Let's say I have this program:
const int width = 4;
void test(int&){}
int main() {
test(width);
}
This will fail to compile. I notice that constant values ( also enumeration constants ) with names ( such as width ) cannot be passed by reference. Why is that so?
Imagine this:
void test (int& j) { j++; }
If test does change the value of the thing referenced, clearly we can't call it with a const parameter. And if it doesn't, why does it take its parameter by non-const reference?
Passing by reference allows us to change the actual object.
If an object is defined as const, it cannot be changed. That is exactly what const means - it's constant.
Related
I've got a question in my assignment that asks me to evaluate whether the following function call is correct. I'm not sure if a const int can be initialized with a variable of type const int&. I know that a const int can be initialized with another int, for example
int i=3; const int j=i
works perfectly fine, but I'm not sure if the following code is semantically correct (the line const int j=bar(++i);)
int foo (int& i) {return i+=2;}
const int& bar(int &i){ return i+=2;}
int main(){
int i=5;
const int j=bar(++i);
}
Yes it can. You can initialize an object with any value category.
When constructing the type, it will simply call its constructor with the correct overload, such as type(type&& other) or type(type const&).
For trivial types, it's always a copy. So as long as the types are compatible, no matter their value category, it will work.
Initializing a reference is different. You must have an expression with a compatible value category. For example, creating a mutable reference from your function won't work:
int& j = bar(++i); // won't compile, int& cannot be bound to int const&
This is because bar returns a reference to const, thus cannot be bound to a reference to mutable.
As a side note, even though it's an int constant, it is not a compile time constant anymore. Thus you won't be able to use it as an array size or template parameter.
To fix that you'd have to use constexpr, which will guarantee that the value of your variable is available at compile time.
int a = 9;
constexpr int b = a; // Won't work, `a` is a runtime value, `b` is compile time
constexpr int a = 1;
constexpr int b = a + 1; // Works! Both compile time values
A reference is basically just an alias to the object it is bound to (with possible indication of immutability by const). Since bar(++i) returns a reference bound to i, it is the same as if you initialized j by i after bar(++i) call:
bar(++i);
const int j = i;
Const reference just says that you cannot modify the bound object through that reference. But there is nothing in your code that would try doing this.
I am calling a function with the signature
void setValue(int& data)
I would like to pass a literal number to it:
setValue(1);
But I get:
error: invalid initialization of non-const reference of type 'int&' from an rvalue of type 'int'
Is there a way I can make this work without changing the function (it's in a library) and without assigning each literal value to a variable?
Assuming setValue does not actually modify its argument and just has a wrong signature which you cannot change, here is an approach which is not thread-safe among other things:
#include <iostream>
void setValue(int &i)
{
std::cout << "i = " << i << std::endl;
}
int& evil(int i)
{
static int j;
j = i;
return j;
}
int main()
{
setValue(evil(1));
setValue(evil(2));
}
When you declare the argument as being an int&, you are saying that the function called can change the value and the caller will see the change.
So it is no longer valid to pass a literal value then because how could the function possibly change the given value of a literal?
If you don't want the setValue to be able to change the given value, make the argument either be an int or const int&. And if you do want the setValue function to be able to change the value, then the caller must declare a non-const variable to hold the int and pass in that.
Can I change something at the call site to make it work
The problem with your code is that you declared your function to expect a reference, which means the compiler has to prepare the code to allow the function to change whatever you pass into it at the call site. So yes, sure, you can declare a variable, set it to 1 and call your function with it.
Contrast this with a constant reference in the declaration, where the compiler knows you won't change it inside the function, and then you can pass a literal in without issues. In fact, any logical, thought out design will make setters accept constant parameters because it won't change them, it will just store a possibly processed value in its state.
The answer to „what do I do if a library has a bad interface and I can't change it“ is usually „write a wrapper“. Assuming this is a method of some class BadLibraryClass, you could do something like:
class Wrapper {
public:
BadLibraryClass inner;
setValue(int i) {
inner.setValue(i); // i is an lvalue
}
};
This is just a crude example. Perhaps inner is better off being a pointer, a reference or even a smart pointer. Perhaps you want a conversion operator to BadLibraryClass. Perhaps you can use inheritance to expose other methods of BadLibraryClass.
