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Function parameter semantics (with nested functions)


Function parameter semantics (with nested functions)

By : Prabhanjan
Date : November 22 2020, 02:59 PM
wish help you to fix your issue The expression lambda y: y + x creates a new function, and this function is returned by do_something to its caller. In this function, x is a free variable, and the closure saves a reference to the parameter to the do_something function that was used to create it. So when you call
code :
f = do_something(4)
f = lambda y: y + 4
def f(y):
    return y + 4 


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Best Practise and Semantics of namespace nested functions and the use of extern "C"

Best Practise and Semantics of namespace nested functions and the use of extern "C"


By : swap0
Date : March 29 2020, 07:55 AM
I wish did fix the issue. This is for MSVC.
The namespace itself is not name-mangled, but the name of the namespace is incorporated in to the function's (or object's) name when name mangling occurs. This process is undocumented, but described here.
code :
complex sqrt(complex); // C + + linkage by default 
extern "C" { double sqrt(double); // C linkage } 
#ifndef MAIN_API
#   define MAIN_API __declspec(dllexport)
#endif

namespace x
{
    extern "C" MAIN_API void foo();
};

namespace y
{
    extern "C" MAIN_API void foo();
};
#include <cstdlib>
#include <iostream>
using namespace std;
#define MAIN_API __declspec(dllexport)
#include "main.h"

void x::foo()
{
    cout << "x::foo()\n";
}

void y::foo()
{
    cout << "y::foo()\n";
}

int main()
{
}
#ifndef MAIN_API
#   define MAIN_API __declspec(dllexport)
#endif

namespace x
{
    extern "C" MAIN_API void x_foo();
};

namespace y
{
    extern "C" MAIN_API void y_foo();
};
Static member function semantics when calling virtual functions

Static member function semantics when calling virtual functions


By : omid seedeqi
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Neither. You'll get a compiler error. You can't call non-static methods or access non-static members from a static method.
code :
//...
virtual void VirtualFunction( void ) = 0;

static void CallsVirtual( void ) {
    VirtualFunction();   // Nope!
}
//...
Move semantics when sending object as function's parameter

Move semantics when sending object as function's parameter


By : acaaca todtod
Date : March 29 2020, 07:55 AM
it should still fix some issue An rvalue reference is (surprise:) a reference indeed.
You can move from it, but std::move does not move.
code :
void foo(X&& x)
{
    X mine(std::move(x));

    // x will not be affected anymore
}
void foo(X&& x)
{
    x.stuff();
    x.setBooDitty(42);
}
What are the semantics of adding the parameter in LHS of function definition in haskell?

What are the semantics of adding the parameter in LHS of function definition in haskell?


By : rosy
Date : March 29 2020, 07:55 AM
I hope this helps you . Your last point is perhaps the most important one - fib1 and fib2 are indeed eta-equivalent, and the compiler is completely free to transform one into the other. Since this transformation has no effect on denotational semantics, only operational semantics, it is typically referred to as an optimization, if it "improves" the operational semantics of the program. The trouble is that it is hard to tell what "improves" really means.
The wiki page states
code :
fib3 :: Int -> Integer 
fib3 =
   let fib' 0 = 0
       fib' 1 = 1
       fib' n = fib1 (n - 1) + fib1 (n - 2)
    in \x -> map fib' [0 ..] !! x
fib1 :: Int -> Integer 
fib1 = ..

fib2 :: Int -> Integer 
fib2 x = ..
>:set +s
>fib2 32
2178309
(10.86 secs, 6,063,137,456 bytes)
>fib1 32
2178309
(0.00 secs, 0 bytes)
Why can I force reference move semantics for `&self` parameter of method, but not function parameters?

Why can I force reference move semantics for `&self` parameter of method, but not function parameters?


By : CWade3051
Date : March 29 2020, 07:55 AM
this one helps. I have two versions of a function that are intended to do the same thing. , You can fix it like this:
code :
v_ref = {
    // move `v_ref` to a new variable which will go out of scope by the end of the block
    let r = v_ref;
    // Mutating & slice splitting moved to function
    let (v_l, v_r) = example2_inner(r);

    match len % 2 {
        0 => v_l,
        _ => v_r,
    }
};
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