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          


        

Answer 1

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 

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::value << std::endl; // false
std::cout << std::is_array::value << std::endl; // false
std::cout << std::is_array::value << std::endl; // true
std::cout << std::is_same::value << std::endl; // true
std::cout << std::is_same::value << std::endl; // false