size_t文字？ [英] size_t literals?
问题描述
嗨！
在size_t为64位的机器上，unsigned long为32位，
是否构造了一个size_t文字，表示2 ^ 32？打字
size_t x = 4294967296UL;
抱怨价值对于未签名的长期来说太大了
（显然，一个人太大了。
非标准足够的UI64
编译器也不承认。
我应该构造值为4294967295UL然后增加
size_t变量？
TIA，
- J.
Hi!
On a machine where size_t is 64-bit, unsigned long is 32-bit, how
does one construct a size_t literal that says 2^32? Typing in
size_t x = 4294967296UL;
complains about the value being too large for unsigned long
(obviously, it''s too large by one).
The nonstandard suffices "UI64" are also not recognized by the
compiler.
Should I construct with a value of 4294967295UL and then increment
the size_t variable?
TIA,
- J.
推荐答案
ja ************ @ gmail.com 写道：
在size_t为64位的机器上，无符号长整数是32位，如何构造一个size_t字面值为2 ^ 32的
？打字
size_t x = 4294967296UL;
抱怨价值对于未签名的长期来说太大了
（显然，一个人太大了。
非标准足够的UI64
编译器也不承认。
我应该构造值为4294967295UL然后增加
size_t变量？
On a machine where size_t is 64-bit, unsigned long is 32-bit, how
does one construct a size_t literal that says 2^32? Typing in
size_t x = 4294967296UL;
complains about the value being too large for unsigned long
(obviously, it''s too large by one).
The nonstandard suffices "UI64" are also not recognized by the
compiler.
Should I construct with a value of 4294967295UL and then increment
the size_t variable?
由于''size_t''的表示不是由语言决定的，
你做的实现允许。如果''size_t''是64位，
那么你需要找到能满足你需要的类型。也许
一双。也许是一个未签名的长多？
size_t x = 4294967296ULL;
另一种可能性是得到2 ^ 16的值并将其平方：
size_t x =（size_t）（1 << 16）*（size_t）（1 << 16）;
和希望编译器能够简化它并使其成为
a值而不是运行时表达式。
V
-
请在通过电子邮件回复时删除资金''A'
我没有回复最热门的回复，请不要问
Since the representation of ''size_t'' is not dictated by the language,
you do what your implementation allows. If your ''size_t'' is 64 bits,
then you need to find the type that would do what you need. Perhaps
a double. Maybe an unsigned long long?
size_t x = 4294967296ULL;
Another possibility is to get 2^16 value and square it:
size_t x = (size_t)(1 << 16) * (size_t)(1 << 16);
and hope that the compiler will be able to simplify it and make it
a value instead of a run-time expression.
V
--
Please remove capital ''A''s when replying by e-mail
I do not respond to top-posted replies, please don''t ask
2008-02-28 09:32:01 -0800， ja ************ @ gmail.com 说：
On 2008-02-28 09:32:01 -0800, ja************@gmail.com said:
>
嗨！
在size_t为64位的机器上，无符号长整数为32位，
如何构造size_t文字那说2 ^ 32？打字
size_t x = 4294967296UL;
抱怨价值对于未签名的长期来说太大了
（显然，一个人太大了。
>
Hi!
On a machine where size_t is 64-bit, unsigned long is 32-bit, how
does one construct a size_t literal that says 2^32? Typing in
size_t x = 4294967296UL;
complains about the value being too large for unsigned long
(obviously, it''s too large by one).
大概这是一个警告。编译器有义务将
值转换为合适的类型，尽管有后缀，并且您的代码应该按照您的预期工作。
Presumably this was a warning. The compiler is obliged to convert the
value to a suitable type, despite the suffix, and your code should
work as you expect it to.
>
非标准足以满足UI64的要求。
编译器也无法识别。
>
The nonstandard suffices "UI64" are also not recognized by the
compiler.
未来的标准将使用ULL或任何其他组合的案件用于那些
三个字母。
The future standard will use ULL, or any other mix of cases for those
three letters.
>
我应该构造值为4294967295UL然后增加
size_t变量吗？
>
Should I construct with a value of 4294967295UL and then increment
the size_t variable?
No.
-
Pete
Roundhouse咨询有限公司（ www.versatilecoding.com ）
的作者
标准C ++库扩展：教程和参考
（ www.petebecker.com/tr1book ）
No.
--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)
ja ************ @ gmail.com 在新闻中写道：e6da1a2b-2561-4623-89a7-8303394cb7f9
@ k2g2000hse.googlegroups.com:
ja************@gmail.com wrote in news:e6da1a2b-2561-4623-89a7-8303394cb7f9
@k2g2000hse.googlegroups.com:
>
嗨！
在机器上size_t是64位，unsigned long是32位，怎么用
构造一个size_t文字，说2 ^ 32？打字
size_t x = 4294967296UL;
抱怨价值对于未签名的长期来说太大了
（显然，一个人太大了。
非标准足够的UI64
编译器也不承认。
我应该构造值为4294967295UL然后增加
size_t变量？
TIA，
- J.
>
Hi!
On a machine where size_t is 64-bit, unsigned long is 32-bit, how
does one construct a size_t literal that says 2^32? Typing in
size_t x = 4294967296UL;
complains about the value being too large for unsigned long
(obviously, it''s too large by one).
The nonstandard suffices "UI64" are also not recognized by the
compiler.
Should I construct with a value of 4294967295UL and then increment
the size_t variable?
TIA,
- J.
即将成为标准说明符怎么样？ ULL？否则，数学
方法应该有效。
joe
What about the soon to be standard specifier of ULL? Otherwise, the math
method should work.
joe
这篇关于size_t文字？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
