Why can we declare an uninitialized variable in the case of a switch?

int i;
i = 2;
switch(i)
{
    case 1: 
        int k;
        break;
    case 2:
        k = 1;
        cout << k << endl;
        break;
}

I don't know why the code above works.

Here, we can never go into case 1 but why we can use k in case 2?

There are actually 2 questions:

1. Why can I declare a variable after case label?

It's because in C++ label has to be in form:

N3337 6.1/1

labeled-statement:

...

• attribute-specifier-seqopt case constant-expression : statement

...

And in C++ declaration statement is also considered as statement (as opposed to C):

N3337 6/1:

statement:

...

• declaration-statement

...

2. Why can I jump over variable declaration and then use it?

Because: N3337 6.7/3

It is possible to transfer into a block, but not in a way that bypasses declarations with initialization. A program that jumps (The transfer from the condition of a switch statement to a case label is considered a jump in this respect.)

from a point where a variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has scalar type, class type with a trivial default constructor and a trivial destructor, a cv-qualified version of one of these types, or an array of one of the preceding types and is declared without an initializer (8.5).

Since k is of scalar type, and is not initialized at point of declaration jumping over it's declaration is possible. This is semantically equivalent:

goto label;

int x;

label:
cout << x << endl;

However this wouldn't work if x was initialized at point of declaration:

 goto label;

    int x = 58; //error, jumping over declaration with initialization

    label:
    cout << x << endl;