C while statement in switch clause

This question comes from my C language exam. It asks for the result.

#include <stdio.h>

int main()
{
    int a = 4, b = 30;
    printf("%d\n", a);
    
    switch(a){
        case 1: a = a*10; break;
        case 2: a = a*10; break;
        case 3: a = a*10;
        while (a<b) case 4: {a = a+5;} break; //???
        case 5: a = a*10; break;
        default:
            b = a*10;break;
    }
    printf("%d\n", a);

    return 0;
}

The result is:

>4 <br/>
>34

I know the basic idea of while statement and switch statement, but I don't understand what's going on in the fourth line of the switch block.

My guess is that the while condition is tested 7 times until variable a reaches 34 (4<30, 4+5+5+5+5+5+5 = 34). But how about the 'case 4' condition? Shouldn't this condition be tested before adding 5 to variable a? And how about the 'break' after it? I'm totally confused.

Thanks!

while (a<b) case 4: {a = a+5;} break; //???

case 4 is just a label.

This is equivalent to the below:

while (a<b) 
some_label:
{
   a = a+5;
}
break; //This will break out of the surrounding switch.

while (a<b) case 4: {a = a+5;} break; 

This line of code is within a switch statement. The first thing that the switch does is to jump to the case label 4.

But this case label is within a while loop. At the end of the a=a+5 instruction, the program goes "back" to the while instruction.

What the program does is performing a goto inside the while loop.

goto statements are discouraged, unless when within switch blocks with perfectly legible constructs like:

  switch(a)
  {
    case 1: a = a*10; break;
    case 2: a = a*10; break;
    case 5: a = a*10; break;
    default:
        b = a*10;break;
}

If you're trying to be smart, omitting the break, inserting instructions between the break and the next case ... you have to be aware of what you're doing. And someone else reading the code may have a hard time to interpret it.

That also implies not jumping in the middle of a loop construct or strange things like that.

If you encounter such code in real life, rewrite it. The equivalent would use a do ... while instruction (the first test is skipped, we can enter the loop even if a<b:

case 4: do {a = a+5;} while (a<b); break; 

You get the same result with this code, except that it compiles without warnings and everyone understands what it does.

As you said, while (a < b) each time it will repeat case 4 and add 5 to a, by the time a is more than 30, it will break out of the while loop and switch scope. Then it will print the variable which is now equal to 34.