Stupid question.. sorry..

**kwikness** · 08-18-2006

what's wrong with this statement? I keep getting parse errors.

Note: The variable q1 refers to a 'char' variable, declared as follows:

Code:

 char q1 = 'y';

Code:

if  (q1=='y') or (q1=='Y')
   
printf("Correct!");

i've also tried adding braces..

Code:

if  (q1=='y') or (q1=='Y')
{
   
printf("Correct!");
}

and i've even tried without the parenthesis..

Code:

if  q1=='y' or q1=='Y'
   
printf("Correct!");

I just had this working!! I messed around with it a little, re-wrote it, and now I'm getting parse errors. Is it something i'm not seeing?

Anyone know of a few code examples that use conditionals with operators? (and, or, not)

**whiteflags** · 08-18-2006

The keyword or is a readable alternative to the logical OR operator defined in <iso646.h>
You should only consider using it if you can't type this:

if ( q1=='y' || q1=='Y' )

or even

#include <ctype.h>

if ( tolower(q1) == 'y' )

**kwikness** · 08-18-2006

Thanks citizen, the || operator worked fine.

I could have sworn I compiled it using "or", and all I included was the <stdio.h>,
maybe i'm too many beers deep..

**whiteflags** · 08-18-2006

Compilers should warn you about implicit declarations of things you're using, when you hadn't included the proper header file. Sometimes the compiler is smart and the implicit declaration is correct, but it's a good thing to strive for warning-free compilation.

**quzah** · 08-18-2006

Originally Posted by kwikness

I could have sworn I compiled it using "or", and all I included was the <stdio.h>,

You could have been compiling as C++. Quoting 'C: A Reference Manual, 5th Edition', page 333:

Amendment 1 to C89 added the header file iso646.h, which contains definitions of mac-
ros that can be used in place of certain operator tokens. Those tokens could be inconvenient
to write in a restricted source character set (such as ISO 646). In C++, these identifiers are keywords.

The macros in question are:

and, and_eq, bitand, bitor, compl, not, not_eq, or, or_eq, xor, and xor_eq

So it is possible that had you compiled with C++, as many do here, it let you use your keyword or without needing any additional headers.

Quzah.

**Salem** · 08-19-2006

> if (q1=='y') or (q1=='Y')
You need () round the whole thing as well

if ( (q1=='y') || (q1=='Y') )

**Xero** · 08-19-2006

Originally Posted by Salem

> if (q1=='y') or (q1=='Y')
You need () round the whole thing as well

if ( (q1=='y') || (q1=='Y') )

Totally agreed to him.

**kwikness** · 08-19-2006

Cool. Thanks for your help guys.

**kwikness** · 08-19-2006

Ok, here's another one if anyone is willing to help. The source compiles fine, but when I execute the program, I get a whole bunch of errors.. I believe the bug has to be somewhere in the "lucky number" part of the program.

Code:

#include <stdio.h>
main()
{
int randomNum;
srand(time());

randomNum = (rand() %4) + 1;

printf("\n\n\nFortune cookie say..\n");

switch (randomNum)
{

case 1:
printf("\nHim with little shovel, not do big dig.\n");
break;

case 2:
printf("\nBoiled rice better then fried rice.\n");
break;

case 3:
printf("\nIf ghost scare you in dream, wake up.\n");
break;

case 4:
printf("\nConfucious is philosopher, not chef.\n");
break;

 }
printf("\n\nYour lucky number is...\n");
printf("%d ", (rand() % 49) +1);

}

**Salem** · 08-19-2006

> srand(time());
You need to include stdlib.h and time.h
Then you'll see time() needs a parameter.

> main
int main
at the start, and finish with
return 0;

> printf("%d ", (rand() % 49) +1);
rand() gives you a new result each time, try
printf("%d ", randomNum);

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