如何使用size_t类型的最大大小？ [英] How to use maximum size of size_t type ?
问题描述
嗨：
我对C编程有点新意，我正在尝试使用
为malloc写一个包装器。由于malloc需要
size_t作为参数，我希望能够
看看我的函数收到的参数是否小于
或者等于max size_t类型。
所以：
//我的包装函数mallocstr
char * mallocstr（size_t num）
{
// TEST
if（CODE HERES TO TEST FOR num< = max size of size_t）{
printf（一些调试错误消息\ n）;
返回NULL;
}
// END TEST
return（char *）malloc（num）;
}
现在，忽略malloc将返回NULL的事实
不成功，我可能不需要测试。如果我想以上述方式构建我的代码，那么是否需要
任何_way_这样做？
IE ，替换CODE HERE ...与一些
测试相符，可以告诉我我的论点小于
等于size_t类型的最大大小。
我在gcc系统上，想要便携式地做这件事。
据我所知，没有gcc定义的宏
或任何值limit.h告诉我最大尺寸
size_t。
也许我错过了什么，但这看起来更难了
比它应该的好。
祝你好运，
--j
Hi:
I am a bit new to C programming and am trying to
write a wrapper around malloc. Since malloc takes
size_t as it''s parameter, I want to be able to
see if my function recieved an argument less than
or equal to the max size_t type.
So:
//my wrapper function mallocstr
char * mallocstr(size_t num)
{
//TEST
if (CODE HERE TO TEST FOR num <= max size of size_t) {
printf("some debug error message\n");
return NULL;
}
//END TEST
return (char *) malloc(num);
}
Now, ignore the fact that malloc will return NULL if
unsucessful and I may not need the test. If I _do_
want to structure my code in the above fashion, is there
any _way_ to do it ?
I.E., replace the "CODE HERE..." line with some
test that can tell me with my argument was less than
of equal to the max size of the size_t type.
I am on a gcc system and would like to do this portably.
As far as I can tell, there is no gcc defined macro
or any value in limit.h that tells me the max size of
size_t.
Maybe I am missing something, but this appears tougher
than it should be.
Best regards,
--j
推荐答案
javadesigner写道：
javadesigner wrote:
嗨：
我对C编程有点新意，我正在尝试编写一个包装器在malloc附近。由于malloc将
size_t作为参数，我希望能够看到我的函数是否收到了小于
或等于max size_t类型的参数。
所以：
//我的包装函数mallocstr
char * mallocstr（size_t num）
{
// TEST
if（CODE这里测试num< = max size of size_t）{
printf（一些调试错误消息\ n）;
返回NULL;
}
//结束测试
返回（char *）malloc（num）;
}
现在，忽略如果
不成功则malloc将返回NULL的事实，我可能不需要考试。如果我想以上述方式构建我的代码，是否有任何_WAY>这样做？
I.E.，替换CODE HERE ...一些
测试可以告诉我我的论点小于
等于size_t类型的最大大小。
我在gcc系统上并希望为了便于这样做。
据我所知，没有gcc定义的宏
或limit.h中的任何值告诉我
size_t的最大大小。
也许我错过了一些东西，但这看起来比它应该更加坚固。
Hi:
I am a bit new to C programming and am trying to
write a wrapper around malloc. Since malloc takes
size_t as it''s parameter, I want to be able to
see if my function recieved an argument less than
or equal to the max size_t type.
So:
//my wrapper function mallocstr
char * mallocstr(size_t num)
{
//TEST
if (CODE HERE TO TEST FOR num <= max size of size_t) {
printf("some debug error message\n");
return NULL;
}
//END TEST
return (char *) malloc(num);
}
Now, ignore the fact that malloc will return NULL if
unsucessful and I may not need the test. If I _do_
want to structure my code in the above fashion, is there
any _way_ to do it ?
I.E., replace the "CODE HERE..." line with some
test that can tell me with my argument was less than
of equal to the max size of the size_t type.
I am on a gcc system and would like to do this portably.
As far as I can tell, there is no gcc defined macro
or any value in limit.h that tells me the max size of
size_t.
Maybe I am missing something, but this appears tougher
than it should be.
这是没有意义的。
你可以在函数中测试的任何值，
将在size_t的范围内。
适合size_t对象的最大值，
是（（size_t）-1）。
C99中有一个SIZE_MAX宏。
-
pete
It''s pointless.
Any value that you can test within the function,
will be within the range of size_t.
The largest value that will fit in a size_t object,
is ((size_t)-1).
There''s a SIZE_MAX macro in C99.
--
pete
问候。
文章< bu ******* ****@netnews.upenn.edu>，javadesigner写道：
Greetings.
In article <bu***********@netnews.upenn.edu>, javadesigner wrote:
我在gcc系统上，并希望这样做便携。
目前为止我可以说，没有gcc定义的宏
或limit.h中的任何值告诉我
size_t的最大大小。
也许我错过了什么，但是这看起来比它应该更加坚固。
I am on a gcc system and would like to do this portably.
As far as I can tell, there is no gcc defined macro
or any value in limit.h that tells me the max size of
size_t.
