What is the name of this operator: “-->”?
29 
After reading "Hidden Features and Dark Corners of C++/STL" on comp.lang.c++.moderated, I was completely surprised that it compiled and worked in both Visual Studio 2008 and G++ 4.4. The code:

I'd assume this is C, since it works in GCC as well. Where in the standard is this defined, and where did it come from?
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Aileen MacGyver Created at: 2013-11-13 17:07:48 UTC By Aileen MacGyver 
Behold the confusion that two little parenthesis could have prevented :) +1 for being disciplined enough to say WTF and ask :) - Oma Barton
Or even just proper spacing... I don't think I've ever seen a space between the variable and either ++ or -- before... - Katharina Becker
This "goes to" operator can be reverted ( 0 <-- x ). And also there is a "runs to" operator ( 0 <---- x ). Geez, the funniest thing I've ever heard of c++ syntax =) +1 for the question. - Shanon Runolfsdottir
Funnily enough, although the interpretation is very wrong, it does describe what the code does correctly. :) - Kayleigh Beer
Imagine the new syntax possibilities: #define upto ++<, #define downto -->. If you're feeling evil, you can do #define for while( and #define do ) { (and #define done ;}) and write for x downto 0 do printf("%d\n", x) done Oh, the humanity... - Mr. Ryann Dietrich
16 Answers
0 ✓ 
Its combination of two operators.First "--" is for decrementing the value and ">" for checking greater than value.

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


output will be:

9 8 7 6 5 4 3 2 1 0
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0 
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.
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0 
That's not an operator -->. That's two separate operators, -- and >.

Your condition code is decrementing x, while returning xs original (not decremented) value, and then comparing the original value with 0 using the > operator.

To better understand, the statement could be as follows:

while( (x--) > 0 )
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0 
That's a very complicated operator, so C++ Standard committee even placed its description in two different parts of Standard. 

Joking aside, they are two different operators: -- and > described respectively in 5.2.6/2 and 5.9 of the C++03 Standard.
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0 
It's equivalent to

while (x-- > 0)
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0 
It's

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


Just the space make the things look funny, -- decrements and > compare.
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0 
while( x-- > 0 )


is how that's parsed.
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0 
In one book I read 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 biggest token: 

token 1: x
token 2: --
token 3: >
token 4: 0
conclude: x-- > 0


Same as:

a-----b


After parse

token 1: a
token 2: --
token 3: --
token 4: -
token 5: b
conclude: (a--)-- - b


I hope it helps to understand the complicated expression ^^
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0 
This is exactly the same as

while (x--){
   printf("%d ", x);
}


for non-negative numbers
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0 
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(void)
{
     int x = 10;
     while( --x> 0 ) // x goes to 0
     {
       printf("%d ", x);
     }
     return 0;
}


that output is :

9 8 7 6 5 4 3 2 1
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0 
There is a space missing between -- and >. x is post decremented, that is, decremented after checking the condition x>0 ?.
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0 
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;
}


is a different way of writing the same thing.

It is nice that the original looks like "while x goes to 0" though.
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0 
-- is the decrement operator and > is the greater-than operator.

The two operators are applied as a single one like -->.
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0 
Actually x is post-decrementing and with that condition is being checked. its not -->, its (x--) > 0

Note: value of x is changed after condition is checked because it post-decrementing. Some similar cases can also occur e.g.

-->    x-->0
++>    x++>0
-->=   x-->=0
++>=   x++>=0
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0 
It is to prevent confusion (obfuscation, perceived cleverness etc) such as this, that static analysis checkers (eg) MISRA seeks to prevent assignment operations (including ++ and -- as well as the obvious =) in conditional statements.
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0 
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.
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