！！， 它是什么？ [英] !!, what is it?
问题描述
你好，
我在一些代码中偶然发现了这个：
n = !! x;
我已经运行它了''n''总是0或1.是''!!''是一个操作员（它是什么？
叫做？）是标准/便携式吗？
-
benkibbey ::: bjk AT arbornet DOT org
" bjk当然 <使用-作者提供地址报头@ [127.1] GT;写在
消息新闻：20 ******************** @ cpe-66-1-217-9.fl.sprintbbd.net ...你好，
我在一些代码中偶然发现了这个：
n = !! x;
我已经运行了它并且''n''总是0或1.是''!!''是一个运算符（它叫什么
？）并且它是标准的/便携式？
n = !! x;
应相当于：
n =！（！x） ;
甚至：
n =！（！（x））;
如果x非零，则n应为1 ，如果x为零，则为0。
有了这个主意吗？
Alex
当然是bjk <使用-作者提供地址报头@ [127.1] GT;写在
消息新闻：20 ******************** @ cpe-66-1-217-9.fl.sprintbbd.net ...你好，
我在一些代码中偶然发现了这个：
n = !! x;
我已经跑了，''''总是0或1.
是的，因为那总是由于！ （逻辑不是）
运营商。
是''!!''运营商
不，！是一个运营商。它的结果总是为零或一。
它只是在上面的代码中被使用了两次。例如
！0 == 1
！1 == 0
！42 == 0;
!! 42 == 1 / *因为！42 == 0，然后！（！42）== 1
（因为！0 == 1）* /
等
（它的名字是什么？）
逻辑没有。
并且是标准/便携式吗？
是的。
int i = 42;
int j =！i; / *用0 * /
int k初始化''j''; = !!我; / *用1 * /
int m =！j初始化''k''; / *初始化''m''用1 * /
逻辑
非运算符的'双重应用''的典型原因是将非零值（在逻辑上将
总是计算为''true''）转换为特定的''true''值（在
中，这种情况为一（1））。需要这样做的原因取决于应用程序上的
。这是一个简单的人为例子：
#include< stdio.h>
/ *零值无效，任何非零值有效* /
void validate（int value）
{
static const char * msg [] = {" invalid"，" valid"};
printf（" value％d is％s \ n"，msg [ !! i]）;
/ *注意!! 0总是正好0 * /
}
int main （）
{
验证（42）;
返回0;
}
IOW，！42 == 0，!! 42 == 1，!! 42！= 42
如果有的话需要其他特定的真实价值，人们只需要对价值1执行更多算术（例如，添加
一些价值）。
-Mike
bjk当然是ha scritto：你好，
我在一些代码中偶然发现了这个：
n = !! x;
我'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''' >
这意味着''不是x''，如果x为0则评估为0或如果x不为0则评估为1.
-
Devaraja（Xdevaraja87 ^ gmail ^ c0mX）
Linux注册用户＃338167
http://counter.li.org
hello,
I''ve stumbled upon this in some code:
n = !!x;
I''ve ran it and ''n'' is always either 0 or 1. Is ''!!'' an operator (what''s it
called?) and is it standard/portable?
--
b e n k i b b e y ::: bjk AT arbornet DOT org
解决方案"bjk of course" <Use-Author-Supplied-Address-Header@[127.1]> wrote in
message news:20********************@cpe-66-1-217-9.fl.sprintbbd.net...hello,
I''ve stumbled upon this in some code:
n = !!x;
I''ve ran it and ''n'' is always either 0 or 1. Is ''!!'' an operator (what''s it called?) and is it standard/portable?
n = !!x;
should be equivalent to:
n = !(!x);
or even:
n = !(!(x));
And n should be 1 if x is nonzero, 0 if x is zero.
Got the idea?
Alex
"bjk of course" <Use-Author-Supplied-Address-Header@[127.1]> wrote in
message news:20********************@cpe-66-1-217-9.fl.sprintbbd.net...hello,
I''ve stumbled upon this in some code:
n = !!x;
I''ve ran it and ''n'' is always either 0 or 1.
Yes, because that''s always the result of the ! ("logical not")
operator.
Is ''!!'' an operator
No, ! is an operator. It''s result is always either zero or one.
It is simply being used twice in the above code. E.g.
!0 == 1
!1 == 0
!42 == 0;
!!42 == 1 /* because !42 == 0, then !(!42) == 1
(because !0 == 1) */
etc.
(what''s it
called?)
"Logical not".
and is it standard/portable?
Yes.
int i = 42;
int j = !i; /* initializes ''j'' with 0 */
int k; = !!i; /* initializes ''k'' with 1 */
int m = !j; /* initializes ''m'' with 1 */
The typical reason for this ''double application'' of the logical
not operator is to transform a nonzero value (which logically
always evaluates to ''true'') into a specific ''true'' value (in
this case one (1). The reason for needing to do this depends
upon the application. Here''s a simple contrived example:
#include <stdio.h>
/* zero value is invalid, any nonzero value is valid */
void validate(int value)
{
static const char *msg[] = {"invalid", "valid"};
printf("value %d is %s\n", msg[!!i]);
/* note that !!0 always is exactly 0 */
}
int main()
{
validate(42);
return 0;
}
IOW, !42 == 0 , !!42 == 1 , !!42 != 42
If some other specific ''true'' value were needed, one would
simply perform more arithmetic upon the value one (e.g. add
some value).
-Mike
bjk of course ha scritto:hello,
I''ve stumbled upon this in some code:
n = !!x;
I''ve ran it and ''n'' is always either 0 or 1. Is ''!!'' an operator (what''s it
called?) and is it standard/portable?
It means ''not not x'', evaluating to 0 if x is 0 or to 1 if x is not 0.
--
Devaraja (Xdevaraja87^gmail^c0mX)
Linux Registerd User #338167
http://counter.li.org
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