无法理解break

Question

Could someone please take the time and try to explain how does this code work, don't understand the break part and how the answer at the end is 0.

int a=3, x;
switch(a==5){
   case 0: x=0;break;
   case 1: x=1;break;
   case 3: x=3;break;
   case 5: x=5;break;
   default: x=7;
}
printf("%d",x);

Answer 1

Without break , execution would continue from each case to the next (fallthrough) and every path would eventually end up at default and assign 7 to x .

Also note that you have a bug: switch(a==5) should be switch(a) .

Answer 2

a==5 returns a true or false value (0 or 1). In this example, most of the cases will not be triggered. Change switch(a==5) to switch(a)

This means that in your example, case 0 is triggered, because a==5 results in false (0).

Answer 3

Since a==5 is false, because a is 3, it returns 0. That's why case 0: is being triggered.

break; causes the code to stop, and without it the code would continue until default: .