Why isn't the “noexcept” specifier part of the function type?

问题

I don't get it why? I don't think compatibility should be a problem as functions declared without the specifier actually have it implicitly defined to false. If it's about name mangling - can we just suppose that old one (existing) will imply noexcept(false) and add another new symbol to the mangling for noexcept(true).

This is going to be useful when working with templates as now comparing function type and noexcept specifier should be done seperatly. What I basically mean is this:

int func() noexcept(true), func_1() noexcept(false);

decltype(func) == decltype(func_1); //this now equals true

But on the other hand if we had function assignment by either using pointer or reference then - the noexcept specifier is checked as if it was part of the type:

int (&refFunc)() noexcept(true) = func_1; //target exception specification is not superset of source

int (&refFunc)() noexcept(true) = func; //ok

So for now implementing full function matching should be done by both performing type and noexcept check which is kinda complex:

decltype(func) == decltype(func_1) && noexcept(func()) == noexcept(func_1()); //this now equals false

Imagine if functions have got parameters:

int func(int, double) noexcept(true), func_1(int, double) noexcept(false);

decltype(func) == decltype(func_1) && noexcept(func(int{}, double{})) == noexcept(func_1(int{}, double{})); //this now equals false

回答1:

Starting with C++17, the noexcept specifier is a part of the function type.

Reference

The noexcept-specification is a part of the function type and may appear as part of any function declarator. (since C++17)

回答2:

C/C++03 compatibility

One of most important C++ basic ideas is backward compatibility with C language and older C++ versions. And it works in most cases. Adding exceptions to function specifiers would deny this idea. There was no noexcept in C++03 and C and it would cause problems with function pointers and so on.

Low level

Let's have a think how function work in low level. They are basically jump with saved (on the stack) return address. They pass arguments and return values also via stack (not always, but let's simplify a bit). So, when you declare a function you actually tell how many bytes function should take from stack and how many it leaves on it. Function declaration is most of all telling the program what to EXPECT NORMALLY from function. Now, knowing that, do exceptions change anything in the information normally passed to/from function? They don't. And I think this is major reason why exceptions aren't part of type.

Edit:

In C++17 noexcept actually became part of the type system so you cannot do this:

void (*p)();
void (**pp)() noexcept = &p; //error here

The rationale behind this decision is, from what I know, allowing better optiization.

来源：https://stackoverflow.com/questions/29492885/why-isnt-the-noexcept-specifier-part-of-the-function-type