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

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.