Getting a dangling pointer by returning a pointer from a local C-style array

Question

I am a bit confused by the following code:

#include 

const char* f()
{
    const char* arr[]={\"test\"};
    return arr[0];
}

int main()
{
             


        

Answer 1

The array arr has local storage duration and will disappear at the end of scope. The string literal "test" however is a pointer to a static storage location. Temporarily storing this pointer in the local array arr before returning it doesn't change that. It will always be a static storage location.

Note that if the function were to return a C++ style string type instead of a C style const char *, the additional conversion/bookkeeping would likely leave you with a value limited in lifetime according to C++ temporary rules.

Answer 2

No, it's not UB.

This:

const char* f()
{
    const char* arr[]={"test"};
    return arr[0];
}

Can be rewritten to the equivalent:

const char* f()
{
    const char* arr0 = "test";
    return arr0;
}

So we're just returning a local pointer, to a string literal. String literals have static storage duration, nothing dangles. The function really is the same as:

const char* f()
{
    return "test";
}

---

If you did something like this:

const char* f() {
    const char arr[] = "test"; // local array of char, not array of char const*
    return arr;
}

Now that is UB - we're returning a dangling pointer.