How do I pass a constant to pointer argument in C?

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lets say we have a function

C

int func(int* p)
{
....
}

if i want to pass variable k=15 i do:

C

x= func(&k); //and all works fine

what if if i want to pass just 15?(for the code to be more readable is some cases)

C

x=func(15);
x=func((int*)15);
x=func(&15);

these doesnt work or function returns wrong result? is it possible?

What I have tried:

x=func(15);
x=func((int*)15);
x=func(&15);

Posted 2-Jan-18 12:13pm

Member 13603933

Updated 14-Sep-22 4:56am

CPallini

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Peter_in_2780 · Answer 1 · 2018-01-02T18:44:00

Solution 1

To pass a pointer (the address of something), you need the "something" to exist in an identifiable memory location.
the simplest way is

C++

const k = 15;
x = func(&k);

For readability, you can adopt a naming convention for constants, suck as calling it k15.

Posted 2-Jan-18 18:44pm

Peter_in_2780

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CPallini 3-Jan-18 2:33am

5.

Tasin1010 · Answer 2 · 2022-09-13T21:01:00

<pre>
#include<iostream>
using namespace std;

void add(const int a)
{
    int y = 10 + a;
    cout<<"value of y : "<<y<<endl;
    //a = a + 4; can't modify
}

void display(const int& p)
{
    //p = p + 3; can't modify the value as using const
    cout<<"value of x : "<<p<<endl;
}

void display2(int const *p)
{
    cout<<"value of p means address of y : "<<p<<endl;
}

int main()
{
    int x = 10;
    add(x);//call by value

    int y =3;
    display(y);//call by reference

    display2(&y);//call by pointer
    cout<<"address of y is : "<<&y<<endl;

    return 0;
}