Two options:
Use the result of assignment:
static int _data;
void myCall() {
setValue((_data = 3));
}
Write a wrapper:
struct setValueW {
int _data;
// constructor
setValueW(int _data) : _data(_data) {
setValue(_data);
}
// if you want to call it again
void operator()() {
setValue(_data);
}
};
void myCall2() {
setValueW(3);
}
AFAIK, references keeps the addresses of the variable. 1 is not variable. It is temporary.
Take a look this article(this is a quote from this site)
c++11 introduced a new kind of reference variable -- an r-value reference
To declare one, use && after a type
int & // type designation for an L-value reference
int && // type designation for an R-value reference
L-value references can only refer to L-values
R-value references can reference to R-values (temporaries)
int x, y, z; // regular variables
int & r = x; // L-value reference to the variable x
int & r2 = x + y; // This would be ILLEGAL, since x + y is an R-value
int && r3 = x + y; // LEGAL. R-value reference, referring to R-value
So you can use (But this is not useful. It may be more useful if you write this in plain without rvalue or lvalue.):
void setValue(int&& data)
setValue(1);
Or you can use that:
void setValue(int& data)
int a = 11;
setValue(a);
Don't forget for second example. If you change the value of data parameter. You will have change the a variable value.
No, you can't.
An lvalue reference like that binds to a variable (roughly speaking).
Your literal is not such a thing. It never had a name, and may not even have a home in memory.
Your two options are the two things you ruled out, I'm afraid.
For what it's worth, this is not your fault: that is a rather poor setter. It should take const int& (which will automatically create a nice temporary variable for you out of the literal!), or even just const int.
Why these definitions are all ok:
int func(int p=255) {
return p;
}
int func1(const int &p=255) {
return p;
}
but this definition:
int func2(int &p=255) {
return p;
}
leads to compile error ?
What is the logic behind it ?
Taking arguments by reference means, you dont work with your local copy of the variable, but with a variable already defined in the scope of the calling function.
While your first example makes sense (you have a local variable p that you can fill with a default value) the second example is a bit more tricky: Usually when using references you expect the variable to have an address, since you want to modify it. For const-refernces, the compiler will still allow you to pass a literal, even if something like "reference to a literal" makes no sense at all.
In the third case the compiler expects you to modify p. But what part of the memory should this modification affect? "255" has no address - therefore it cant be used as a reference.
If you want to have a more detailed explanation, you should probably look for keywords like "rvalue" and "lvalue".
The attempted function definition
auto func2( int& p = 255 )
-> int
{ return p; }
… fails because you can't bind an rvalue to a reference to non-const. Basically that rule is because a simple value like 255 isn't modifiable. While the reference can be used to modify.
One simple solution is to express the default as a separate overload:
auto func2( int& p )
-> int
{ return p; }
auto func2()
-> int
{
int scratchpad = 255;
return func2( scratchpad );
}
A non-const reference must be bound to lvalue (i.e. its address could be got). 255 (i.e. an int literal) is not a lvalue, so int &p=255 fails.
A const reference could be bound to rvalue, and for this case, a temporary int will be created and initialized from 255. The temporary int's lifetime will be the same as the const reference.
int func(int p=255) {
return p;
}
p here is copied by value, and it is defined to exist in the scope of func.
int func2(int &p) {
return p;
}
// e.g. use:
int value = 10;
func2(value); // func2 *could* modify value because it is passed by non-const reference
In this case the compiler here expects p to have a name somewhere in memory (i.e. lvalue), so it can possibly write to it within func2. Passing by non-const reference allows you to modify the variable used in the function call. Since p must belong to someone else somewhere since it can be modified, you can't assign a default value to it.
But what about the const-reference case? Here, the compiler is smart enough to know that p can never be written to since it is const, so it doesn't need to have a name in memory to write to. In cases of a literal being passed (e.g. 255), it (behind the scenes) essentially creates a temporary and passes that temporary variable to the function.
int func1(const int &p=255) {
return p;
}
func1(10);
// Behind the scenes, the compiler creates something along these lines
// since it can never be modified.
const int some_temporary = 10;
func1(some_temporary);
suppose I have a function which accept const reference argument pass,
int func(const int &i)
{
/* */
}
int main()
{
int j = 1;
func(j); // pass non const argument to const reference
j=2; // reassign j
}
this code works fine.according to C++ primer, what this argument passing to this function is like follows,
int j=1;
const int &i = j;
in which i is a synonym(alias) of j,
my question is: if i is a synonym of j, and i is defined as const, is the code:
const int &i = j
redelcare a non const variable to const variable? why this expression is legal in c++?
