error C2361: initialization of 'found' is skipped by 'default' label

Question

Possible Duplicate:
Why can't variables be declared in a switch statement?

I have a strange error in my code below:

char choice=Getchar();
switch(choice)
{
case 's':
    cout<<" display tree ";
    thetree->displaytree();
    break;

case 'i':
    cout<<"  enter value to insert "<<endl;
    cin>>value;
    thetree->insert(value);
    break;
case 'f' :
    cout<< "enter value to find ";
    cin>>value;
    int found=thetree->find(value);
    if(found!=-1)
        cout<<" found  =  "<<value<<endl;
        else
            cout<< " not found " <<value <<endl;
        break;
default:
    cout <<" invalid entry "<<endl;;
    }

Visual Studio 2010 compiler says that:

1>c:\users\daviti\documents\visual studio 2010\projects\2-3-4\2-3-4\2-3-4.cpp(317): error C2361: initialization of 'found' is skipped by 'default' label
1>          c:\users\daviti\documents\visual studio 2010\projects\2-3-4\2-3-4\2-3-4.cpp(308) : see declaration of 'found'

I think that I have correctly written break and default statements, so where is the error?

This is only an exact duplicate if you already know the answer to the problem. The cryptic "error C2361: initialization of 'found' is skipped by 'default' label" does not necessarily lead you to the question 'Why can't variables be declared in a switch statement?' — Tim MB, Oct 17 '12 at 16:08
i got today the same problem :) i am not a c++ professional and i couldn't know that declaring a pointer in a 'case' without curly braces is not allowed. so just a thought, if you know the answer or solution and want to share it then please share, but stop acting here like a smart ass. — Constantin, Mar 07 '13 at 15:29
@Constantin agree with 'smart ass' thing - but whom you are referencing to? :) — Alexander, Sep 24 '19 at 12:23

score 66 · Accepted Answer · answered Apr 30 '12 at 09:23

You need to either enclose your case 'f': with a scoped brace:

case 'f' :
{  
    cout<< "enter value to find ";
    cin>>value;
    int found=thetree->find(value);
    if(found!=-1)
        cout<<" found  =  "<<value<<endl;
    else
        cout<< " not found " <<value <<endl;
    break;
}

or place the declaration of found outside of the switch

score 25 · Answer 2 · answered Apr 30 '12 at 09:27

25

The semantics of a switch are those of a goto: cases don't introduce a new scope. So found is accessible in your default: case (although you don't actually access it). Jumping over a non-trivial initialization is illegal, so your code becomes illegal.

Given the complexity of your case 'f':, the best solution is probably to factor it out into a separate function. Failing that, you can put the entire case in {...}, creating a separate scope, or forgo the initialization, writing:

int found;
found = thetree->find(value);

(I mention this for completeness. It is not the solution I would recomment.)

answered Apr 30 '12 at 09:27

James Kanze

150,581
18
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9

Upvote for an actual explanation. – Fulluphigh Aug 10 '16 at 20:17
"`case`s don't introduce a new scope" - this sentence explains everything. – Bojan Komazec Nov 09 '17 at 12:14
Not really, since if you have all usages below the definition there should be no issue regardless of scope, but my compiler still complains – Ferazhu Apr 13 '23 at 18:16

score 7 · Answer 3 · answered Apr 30 '12 at 09:17

7

You need to declare the internal variables of switch's case within curly braces. i.e.

case 'f' :
{
    ...
    int found=thetree->find(value);
    ...
}

answered Apr 30 '12 at 09:17

iammilind

68,093
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error C2361: initialization of 'found' is skipped by 'default' label

3 Answers3