What is the difference between temporary variable and constant in C++?

Question

Allow me to post my code first:

void Fun(short &s){}
void FunCon(const short &s){}

int main()
{
  int iTest = 20;//note: iTest is int but parameter of Fun is "short &"
  Fun(iTest);//error, I know there is a temp variable(typecast)
  FunCon(iTest);//ok
  return 0;
}

I know Fun(iTest);will generate a new temp variable(typecast), but I wonder if the temp variable is a constant?

If No: Why can't I pass the temp variable to short &

If Yes: I have another code like this:

class MyObject{
  public :
  void F(){}
};

MyObject MOCreator(){
  return MyObject();
}

int main()
{
   MOCreator().F();//OK
   return 0;
}

If temp variable that returned by MOCreator() is constant, why the temp variable can call non-const member function?

My questions are：

1) What is the difference between temporary variable and constant in C++?

2) There is a sentence in Thinking in C++(page 507). Is the sentence right? and why?:

Temporary objects are automatically const

I was asked for a simple question by someone, and I encounter more questions on my way to solving the question. I do know they may be a very common questions,and I am searching for a long time on net. I also got many different answers. But I'm more confused about it now.

Thanks in advance.

songyuanyao · Accepted Answer · 2016-03-31T05:47:49.510

4

I wonder if the temp variable is a constant?

No, it's just a temporary variable which will be destroyed at the end of the full expression in which it was created.

Why can't I pass the temp variable to short &

A temporary can't be bound for a non-const reference. But it could be bound for const reference, and the lifetime of the temporary will be extended to match the lifetime of the reference. That's why FunCon(iTest); is fine.

why the temp variable can call non-const member function?

It's fine. The only special thing is the temporary variable will be destroyed at the end of the full expression.

Is the sentence right? and why?
Temporary objects are automatically const

No. Unless you explicitly declare it to be const.

edited Mar 31 '16 at 05:47

answered Mar 31 '16 at 05:25

songyuanyao

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"Temporary objects are automatically const" Is the sentence right? – Joe Mar 31 '16 at 05:33
@Joe I don't know the context of it, I'll say it's wrong only for this single sentence. – songyuanyao Mar 31 '16 at 05:36
@Joe See [Are temporary objects in C++ const indeed?](http://stackoverflow.com/q/4701558/3309790). – songyuanyao Mar 31 '16 at 05:48

Miles Budnek · Answer 2 · 2016-03-31T05:36:37.367

1

Anonymous temporaries are rvalues, which can only be bound to an rvalue reference or a const lvalue reference. So your temporary short can be bound to a short&& or a const short&.

They are not const; the main purpose of rvalue references is to alter them by moving expensive-to-copy resources out of them and into a new object.

edited Mar 31 '16 at 05:36

answered Mar 31 '16 at 05:29

Miles Budnek

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score 1 · Answer 3 · answered Mar 31 '16 at 05:39

Given the following:

void f(short& s) {
   s = 42;
}

Now consider this simple use case:

short s = 0;
f(s);
std::cout << s << "\n";

This will print 42 - calling f had the side effect of modifying s.

But now consider the following, if it were legal:

int i = 0;
f(i);
std::cout << i << "\n";

This would print 0 because what you would be modifying in f is the temporary, with the original i untouched.

To prevent this behavior, you can only pass a temporary to a const reference or an rvalue reference.

What is the difference between temporary variable and constant in C++?

3 Answers3