do while loop

[jlharrison]

Whats wrong with this loop?
I cant proceed with the program even when S or R is input.
I can exit if I take the || out but I need to repeat untill I get a R or a S.

do
		{
		printf ("Enter R for Required or S for Suggested: ");
		while (getchar() != '\n');
		reqOrSugg = getchar();
		printf ("You entered %c\n", reqOrSugg);
		}
while (reqOrSugg != 'R' || 'S');

[Dave_Sinkula]

Methinks you want this:

while ( reqOrSugg != 'R' && reqOrSugg != 'S' );

[cwr]

The reason you have to do what Dave_Sinkula said and cannot do

while (foo != 'A' || 'B')

is because that expression is effectively

foo != ('A' || 'B')

Since 'A' is non-zero, it becomes

foo != 1

This is clearly not what you want.

[jlharrison]

Thanks Again Guys

[Dave_Sinkula]

The reason you have to do what Dave_Sinkula said and cannot do

while (foo != 'A' || 'B')

is because that expression is effectively

foo != ('A' || 'B')

Since 'A' is non-zero, it becomes

foo != 1

This is clearly not what you want.

Doesn't the != bind more tightly than the ||, thus making the equivalent like this?

(foo != 'A') || 'B'

The second part will always be true.

[cwr]

Doesn't the != bind more tightly than the ||, thus making the equivalent like this?

(foo != 'A') || 'B'

Yes of course, apologies, thanks for pointing this out.

When I made the post I quickly and sleepily looked at the list below and saw |= as !=

Operator                             Associativity
--------                             -------------
() [] -> .                           left to right
! ~ ++ -- - (type) * & sizeof        right to left
* / %                                left to right
+ -                                  left to right
<< >>                                left to right
< <= > >=                            left to right
== !=                                left to right
&                                    left to right
^                                    left to right
|                                    left to right
&&                                   left to right
||                                   left to right
?:                                   right to left
= += -= *= /= %= <<= >>= &= ^= |=    right to left
,                                    left to right

[jlharrison]

I am really new to C and my instructor does not do a great job explaining things. Wondering if one of you can clarify this for me.

I need to keep asking for input untill I get a R or S.

while ( reqOrSugg != 'R' && reqOrSugg != 'S' );

This looks to me like repeat the loop untill the reqOrSugg is equal to BOTH R and S. How can this work because the variable reqOrSugg can not be R and S? This cant be the case because it is what made the program work.

while ( reqOrSugg != 'R' || reqOrSugg != 'S' );

This looks like the code that should work to me because reqOrSugg can be equal to EITHER R or S. Only one needs to be satisfied to exit the loop...

I guess there is something I dont understand aboutthe and(&&) and or(||) operators so could one of you explain this?

[skripti]

while ( reqOrSugg != 'R' || reqOrSugg != 'S' );
Should Be
while ( reqOrSugg != 'R' || reqOrSugg != 'S' ) <- colon not needed for loops

James

[Dave_Sinkula]

Throw a couple of values at the expression like 'R', 'S', and 'T'. Mentally do the comparisons and get your result. Or write a little bit of code to help.

#include <stdio.h>

void foo(char reqOrSugg)
{
   printf("'%c' != 'R' && '%c' != 'S' => ", reqOrSugg, reqOrSugg);
   printf("%d && %d = %d\n", reqOrSugg != 'R', reqOrSugg != 'S',
          reqOrSugg != 'R' && reqOrSugg != 'S');
}

void bar(char reqOrSugg)
{
   printf("'%c' != 'R' || '%c' != 'S' => ", reqOrSugg, reqOrSugg);
   printf("%d || %d = %d\n", reqOrSugg != 'R', reqOrSugg != 'S',
          reqOrSugg != 'R' || reqOrSugg != 'S');
}

int main (void)
{
   foo('R');
   foo('S');
   foo('T');
   bar('R');
   bar('S');
   bar('T');
   return 0;
}

/* my output
'R' != 'R' && 'R' != 'S' => 0 && 1 = 0
'S' != 'R' && 'S' != 'S' => 1 && 0 = 0
'T' != 'R' && 'T' != 'S' => 1 && 1 = 1
'R' != 'R' || 'R' != 'S' => 0 || 1 = 1
'S' != 'R' || 'S' != 'S' => 1 || 0 = 1
'T' != 'R' || 'T' != 'S' => 1 || 1 = 1
*/

[Dave_Sinkula]

while ( reqOrSugg != 'R' || reqOrSugg != 'S' ) <- colon not needed for loops

Look up a do...while.

[ZuK]

 // while var is not 'R' and var is not 'S'
while ( reqOrSugg != 'R' && reqOrSugg != 'S' )

or

 // while not ( var is 'R' or var is  'S' )
while ( !( reqOrSugg == 'R' ||  reqOrSugg == 'S' ))

does the same ( break the loop if either 'R' or 'S' is entered )

Kurt

[dwks]

Remember, the loop is running until either S or R is entered. So you want the loop to continue while the input was not S and the input was not R. If you used or, the loop would run forever, since to end the loop, the input would have to be S and R at the same time, which is impossible.

while ( reqOrSugg != 'R' || reqOrSugg != 'S' ) <- colon not needed for loops

And it's a semi-colon, too.