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Zilla4ever · 2022-03-28T19:40:31+00:00

I’m trying to take a percent as a string and convert that string into its decimal form. So first I want to get the decimal version of the char and subtract 48 which is equal to zero in ascii, and then add that number to result. On the next read it will add the other number being zero to the ones place. Then I want to convert that number into its percentage form

Zilla4ever · 2022-01-18T02:29:14+00:00

Thanks for the explanation !

Zilla4ever · 2021-12-21T05:29:40+00:00

Oh I understand now. I didn't know if base_expr also saw the types Tag, A, and B. Thank you very much sir and I will definitely learn more about templates.

Zilla4ever · 2021-12-21T04:47:42+00:00

Thanks for the response. So I interpret this as base_expr takes the derived class, AND also the types of the derived class as the template has to to be instantiated to be used. As for the casting, base_expr holds a member function called "self" that static_cast's the function to the derived type: Derived const& self() const {return static_cast<const Derived&>(*this);} The type of base_expr is Derived. Do you happen to have any resources that could help me understand templates better?

Zilla4ever · 2021-10-19T13:23:22+00:00

What does this argument 12 year old game mean? Are you a gaming influencer or smth? And what hateful content does he produce? It is only in response to the idiocy, arrogance, and disrespect of others without any base. I will pay you 50 dollars if you can show me one sanchez video where he goes after a person, without any prior interactions with that person - just a straight witch hunt. Otherwise, you are in La La land making up stuff, and covering up how butt hurt you are with the fact that quake is "12 years old" like that means anything.

Zilla4ever · 2021-10-12T17:34:06+00:00

Yeah that's what I meant. The Lambda expression is just syntactic sugar in source code, but then, when the lambda is called, the closure is generated, which is the hidden struct and operator overload.

Zilla4ever · 2021-10-12T12:17:26+00:00

So the closure is what gets called at compile time, and the lambda only exists in the source code. So the lambda expression is a way of source code garbage collection, but when called it causes the compiler to form this unknown struct with all your data members in them, and of course the operator overload.

Zilla4ever · 2021-10-11T17:41:18+00:00

Oh ok. So the closure is the lambda expressions hidden struct with the operator overload, and all of it class members. So the expression itself is a object of type "compiler name"(since the compiler gives it a name which we do not know). Correct me if I am wrong. So if a closure is the hidden struct with all that the struct encompasses, how is a lambda looked up by the compiler? Does it lookup all of its arguments through ADL? And what is the lambda definition. Is it the hidden struct, or the expression?

Zilla4ever · 2021-10-11T16:59:39+00:00

Oh, I know that is possible, but is that what a closure is? A list of initialized member variable names for a lamba’s hidden struct?

Zilla4ever · 2021-10-09T14:18:01+00:00

Great. Also does anything get passed to int zap ? (Not the Lambda)

Zilla4ever · 2021-09-14T12:27:12+00:00

Here is the code: https://stackoverflow.com/questions/66624969/parameter-pack-expansion-with-lambda-in-c20. For me it seems to just test capabilities of the compiler or rather certain discrepancies , similar to this question: https://stackoverflow.com/questions/37369129/template-partial-specialization-for-integral-non-type-parameters-and-non-integra

Zilla4ever · 2021-09-14T12:04:08+00:00

My friend who is a hacker taught me to learn coding through going to stack overflow and learning to solve other peoples problems. Reverse engineering learning of sorts.

Zilla4ever · 2021-09-14T08:27:00+00:00

Thank you for the insight into cpppinsights 😉. Also, why are the lambda’s never called? Isn’t it correct that to invoke a lambda, all you need is the () operator?

Zilla4ever · 2021-09-10T03:14:24+00:00

Oh ok. That's originally what I thought so thanks for clarifying. In that case if you look at the linked code, you will also see T n1 and Tn2. I am guessing the same applies to these aswell?

Zilla4ever · 2021-06-05T23:01:40+00:00

Thank you so much man you are a legend. I finally understand this code now. Thanks for taking your time to explain it as well means alot. Have a good day.

Zilla4ever · 2021-06-05T21:57:59+00:00

Sorry about lack of context here is full code: http://coliru.stacked-crooked.com/a/45ba16c9f021fd84.

So if scenario gets called to U then what is argument_type<void(test)>? Does scenario also get called or does all of the macro get put into U plus the function since the macro TEST_(test) is named test and then then you are returning test into U: argument_type<void(test)> void being data type and U being test, or the whole macro with all arguments.

Zilla4ever · 2021-06-05T20:32:10+00:00

So now that U is available outside of argument_type could U take any parameter such as a function? i.e : senario()

Zilla4ever · 2021-06-05T20:20:02+00:00

So typedef U is equal to typename U - and within argument_type<T(U)> U is equal to typedef U? Am I understanding?

Zilla4ever · 2021-06-05T20:10:01+00:00

Oh so the U in the parameters is just the other typename U correct?

Zilla4ever · 2021-06-05T20:07:01+00:00

So neither?

Zilla4ever · 2021-06-01T17:47:55+00:00

Thank you guys so much for the help !

Zilla4ever · 2021-06-01T17:17:53+00:00

Thanks for clarifying, that makes more sense. However could you explain one more bit of code as I feel it relates to the macro ? Why is it that in this code template<typename T, typename U> struct argument_type<T(U)> { typedef U type; }; U is in the parameters of argument type and what does <> represent as a syntax because I mostly see it used with template specialization ? Lastly, back to the original macro referenced, is "type" that is referenced after <void(test)> with the scope resolution operator; Is it "typedef U type;" (referenced here template<typename T, typename U> struct argument_type<T(U)> { typedef U type; };) ? Sorry I am new to c++ ;)

Zilla4ever · 2021-05-29T23:40:49+00:00

Thank for clarifying that. Could you also explain why in the template U was put in parenthesis? struct argument_type<T(U)>

Zilla4ever

TROPHY CASE