[Solved] C language, explain this code [closed]

Question

Reason for specific output. Since you don not have a break; for switch conditions you fall through all the switch cases from the first match found

From this tutorial,

When the variable being switched on is equal to a case, the statements following that case will execute until a break statement is reached.

If no break appears, the flow of control will fall through to subsequent cases until a break is reached.

switch (i % 3) {
     case 0:  printf("zero, "); // <= No break so once this get match all the below will get execute. (Till a break is reached)
     case 1:  printf("one, ");
     case 2:  printf("two, ");
     default: printf("what? ");
}

So in your case for i=0 to i=4, following happens,

When i=1 you get i%3 will match with case 1 and output will be one,two,what?.

When i=2 you get i%3 will match with case 2 and output will be two,what?

When i=3 you get i%3 will match with case 0 and output will be zero,one,two,what?

When i=4 you get i%3 will match with case 1 and output will be one,two,what?

And note that default is the case given to fulfill when no case is met. but in your case since you do not have break , this too get executed ans results for what?.

And what puts() does is , simply put a string to standard output. In your case puts(” “) will put a space. And note that puts() will append a newline at the end.

Accepted Answer