I would like to use parameter in C++ to store back whatever value/object.
In this example, I try to store the value from the global variable as a simplified example.
This code doesn't work,
int value = 20;
void returnPointer2(int* hello)
{
hello = &value;
}
// It changes nothing
int value2 = 100;
returnPointer2(&value2);
cout << value2 << endl;
as I needed double pointer.
void returnPointer3(int** hello)
{
*hello = &value;
}
int* vp2 = new int();
*vp2 = -30;
returnPointer3(&vp2);
cout << *vp2 << endl; // expects 20
I reminded of the reference, and I can use pointer reference to get the same result.
void returnPointer4(int* & hello)
{
cout << "value : " << value;
hello = &value;
}
int* vp3 = new int();
*vp3 = -130;
returnPointer4(vp3); // also expects 20, but much simpler to use
cout << "better : " << *vp3 << endl;
I tried with double &, and it compiles.
void returnPointer5(int&& hello)
{
cout << "value : " << value;
hello = value;
}
However, it doesn't compile with the input of integer variable.
int vp4 = 123;
returnPointer5(vp4); // also expects 20, but even more simpler to use
cout << "best : " << vp4 << endl;
This is an error message.
pointer_return.cpp:31:6: error: initializing argument 1 of 'void returnPointer5(int&&)'
void returnPointer5(int&& hello)
I happened to know about move, and it works with this code.
int vp4 = 123;
returnPointer5(move(vp4)); // also expects 20, but much simpler to see
cout << "best : " << vp4 << endl;
What's the magic/logic behind this move function?
There is a lot of stuff getting mixed in here, but to keep it simple I'll address your root question.
&& is nothing at all like **.
&& is an rvalue reference, while ** is a pointer to a pointer.
As a second point, you are declaring in your function name what you want to do: returnPointer4.
You want to have a pointer to an integer returned back. int*& is the correct syntax for having a reference to a pointer.
Reading over your question again, why don't you use the following:
int& returnGlobalReference() {
return value;
}
Then in your other function:
int& value2 = returnGlobalReference();
The first attempt makes the classic mistake of passing a pointer by value, modifying its address in the function and expecting what it points to to change.
As mentioned in the comments,
void returnPointer2(int* hello)
{
hello = &value; // don't do this, it only modifies what the
// pointer hello, which resides in the stack, points to
*hello = value; // do this instead. even though hello still resides in the
// stack, you're modifying the location that hello points to,
// which was your original intention
}
why do you want to pass pointers however? is the static variable not available when you call the function? (perhaps, different files?)
The magic of std::move is:
The actual declaration for std::move is somewhat more involved, but at its heart, it's just a static_cast to an rvalue reference.
Taken from here.
As Jeffery Thomas already said, a && is not a reference to a reference, but a reference to a rvalue.
Related
I can pass any argument by reference and the changes to that variable in any of the function calls are reflected in other function calls too.
Likewise if I want to keep the address stored in a pointer consistent across all the function call such that it behaves like we are passing a pointer by reference.
You can pass pointers by reference too, as suggested in a comment.
#include <iostream>
void f(int * & p) {
std::cout << p << std::endl;
}
int main() {
int x;
int * p = &x;
std::cout << p << std::endl; // This prints the same...
f(p); // ... as this.
}
I am thinking of this solution, but not sure if it is correct.
Type* myObjP;
void setSmth(Type& toBePassed) {
myObjP = *toBePassed;
};
Syntactically speaking, you would need to use:
void setSmth(Type& toBePassed) {
myObjP = &toBePassed; // Addressof operator rather than dereference operator.
}
Word of Caution
When using such a mechanism, you'll have to be very careful when you dereference myObjP. If the object passed to setSmth is not alive, dereferencing myObjP will cause undefined behavior.
There are more robust mechanisms in the language but it's hard to suggest anything without additional context.
toBePassed is a reference, not a pointer. So you actually need the & operator (address of operator), not the * operator (dereference operator):
myObjP = &toBePassed;
Note: You're not really storing a reference in a pointer. You're just storing a regular pointer to the object that is being referenced.
