What is the name of this operator: "-->"?

Question

After reading this post on comp.lang.c++.moderated, I was completely surprised that it compiled and worked in both VS 2008 and G++ 4.4. The code:

#include <stdio.h>
int main()
{
     int x = 10;
     while( x --> 0 ) // x goes to 0
     {
       printf("%d ", x);
     }

} 

Where in the standard is this defined, and where did it come from?

I'd assume C, since it works in GCC as well, but I put C++ on there just in case C++ has more to mention on it. On a more subjective note, I've never heard of this before, had anybody else? Is it worth using?

Edit

Because people keep taking this the wrong way, yes, I knew the answer and merely thought it would be a fun question. :P Kirill was the first to call me out, after I left a few "egging on" comments.

You sillies.

Answer 1

That's not an operator -->. That's two separate operators, -- and > You're post decrementing x and then comparing x and 0 with the > operator

Answer 2

I think it's equivalent to:

while( x-- > 0 )

Answer 3

It's

#include <stdio.h>
int main()
{
     int x = 10;
     while( x-- > 0 ) // x goes to 0
     {
       printf("%d ", x);
     }

}

Just the space make the things look funny, -- decrements and > compare.

Answer 4

while( x-- > 0 )

is how that's parsed.

Answer 5

That's a very complicated operator, so C++ Standard committee even placed its description in two different parts of Standard. You could read about it in 5.2.6/2 and 5.9 in C++03 Standard.

Answer 6

Anyway, we have a "goes to" operator now. "-->" is easy to be remembered as a direction, and "while x goes to zero" is meaning-straight.

Furthermore, it is a little more efficient than "for (x = 10; x > 0; x --)" on some platforms.

Answer 7

That is great, you could also do x++<10 which would grow till it hit 10, but isn't nearly as sneaky.

I am not sure what we could call ++< It looks like it should be some emotion icon.

Answer 8

Equally confusing is the rises to operator:

while( x ++< 100 )

(Edit: Fixed error caused by Murphy's law.)

Answer 9

Also:

(x-->0)  (x--<0)  (--x>0) 
(--x<0)
(x++>0)
(x++<0)
(++x>0) 
(++x<0)

Also, mix these expressions with the = operator, like this: (x--=>). This is really nice.

Answer 10

As a side note, the opposite of the --> operator

while (x --> 0)

i.e. the <-- operator

while (0 <-- x)

also works, though in a slightly different manner

Answer 11

   #include <stdio.h>

   int main() {
         int x = 10;
         int y = ???;
         int z = ???;
         while( x -->> y -->> z )
         {
           printf("%d ", x);
         } 
    }

outputs:

9 8 7

What is the value of y and z?

Answer 12

Such syntax trick was used early in C and widely used in present time. Look here, for instance (sqlite3 in the list).

Answer 13

The usage of --> has historical relevance. Decrementing was (and still is in some cases), faster than incrementing on the x86 architecture. Using --> suggests that x is going to 0, and appeals to those with mathematical backgrounds.

Answer 14

This code first compare x and 0 then decrement x . [ also say in first answer : You're post decrementing x and then comparing x and 0 with the > operator ] see the output of this code :

9 8 7 6 5 4 3 2 1 0

we know first compare and then decrement by see 0 in output .

if we want to first decrement and then compare use this code :

#include <stdio.h>
int main()
{
     int x = 10;
     while( --x> 0 ) // x goes to 0
     {
       printf("%d ", x);
     }

}

that output is :

9 8 7 6 5 4 3 2 1

Answer 15

there is a space missing between -- and > .. x is post decremented ie decremented after checking the condition X>0 ?

Answer 16

This is exactly the same as

while (x--)

Answer 17

My compiler will print out 9876543210 when I run this code.

#include <iostream>
int main()
{
    int x = 10;
    while( x --> 0 ) // x goes to 0
    {
        std::cout << x;
    }
}

As expected. The while( x-- > 0 ) actually means while( x > 0). The x-- post decrements x.

while( x >= 0 ) {
    x--;
    std::cout << x;
}

could be written to do the same thing.

It is nice that it looks like while x goes to 0 though.

Answer 18

cool_me5000: "The result is undefined when a variable is referenced within an expression that increments or decrements it" There is nothing wrong with the code above, well, except that it looks funny. If what your are saying was true, they'd be no need in having both i++ and ++i in the language. You are probably thinking about the case such as a[i++] = i++ which really is problematic.

Answer 19

-- is decrement operator and '>' greater then operator, is two operators applied as a single one like -->

Answer 20

in one book i read before they said ( i don't remember correctly what is book ): Compiler try to parse expression to bigest token by using left right rule: In this case is:

x-->0

Parse to bigest token:

token 1: x--
token 1: >
token 2: 0
conclude: x-- > 0

Same as:

a-----b

After parse

token 1: a--
token 2: -
token 3: --b
conclude: a-- - --b

Hope it help to understand complicate expression ^^

Answer 21

Utterly geek, but I will be using this:

#define as ;while
int main(int argc, char* argv[]){
 int n = atoi(argv[1]);
 do printf("n is %d\n", n) as ( n --> 0);
 return 0;
}