C++: pass reference as parameter but that function doesn't accept reference as parameter

Question

I have this code snippet from here. Pls have a look at the file spec.cc.

Eval parse_recursive_descent (Grammar g, SymbolString input) {
    return
        parse_recursive_descent
            (g, Stack (SymbolString {g.start}, input, Path ())) ;
}

// MAIN ////////////////////////////////////////////////////////////////////////

int main () {
    Input in = read_input (std::cin) ;
    Eval  ev = parse_recursive_descent (grammar (in), string (in)) ;
    print (make_output_string (derivations (ev), accept (ev))) ;
    return 0 ;
}

The prototype of function grammar is: Grammar& grammar (Input& in) ;. So it returns a reference.

But parse_recursive_descent doesn't have the g parameter as reference. The question is: the g in function parse_recursive_descent is a reference or a value?

Answer 1

It's a value, copy-constructed from the reference you pass as argument.

The general idea is that all the parameters of a function are variables (of the specified type) local to the function. When you pass an argument, it is used as the initializer to construct the corresponding parameter.

Answer 2

基本的是你可以为同一类型的变量分配一个引用，如：

int i = 10;
int& j = i;
int k = j;//here j is a reference and value from j is copied to k

在这里，正如您所说，函数grammar返回类型为Grammar的引用。现在，此返回值将作为参数发送到函数parse_recursive_descent（到Grammar g）。因此，只需将Grammar类型的变量分配给g即可。

Grammar g = variable of type Grammar; //this is a value copy

因此，g只是一个值而非参考。

*复制发生在Grammar类的复制构造函数中。

Answer 3

Eval parse_recursive_descent(Grammar g, SymbolString input)

g是Grammar类型的值。你知道，因为这就是宣言所说的。

当你调用函数时，如果传递的类型与函数期望的类型不完全匹配，编译器会生成代码以使它们匹配。所以在通话中

parse_recursive_descent(grammar(in), string(in));

编译器将grammar(in)返回的引用转换为Grammar类型的对象，并将其传递给parse_recursive_descent。为了使对象为Grammar类型的对象留出空间并使用复制构造函数，传递grammar(in)返回的引用。你可以做类似的事情：

Grammar temp(grammar(in));
parse_recursive_descent(temp, string(in));

如果你这样做的唯一区别是你将有一个命名对象（temp）而不是编译器在你的实际代码中创建的（未命名的）临时对象