C++ char*[] to char** conversion

Question

I have this simple code that compiles without errors/warnings:

void f(int&, char**&){}

int main(int argc, char* argv[])
{
    f(argc, argv);
    return 0;
}

And next similar code that doesn't compile:

void f(int&, char**&){}

int main()
{
    int argc = 2;
    char* argv[] = { "", "", nullptr };
    f(argc, argv); 
    //@VS2013 error: cannot convert argument 2 from 'char *[3]' to 'char **&'
    //@GCC error: invalid initialization of non-const reference of type 'char**&' from an rvalue of type 'char**'
    return 0;
}

Why char*[] can be converted to char**& in the first sample and can't be converted in the second sample? Does it matter if the size is known at compile time?

EDIT: I think there are 2 conversions needed in the second case, and only one implicit conversion can be done by compiler.

This code compiles fine:

void f(int&, char**&){}

int main()
{
    int argc = 2;
    char* temp[] = { "", "", nullptr };
    char** argv = temp;
    f(argc, argv);
    return 0;
}

Answer 1

The type of temp in

char* temp[] = { "", "", nullptr };

is not char*[], it's

char*[3]

The latter can't be implicitly converted to `char**'.

In main, the type of argv is an unbound char* array which is equivalent to char**

I admit, it's confusing :)

Answer 2

Jefffrey's comment references the standard, here it is:

4.2 Array-to-pointer conversion [conv.array]

An lvalue or rvalue of type “array of N T” or “array of unknown bound of T” can be converted to a prvalue of type “pointer to T”. The result is a pointer to the first element of the array.

And a prvalue is:

A prvalue ("pure" rvalue) is an expression that identifies a temporary object (or a subobject thereof) or is a value not associated with any object.

You cannot bind a non-const reference to a temporary.

int& i = int(); // error

char* argv[] = { "", "", nullptr };
// the result of the conversion is a prvalue
char**& test = argv; // error

Therefore the following code will happily compile:

#include <iostream>

void f(int& argc, char** const& argv){
    std::cout << argv[0] << std::endl; // a
}

int main()
{
    int argc = 2;
    char* argv[] = { "a", "b", nullptr };
    f(argc, argv); 
    return 0;
}

---

One important thing I glazed over is pointed out in Kanze's comment.

In the first example provided in the OP, char* argv[] and char** argv are equivalent. Therefore, there is no conversion.

std::cout << std::is_array<decltype(argv)>::value << std::endl; // false
std::cout << std::is_array<char**>::value << std::endl; // false
std::cout << std::is_array<char*[]>::value << std::endl; // true
std::cout << std::is_same<decltype(argv), char**>::value << std::endl; // true
std::cout << std::is_same<decltype(argv), char*[]>::value << std::endl; // false

Answer 3

Because despite appearances, the second argument to main has type char**. When used as the declaration of a function argument, a top level array is rewritten to a pointer, so char *[] is, in fact, char**. This only applies to function parameters, however.

A char*[] (as in your second case) can convert to a char**, but the results of the conversion (as with any conversion) is an rvalue, and cannot be used to initialize a non-const reference. Why do you want the reference? If it is to modify the pointer, modifying the char** argument to main is undefined behavior (formally, in C, at least—I've not checked if C++ is more liberal here). And of course, there's no way you can possibly modify the constant address of an array. And if you don't want to modify it, why use a reference?