Programming in C

300 Chapter 13 The Preprocessor

defined as 1 , the preceding statement has the effect of assigning 1 to gameOver.The pre-

processor statement

#define NO 0

defines the name NOand makes its subsequent use in the program equivalent to specify-

ing the value 0 .Therefore, the statement

gameOver = NO;

assigns the value of NOto gameOver, and the statement

if ( gameOver == NO )

...

compares the value of gameOveragainst the defined value of NO.Just about the only

place that you cannotuse a defined name is inside a character string; so the statement

char *charPtr = "YES";

setscharPtrpointing to the string "YES"and not to the string "1".

A defined name is nota variable.Therefore, you cannot assign a value to it, unless the

result of substituting the defined value is in fact a variable.Whenever a defined name is

used in a program, whatever appears to the right of the defined name in the #define

statement gets automatically substituted into the program by the preprocessor. It’s analo-

gous to doing a search and replace with a text editor; in this case, the preprocessor

replaces all occurrences of the defined name with its associated text.

Notice that the #definestatement has a special syntax:There is no equal sign used to

assign the value 1 to YES. Furthermore, a semicolon does notappear at the end of the

statement. Soon, you will understand why this special syntax exists. But first, take a look

at a small program that uses the YESand NOdefines as previously illustrated.The function

isEvenin Program 13.1 simply returns YESif its argument is even and NOif its argument

is odd.

Program 13.1 Introducing the #defineStatement

#include <stdio.h>

#define YES 1

#define NO 0

// Function to determine if an integer is even

int isEven (int number)

{

int answer;

if ( number % 2 == 0 )

answer = YES;