“1u”的含义 [英] meaning of "1u"
问题描述
这应该是一个简单的...1u的含义是什么，例如：
#define DUMMY（1U<< 31）
而不仅仅是：
#define DUMMY（1<< 31）
this should be an easy one ... what is the meaning of "1u", for instance:
#define DUMMY (1U<<31)
as opposed to just:
#define DUMMY (1<<31)
推荐答案
df****@earthlink.net （godfather2）写在
新闻：d1 ************************** @ posting.google.c om：
df****@earthlink.net (godfather2) wrote in
news:d1**************************@posting.google.c om:
这应该是一个简单的...1u的意思是什么，对于
实例：
#define DUMMY（1U<< 31）
而不仅仅是：
#define DUMMY（1<< 31）
this should be an easy one ... what is the meaning of "1u", for
instance:
#define DUMMY (1U<<31)
as opposed to just:
#define DUMMY (1<<31)
前者迫使C将31推广到unsigned int匹配unsigned
int 1U。这意味着生成的表达式将具有
unsigned int（或unsigned long取决于平台）的类型，而不是
签名的替代品，这可能不需要。
-
- 马克 - >
-
The former forces C to promote 31 to an unsigned int to match the unsigned
int 1U. This means that the resulting expression will have a type of
unsigned int (or unsigned long depending upon the platform) instead of the
signed alternative which one may not want.
--
- Mark ->
--
Mark A. Odell <无**** @ embeddedfw.com>写道：
"Mark A. Odell" <no****@embeddedfw.com> writes:
df****@earthlink.net （godfather2）写道
新闻：d1 ************************** @ posting.google.c om：
df****@earthlink.net (godfather2) wrote in
news:d1**************************@posting.google.c om:
这应该是一个简单的...1u的含义是什么，对于
实例：
#define DUMMY（1U<< 31）
而不仅仅是：
#define DUMMY（1<<< 31）
this should be an easy one ... what is the meaning of "1u", for
instance:
#define DUMMY (1U<<31)
as opposed to just:
#define DUMMY (1<<31)
前者迫使C将31推广为无符号int匹配unsigned
int 1U。这意味着生成的表达式将具有
unsigned int（或unsigned long取决于平台）的类型，而不是
签名的替代，这可能是您不想要的。
The former forces C to promote 31 to an unsigned int to match the unsigned
int 1U. This means that the resulting expression will have a type of
unsigned int (or unsigned long depending upon the platform) instead of the
signed alternative which one may not want.
不，表达式将具有unsigned int类型，无论平台是否为
。值1总是适合无符号。另一方面，如果平台的无符号整数小于32位
宽，则1U << 31未定义。
-
如果行为未定义，这将是一个更好的未定义行为示例
。
- -Michael Rubenstein
No, the expression will have a type of `unsigned int'' regardless
of the platform. The value 1 always fits in an unsigned. On the
other hand, if the platform''s unsigned ints are less than 32 bits
wide, then 1U<<31 is undefined.
--
"It would be a much better example of undefined behavior
if the behavior were undefined."
--Michael Rubenstein
" godfather2" < DF **** @ earthlink.net>在消息中写道
news：d1 ************************** @ posting.google.c om ...
"godfather2" <df****@earthlink.net> wrote in message
news:d1**************************@posting.google.c om...
这应该是一个简单的...1u的含义是什么，例如：
#define DUMMY（1U<< 31）
而不仅仅是：
#define DUMMY（1<< 31）
this should be an easy one ... what is the meaning of "1u", for instance:
#define DUMMY (1U<<31)
as opposed to just:
#define DUMMY (1<<31)
在32平台上-bit int'，前者是unsigned int（0x80000000），
后者是int（-2147483648）。
相同位模式。
我读C99标准的方式，后者将导致未定义的
行为。
" E1的结果<< ; E2是E1左移E2位位置;腾出的位
充满了
零。如果E1具有无符号类型，则结果的值为E1 *（2 ** E2），
减少模数
比结果中可表示的最大值多一个类型。如果E1已经签署
类型和非负值，并且E1 *（2 ** E2）在结果中可以表示
类型，那么那是
结果值;否则，行为未定义。
Carsten
On a platform with 32-bit int''s, the former is an unsigned int (0x80000000),
the latter is an int (-2147483648).
Same bit pattern.
The way I read the C99 Standard, the latter will result in undefined
behavior.
"The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits
are filled with
zeros. If E1 has an unsigned type, the value of the result is E1 * (2**E2),
reduced modulo
one more than the maximum value representable in the result type. If E1 has
a signed
type and nonnegative value, and E1 * (2**E2) is representable in the result
type, then that is
the resulting value; otherwise, the behavior is undefined."
Carsten
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