In your second example, the arguments are the same if:
1. Arguments are constexpr and implement a constexpr operator==
2. Are trivial and their byte layout is known at compile time
3. Exceptions for standard containers that can but don't have to make them
the same, implementation defined.

On Tue, 4 Mar 2025, 12:43 Tymi, <tymi.cpp_at_[hidden]> wrote:

> Hi Ell,
>
> Well, if the function arguments change (as in your 1st example), then they
> are not the same.
> In your second example, it could be implementation defined, so, I don't
> know.
>
> On Tue, 4 Mar 2025, 08:07 Tymi, <tymi.cpp_at_[hidden]> wrote:
>
>> Some functions have no side effects, and while they can be often
>> optimised, the compiler is not aware of that.
>> Consider this example:
>> ```cpp
>> template <std::arithmetic T> constexpr T(const T a, const T b) { return a
>> + b; }
>> ```
>> That function could be declared as pure, or in other words, produces no
>> side effects.
>>
>> I think this is a great way to optimise frequent calls, such as a pure
>> function's result can be cached for the same arguments provided.
>>
>> Diagnostics: if a pure function has side effects, the compiler shall
>> issue a warning and calling that function is undefined.
>>
>> pure member function shall be declared as const, otherwise, the program
>> is ill-formed.
>>
>>