C++ array as function parameter

Question

using Data = char[10];
void f(Data x)
{
    x = nullptr; // this compiles
}
void g(Data &x)
{
    x = nullptr; // this does not compile, msvc complain "expression must be a modifiable lvalue
}

I am confused why the assignment expression in f compiles, but does not compile in g. I expect that both assignment will fail since array type is not modifiable.

Any reference to c++ standard will be appreciated.

Possible duplicate of When a function has a specific-size array parameter, why is it replaced with a pointer? — user7860670
– user7860670, Commented Nov 20, 2017 at 7:26

StoryTeller - Unslander Monica · Accepted Answer · 2017-11-20 07:27:40Z

2

This has everything to do with function parameter type adjustment ([dcl.fct]/5):

After determining the type of each parameter, any parameter of type “array of T” or of function type T is adjusted to be “pointer to T”.

Since a type alias is equivalent to the type it names, in the first case the type is determined as char[10] and adjusted to char*. You can modify a pointer.

In the second case, you form a char(&)[10]. It's a reference to an array, and arrays cannot be assigned to, even via a reference.

answered Nov 20, 2017 at 7:27

StoryTeller - Unslander Monica

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2 Comments

xdot Over a year ago

So the standard acts different between type is "array of T" and type is "reference to array of T", I might just accept that although it seems weird to me...

StoryTeller - Unslander Monica Over a year ago

@xdot - That array type pulled in from C are weird. I concur with the comment on your post. std::array is better for such things. It's an actual value type.

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