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Function pointers with default parameters in C++


Function pointers with default parameters in C++

By : user20276
Date : November 21 2020, 07:01 PM
To fix this issue How does C++ handle function pointers in relation to functions with defaulted parameters? , Both foo() and bar() can only be assigned to func_ptr2.
§8.3.6/2:
code :


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combining similar functions into one common function involving passing function pointers as parameters

combining similar functions into one common function involving passing function pointers as parameters


By : user1082125
Date : March 29 2020, 07:55 AM
it fixes the issue The simplest way to make your code a bit more generic is the following :
code :
template<typename ComparisonType>
double* NeedleUSsim::FindIdx(double* containerBegin, double* containerEnd, ComparisonType comparison) {
    double* p = std::find_if(containerBegin, containerEnd, comparison);
    double* idx = 0;
    while(p != containerEnd)
    {
        idx = p - containerBegin;
        p = std::find_if(p+1, containerEnd, comparison);
    }
    return idx;
}

void NeedleUSsim::FindIdxRho()
{
  searchTmp = &ninfo->rho;
  double* idx = FindIdx(tplRho_deg, tplRho_deg+sampleDim[2], &NeedleUSsim::GreaterThanOrEqualTo);
  if( idx != 0 )
  {
    idxL = idx;
  }

}

void NeedleUSsim::FindIdxDepth()
{
  searchTmp = &ninfo->l;
  double* idx = FindIdx(tplL, tplL+sampleDim[1], &NeedleUSsim::LessThanOrEqualTo);
  if( idx != 0 )
  {
    idxRho = idx;
  }
}
golang pointers on pointers as function parameters

golang pointers on pointers as function parameters


By : Kunal
Date : March 29 2020, 07:55 AM
To fix this issue An interface type is simply a set of methods. Notice that the members of an interface definition do not specify whether or not the receiver type is a pointer. That is because the method set of a value type is a subset of the method set of its associated pointer type. That's a mouthful. What I mean is, if you have the following:
code :
type Whatever struct {
    Name string
}
func (w *Whatever) Foo() {
    ...
}

func (w Whatever) Bar() {
    ...
}
type Grits interface {
    Foo()
    Bar()
}
package main

import "fmt"

type Fruit struct {
    Name string
}

func (f Fruit) Rename(name string) {
    f.Name = name
}

type Candy struct {
    Name string
}

func (c *Candy) Rename(name string) {
    c.Name = name
}

type Renamable interface {
    Rename(string)
}

func Rename(v Renamable, name string) {
    v.Rename(name)
    // at this point, we don't know if v is a pointer type or not.
}

func main() {
    c := Candy{Name: "Snickers"}
    f := Fruit{Name: "Apple"}
    fmt.Println(f)
    fmt.Println(c)
    Rename(f, "Zemo Fruit")
    Rename(&c, "Zemo Bar")
    fmt.Println(f)
    fmt.Println(c)
}
Create template function for function that accepts function pointers as parameters for many types of functions

Create template function for function that accepts function pointers as parameters for many types of functions


By : Oleks
Date : March 29 2020, 07:55 AM
may help you . It's not clear what you are trying to achieve, but here is an example how your code may me translated into valid C++. First, let's define your Func with maximum number of arguments:
code :
template<typename T1, typename T2>
void Func(T1 par1, T2 par2) {
        std::cout << par1 << " " << par2 << std::endl;
}
template<typename ...T>
void FuncFunc(const std::function<void(T...)>& fPtr) {
        std::string funcVar = "Foo!";
        fPtr(funcVar);
}
template<typename F>
void GuncFunc(const F& fPtr) {
        std::string funcVar = "Foo!";
        fPtr(funcVar);
}
auto Func1 = std::bind(Func<float, std::string>, 3.14f, std::placeholders::_1);
auto Func2 = std::bind(Func<std::string, int>, std::placeholders::_1, 42);
FuncFunc(std::function<void(std::string)>(Func1));
FuncFunc(std::function<void(std::string)>(Func2));
return 0;
// Using std::bind
GuncFunc(std::bind(Func<float, std::string>, 3.14f, std::placeholders::_1));
GuncFunc(std::bind(Func<std::string, int>, std::placeholders::_1, 42));

// Or using lambdas:
GuncFunc([](std::string s){Func<float, std::string>(3.14f, s);});
GuncFunc([](std::string s){Func<std::string, int>(s, 42);});
Passing a function with templated parameters as an argument, with function pointers vs. std::function

Passing a function with templated parameters as an argument, with function pointers vs. std::function


By : Mithrandir2112
Date : March 29 2020, 07:55 AM
Hope that helps Template argument deduction does not work like that.
Template argument deduction is a pattern match. Is example an object of type std::function&)> for some type Targ?
C++(11) Default arguments for functions as template parameters, or as function pointers

C++(11) Default arguments for functions as template parameters, or as function pointers


By : stoyan1995
Date : March 29 2020, 07:55 AM
should help you out Yes There is a Good way. Function Object. I highly recommend you to see this link.
http://www.stanford.edu/class/cs106l/course-reader/Ch13_Functors.pdf
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