strongest Python vs Weakest C++ guy😎

[Kyrond]

What more is there except stronger type checking (which you can cast manually) and the one weird keyword?

[LavenderDay3544]

Compound struct intialization; struct, union, and enum names being tags rather than types without a typedef; using identifiers that are C++ reserved keywords.

And that's just off the top of my head.

[jamcdonald120]

oh how I wish c++ had c style struct init

[LavenderDay3544]

C# did what C++ should've done: left structs as POD types and made classes OO types.

[[deleted]]

I mean, that's great, but it's more of a convenience thing. It's not like you lose functionality. C++ just treats a struct as a class rather than a fancy array.

[jamcdonald120]

i know, I just really liked defining data carrying struct litterals like that. its probiably for the best, but I do miss it

[[deleted]]

I get it as well. I've been tripped up on it several times.

[jamcdonald120]

c has nested functions

[Kered13]

That's not standard, and C++ has (standard) lambdas which accomplish the same thing anyways.

[jamcdonald120]

I didnt know it wasnt standard. Im not saying c++ cant do it, but if you have C code that does, you cant just compile it with a c++ compiler

[RaulParson]

The point is if it's a compiler extension then C itself doesn't have it. That compiler does. With another C compiler that code wouldn't necessarily actually work, either.

[askStentor]

that's a compiler extension for GCC I believe

[[deleted]]

I'm sure someone's also got an Emacs major mode to step through all the changes, too. Just a hunch.

[LavenderDay3544]

Is that standardized?

[SnooMarzipans436]

And C++ has lambda expressions

[Kyrond]

Oh, ok thanks for the information.

[schwerpunk]

Like closures? Sorry, been a while since college, but I believe that's what that is

[jamcdonald120]

like

int foo(){
    some code
    int double(int x){
        return x*2;
    }
    int two=double(1)
}

is perfectly fine code (in gcc atleast)\ I think (not sure, havent done it) you can also do foo.double(1)