Newbie need help in repeating a program after user input

This is a discussion on Newbie need help in repeating a program after user input within the C Programming forums, part of the General Programming Boards category; Hi,i am currently having problem with the following code. I want the program to ask the user whether he wants ...

  1. #1
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    Newbie need help in repeating a program after user input

    Hi,i am currently having problem with the following code. I want the program to ask the user whether he wants to repeat himself at the end of the program.This is what i come up with and i am not sure why it doesn't work.Hope u all can help me.THANK!
    Code:
    #include <stdio.h>
    
    int main(void){
    int age;
    char answer;
    
    do{
        printf("Please enter a number between 1 and 100.\n");
        scanf("%d", &age);
    if (age>= 1 && age<= 20 )
        printf("you are very young\n");
    else if (age> 20 && age<= 50 )
        printf("you are old\n");
    else {
        printf("u are very old\n");
    }
        printf("Do u want to continue?\n");
        scanf("%c",&answer);
        printf("\n");
    
    }while (answer ='Y');
    
    }

  2. #2
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    Do you know the difference between = and ==?


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    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

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    Quote Originally Posted by matsp View Post
    Do you know the difference between = and ==?


    --
    Mats
    Sorry, am not really that that sure when should use = and ==, i only know i have to use = when declaring variable?

  4. #4
    Woof, woof! zacs7's Avatar
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    Then you probably didn't write
    if (age>= 1 && age<= 20 )
    did you?

    Have a look at the tutorials on this website, or the list of books at the top of the C board.

  5. #5
    Registered User Dogmasur's Avatar
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    I am fairly new to programming myself, but I noticed something about your code that I thought I would comment on.

    What if the user answers with yes instead of 'Y'? Maybe you should clarify that they should input 'Y' or 'N'.
    "The art of living is more like wrestling than dancing." - Marcus Aurelius

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    Whilst nitpicking:
    Code:
        printf("Please enter a number between 1 and 100.\n");
    Would it not be better to ask for "enter your age", rather than "enter a number ..." - if the user knows what the number is used for, it will help them input something that is meaningfull. If I get a prompt to enter a number 1-100, I may well enter 42 - but that's not my age [it's close, and the output would actually be the same - but still].

    And to explain:
    = assigns a value. E.g. a = 7;
    == compares a value. E.g. a == 1 is true if a has the value 1.
    Your while-condition sets answer to 'Y'. And because the way that C works, it then says "is the value we set answer to non-zero" - yes, then continue the loop. So I expect that your program loops forever, whatever the user inputs.

    Your main should also end with "return 0;" to indicate that you successfully finished.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

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    Quote Originally Posted by matsp View Post
    Whilst nitpicking:
    Code:
        printf("Please enter a number between 1 and 100.\n");
    Would it not be better to ask for "enter your age", rather than "enter a number ..." - if the user knows what the number is used for, it will help them input something that is meaningfull. If I get a prompt to enter a number 1-100, I may well enter 42 - but that's not my age [it's close, and the output would actually be the same - but still].

    And to explain:
    = assigns a value. E.g. a = 7;
    == compares a value. E.g. a == 1 is true if a has the value 1.
    Your while-condition sets answer to 'Y'. And because the way that C works, it then says "is the value we set answer to non-zero" - yes, then continue the loop. So I expect that your program loops forever, whatever the user inputs.

    Your main should also end with "return 0;" to indicate that you successfully finished.

    --
    Mats
    Thanks everybody for the help!
    I dun really understand what u mean by this
    " And because the way that C works, it then says "is the value we set answer to non-zero" - yes, then continue the loop."
    Do u mean that the C will only not loop if the answer is zero?
    I knew there is something wrong with my while function but i still dun really understand why doesn't compare correctly to my input.
    Thanks!

  8. #8
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    Quote Originally Posted by dragee View Post
    Thanks everybody for the help!
    I dun really understand what u mean by this
    " And because the way that C works, it then says "is the value we set answer to non-zero" - yes, then continue the loop."
    Do u mean that the C will only not loop if the answer is zero?
    I knew there is something wrong with my while function but i still dun really understand why doesn't compare correctly to my input.
    Thanks!
    In C, any condition (as in, if, while and similar), the determination of whether it is true or false is "if it's not zero, it's true".

    So if you do "if (x = y)" then we are assiging x with the value of y, then checking if the "result" is zero - so if y is zero, then x is zero and the "result" is zero - that is, false. If y is not zero, then the whole thing is true.

    Same with your "answer = 'Y'" - you are setting the variable answer to 'Y'. Since 'Y' is not the value zero, so the loop continues. (In ASCII/ANSI, 'Y' is 89 - there are other values that may generate the letter Y on some other computers, I'm 99.999% sure that there is not a single one where the letter 'Y' is zero - because zero is very very often used as a special character, such as marking the end of strings and such).

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

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