You can

hm

Let's switch to Rose

~~merryrose~~

my command will fuck up the saved then

But I need Marie to be up so that I can see the saved stuff

Marie Rose .. now I see what you did there

/save howtoask

Yas! Added howtoask. Get it with /get howtoask, or #howtoask

/saved

/save howtoask Try phrasing your question like this: 1. Tell us the problem 2. Tell us what you have done so far. 3. Tell us how you got to this point 4. Paste code in pastebin or something similar 5. Wait for beautiful answers

Yas! Added howtoask. Get it with /get howtoask, or #howtoask

/saved

Notes in chat: - cpp - cppbook - freeprogrammingbooks - goodcodingmentality - goodgoogling - googleit - hownottoask - howtoask - ide - learn - meta - noendl - offtopic - ot - projects

#howtoask

It's slow af

I will just search them all in the chat and do it

y

y

We must switch to Rose now

#howtoask

Try phrasing your question like this: 1. Tell us the problem 2. Tell us what you have done so far. 3. Tell us how you got to this point 4. Paste code in pastebin or something similar 5. Wait for beautiful answers

k

/saved

/saved

Notes in chat: - cpp - cppbook - freeprogrammingbooks - goodcodingmentality - goodgoogling - googleit - hownottoask - howtoask - ide - learn - meta - noendl - offtopic - ot - projects

Hellow here

Notes in chat: - cpp - cppbook - freeprogrammingbooks - goodcodingmentality - goodgoogling - googleit - hownottoask - howtoask - ide - learn - meta - noendl - offtopic - ot - projects

why this doesn't compile auto lambda = []() { std::cout << "hello "; return lambda; }; but this? ` auto lambda = [](auto dummy) { std::cout << "hello "; return lambda; };

Give us your compiler message!

first one?

Compiler can't deduce the type

Why do you return lambda btw?

Nothing important just trying to get javacript like func()()() syntax

Compiler can't deduce the type

Try specifying the return type

Nothing important just trying to get javacript like func()()() syntax

Lambda isnt available inside of the scope of the lambda, caputer it first!

this doesn't work either auto lambda = [lambda]() { std::cout << "hello "; return lambda; };

show me the compiler warning please

hackerrank.cpp:4:17: warning: capture of variable 'lambda' with non-automatic storage duration auto lambda = [&lambda]() { ^~~~~~ hackerrank.cpp:4:6: note: 'auto lambda' declared here auto lambda = [&lambda]() { ^~~~~~ hackerrank.cpp: In lambda function: hackerrank.cpp:6:12: error: use of 'lambda' before deduction of 'auto' return lambda; ^~~~~~ hackerrank.cpp: In function 'int main(int, const char**)': hackerrank.cpp:11:11: error: void value not ignored as it ought to be lambda()()();

It can't deduce type but the silly thing is how it can deduce when I give it to a dummy parameter?

Show me the whole code

auto lambda = [lambda]() { std::cout << "hello "; return lambda; }; main(int argc, char const *argv[]) { lambda()()(); }

Dafaq why do you have three brackets?

Remove two of em

auto lambda = [](auto) { std::cout << "hello "; return lambda; }; main(int argc, char const *argv[]) { lambda(0)(0)(0); }

this compiles btw

Mhh

Why do you do such fuckeries? 🤔

XD

I was just trying something as I said nothing important but it is really interesting how it a dummy parameter make it be able to deduce the type

Compiler type deduction can be weird

Pretty sure olli knows more about it

why this doesn't compile auto lambda = []() { std::cout << "hello "; return lambda; }; but this? ` auto lambda = [](auto dummy) { std::cout << "hello "; return lambda; };

@ollirz i'm interested as well tbh

Actually I don't think it is possible to write its return type, since it returns itself which makes infinite recursive type name

Actually I don't think it is possible to write its return type, since it returns itself which makes infinite recursive type name

I think that you should stop doing weird stuff xD

I like to try things like these when I am too tired to write actual code 😀

then sleep

I like to try things like these when I am too tired to write actual code 😀

How about you stop trying to shoot yourself in the foot? :D

How about specifying the return value urself?

He can't

Since lambdas aren't a 'thing'

``` #include <functional> std::function< std::function<void()>() > ```

We are talking about lambdas not snd functions

Besides that wouldn't be what he wants to do

Language: cpp Source: #include <iostream> #include <functional> int main () { std::function<std::function<void()>()> lam = [&]() { std::cout << "lam()\n"; return lam; }; lam()(); } Result: lam() lam() Note: cplusplus_gcc assumed, other valid options are cplusplus_clang, visual_cplusplus, you can be more specific next time.

Aww C++

how to compile c++ in linux terminal

after saveing the file with .c extention

/saved

/saved

Notes in chat: - cpp - cppbook - freeprogrammingbooks - goodcodingmentality - goodgoogling - googleit - hownottoask - howtoask - ide - learn - meta - noendl - offtopic - ot - projects

/saved

/saved

Notes in chat: - cpp - cppbook - freeprogrammingbooks - goodcodingmentality - goodgoogling - googleit - hownottoask - howtoask - ide - learn - meta - noendl - offtopic - ot - projects

Guys how to copy text from notepad to excel using concatenate symbol &

I did by using double quotes( "") but it is upto 255 character only

#freeprogrammingbooks

#freeprogrammingbooks

https://github.com/EbookFoundation/free-programming-books/blob/master/free-programming-books.md

#projects

#projects

1. Go to github.com 2. Tap on search. 3. Use advanced search. 4. Go to Advanced Search. 5. Select user's working language. 6. Press the search button. 7. Hover through the repositories and select one of them. 8. Fork it. 9. Do what you want to do. 10. Make a pull request so that the "original owner" can look at the changes and merge them if he feels the need to.

how to compile c++ in linux terminal

g++ filename.cpp

why this doesn't compile auto lambda = []() { std::cout << "hello "; return lambda; }; but this? ` auto lambda = [](auto dummy) { std::cout << "hello "; return lambda; };

@linuxer4fun I canot reproduce this. Neither Clang, G++, nor MSVC compiles it. imho Clang has the best error diagnostic <source>:13:16: error: variable 'lambda' declared with deduced type 'auto' cannot appear in its own initializer See: https://godbolt.org/z/7B30bp

Nothing important just trying to get javacript like func()()() syntax

using a functor such as the following Foo, you can call it and its return value. To do something like f(1)(2)(3)("Hello"); struct Foo { template <class T> const Foo& operator()(const T& t) const { std::cout << t << " "; return *this; } }; void bar() { Foo f; f(1)("Hi there")(3)(3.141f); }

using a functor such as the following Foo, you can call it and its return value. To do something like f(1)(2)(3)("Hello"); struct Foo { template <class T> const Foo& operator()(const T& t) const { std::cout << t << " "; return *this; } }; void bar() { Foo f; f(1)("Hi there")(3)(3.141f); }

are mem_func_t's called functors as well?

What's the definition of those suckers?

What's the definition of those suckers?

Functors? The term is not defined by the standard but is used to describe function objects that can be "called" (defining operator () ) mem_func_t in Freebsd are function pointers (typedef for a function pointer returning int and taking several parameters)

No mem_func_t in std::

No mem_func_t in std::

std::mem_fn ? or std::mem_fun_t ? Although the exact return type of std::mem_fn is not specified it return a (stateful) function object std::mem_fun_t is a function object (has been deprecated in C++ 11 and finally removed in C++17)

ye,, those suckers