C++ 中的“-->”运算符是什么？
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What is the "-->" operator in C++?
提问by GManNickG
After reading Hidden Features and Dark Corners of C++/STLon comp.lang.c++.moderated, I was completely surprised that the following snippet compiled and worked in both Visual Studio 2008 and G++ 4.4.
看完后隐藏的功能和C ++ / STL的暗角上comp.lang.c++.moderated，我完全意外的是，下面的代码片断编译并在两个Visual Studio 2008和G ++ 4.4的工作。
Here's the code:
这是代码：
#include <stdio.h>
int main()
{
int x = 10;
while (x --> 0) // x goes to 0
{
printf("%d ", x);
}
}
Output:
输出：
9 8 7 6 5 4 3 2 1 0
I'd assume this is C, since it works in GCC as well. Where is this defined in the standard, and where has it come from?
我认为这是 C，因为它也适用于 GCC。这在标准中在哪里定义，它来自哪里？
采纳答案by Charles Salvia
-->is not an operator. It is in fact two separate operators, --and >.
-->不是运营商。它实际上是两个独立的运算符，--和>。
The conditional's code decrements x, while returning x's original (not decremented) value, and then compares the original value with 0using the >operator.
条件的代码递减x，同时返回x的原始（未递减）值，然后将原始值与0使用>运算符进行比较。
To better understand, the statement could be written as follows:
为了更好地理解，该语句可以写成如下：
while( (x--) > 0 )
回答by unsynchronized
Or for something completely different... xslides to 0.
或者对于完全不同的东西......x滑动到0.
while (x --\
\
\
\
> 0)
printf("%d ", x);
Not so mathematical, but... every picture paints a thousand words...
不是那么数学，但是……每张图片都描绘了一千个字……
回答by Kirill V. Lyadvinsky
That's a very complicated operator, so even ISO/IEC JTC1 (Joint Technical Committee 1)placed its description in two different parts of the C++ Standard.
这是一个非常复杂的运算符，因此即使是ISO/IEC JTC1（联合技术委员会 1）也将其描述放在 C++ 标准的两个不同部分中。
Joking aside, they are two different operators: --and >described respectively in §5.2.6/2 and §5.9 of the C++03 Standard.
顺便说一句，它们是两个不同的运算符：--和>分别在 C++03 标准的 §5.2.6/2 和 §5.9 中描述。
回答by Jay Riggs
It's equivalent to
它相当于
while (x-- > 0)
x--(post decrement) is equivalent to x = x-1so, the code transforms to:
x--(post decrement) 等价于x = x-1so，代码转换为：
while(x > 0) {
x = x-1;
// logic
}
x--; // The post decrement done when x <= 0
回答by doc
xcan go to zero even faster in the opposite direction:
x可以在相反方向更快地归零：
int x = 10;
while( 0 <---- x )
{
printf("%d ", x);
}
8 6 4 2
8 6 4 2
You can control speed with an arrow!
你可以用箭头控制速度！
int x = 100;
while( 0 <-------------------- x )
{
printf("%d ", x);
}
90 80 70 60 50 40 30 20 10
90 80 70 60 50 40 30 20 10
;)
;)
回答by RageZ
It's
它是
#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 >compares.
只是空间让事情看起来很有趣，--减少和>比较。
回答by Matt Joiner
The usage of -->has historical relevance. Decrementing was (and still is in some cases), faster than incrementing on the x86 architecture. Using -->suggests that xis going to 0, and appeals to those with mathematical backgrounds.
的使用-->具有历史意义。在 x86 架构上，递减（在某些情况下仍然如此）比递增更快。使用-->表明这x将会0，并吸引那些具有数学背景的人。
回答by Grumdrig
while( x-- > 0 )
is how that's parsed.
这是如何解析的。
回答by Escualo
Utterly geek, but I will be using this:
完全的极客，但我将使用这个：
#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;
}
回答by NguyenDat
One book I read (I don't remember correctly which book) stated: Compilers try to parse expressions to the biggest tokenby using the left right rule.
我读过的一本书（我记不清是哪本书了）说：编译器尝试使用左右规则将表达式解析为最大的标记。
In this case, the expression:
在这种情况下，表达式：
x-->0
Parses to biggest tokens:
解析为最大的令牌：
token 1: x
token 2: --
token 3: >
token 4: 0
conclude: x-- > 0
The same ruleapplies to this expression:
同样的规则适用于这个表达式：
a-----b
After parse:
解析后：
token 1: a
token 2: --
token 3: --
token 4: -
token 5: b
conclude: (a--)-- - b
I hope this helps to understand the complicated expression ^^
我希望这有助于理解复杂的表达^^