The reference is const, not the object. It doesn't change the fact that the object is mutable, but you have one name for the object (j) through which you can modify it, and another name (i) through which you can't.
In the case of the const reference parameter, this means that main can modify the object (since it uses its name for it, j), whereas func can't modify the object so long as it only uses its name for it, i. func could in principle modify the object by creating yet another reference or pointer to it with a const_cast, but don't.
const int &i = j;
This declares a reference to a constant integer.
Using this reference, you won't be able to change the value of the integer that it references.
You can still change the value by using the original variable name j, just not using the constant reference i.
I know that if you write void function_name(int& a), then function will not do local copy of your variable passed as argument. Also have met in literature that you should write void function_name(const int & a) in order to say compiler, that I dont want the variable passed as argument to be copied.
So my question: what is the difference with this two cases (except that "const" ensures that the variable passed will not be changed by function!!!)???
You should use const in the signature whenever you do not need to write. Adding const to the signature has two effects: it tells the compiler that you want it to check and guarantee that you do not change that argument inside your function. The second effect is that enables external code to use your function passing objects that are themselves constant (and temporaries), enabling more uses of the same function.
At the same time, the const keyword is an important part of the documentation of your function/method: the function signature is explicitly saying what you intend to do with the argument, and whether it is safe to pass an object that is part of another object's invariants into your function: you are being explicit in that you will not mess with their object.
Using const forces a more strict set of requirements in your code (the function): you cannot modify the object, but at the same time is less restrictive in your callers, making your code more reusable.
void printr( int & i ) { std::cout << i << std::endl; }
void printcr( const int & i ) { std::cout << i << std::endl; }
int main() {
int x = 10;
const int y = 15;
printr( x );
//printr( y ); // passing y as non-const reference discards qualifiers
//printr( 5 ); // cannot bind a non-const reference to a temporary
printcr( x ); printcr( y ); printcr( 5 ); // all valid
}
So my question: what is the difference
with this two cases (except that
"const" enshures that the variable
passes will not be changed by
function!!!)???
That is the difference.
You state the difference right. You may also formulate it as:
If you want to specify that the function may change the argument (i.e. for init_to_big_number( int& i ) by specifying the argument by (variable) reference. When in doubt, specify it const.
Note that the benefit of not copying the argument is in performance, i.e. for 'expensive' objects. For built-in types like int it makes no sense to write void f( const int& i ). Passing the reference to the variable is just as expensive as passing the value.
There is a big difference in terms of parameter they could operate on,
Say you have a copy constructor for your class from int,
customeclass(const int & count){
//this constructor is able to create a class from 5,
//I mean from RValue as well as from LValue
}
customeclass( int & count){
//this constructor is not able to create a class from 5,
//I mean only from LValue
}
The const version can essentially operate on temporary values and non constant version could not operate on temporary, you would easily face issue when you miss out const where it is needed and use STL, but you get weired error telling it could not find the version that takes temporary. I recommend use const where ever you can.
They are used for different purposes. Passing a variable using const int& ensures you get the pass-by-copy semantics with much better performance. You are guaranteed that the called function (unless it does some crazy things using const_cast) will not modify your passed argument without creating a copy. int& is used when there are generally multiple return values from a function. In that case these can be used hold the results of the function.
I would say that
void cfunction_name(const X& a);
allows me to pass a reference to temporary object as follows
X make_X();
function_name(make_X());
While
void function_name(X& a);
fails to achieve this. with the following error
error: invalid initialization of non-const reference of type 'X&' from a temporary of type 'X'
leaving out the performance discussion, let the code speak!
void foo(){
const int i1 = 0;
int i2 = 0;
i1 = 123; //i gets red -> expression must be a modifiyble value
i2 = 123;
}
//the following two functions are OK
void foo( int i ) {
i = 123;
}
void foo( int & i ) {
i = 123;
}
//in the following two functions i gets red
//already your IDE (VS) knows that i should not be changed
//and it forces you not to assign a value to i
//more over you can change the constness of one variable, in different functions
//in the function where i is defined it could be a variable
//in another function it could be constant
void foo( const int i ) {
i = 123;
}
void foo( const int & i ) {
i = 123;
}
using "const" where it is needed has the following benefits:
* you can change the constness of one variable i, in different functions
in the function where i is defined it could be a variable
in another function it could be constant value.
* already your IDE knows that i should not be changed.
and it forces you not to assign a value to i
regards
Oops