What does operator <- do in c++?

Question

Reading really long functions can be very fun!

int crazylongfun()
{
  int x = -1;
  foo b = -42;

//... 1000 lines below

  if( b<-x)
  {
    std::printf("How did I get here there is no left arrow operator?\n");
  }

  return 0;    
}

Looking at foo definition

struct foo
{
  int  x;

  foo(int a) : x(a) {}

  operator int() const
  {
    return x;
  }
};

This compiles just fine and produces the desired output. What is the mechanism that allows this?

see also "tadpole operator": http://blogs.msdn.com/b/oldnewthing/archive/2015/05/25/10616865.aspx — Trevor Hickey, Oct 22 '15 at 23:10

score 5 · Answer 1 · answered Oct 22 '15 at 23:06

5

It's simply, "less than negative X".

if (b < -x)

After 1,000 lines I'd be kinda cross-eyed, too!

answered Oct 22 '15 at 23:06

Graham Perks

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score 3 · Accepted Answer · edited May 23 '17 at 12:31

Arguably implicit conversion operators in c++ are controversial, e.g. Are implicit conversions good or bad in modern C++?. When they are defined there can be very interesting syntactical situations to the unaware programmer. The above is just an example.

Here is how this works:

Struct foo has a conversion operator defined and therefore it is convertible to int which causes it to be implicitly converted and compared to the local variable x for less than minus x.

The code is actually:

if(int(b)<(-x))

i.e. There is no left arrow operator <- in c++ .

score 3 · Answer 3 · answered Oct 22 '15 at 23:16

What does operator <- do in c++?

It's actually not an operator (in contrast to ->, which is one).

As mentioned in the other answers, there are in fact two operator functions applied:

Unary operator-() applying a negative sign to the value
Less comparison operator<()

What does operator <- do in c++?

3 Answers3