You can try it.
you should use & operator for pass address to pointer.
double* myObjP;
void setSmth(double& toBePassed)
{
myObjP = &toBePassed;
};
int main()
{
double testValue{8.0};
setSmth(testValue);
if (testValue == *myObjP)
{
cout << "We store a reference to an object in the pointer\n";
cout << "testValue ==> " << testValue << " " << "*myObjP ==> " << *myObjP<<std::endl;
}
std::getchar();
}
This question already has answers here:
When I change a parameter inside a function, does it change for the caller, too?
(4 answers)
Closed 4 years ago.
I know call by pointers in which we pass address of variables.Something like this:
void swap(int *x, int *y)
{
int temp;
temp = *x; /* save the value at address x */
*x = *y; /* put y into x */
*y = temp; /* put x into y */
return;
}
swap(&a, &b);
And also call by reference,in both of these methods the changes made in the function are reflected in actual arguments' variable.
But why are actual parameters passed in this case of call not changed:
#include <iostream>
using namespace std;
void foo(int* c){
c=c+1;
}
int main()
{
int a=5;
int *c=&a;
cout<<&c<<endl; //0x7ffe1a74f3b0
foo(c);
cout<<*c<<endl;//5
cout<<&c<<endl;//0x7ffe1a74f3b0
}
Here c passed to foo() is address of a.So how this is call by value.
Here c should have printed garbage value according to me.Please explain what has happened here.
And also call by reference, in both of these methods the changes made in the function are reflected in actual arguments' variable.
There is an important difference, though: the changes are always made to whatever is referenced/pointed to, never to the reference/pointer itself (modifying a reference is impossible in general).
That is why assigning c a new value inside foo has no effect on c outside foo: the pointer passed to a function is copied.
If you need to modify the pointer, you need to add another level of dereference by passing a pointer reference or a pointer to a pointer.
Following on from comments, the variable c defined in function main is a different variable to the parameter c of function foo. If you want foo to be able to modify main's c, that is modify the address that c's pointer type holds, then you need to pass either a reference or pointer to c to the function instead.
Here is an example that shows the difference between passing c by value (as int *), or by reference (as int ** or int *&). Don't be fooled by the fact that int * is a pointer type, that means that it can receive an int by reference or an int * by value. And since main's c is int * rather than int, main c is being passed by value.
Note the differences in how the functions are called (whether c needs the address operator & in the function call) and the outcome of each function.
#include <iostream>
using namespace std;
void foo_int_ptr(int* c)
{
c=c+1;
}
void foo_int_ptr_ptr(int** c)
{
*c=*c+1;
}
void foo_int_ptr_ref(int*& c)
{
c=c+1;
}
int main()
{
int a=5;
int *c=&a;
cout << "&c=" << &c << ", c=" << c << ", *c=" << (c==&a ? std::to_string(*c) : std::string("INVALID PTR")) << endl;
foo_int_ptr(c);
cout << "&c=" << &c << ", c=" << c << ", *c=" << (c==&a ? std::to_string(*c) : std::string("INVALID PTR")) << endl;
foo_int_ptr_ptr(&c);
cout << "&c=" << &c << ", c=" << c << ", *c=" << (c==&a ? std::to_string(*c) : std::string("INVALID PTR")) << endl;
foo_int_ptr_ref(c);
cout << "&c=" << &c << ", c=" << c << ", *c=" << (c==&a ? std::to_string(*c) : std::string("INVALID PTR")) << endl;
}
Output:
&c=0x7e02d81808b8, c=0x7e02d81808ac, *c=5
&c=0x7e02d81808b8, c=0x7e02d81808ac, *c=5
&c=0x7e02d81808b8, c=0x7e02d81808b0, *c=INVALID PTR
&c=0x7e02d81808b8, c=0x7e02d81808b4, *c=INVALID PTR
there is a mistake in your thinking about this ..
int *c = &a;
this doesn't mean that c "contains" address of a, this means that c is a pointer TO the address of a. Passing a pointer to foo() will not do anything.
