Thread: Unable to accept input

  1. #1
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    Question Unable to accept input

    Hell all, I have never experienced this issue before. I am unable to accept input for
    Code:
    divisor
    in the following code block:
    Code:
    #include<stdio.h>
    void divide(int dividend, int divisor, int *quotient, int *remainder)
    {
      *quotient = dividend / divisor;
      *remainder = dividend % divisor;
    }
    
    int main()
    {
      double dividend, divisor, quotient = 0;
      int remainder = 0;
      printf("Enter Dividend: ");
      scanf("&d", &dividend);
      printf("Enter Divisor:");
      scanf("&d", &divisor);
    
    }
    I have tried to but a blank space and a "\n" before %d on the divisor input but to no avail. This is very odd. The terminal accepts the value for dividend, prints "Enter divisor: " and just terminates like so:
    Code:
    Enter Dividend: 6
    Enter Divisor:
    C:\cProg>
    Any help appreciated.
    Last edited by Salem; 06-20-2021 at 12:51 AM. Reason: Removed crayola

  2. #2
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    Code:
    scanf("&d", &dividend);
    You likely wanted
    Code:
    scanf("%d", &dividend);
    Notice I used "%d" instead of "&d".

    Tim S.
    "...a computer is a stupid machine with the ability to do incredibly smart things, while computer programmers are smart people with the ability to do incredibly stupid things. They are,in short, a perfect match.." Bill Bryson

  3. #3
    Registered User rstanley's Avatar
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    On top of that, your divide() function takes ints and int pointers, however, in main() they are declared as doubles, except for the int reminder.

    You need to be consistent.

  4. #4
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    Quote Originally Posted by stahta01 View Post
    Code:
    scanf("&d", &dividend);
    You likely wanted
    Code:
    scanf("%d", &dividend);
    Notice I used "%d" instead of "&d".

    Tim S.
    My God after countless hours of coding, this is the biggest brain freeze I have experienced since...thank you kind sir.

  5. #5
    and the hat of int overfl Salem's Avatar
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    Compiling with warnings enabled would have told you something was wrong from the start, without the "hours of coding".

    Code:
    $ cat foo.c
    #include<stdio.h>
    void divide(int dividend, int divisor, int *quotient, int *remainder)
    {
      *quotient = dividend / divisor;
      *remainder = dividend % divisor;
    }
    
    int main()
    {
      double dividend, divisor, quotient = 0;
      int remainder = 0;
      printf("Enter Dividend: ");
      scanf("&d", &dividend);
      printf("Enter Divisor:");
      scanf("&d", &divisor);
    
    }
    $ gcc -Wall foo.c
    foo.c: In function ‘main’:
    foo.c:13:9: warning: too many arguments for format [-Wformat-extra-args]
       scanf("&d", &dividend);
             ^
    foo.c:15:9: warning: too many arguments for format [-Wformat-extra-args]
       scanf("&d", &divisor);
             ^
    foo.c:11:7: warning: unused variable ‘remainder’ [-Wunused-variable]
       int remainder = 0;
           ^
    foo.c:10:29: warning: unused variable ‘quotient’ [-Wunused-variable]
       double dividend, divisor, quotient = 0;
                                 ^
    $
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  6. #6
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    Shoudn't be %f instead of %d?

  7. #7
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    Quote Originally Posted by flp1969 View Post
    Shoudn't be %f instead of %d?
    Nope, everything should be an int here, especially since % is used.

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