold Abelson

  11. #11
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    Code:
    //A program that calculates the reduced mass of diatomic molecules.
    //The following libraries will be used.
    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    
    Manual_Input()
    {
    
    }
    
    Textfile_Input()
    {
    
    }
    
    int main()
    {
    int selection;//Defines the variable used to decide on a keyboard or text file input.
    selection=4;//Sets the value of the variabe selection so it can be used for the do while loop.
    
    do {
        printf("A program to calculate the reduced mass of diatomic molecules.\n");//Intro Message.
        Data_input://Allows the program to loop if the user selects a value other than 1,2, or 3.
        printf("Input data from the keyboard, from a text\nfile, or close the program?\n");
        printf("1. From the keyboard.\n");printf("2. From a text file.\n");printf("3. Close program.\n");printf(":");
        scanf("%d",&selection);//Allows the user to choose whether or not to input from a text file or the keyboard.
    
        if (selection==1)
            {
                printf("Hello");
            }
        if (selection==2)
            {
                printf("how are you?");
            }
        if (selection!=1 || selection!=2 || selection!=3)
            {
                printf("Invalid Option, please choose again. ");
            }
    
    } while (selection!=1 || selection!=2 || selection!=3);
    return 0;
    }
    I've changed the code to this, and also set the value of the selection variable so it has a different value to that of 1, 2 or 3.

    Code:
    int selection;//Defines the variable used to decide on a keyboard or text file input.
    selection=4;//Sets the value of the variabe selection so it can be used for the do while loop.
    However the loop doesn't seem to be working. Even when I set the value to 1, 2 or 3 the loop repeats.

  12. #12
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    Quote Originally Posted by Salem View Post
    > if (selection!=1 || selection!=2 || selection!=3)
    You want && here, not ||

    At the moment, you have say
    selection = 1
    which means you have
    if ( false || true || true )

    But then if you have
    selection = 2
    you have
    if ( true || false || true )

    Your if statement as written is always true.
    Never mind, I maybe misunderstood how if statements work. This is now working. Thank you very much.

  13. #13
    SAMARAS std10093's Avatar
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    Read again Salem's post

    Also when you say doesn't work, specify what is happening

    //you just read it :P
    Code - functions and small libraries I use


    It’s 2014 and I still use printf() for debugging.


    "Programs must be written for people to read, and only incidentally for machines to execute. " —Harold Abelson

  14. #14
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by Alex Saunders View Post
    Never mind, I maybe misunderstood how if statements work. This is now working. Thank you very much.
    No, it appears that you misunderstood boolean logic, not if-statements.
    My homepage
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