Help me understand how the program works.

I take it from the book of Steven Pratt “Program language C” I find on page № 289 exercise №6.

It's a program, her code here:

#include <stdio.h>
#include <stdlib.h>
int main(void)
{
    int i = 0;
    while (i<3) {
        switch (i  ) {
            case 0 : printf ("fat ");
            case 1 : printf ("hat ");
            case 2 : printf ("cat ");
            default : printf ("Oh no ");
        }
        putchar('\n');
    }
    return 0;
}

Please correct me if I am not right at all. Right now, I would like to explain how it works. Firstly to work the function “while” three times.

The variable “i” has first time value 0. Then during the function "while" work, the variable has to be added three times three times.

As a result of program work, I had this result:

fat hat cat oh no
hat cat oh no
cat oh no

But I can't understand how I've got such a result. I am used to IDE CodeBlock, and my compiler works well. I changed the program code many times to understand how it worked at all. I changed the operatore "case".

Right now I changed the program kode so:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
  int i = 0;
  while (i<2) {
    switch (i  ) {
    case 1 : printf ("fat ");
    case 2 : printf ("hat ");
    case 3 : printf ("cat ");
    default : printf ("Oh no ");
  }
  putchar('\n');
  }
  return 0;
}

After I got such a result:

oh no
fat hat cat oh no

But I can’t understand why the program works so? In my opinion the program must give such a result at all:

fat
Oh no

I think result or program work, must be only so. Why the program work so?

CodePudding user response：

Your code should look like this:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
  int i = 0;
  while (i<2) {
    switch (i  ) {
     case 1 : printf ("fat "); 
              break;
     case 2 : printf ("hat "); 
              break;
     case 3 : printf ("cat "); 
              break;
     default : printf ("Oh no "); 
              break;
    }
    putchar('\n');
  }
 return 0;
}

CodePudding user response：

"Why the program work so?"

In short it is because of the way a post-increment variable works, and that the book's implementation of the switch statement, although legal is non-typical, failing to limit execution flow to one case per iteration.

First, the form of the switch statement out of the book is not typical. I would expect that some of the text in that section of the book should eventually include use of the break statement placed between each of the case lines

In the book's original example, when switch is called on the first loop, i == 0, resulting in this execution flow:

`case 0:` then increments `i` to `1`, then outputs `fat`,   
`case 1:` outputs `hat`,   
`case 2:` outputs `cat`,  
`default` outputs `Oh no`    

The 2nd iteration of the while loop, with i == 1, again post-incremented, starts at the 2nd case

`case 1:` `i` is incremented to `2`, then outputs `hat`,   
`case 2:` outputs `cat`,   
`default` outputs `Oh no`  

With i == 2 the third iteration follows:

`case 2:` i is incremented to `3` outputs `cat`,   
`default` outputs `Oh no`   

In your modified code, the 0 case is removed. Because i is initialized to 0 execution flow jumps immediately to the default case. And you can follow the rest.

Note, in the text of your book you will likely learn that the switch statement is commonly used in conjunction with the break statement, ensuring only one case is executed per swtich selection:

int main(void)
{
    int i = 0;
    while (i<3) {
        switch (i  ) {
            case 0 : printf ("fat ");
              break;// brings execution flow directly to the closing }
            case 1 : printf ("hat ");
              break;// brings execution flow directly to the closing }
            case 2 : printf ("cat ");
              break;// brings execution flow directly to the closing }
            default : printf ("Oh no ");
        }
        putchar('\n');
    }
    getchar();
    return 0;
}