What are the serious uses of returning the this pointer from a function?

Question

Following is a fancy code:

class B
{
    private:
        int sum;

    public:
        B ()
        {
            sum = 0;
        }

        B& add (int number)
        {
            sum =+ number;
            return *this;
        }
};

int main ()
{
    B obj;
    obj.add (1).add (2).add (3). add (4);
}

Fine, what are the "serious" uses of returning the this pointer from a function call?

What really is serious is that you are returning `*this` from a function declared to return `void` :-) — hochl, Mar 13 '12 at 16:25
Operators will often return a reference to the object in question, to allow chaining, so `return *this`. Much as in your example. This *is* a "serious" use. — BoBTFish, Mar 13 '12 at 16:25

score 4 · Accepted Answer · answered Mar 13 '12 at 16:26

4

An example would be;

class basic_ostream
    : ...
{
    basic_ostream& operator<<(bool n);

You really want to return this to be able to chain to;

std::cout << boolValue << std::endl;

answered Mar 13 '12 at 16:26

Joachim Isaksson

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Shouldn't this be `basic_ostream& operator<<(basic_ostream& osr, bool n)` ? – hochl Mar 13 '12 at 16:27
2

If you are creating an `operator<<` for your own class, yes, as it won't be a member function. But for the standard ones it is probably a member function, so a relevant example. – BoBTFish Mar 13 '12 at 16:30
@hochl It's a member of basic_ostream actually, added a skeleton class header to clarify. – Joachim Isaksson Mar 13 '12 at 16:32
"The defense rests, your honor." – hochl Mar 13 '12 at 16:36

score 2 · Answer 2 · answered Mar 13 '12 at 16:31

2

mycode $ fgrep -r 'return *this' /usr/include/c++/4.4.3 | wc -l
592
mycode $

answered Mar 13 '12 at 16:31

n. m. could be an AI

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@Anisha: the point is that there are 592 uses of `return *this` in GCC's library headers. Presumably all or most of these are "serious", so you can look at any of them for examples. – Steve Jessop Mar 13 '12 at 16:38

score 2 · Answer 3 · edited May 23 '17 at 12:15

2

One use of this is for the Named Parameter Idiom. It depends on method chaining.

class Person;

class PersonOptions
{
  friend class Person;
  string name_;
  int age_;
  char gender_;

public:
   PersonOptions() 
   : age_(0), gender_('U')
   {}

   PersonOptions& name(const string& n) { name_ = n; return *this; }
   PersonOptions& age(int a) { age_ = a; return *this; }
   PersonOptions& gender(char g) { gender_ = g; return *this; }
};

class Person
{
  string name_;
  int age_;
  char gender_;

public:
   Person(const PersonOptions& opts) 
   : name_(opts.name_), age_(opts.age_), gender_(opts.gender_)
   {}
};
Person p = PersonOptions().name("George").age(57).gender('M');
Person p = PersonOptions().age(25).name("Anna");

edited May 23 '17 at 12:15

Community

1
1

answered Mar 13 '12 at 16:34

Fred Larson

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@MooingDuck: Thanks for covering my laziness. ;v) – Fred Larson Mar 13 '12 at 17:03
middle click, CTRL+C, select this tab, CTRL+V? – Mooing Duck Mar 13 '12 at 17:08

score 1 · Answer 4 · edited Mar 13 '12 at 17:22

The initial reason wass for chaining mathematical operations, like this:

class mynumberclass {
    int internal;
public:
    mynumberclass(int);
    mynumberclass operator+(const mynumberclass&) const;
    mynumberclass operator-(const mynumberclass&) const;
    mynumberclass operator*(const mynumberclass&) const;
    mynumberclass operator/(const mynumberclass&) const;
    mynumberclass operator%(const mynumberclass&) const;
    mynumberclass& operator+=(const mynumberclass&);
    mynumberclass& operator-=(const mynumberclass&);
    mynumberclass& operator*=(const mynumberclass&);
    mynumberclass& operator/=(const mynumberclass&);
    mynumberclass& operator%=(const mynumberclass&);
};

int main() {
    mynumberclass a(3);
    mynumberclass b(4);
    mynumberclass c = (a * b + b) / 2; //this chains 3 of the above operators
}

without chaining, that code would have to look like this:

int main() {
    mynumberclass a(3);
    mynumberclass b(4);
    mynumberclass c(a);
    c *= b;
    c += b;
    c /= 2;
}

FredLarson also mentions the Named Parameter Idiom, which is certainly an awesome thing you can use chaining for.

You for sure don't want dyadic operators modifying their left operand. You return by value from them. — AProgrammer, Mar 13 '12 at 16:59
@AProgrammer: Never seen the word "dyadic" before, but I realized what you meant anyway. Fixed. Thanks! — Mooing Duck, Mar 13 '12 at 17:01

score 0 · Answer 5 · answered Mar 13 '12 at 16:25

0

The first thing that comes to mind is chaining functions to perform operations on the same object. jQuery, although not C++, has shown that this can be a very useful paradigm.

answered Mar 13 '12 at 16:25

Scott M.

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What are the serious uses of returning the this pointer from a function?

5 Answers5