[PS: your clock is off by a month]
Chris wrote:
> what am i doing wrong here. it compiles and works but after you enter in
> the first value it just stops. /*
> if ( ( j = read_x1( x1 ) ) == 'E' );
> return EXIT_FAILURE;
[snip]
> float read_x1( float x1 )
>
> {
>
> printf( "\nenter x1: " );
>
> if ( scanf( "%f" , &x1 ) != 1 )
> {
> printf( "\nFailed to read entered value\n" );
> x1 = 'E';
> return x1;
> }
>
> return x1;
>
> }
1. This (and the x2/y1/y2 versions) won't work. C uses call-by-value,
not call-by-reference. You would use either
float read_x1( void )
or
void read_x1(float *px1)
instead.
2. there isn't any point having four (nearly) identical functions,
which differ only in the prompt. Use something like:
float read_float(const char *prompt)
{
float x;
printf( "enter %s: ", prompt );
fflush( stdout );
if ( scanf( "%f" , &x ) != 1 )
{
fprintf( stderr, "Failed to read entered number\n" );
return NAN; /* IEEE Not-A-Number */
}
return x;
}
3. However, the reason why your program is failing is that you are
unconditionally returning after the call to read_x1(). The indentation
doesn't match the presence of the semicolons; you wrote:
if ( ( j = read_x1( x1 ) ) == 'E' );
return EXIT_FAILURE;
but it would be more accurate to write
if ( ( j = read_x1( x1 ) ) == 'E' )
;
return EXIT_FAILURE;
I.e. you are conditionally executing an empty statement, then
unconditionally returning. Remove the semicolon, i.e.
if ( ( j = read_x1( x1 ) ) == 'E' ) /* no semicolon here */
return EXIT_FAILURE;
4. This has nothing to do with Linux. The above is all vanilla ANSI C,
and would apply equally to MS-DOS or VMS.
5. I think that you need to get a book on C.
--
Glynn Clements <[EMAIL PROTECTED]>
