How to initialize the reference member variable of a class? - c++

Consider the following code C++:
#include<iostream>
using namespace std;
class Test {
int &t;
public:
Test (int &x) { t = x; }
int getT() { return t; }
};
int main()
{
int x = 20;
Test t1(x);
cout << t1.getT() << " ";
x = 30;
cout << t1.getT() << endl;
return 0;
}
It is showing the following error while using gcc compiler
est.cpp: In constructor ‘Test::Test(int&)’:
est.cpp:8:5: error: uninitialized reference member ‘Test::t’ [-fpermissive]
Why doesn't the compiler directly call the Constructor?

That is because references can only be initialized in the initializer list. Use
Test (int &x) : t(x) {}
To explain: The reference can only be set once, the place where this happens is the initializer list. After that is done, you can not set the reference, but only assign values to the referenced instance. Your code means, you tried to assign something to a referenced instance but the reference was never initialized, hence it's not referencing any instance of int and you get the error.

My compiler emits this error:
error C2758: 'Test::t' : must be initialized in constructor base/member initializer list
And that's exactly what you must do. References must be initialized in the initializer list:
#include<iostream>
using namespace std;
class Test {
int &t;
public:
Test (int &x) : t(x) { } // <-- initializer list used, empty body now
int getT() { return t; }
};
int main()
{
int x = 20;
Test t1(x);
cout << t1.getT() << " ";
x = 30;
cout << t1.getT() << endl;
return 0;
}
Explanation:
If the reference is not in the initiliazer list, it's next to impossible for the compiler to detect if the reference is initialized. References must be initialized. Imagine this scenario:
Test (int &x, bool b)
{
if( b ) t = x;
}
Now it would be up to the caller of the constructor to decide if valid code was generated. That cannot be. The compiler must make sure the reference is initialized at compile time.

Related

Why is object member value changing between 2 getter calls

I am getting unexpected value in second getter call which looks wrong to me, any specific reason for this happening?
#include<iostream>
using namespace std;
class Test {
public:
int &t;
Test (int x):t(x) { }
int getT() { return t; }
};
int main() {
int x = 20;
Test t1(x);
cout << t1.getT() << " ";
cout << t1.getT() << endl;
return 0;
}
The problem here is that your code results in undefined behavior.
The constructor of Test does not take a reference to an int but a copy, and due to int x only being a temporary copy which is not guaranteed to live until your second function call you will end up with undefined behavior.
You would have to change your constructor to the following to make the code work properly:
Test(int& x) : t(x) {}
Now the reference you're working with in Test will be the same x as defined in main

set default parameter to a member variable [duplicate]

This question already has answers here:
How to use a member variable as a default argument in C++?
(4 answers)
Closed 1 year ago.
I tried to set a default parameter to a member variable, it gave me this bug.
[cquery] invalid use of non-static data member 'num'
code
#include <iostream>
class Test{
private:
int val = 0;
public:
void print_num(int num = val){
std::cout << num << '\n';
}
}
int main(){
Test test;
test.print_num();
return 0;
}
Despite it being quite obvious what this would mean, you just can’t. The usual workaround is to provide an overload that (implicitly) uses this to get the member value.
In order to set a default value for a function parameter to a class member, the member must be static.
Citing the documentation,
Non-static class members are not allowed in default arguments
However, given the context of your code sample I doubt this is what you want to do (as every instance of Test class will point to the same val. Citing the docs:
Static members of a class are not associated with the objects of the class: they are independent variables
If you make val static in your example, and have multiple instances of Test in use by your code, you can (and very likely will) have some unexpected behavior. Consider:
#include <iostream>
class Test{
private:
static int val;
public:
void print_num(int num = val){
std::cout << num << '\n';
}
void set_val(int num) {
val = num;
}
}
int Test::val = 0;
int main(){
Test test1;
test1.set_val(1);
Test test2;
test2.set_val(2);
test1.print_num(); // results in "2"
return 0;
}
A better alternative would be to pass a pointer to your function like this:
#include <iostream>
class Test{
private:
int val = 0;
public:
void print_num(int* numptr = nullptr){
int num = (numptr ? *numptr : val);
std::cout << num << '\n';
}
}
int main(){
Test test;
test.print_num();
return 0;
}
Or, as described by Davis Herring, use an overload:
#include <iostream>
class Test{
private:
int val = 0;
public:
void print_num(int num){
std::cout << num << '\n';
}
void print_num() {
return print_num(val); // return is irrelevant here, but my preferred coding style
}
}
int main(){
Test test;
test.print_num();
return 0;
}
Edited to reflect the comment and example.
Your function definition doesn't actually know what val is because it doesn't actually live inside of your class. The compiler is actually making a function that contains your class as a parameter and abstracts away all of that. You'll want to set num to val within the function body.

How can i increase value of x in function?