From what I understand, this should work:
const char* x = "x";
std::cout << x << std::endl;
Passing x into this function:
void myClass::passAsVoid(void* v) {
std::cout << (const char*)v << std::endl;
}
The first example prints "x";
The second example prints "\350\224A";
I want to learn what's going on, and the correct approach to do this!
The actual code:
float delay = 1;
std::string txt = "random filler text that is not lorum ipsum";
for (int i = 0; i < txt.length(); ++i) {
const char* x = "x";
std::cout << x << "code1" << std::endl;
CCSequence* seq = CCSequence::create(CCDelayTime::create(i*delay),
CCCallFuncND::create( this, callfuncND_selector(OverWorldView::setString), (void*)x ),
NULL);
this->runAction(seq);
}
Callback function:
void OverWorldView::setString(void* x) {
std::cout << (const char*)x << "code2" << std::endl;
label1->setString( (const char*)x );
}
I'm using cocos2dx 2.1.4
It looks like you are not using the API correctly.
The documentation says you need to use a SEL_CallFuncND type callback, which receives two arguments, not one.
For everyone's convenience, callfuncND_selector is a macro that hides a static_cast, or probably even a C-style cast (could not find other API versions online) which lets you use just about anything as a callback without getting any compilation errors. Pure joy.
This won't even compile because you're passing a const pointer to a function which accepts a non-const pointer. If you can't change the type of the callback then instead of passing a string literal, pass a character array.
char buffer[100];
callback(buffer); //buffer decays to char* which is implicitly converted to void*
If you can change the type of the callback though then modify the parameter to be const void*.
Can someone tell why test(2) object is destroyed after test_method() call?
#include<iostream>
#include<string>
using namespace std;
class test
{
int n;
public:
test(int n) : n(n)
{
cout << "test: " << n << endl;
}
~test()
{
cout << "~test: " << n << endl;
}
test & test_method()
{
cout << "test_method: " << n << endl;
return *this;
}
};
int main(int argc, const char *argv[])
{
cout << "main start" << endl;
const test &test1 = test(1);
const test &test2 = test(2).test_method();
cout << "main end" << endl;
}
Output is:
main start
test: 1
test: 2
test_method: 2
~test: 2
main end
~test: 1
test(2).test_method() returns a reference, which is bound to test2, and then the object to which it refers is destroyed at the end of the full expression, since it is a temporary object. That should not be a surprise.
The real surprise is that test1 remains a valid reference, because it is directly bound to a temporary, and binding a temporary to a reference extends the lifetime of the temporary to that of the reference variable.
You only have to note that in the test(2) case, the temporary object isn't bound to anything. It's just used to invoke some member function, and then its job is done. It doesn't "babysit" member functions, or in other words, lifetime extension isn't transitive through all possible future references.
Here's a simple thought experiment why it would be impossible to actually have "arbitrary lifetime extension":
extern T & get_ref(T &);
{
T const & x = get_ref(T());
// stuff
// Is x still valid?
}
We have no idea if x remains valid beyond the first line. get_ref could be doing anything. If it's implemented as T & get_ref(T & x) { return x; }, we might hope for magic, but it could also be this:
namespace { T global; }
T & get_ref(T & unused) { return global; }
It's impossible to decide within the original translation unit whether anything needs to be extended or not. So the way the standard has it at present is that it's an entirely trivial, local decision, just made when looking at the reference declaration expression, what the lifetime of the temporary object in question should be.
Because the C++ standard requires this behavior. Give the object a name if you want it to persist. It will persist as long as the name.
Edit: You your example, test1 is the name that you gave to the first object, whereas the second object has obtained no name at all, and so it does not outlast evaluation of the expression.