Trouble with receiving input

This is a discussion on Trouble with receiving input within the C Programming forums, part of the General Programming Boards category; I am having trouble with receiving input from the user in my programs. I have a situation where the user ...

  1. #1
    Registered User BellosX's Avatar
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    Angry Trouble with receiving input

    I am having trouble with receiving input from the user in my programs. I have a situation where the user has to input whether or not they are ready to continue, and when it gets to there, it immediately follows on as if they said yes. When i went to debug it, i got my program to print out the variable which contained the input, and i found it was 10. If i'm not mistaken, 10 is the numeric value of the newline character (\n). I was wondering if there is anyway i can bypass my program spitting 10 when there is a scanf() or a getc() / getchar() function.

    Thanks

  2. #2
    Ethereal Raccoon Procyon's Avatar
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    What to do depends on the functions and variables you're using to collect your input. Do you think you could post that part of the code?

  3. #3
    Guest Sebastiani's Avatar
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    Yes post the code!!!
    Code:
    bool fun(bool value)
    {
        return std::pow(std::exp(1), std::complex<float>(0, 1) 
        * std::complex<float>(std::atan(1)*(1 << (value + 2))))
        .real() > 0;
    }

  4. #4
    and the hat of wrongness Salem's Avatar
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    > If i'm not mistaken, 10 is the numeric value of the newline character (\n).
    This is correct.
    Welcome to the strange world of scanf - a wonderful function for converting user input, but has way too many traps for the unwary.

    The thing you've got to remember about scanf is that it only uses the minimum number of characters possible to satisfy the current conversion requirement (even when it fails).

    If this is your scanf call
    &nbsp; int res = scanf( "%d", &my_int );

    An input buffer of "4\n" would have res=1 (the number of sucessful conversions and assignments), my_int = 4 (the number you typed in), and "\n" left on the input stream.

    An input buffer of "Four\n" would result in res=0, my_int unmodified, and "Four\n" still in the input buffer.

    Code:
    #include <stdio.h>
    
    void flush_input ( FILE *fp ) {
        int ch;
        while ( (ch=fgetc(fp)) != EOF && ch != '\n' );
    }
    
    int main ( ) {
        int res, my_int;
        printf( "Type in a number > " );
        fflush( stdout );   /* the only correct use of fflush is on output streams */
        res = scanf( "%d", &my_int );
        if ( res == 1 ) {
            printf( "You typed in %d\n", my_int );
        } else {
            printf( "That wasn't a number\n" );
        }
        printf( "Press return to finish >" );
        fflush( stdout );
        flush_input( stdin );
        getchar();
        return 0;
    }
    The better way of coping with this is to use fgets and sscanf.
    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.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

  5. #5
    Registered User BellosX's Avatar
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    I've worked it out now, thanks and sorry i didn't respond to your requests i have been very busy lately
    Thanks Salem for your advice!

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