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Using pointers as parameters


Using pointers as parameters

By : user2185234
Date : November 21 2020, 04:01 AM
Does that help Your code is not doing what you think it does. To understand, try to compile this:
code :
void foo(int* x) {}

int main() { 
    int x = 42;
    foo(x);
}
using namespace std;
 swap(a, b); 
void swap(int& x, int& y){
    int temp;
    temp = x; 
    x = y; 
    y =temp;
}


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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)
}
Pointers as parameters in C++

Pointers as parameters in C++


By : v22
Date : March 29 2020, 07:55 AM
wish helps you EDIT (based on comments down here where poster says it's called actually with the address of an int: foo((u_char*)&count) ):
code :
void foo(u_char *ct)
{
   int *pcounter = (int *)ct;  // change pointer back to an int *
   printf("Count: %d\n", *pcounter);
   (*pcounter)++;  // <<-- brackets here because of operator precedence.
}
void foo(u_char *ct)
{
   printf("Count: %d\n", (*(int *)ct)++);
}
Using pointers as function parameters

Using pointers as function parameters


By : neha
Date : March 29 2020, 07:55 AM
this one helps. I want to do a function that uses pointers a s parameters and return one of the pointers, is it possible? , Assuming this worked (it doesn't compile, rightly so):
code :
 total=x+y;
*total = *x + *y; 
Why should I prefer references on smart pointers over smart pointers as parameters in C++

Why should I prefer references on smart pointers over smart pointers as parameters in C++


By : talaysa
Date : March 29 2020, 07:55 AM
wish of those help The possible reasons are:
Performance. It should be faster to pass a reference (one CPU register) rather than a smart pointer by value. There is something wrong with your performance tests. Saving stack space. A smart-pointer passed by value takes more space on the stack than a reference.
Can i use pointers in and with template parameters?

Can i use pointers in and with template parameters?


By : user3552962
Date : March 29 2020, 07:55 AM
To fix the issue you can do I'm trying to do someting like this: ,
Can i use pointers in and with template parameters?
code :
std::map<const FromType*, ToType*> Data;
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