switch case do while nesting

Question

#include<stdio.h>
int main()
{
    int n=0, i=2;
    switch(1)
    {
    case 0:do
           {
    case 1: n++;
    case 2: n++;
           }while(--i > 0);
    }
    printf("n = %d",n);
}

I was expecting output for above code to be 0, as case 1 and case 2 are inside a do while which is inside case 0. Switch is testing on value 1, so case 0 will never be executed and hence neither case 1 or 2.

value of n coming to be 4. Any explanation ?

Each case needs to be an atomic code block - you cannot nest cases within other cases. — Brian Driscoll, Oct 15 '14 at 16:14
@BrianDriscoll: what's that supposed to mean? Switch cases aren't required to be blocks. — Fred Foo, Oct 15 '14 at 16:16
http://stackoverflow.com/questions/514118/how-does-duffs-device-work — mafso, Oct 15 '14 at 16:18
This is perfectly value C, if obfuscated and P45-inducing on the grounds of unmaintainability. Interesting question though, +1. — Bathsheba, Oct 15 '14 at 16:20
@BrianDriscoll That is incorrect. If a case statement has no break then execution will fall through. — Code-Apprentice, Oct 15 '14 at 16:23
I think a `break` within the `do` loop will exit the loop rather than the `switch`. — Bathsheba, Oct 15 '14 at 16:24
@Bathsheba: Yes, a `break` refers to the nearest enclosing `switch`, `while`, `do`, or `for` statement. Whereas a `continue` refers to the nearest enclosing `while`, `do`, or `for` statement. (Could be IOCCC fodder.) — Keith Thompson, Oct 15 '14 at 17:18
@Code-Apprentice falling through is not the same as nesting. — Brian Driscoll, Oct 15 '14 at 17:57
@BrianDriscoll The OP is about nesting a do while loop in a switch not about nesting cases within other cases. — Code-Apprentice, Oct 15 '14 at 18:11
possible duplicate of [Mixed 'switch' and 'while' in C](http://stackoverflow.com/questions/25592451/mixed-switch-and-while-in-c) — midor, Oct 15 '14 at 18:55

score 7 · Answer 1 · answered Oct 15 '14 at 16:15

7

Your code jumps into the middle of a loop. From that point on, it ignores the cases of the switch statement, so it does n++, n++, check condition, loop, etc. It helps if you think of a case as a label, and switch as a goto:

    int n=0, i=2;

    if (1 == 0)
        goto label0;
    else if (1 == 1)
        goto label1;
    else if (1 == 2)
        goto label2;

label0:
    do {
        puts("loop");
label1:
        n++;
label2:
        n++;
    } while (--i > 0);

Like your original code, this simply skips the part of the loop body before label1 (the puts), then continues as if it had entered the loop in the normal way.

answered Oct 15 '14 at 16:15

Fred Foo

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@larsmans : With your explanation that case are label - wouldn't it cause problem for nested switch, if both outer and inner switch has same case statements. E.g. if I put following in case 0 switch (i) { case 1: printf("here1"); case 2: printf("here2"); } now label1 and label2 should conflict as they have a function scope. Please elaborate more on this. FYI - I don't get any error though for doing above. – Sagrian Oct 15 '14 at 16:46
1

@Sagrian: A `case ...:` label is associated with the nearest enclosing `switch` statement. There's no ambiguity. – Keith Thompson Oct 15 '14 at 17:16

switch case do while nesting

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