Why can't we static cast a char pointer to an int pointer?

Question

So i was reading the learncpp tutorial, and i was trying to print the address of a char pointer, because when you do :

char c{'i'}; cout << &c << endl;

It will just print the string "i", because i guess the compiler identify a char pointer to a char array, which is a string (however i don't understand why it's just printing "i", because there is not '\0' after the character 'i').

Anyway, i then tried to static_cast the char pointer into an int pointer so that is will print the address :

char c{'i'}; cout << static_cast<int*>(&c) << endl;

But this doesn't even compile, i have an error "invalid static_cast from the type << char* >> to the type << int* >>".

Finally, i simply tried a C-type cast like this :

char c{'i'}; cout << (int*)(&c) << endl;

And it works, it prints an address.

So my questions are :

why the static_cast doesn't work ? Is there any way to make a valid "C++ style cast" from char* to int* ?
in the first example, when I "cout << &c", why does it only print "i", while i haven't put any '\0' after the character 'i' ?

Thanks

static_cast is not the same as a c cast ((T*)). A c cast will cast just about anything to anything else regardless of type safety. static_cast only allows some of the safer conversions. — François Andrieux
– François Andrieux, Commented Dec 24, 2019 at 17:14

Cruz Jean · Accepted Answer · 2019-12-24 17:29:28Z

Firstly, the static_cast<int*>(&c) will fail because static_cast only allows a very specific subset of type conversions. For instance, you can static_cast between numeric types, or between two class types which are related by inheritance (i.e. one inherits the other at some point in the inheritance hierarchy).

What you're doing is simply trying to "reinterpret" the address as an address to a different type. The reason being that C++ std::ostream overloads operator<< for const char*. If you only want to print the address itself, standard convention is to convert to void* via static_cast<void*>(&c) or just (void*)&c if you're ok with C-style casts. void* is basically a pointer with no type information, which makes it perfect if all you want to do is print out the value of the address itself.

The reason the C-style case compiles is because a C-style cast employs very ... shall we say risky ... methods to perform the conversion. It's allowed to perform const_cast, reinterpret_cast, and static_cast - whatever is necessary - in order to perform the cast you specified. This behavior is detailed on cppreference.com.

As for why the example of printing the char* directly only prints a single character despite not being terminated: undefined behavior is undefined. Apparently, it just so happens that (with this compiler) the executable happens to have a zero (null) byte immediately following it, wherever it may be. The compiler might even realize you assign a value but never modify it, so it might not even be in the stack (it might be in the data segment of the executable). The moral of the story is just not to invoke undefined behavior.

Ryan M · Accepted Answer · 2022-07-21 17:50:49Z

Let me give an example. Printing address is same as printing pointer. Such as

Int* a; cout<<a;// print the address Lets look another Int a; cout<<&a; // print the address, after casting into int*

Both are same.

Now take similar case for char, and look what happen

char a = 'a' cout<<&a; // it is printing value of *a means char* a

So that's why &a prints the value of char rather than address, because printing the address of char, after casting, char becomes the char* which is the data type called char pointer.

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