int a = -2147483648; [英] int a = -2147483648;
问题描述
警告C4146：一元减号运算符应用于无符号类型，结果仍然是
未签名。
？
" Skybuck Flying" < sp ** @ hotmail.comwrote in message
news：f6 ********** @ news6.zwoll1.ov.home.nl ...
< blockquote class =post_quotes>
>
警告C4146：一元减号运算符应用于无符号类型，结果仍然是
unsigned。
？
有符号整数的值太大（检查整数边界）
Skybuck飞行说：
>
警告C4146：一元减号运算符应用于无符号类型，结果
仍然没有签名。
？
警告CLC123：长期的巨魔返回到comp.lang.c - 在撰写回复之前建议
归档检查。
-
Richard Heathfield< http://www.cpax.org.uk>
电子邮件：-www。 + rjh @
谷歌用户：< http://www.cpax.org.uk/prg/writings/googly.php>
Usenet是一个奇怪的放置" - dmr 1999年7月29日
On Sun，2007年7月8日08:03:23 +0200，Skybuck Flying写道：
>
警告C4146：一元减号运算符应用于无符号类型，结果仍然是
无符号。
？
我猜你有32位整数和2'补码。
C99基本原理（6.4.4.4。）说 ;尽管一般规则
，文字常数是非负的，3 ......脚注
是3注意-4是表达式：一元减去操作数4.
INT_MAX是0x7FFFFFFF，所以十进制常数为0x80000000不是
适合签名的int，如果INT_MAX == LONG_MAX，甚至不是
签名长。在C89中，这将导致它具有无符号类型。
否定它会产生UINT_MAX + 1 - 0x80000000，即0x80000000，
不适合签名的int。
尝试使用：
long a = -2147483647 - 1，例如MSVC #define''s INT_MIN this
way（但带括号）周围）。请注意，在C89中没有类型是
*必需*才能保持该值，因此程序不会随处可移动。但是使用long会使它即使在具有16位int的系统上也可以移植，只要它们使用两个补码。
（如果你的意思是INT_MIN。如果你的意思是。你有C99也许你想要
使用INT32_MIN并声明一个为int32_t，这些是在
< stdint.h>中定义的。）
-
Army1987（将NOSPAM替换为电子邮件）
永远不要归咎于可以充分解释的恶意
愚蠢。 - R. J. Hanlon（？）
warning C4146: unary minus operator applied to unsigned type, result still
unsigned.
?
"Skybuck Flying" <sp**@hotmail.comwrote in message
news:f6**********@news6.zwoll1.ov.home.nl...>
warning C4146: unary minus operator applied to unsigned type, result still
unsigned.
?
too large value for signed integer (check integer boundary)
Skybuck Flying said:
>
warning C4146: unary minus operator applied to unsigned type, result
still unsigned.
?warning CLC123: longstanding troll returns to comp.lang.c - suggest
archive check before composing reply.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
On Sun, 08 Jul 2007 08:03:23 +0200, Skybuck Flying wrote:
>
warning C4146: unary minus operator applied to unsigned type, result still
unsigned.
?I guess you have 32-bit ints with 2''s complement.
The C99 rationale (6.4.4.4.)says "Notwithstanding the general rule
that literal constants are non-negative, 3..." where the footnote
is "3 Note that -4 is an expression: unary minus with operand 4.".
INT_MAX is 0x7FFFFFFF, so a decimal constant of 0x80000000 doesn''t
fit in a signed int, and if INT_MAX == LONG_MAX, not even in a
signed long. In C89 this would cause it to have unsigned type.
Negating it yields UINT_MAX + 1 - 0x80000000 which is 0x80000000,
which doesn''t fit in a signed int.
Try using:
long a = -2147483647 - 1, for example MSVC #define''s INT_MIN this
way (but with parentheses around it). Note that in C89 no type is
*required* to be able to hold that value, so the program won''t be
portable everywhere. But using long will make it portable even on
systems with 16-bit int, provided they use two''s complement.
(Or INT_MIN if you mean that. If you have C99 maybe you want to
use INT32_MIN and declare a as an int32_t, these are defined in
<stdint.h>.)
--
Army1987 (Replace "NOSPAM" with "email")
"Never attribute to malice that which can be adequately explained
by stupidity." -- R. J. Hanlon (?)
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