铸造返回malloc [英] Casting return of malloc
问题描述
我读过一篇旧帖子，里面没有指针，因为malloc会返回
。我想知道原因。
提前致谢，
YTR
在文章< 11 ********************** @ g14g2000cwa.googlegroups .com>，
yt **** @ gmail.com < yt **** @ gmail.com>写道：
我已经阅读了一篇旧帖子，其中没有指出由malloc返回的指针。我想知道原因。
没有必要，它隐藏了错误。为什么额外输入
带有负面利益的东西？
dave
-
Dave Vandervies dj******@csclub.uwaterloo.ca
话虽如此，你的错误往往比大多数人少得多（
可能就是为什么我们采取这种不圣洁的喜悦来试图找到错误的
你是什么意思写）。 --Richard Heathfield in comp.lang.c
yt **** @ gmail .com < yt **** @ gmail.com>潦草地写了下面的内容：
我已经阅读了一篇旧帖子，其中没有指出由malloc返回的指针。我想知道原因。
它不会修复任何东西，但它可能会让编译器认为问题是
已修复当他们真的不是。
更长的解释：
没有适当的malloc（）原型，编译器认为它返回
int。因此，当malloc创建void *值并返回它时，
编译器会隐式将其转换为int。在这种情况下，值已经*已经*分解为
，因此任何投射到任何东西的数量都无济于事。但是，如果你将它转换为指针类型，那么
，编译器会认为*你知道你正在做什么，并且省略警告。因此你会得到破碎的代码
看起来像工作代码。
-
/ - Joona Palaste（pa ** ***@cc.helsinki.fi）-------------芬兰-------- \
\ ------ --------------------------------------------------规则！ -------- /
我们的巫师不喜欢吃我们的话，所以说。
- Sparrowhawk
" YT **** @ gmail.com" < YT **** @ gmail.com>写道：
我读过一篇旧帖子，里面没有指出由malloc返回的指针。我想知道原因。
我不建议投出malloc（）的返回值：
* ANSI C中不要求强制转换。
*转换它的返回值可以掩盖#include
< stdlib.h>的失败，这会导致不确定的行为。
*如果你偶然输入了错误的类型，奇怪的失败可能会导致
结果。
其他一些人不同意，例如PJ Plauger（见文章
< 9s ***************** @nwrddc01.gnilink.net> ;）。
-
"这是一个很棒的答案。
它是偏离主题的，它这是不正确的，它没有回答这个问题。
- 理查德希思菲尔德
Hi,
I have read in one of old posting that don''t cast of pointer which
is returned by the malloc. I would like to know the reason.
Thanks in advance,
YTR
In article <11**********************@g14g2000cwa.googlegroups .com>,
yt****@gmail.com <yt****@gmail.com> wrote:Hi,
I have read in one of old posting that don''t cast of pointer which
is returned by the malloc. I would like to know the reason.
It isn''t necessary, and it hides errors. Why do the extra typing for
something with a negative benefit?
dave
--
Dave Vandervies dj******@csclub.uwaterloo.ca
Having said that, you tend to be a lot less wrong than most people (which
might be why we take such an unholy delight in trying to find errors in
what you write). --Richard Heathfield in comp.lang.c
yt****@gmail.com <yt****@gmail.com> scribbled the following:Hi,
I have read in one of old posting that don''t cast of pointer which
is returned by the malloc. I would like to know the reason.
It won''t fix anything, but it may make the compiler think problems are
fixed when they really aren''t.
Longer explanation:
Without a proper prototype for malloc(), the compiler thinks it returns
int. Thus, when malloc creates a void* value and returns it, the
compiler implicitly converts it to int. The value is *already* broken at
this case, so no amount of casting it to anything will help. However,
if you cast it to a pointer type, the compiler will *think* you know
what you''re doing, and omit a warning. Thus you get broken code that
looks like working code.
--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"We sorcerers don''t like to eat our words, so to say."
- Sparrowhawk
"yt****@gmail.com" <yt****@gmail.com> writes:
I have read in one of old posting that don''t cast of pointer which
is returned by the malloc. I would like to know the reason.
I don''t recommend casting the return value of malloc():
* The cast is not required in ANSI C.
* Casting its return value can mask a failure to #include
<stdlib.h>, which leads to undefined behavior.
* If you cast to the wrong type by accident, odd failures can
result.
Some others do disagree, such as P.J. Plauger (see article
<9s*****************@nwrddc01.gnilink.net>).
--
"This is a wonderful answer.
It''s off-topic, it''s incorrect, and it doesn''t answer the question."
--Richard Heathfield
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