c++ - How to deal with the situation where default is placed in front of the switch statement block

Question

int c,i;
for (int i = 1; i < 3; ++i)
{
    switch (i)
    {
        default: c+=i;
        case 2: c++;break;
        case 4: c+=2;break;
    }
}
printf("%d\n", c);

Why is this code equal to 3? Isn't this the case where default: c =i; is used when i is 1 for the first time, and case 2: c ;break; is used when i is 2 for the second time? Then it ends, and c=2 is output in the end? Why is it 3?

Answer 1

First of all, let’s clarify some points to note in switch:

1.

The

switch statement body consists of a series of case tags and an optional default tag. case Two constant expressions in a statement cannot evaluate to the same value. default tag can only appear once. Marking statements is not a grammatical requirement, but if they are not present, the switch statements are meaningless. The default statement (i.e. the default tag) does not need to appear at the end; it can appear anywhere in the body of the switch statement. case or default tags can only be displayed within switch statements.
Excerpted from: Microsoft Visual Studio 2015 c++ Switch statement official documentation

2.

The above-mentioned case and default themselves are labels, which tell the compiler to start executing backwards when this label is satisfied, and then it will not judge the correctness of other labels until the break statement or switch statement The scope ends.

For this problem

STEP 1 : When i=1, since i!=2&&i!=4, execution starts after the default label. At this time, the statement c+=i;(we now assume that compilation The compiler will help you initialize c to 0. You must know that not all compilers are so friendly) The value of c after execution is 1;
STEP 2: Based on the above 1 and 2, it can be seen that since no break is encountered at this time The statement has not reached the end of the scope of the switch statement (because this default statement is placed first), so it then executes the statements after case 2 backwards (the compiler is no longer full at this time) The case tag is not satisfied) At this time, the statement c++ is executed; after execution, the value of c is 2; when the break statement is encountered, the switch statement is jumped out.
STEP 3: When i=2, since i satisfies case 2, execution starts directly from the statement after the case 2 note. At this time, the statement c++ is executed. After execution, c The value is 3. When the break statement is encountered, the switch statement will be jumped out.
STEP 4: When i=3, jump out of the for loop and output c=3;
(The above process is the conclusion drawn by me using Visual Studio 2015 single-step debugging and combined with the data)

Answer 2

When i=1, it enters default: c=1 and there is no break, so continue to match case 2 and get c=2 break
When i=2, it matches case 2 first and enters case 2: c=3 break
When i=3 The cycle does not hold.
The output c is 3

Answer 3

var c = 0,i;
for (var i = 1; i < 3; ++i)
{
    switch (i) {
        default: c+=i;console.log('defalult:'+c);
        case 2: c++;console.log('case2:'+c);break;
        case 4: c+=2;console.log('case4:'+c);break;
    }
    console.log(i, c);
}
console.log('last:'+c);