C++ Program: - I know that in the function definition, x is not passed so it would get error message but i want to increase in function, so what can i do?
#include <iostream>
using namespace std;
fun(int& p)
{
cout << p;
x++;
}
int main()
{
int x = 15;
int c = 1;
fun(c);
cout << x;
return 0;
}
The value of variable x is undefined for fun() and you should go through a book for basics.
The only method if you just want to manipulate the same variables, is using reference variables as parameters and then you can directly modify its original values.
Refined version of your program for accessing and manipulating the data with functions:
#include <iostream>
void fun(int &, int &);
int main(void) {
int x = 15;
int c = 1;
fun(c, x); // c is printed "1" and x increments with 1
std::cout << x << std::endl; // new value of x prints
return 0;
}
void fun(int & p, int & x) {
std::cout << p << std::endl;
x++; // increments original x
}
Note: Alternatively, you can declare your required variable in global scope by putting them outside of all the functions and underneath the header declaration, so that they'll be visible to the entire program but remember that you must need to use reference for changing the variable values for the whole program.
You can either set x as a global variable or you can also pass it to your function.
fun(int &p, int &x) and then call it from main.

Why this code showing an error lvalue required in line 1? How to resolve it?

#include<iostream>
using namespace std;
class Test
{
private:
int x;
public:
Test(int x = 0) { this->x = x; }
void change(Test *t)
{
this = t; //line 1
}
void print() { cout << "x = " << x << endl; }
};
int main()
{
Test obj(5);
Test *ptr = new Test (10);
obj.change(ptr);
obj.print();
return 0;
}
since we know that this pointer hold the reference of calling object. In line 1 i am trying to change the reference of calling object but it shows an error "lvalue required". Can someone explain this??
You cannot assign a pointer to this pointer, because it's a prvalue.
this pointer is a constant pointer that holds the memory address of the current object.
As a result, this is of type const Test* in your case, so it cannot be assigned to. Doing so (if it was allowed) would effectively allow an object to change its own address in memory, as #Peter mentioned.
Note: const Test* is a pointer to a constant object. The object it points to is constant, not the pointer itself.
PS: this->x = t->x; is probably what you meant to say.
Here you are assigning a pointer(here t) to "this" pointer for a particular object.
"this" pointer is const. pointer that holds the memory address of the current object. You simply can't change the this pointer for an object, since doing this you will practically be changing the location of the object in the memory keeping the name same.
Reference - ‘this’ pointer in C++
#include <iostream>
using namespace std;
class Test
{
private:
int x;
public:
Test(int x=0)
{
this->x = x;
}
void change(Test *t)
{
t->x; //t is a pointer. so make it point to x
}
void print() { cout << "x = " << x << endl; }
};
int main()
{
Test obj(5);
Test obj1(10); //create a new object
Test *ptr = &obj1;//make the pointer point to obj1
obj.change(ptr); //use change() to point to argument of obj1
obj.print(); //print the value of obj now
return 0;
}

Class non-static method pointer to global function

I'm trying to point a class method to a global function, i've seen this
but how i can't do it without instance?.
Consider this:
class x
{
public:
int(x::*GetVal)(int);
};
int RtX(int a)
{
return a * 4;
}
// declaration
int(x::*GetVal)(int) = (int(x::*)(int))&::Rtx; // :: global? // error
int main()
{
x a;
cout << (a.*GetVal)(4) << endl;
}
This returns me the error:
[Error] invalid cast from type 'int ()(int)' to type 'int
(x::)(int)'
x::GetX is a pointer to member. These are deeply complicated beasts, and you can't get them to point to non-member functions. The following code will work:
#include <iostream>
int RtX(int a) // Global non-member function
{
return a * 4;
}
class x
{
public:
int(x::*GetVal)(int);
// This is an instance member function which acts as a proxy and calls the
// the global function
int RtX(int a) { return ::RtX(a); }
};
int main()
{
x a;
a.GetVal =&x.RtX; // Assign the member variable. Could do this in the
// constructor. **NOTE** No casts!
std::cout << (a.*GetVal)(4) << std::endl;
}
If you find yourself reaching for a cast when dealing with function pointers and pointers-to-member-functions, stop - you are almost certainly doing it wrong, and while it will compile, it is very likely not to run properly.
Alternatively, as noted in the comments, use std::function.
#include <iostream>
#include <functional>
int RtX(int a)
{
return a * 4;
}
class x
{
public:
std::function<int(int)> GetVal;
// Initialize GetVal in the constructor.
x() : GetVal(RtX)
{}
// Alternatively, you can initialize the variable in with a default
// initialization. You can also declare the member const if you don't want to
// change it later.
const std::function<int(int)> gv = RtX;
/*
};
int main()
{
x a;
std::cout << a.GetVal(4) << std::endl;
}
Non static member functions need an instance in order to be called. You might consider using a static function member instead and if you also use std::function, you might get a simple code to assign your member function without instance:
#include <iostream>
#include <functional>
int RtX(int a)
{
return a * 4;
}
class x
{
public:
static std::function<int(int)> GetVal;
};
std::function<int(int)> x::GetVal = RtX;
int main()
{
x a;
std::cout << (a.GetVal)(4) << std::endl;
}