empty switch case in C

Question

I'm reading through some books about C. I found the following example to switch case in C, which I try to understand ...

/* caps.c */

   #include <stdio.h>
   #include <ctype.h>

   #define SEEK 0
   #define REPLACE 1

   int main(void)
   {
     int ch, state = SEEK;
     while(( ch = getchar() ) != EOF )
     {
       switch( state )
       {
       case REPLACE:
         switch( ch )
         {
         case ' ':    //leaving case empty will go to default ???
         case '\t':
         case '\n':   state = SEEK;
                      break;
         default:     ch = tolower( ch );
                      break;
         }
         break;
       case SEEK:
         switch( ch )
         {
         case ' ':
         case '\t':
         case '\n':   break;
         default:     ch = toupper( ch );
                      state = REPLACE;
                      break;
         }
       }
       putchar( ch );
     }
     return 0;
   }

It is pretty clear to me that the mode SEEK, and then letters are Capitalized, and then mode is set to REPLACE, then letters are converted to lower. But why empty spaces trigger again SEEK mode ? Is this really the case in my comment ?

Solution 1

This is so-called fall-through behaviour of C switch operator. If you don't have a break at the end of a case region, control passes along to the next case label.

For example, the following snippet

int x = 5;
switch( x ) {
    case 5:
        printf( "x is five!\n" );
    case 2:
        printf( "x is two!\n" );
        break;
    default:
        printf( "I don't know about x\n" );
}

outputs

x is five!
x is two!

Be careful.

Solution 2

In this case what will happen is that the switch will enter at the appropriate case statment, so if ch=' ' at the top then run until it hits a break.

This means that if ch is one of ' ','\t' or '\n' then state will be set to SEEK.

Solution 3

Leaving a case empty does not go to default, it drops through to the next case.

So the piece of code:

case ' ':    // leaving case empty will drop through to `\t` and then `\n`.
case '\t':
case '\n':   state = SEEK;
             break;
default:     ch = tolower( ch );
             break;

will set the state to SEEK for spaces, tabs and newlines, otherwise it will lower-case the character.

What you have here is a little finite state machine with two states.

• In SEEK mode, it will skip over all the white space (tabs, spaces and newlines) and, when it finds a different character, it will upper-case it and switch to REPLACE mode.

• In REPLACE mode, it will lower-case all characters until it finds white space then switch to SEEK mode.

Hence text like:

PaxDiablo is a REALLY cool guy

will become:

Paxdiablo Is A Really Cool Guy

Solution 4

case ' ':    //leaving case empty will go to default ???

First I want to make clear that , if you write case ' ': , you are not leaving the case empty. You are actually writing a case 32: , since ASCII value of space is 32.

In general, whenever you write 'a',or'b' or any charecter constant, there will be an implicit conversion of the charecter specified to its corresponding ASCII value.

Proof: 
 The prototype of isspace() is,
 int isspace ( int c );
 And not as,
 int isspace ( char c );

Comment out the case ' ': statement. Then you will find answer yourself. If I comment out case ' ': ,

 Then on input,
    hello world
    the output will be,
    Hello world   <not expected>
    instead of 
    Hello World  (needed output).

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