use of boost::ref for passing reference to functions that take values

Question

I am confused about the use of boost::ref. I dont understand why any one would want to do the following -

void f(int x)
{
    cout << x<<endl;
    x++;
}

int main(int argc, char *argv[])
{
    int aaa=2;
    f(boost::ref(aaa));
    cout << aaa<<endl;
    exit(0);
}

What is the use of passing a ref to a function. I can always pass by value instead. Also it not that the ref is actually passed. In the above the value of aaa in main remains 2 only.

Where exactly is boost ref useful?

Is it possible to use boost::ref in this scenario. I want to pass iterator refernce to std::sort function. normally the sort works on iterator copies - will boost::ref make it work for references also? (without any changes to std::sort)

Mike Seymour · Accepted Answer · 2013-05-09 12:08:21Z

17

I dont understand why any one would want to do the following

They wouldn't. That's not what boost::ref (or these days std::ref) is for. If a function takes an argument by value, then there's no way to force it to take it by reference instead.

Where exactly is boost ref useful?

It can be used to make a function template act as if it takes an argument by reference, by instantiating the template for the reference (wrapper) type, rather than the value type:

template <typename T>
void f(T x) {++x;}

f(aaa);      cout << aaa << endl; // increments a copy: prints 0
f(ref(aaa)); cout << aaa << endl; // increments "a" itself: prints 1

A common specific use is for binding arguments to functions:

void f(int & x) {++x;}

int aaa = 0;
auto byval = bind(f, aaa);      // binds a copy
auto byref = bind(f, ref(aaa)); // binds a reference

byval(); cout << aaa << endl;   // increments a copy: prints 0
byref(); cout << aaa << endl;   // increments "a" itself: prints 1

Is it possible to use boost:;ref in this scenario. I want to pass iterator refernce to std::sort function. normally the sort works on iterator copies - will boost::ref make it work for references also?

No; the reference wrapper doesn't meet the iterator requirements, so you can't use it in standard algorithms. If you could, then many algorithms would go horribly wrong if they needed to make independent copies of iterators (as many, including most sort implementations, must do).

edited May 9, 2013 at 12:08

answered May 9, 2013 at 11:51

Mike Seymour

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5 Comments

Rohit Over a year ago

oki understand but can you explain how this works. template <class _T> void f(_T x) { x++; } int main(int argc, char *argv[]) { int aaa=2; f<boost::reference_wrapper<int>>(boost::ref(aaa)); cout << aaa<<endl; cin >>aaa; exit(0); } The value printed is 3. Dont understand how x i function became a refernce ?

Mike Seymour Over a year ago

@user2161003: By explicitly writing out the reference_wrapper part (which isn't necessary), you've answered your own question: the template is instantiated with a "reference wrapper" type (returned by ref()), and that acts like a reference to the variable it wraps.

Rohit Over a year ago

but how? and object of reference_wrapper does not have ++ operator. how is x++ working?

Mike Seymour Over a year ago

@user2161003: Using its conversion operator, operator T&(), to convert to a real reference when needed. So ++x becomes ++(x.operator int&()) and increments aaa.

Rohit Over a year ago

ok - i had a hunch that it is beacuse of this. But what i dont understand are the rules of calling this type case operator. If x is a refernce_wrapper the i do understand that the operator will get called if i do something like this - int& a = x. but just doing x++ is also invoking the operator. strange!!!

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