Why is the comparison between const char[] and const char* different?

Question

Suppose I have the following code:

const char str1[] = "asdasd";
const char str2[] = "asdasd";
const char* str3 = "dsadsa";
const char* str4 = "dsadsa";
if (str1 == str2)
{
    cout << "str1 == str2" << endl;
}
if (str3 == str4)
{
    cout << "str3 == str4" << endl;
}

The result is "str3 == str4". Why?

Answer 1

For C++

str3 and str4 are pointing to identical string literals,

The compiler is allowed, but not required, to combine storage for equal or overlapping string literals. That means that identical string literals may or may not compare equal when compared by pointer.

That means whether str3 == str4 is true or false is unspecified, it depends on the implementation, and the one you're using gives true.

On the other hand, str1 and str2 are independent arrays.

String literals can be used to initialize character arrays. If an array is initialized like char str[] = "foo";, str will contain a copy of the string "foo".

So str1 == str2 is guaranteed to be false.

Answer 2

String literal pooling. Compilers tend to merge identical string literals, so there's only one copy in the program. This is allowed but not required; it would be unwise to rely on this happening.