无条件长到长 [英] cast unsigned long to long

问题描述

unsigned long a = ...;

long b =（long）a;


溢出时，b的结果是什么？

谢谢。

解决方案

2005-12-27 ，jeff< xi ***** @ gmail.com>写道：

unsigned long a = ...;
long b =（long）a;

虽然溢出，结果是什么b？
谢谢。

undefined。

jeff说：

unsigned long a = ...;
long b =（long）a;

虽然溢出，b的结果是什么？

a不能溢出。如果a的值超过LONG_MAX，则b的值为

undefined。


-

Richard Heathfield
Usenet是一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk

电子邮件：rjh在上面的域名（但显然放弃了www）

" jeff" < XI ***** @ gmail.com>写道：

unsigned long a = ...;
long b =（long）a;

虽然溢出，b的结果是什么？ / blockquote>


演员阵容是不必要的;声明

long b = a;

是等价的，因为存在隐式转换。 （大多数演员都不需要
。）


如果转换为有符号整数类型溢出，结果

是实现定义的或实现定义的信号是

筹集的 （C99 6.3.1.3p3）。我认为在C99中获得信号的许可是新的b
;在C90中，你只需得到一个实现定义的结果。

（不，它不是未定义的。）


实现定义的意味着你的实现需要

记录它，但是你不应该依赖于它，因为它可能会因为一个实现而不同，所以你的代码

非便携式。


请注意，这与算术溢出时发生的情况不同，

调用未定义的行为签名类型。


-

Keith Thompson（The_Other_Keith） ks * **@mib.org < http://www.ghoti.net/~kst>

圣地亚哥超级计算机中心< *> < http://users.sdsc.edu/~kst>

我们必须做点什么。这是事情。因此，我们必须这样做。

unsigned long a = ...;
long b = (long)a;

while a overflows, what is the result of b?
Thank you.

解决方案

On 2005-12-27, jeff <xi*****@gmail.com> wrote:

unsigned long a = ...;
long b = (long)a;

while a overflows, what is the result of b?
Thank you.

undefined.

jeff said:

unsigned long a = ...;
long b = (long)a;

while a overflows, what is the result of b?

a can''t overflow. If the value of a exceeds LONG_MAX, the value of b is
undefined.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

"jeff" <xi*****@gmail.com> writes:

unsigned long a = ...;
long b = (long)a;

while a overflows, what is the result of b?

The cast is unnecessary; the declaration
long b = a;
is equivalent, since there''s an implicit conversion. (Most casts are
unnecessary.)

If a conversion to a signed integer type overflows, "either the result
is implementation-defined or an implementation-defined signal is
raised" (C99 6.3.1.3p3). I think the permission to raise a signal is
new in C99; in C90, you just get an implementation-defined result.
(No, it''s not undefined.)

"Implementation-defined" means that your implementation is required to
document it, but you shouldn''t depend on this since it''s likely to
vary from one implementation to another, making your code
non-portable.

Note that this is different from what happens on artithmetic overflow,
which invokes undefined behavior for signed types.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

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