Clarification in Section 5.10 of K&R 2

Question

A pattern recognition program must print all lines containing the patter if the input is find pattern. If the input is find -x pattern, the program must print all lines except the lines containing pattern.

// .....
switch(c)
{
case 'x':
    except=1;
    break;
// ......
}

// ......
while(getline(line,MAXLINE)>0)
    {
    line_num++;
    if( (strstr(line,*argv)!=NULL) != except)
        {
        if(number)
            printf("%ld:",linenum);
        printf("%s",line);
        found++;
        }
    }
// ......

In the above code from K&R except can either be 1 or 0. How does if(strstr...) block functions effectively to handle -x ?

Answer 1

The logic is simple. If the pattern is "-x" we should print all lines that do not contain the pattern.

For this pattern except is equal to 1.

So lines that contain the pattern satisfy the condition

strstr(line,*argv)!=NULL

that is this condition will be always equal to 1 if a line contains the pattern.

Thus if except is equal to 1 and the condition strstr(line,*argv)!=NULL is equal to 1 we should skip the pattern.

Otherwise if the condition strstr(line,*argv)!=NULL is not equal to 1 that is if the pattern is not found then the if statement

if( (strstr(line,*argv)!=NULL) != except)

yields true and its compound statement is executed.

On the other hand if except is equal to 0 then to achieve that the condition in the if statement would evaluate to true we need that the condition strstr(line,*argv)!=NULL would be equal to 1.

In fact you can rewrite the if statement

if( (strstr(line,*argv)!=NULL) != except)

the following way

if( ( ( strstr(line,*argv) != NULL ) == 1 && except == 0 ) ||
    ( ( strstr(line,*argv) != NULL ) == 0 && except == 1 ) )

Shortly speaking the if statement does the work if either

1 and 0

or

0 and 1

If either

1 and 1

or

0 and 0

then the if statement will not be executed.

Here 1 and 0 are results of evaluating of the two sub expressions in the if statement.