运算符<-在C ++中做什么？

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?

Answer 1

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

if (b < -x)

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

Answer 2

Arguably implicit conversion operators in c++ are controversial, eg 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))

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

Answer 3

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:

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