Does a variable defined inside a switch/case persists it's value?

Question

I know that it is possible to create, but not initalize, a variable inside a switch-statement. But I wonder if a created variable remains in memory, during subsequent executions of the switch-statement?

I assume the variable named cnt is newley created every time the switch-statement is executed. Therefore the value of cnt is always undefined, until the code inside a case label assigns a value!

#include <iostream>
using namespace std;

int main() {
    int iarr[6] = {0, 1, 1, 1, 0, 1};

    for (int i : iarr) {
        switch (i) {
            int cnt;
            case 0:
                // it is illegal to initalize value, therefore we assign an
                // inital value now
                cout << "Just assign 0 to the variable cnt.\n";
                cout << (cnt = 0) << "\n";
                break;
            case 1:
                cout << "Increment variable cnt.\n";
                cout << ++cnt << "\n";
                break;
        }
    }

    return 0;
}

But at least on my machine and during my tests, the variable cnt defined within the switch-statement persists it's value. I assume this a false positive and my system is (bad luck) accessing always the same memory region?

Output (GCC 4.9):

$ g++ -o example example.cpp -std=gnu++11 -Wall -Wpedantic -fdiagnostics-color && ./example
Just assign 0 to the variable cnt.
0
Increment variable cnt.
1
Increment variable cnt.
2
Increment variable cnt.
3
Just assign 0 to the variable cnt.
0
Increment variable cnt.
1

Thank you

You're invoking undefined behaviour (accessing the value of an uninitialized variable). — Oliver Charlesworth, Jun 21 '14 at 16:46
This is like the fifth UB thing I've seen in the past minute. New record. — chris, Jun 21 '14 at 16:47
possible duplicate of [Why can't variables be declared in a switch statement?](http://stackoverflow.com/questions/92396/why-cant-variables-be-declared-in-a-switch-statement) — aruisdante, Jun 21 '14 at 16:51
Leaving a scope means everything inside it is gone forever no matter the control structure. — diapir, Jun 21 '14 at 16:52
And also this: https://stackoverflow.com/questions/20031147/using-in-a-case-statement-why?rq=1 — aruisdante, Jun 21 '14 at 16:54
Thank you. I swear this "undefined behaviour" question was not timed up with the others. — Peter, Jun 21 '14 at 17:03

Vlad from Moscow · Accepted Answer · 2014-06-21T17:13:41.670

4

The compiler does not initialize the variable explicitly. So it has a value that is stored in the memory allocated for the variable. But according to the C++ Standard the variable is created each time when the control is passed to the switch statement.

In fact nothing prevents the compiler to use the same memory in some other code block included in the range-based for compound statement.

According to the C++ Standard

2 Variables with automatic storage duration (3.7.3) are initialized each time their declaration-statement is executed. Variables with automatic storage duration declared in the block are destroyed on exit from the block (6.6).

edited Jun 21 '14 at 17:13

answered Jun 21 '14 at 16:49

Vlad from Moscow

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Thank you! As note for subsequent readers, you can read more about it in the standard (C++11, Section 6.7, Paragraph 3). – Peter Jun 21 '14 at 17:05
@Peter See my updated post. I included paragraph #2 from my C++ Standard draft. – Vlad from Moscow Jun 21 '14 at 17:15

Does a variable defined inside a switch/case persists it's value?

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