Maybe I am missing something, but this appears tougher
than it should be.
C99（可能是更早的标准;我不记得了）定义SIZE_MAX
宏。最近版本的GCC支持它。
便携式后备是使用表达式（size_t）（ - 1）。
问候，
Tristan
-
_
_V.-o Tristan Miller [ en，（fr，de，ia）]><空间有限
/ |` - '' - = - = - = - = - = - = - = - = - = - = - = - = - = - = - =<> ;在ha句中，所以很难
（7_ \\ http://www.nothingisreal.com/ ><完成你的内容
C99 (and possibly earlier standards; I don''t remember) define the SIZE_MAX
macro. It''s supported by recent versions of GCC.
A portable fallback is to use the expression (size_t)(-1).
Regards,
Tristan
--
_
_V.-o Tristan Miller [en,(fr,de,ia)] >< Space is limited
/ |`-'' -=-=-=-=-=-=-=-=-=-=-=-=-=-=-= <> In a haiku, so it''s hard
(7_\\ http://www.nothingisreal.com/ >< To finish what you
javadesigner< re ******** **********@yahoo.com>写道：
javadesigner <re******************@yahoo.com> wrote:
我对C编程有点新意，并且我试图围绕malloc写一个包装器由于malloc将
size_t作为参数，我希望能够看到我的函数是否收到了小于
或等于max size_t类型的参数。
Erm。如果你的函数需要一个size_t参数...
//我的包装函数mallocstr
char * mallocstr（size_t num）
....它确实如此，那么你无法获得一个不符合size_t的
参数。甚至如果有人试图将更多的数字传递给你的函数，那么它将被调制到
传递给mallocstr（）之前的
大小。如果size_t可以签名，那么
大的参数会调用I-DB，但它需要是无符号的，所以
它总是被传递给你modulo SIZE_MAX。
return（char *）malloc（num）;
除非你忘记了
#include< stdlib.h>，否则没有理由投出malloc（），btw。 malloc（）返回一个void *，它完全有用
因为它可以容纳各种指针并且不需要强制转换。
现在，忽略这个事实如果
不成功，malloc将返回NULL，我可能不需要测试。如果我想以上述方式构建我的代码，是否有任何_way_这样做？
部分。取一个size_t，取一个uintmax_t并检查你的
参数是否为< = SIZE_MAX。因为在C89中都不存在这些，所以如果你没有C99，那么你将需要使用无符号长ans（size_t）-1。
这意味着你的函数不会像
malloc（）那样采用相同的论证。
此外，size_t甚至可能大于unsigned long（至少，我
找不到它为什么不能的原因，这意味着mallocstr（）
实际上比malloc（）实际上有更小的范围。在C99中，使用
uintmax_t，这不会有问题。
据我所知，没有gcc定义的宏
或任何值limit.h告诉我
size_t的最大大小。
I am a bit new to C programming and am trying to
write a wrapper around malloc. Since malloc takes
size_t as it''s parameter, I want to be able to
see if my function recieved an argument less than
or equal to the max size_t type.
Erm. If your function takes a size_t argument...
//my wrapper function mallocstr
char * mallocstr(size_t num)
....which it does, then there''s no way you''re ever going to get an
argument that does _not_ fit in a size_t. Even if someone tries to pass
a larger number to your function, it''s going to be modulated down to
size before being passed to mallocstr(). If size_t could be signed, too
large arguments would invoke I-DB, but it''s required to be unsigned, so
it is always going to be passed to you modulo SIZE_MAX.
return (char *) malloc(num);
There is no reason to cast malloc(), btw, unless you have forgotten to
#include <stdlib.h>. malloc() returns a void *, which is useful exactly
because it can hold all kinds of pointers and doesn''t need casts.
Now, ignore the fact that malloc will return NULL if
unsucessful and I may not need the test. If I _do_
want to structure my code in the above fashion, is there
any _way_ to do it ?
Partially. Instead of a size_t, take a uintmax_t and check whether your
parameter is <=SIZE_MAX. Since neither of those exists in C89, you''ll
have to make do with unsigned long ans (size_t)-1 if you don''t have C99.
This means your function doesn''t take the same kind of argument as
malloc(), however.
Moreover, size_t could even be larger than unsigned long (at least, I
can''t find a reason why it couldn''t), which would mean that mallocstr()
would actually have a smaller range than malloc(). In C99, with
uintmax_t, that would not be a problem.
As far as I can tell, there is no gcc defined macro
or any value in limit.h that tells me the max size of
size_t.
如果你有一个预先C99版本的gcc，但是因为size_t是保证的
是无符号的，分配给无符号类型的值是模数改变的
类型的大小，如果它们超出范围，（size_t）-1会给你
size_t的最大值。当然，如果你首先拿一个
size_t，它仍然没什么用处。
Richard
Not if you have a pre-C99 version of gcc, but since size_t is guaranteed
to be unsigned and values assigned to unsigned types are changed modulo
the size of the type if they''re out of range, (size_t)-1 will give you
the maximum value of a size_t. It''s still not much use if you take a
size_t in the first place, of course.
Richard
这篇关于如何使用size_t类型的最大大小？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！
