是“演员”吗？在“返回”中有必要吗？ [英] Is a "cast" ever necessary in "return"?

问题描述

在职，我已经继承了很多旧的C语言软件。

一个或两个关于什么时候铸造的问题。空指针常量

在幕后清理期间发生了[不需要]

的某些软件。这个问题似乎没有得到解决，

至少不是直接的，在C FAQ中，FAQ 5.2似乎最相关。


参考文献：

* C FAQ 5.2空指针

（包括

空指针常量的转换条件是否需要）

* ANSI C标准6.2.2.3指针

（包括空指针常量）

* ANSI C标准6.6.6.4返回语句

* ANSI C标准7.1.6（包括NULL宏扩展到

a空指针常量）


是否有必要在铸造中使用"返回"声明，一个NULL

（或一个unadorned 0，使用来自C FAQ 5.2的术语）

，在该表中显示为待返回的值;返回"声明？

函数原型是否在范围内是否重要

这个返回的函数是什么？声明出现了？


示例：如果返回下面的陈述出现在C函数中，

将是（char *）必须使用cast吗？

return（char *）NULL;


返回NULL或unadorned 0是否返回在返回中声明

构成一个赋值上下文，因为该术语在C FAQ 5.2中使用？

一般来说，在返回中返回任意值;

语句构成了这样一个赋值语境？

如果确实如此，如果一个朴素的0（不是NULL）
$那就不会有问题b $ b将以返回方式退回。意图是将

解释为整数0而不是（空）指针的声明？


非常感谢...

On-the-job, I have "inherited" a lot of old C language software.
A question or two about when "casting" of null pointer constants
is [not] needed has occurred during behind-the-scenes cleanup
of some of that software. That subject seems not to be addressed,
at least not directly, in the C FAQ where FAQ 5.2 seems most relevant.

References:
* C FAQ 5.2 Null pointers
(Including conditions where "casting" of
null pointer constants is (not) needed)
* ANSI C Standard 6.2.2.3 Pointers
(Including null pointer constant)
* ANSI C Standard 6.6.6.4 The return statement
* ANSI C Standard 7.1.6 (Including NULL macro which expands to
a null pointer constant)

Is it ever necessary to "cast", in a "return" statement, a NULL
(or an "unadorned 0", to use terminology from C FAQ 5.2)
that appears as the to-be-returned value in that "return" statement?
Does it matter whether or not a function prototype is in scope
for the function in which such a "return" statement appears?

Example: If the "return" statement below appeared in a C function,
would the "(char *)" cast ever be necessary?
return (char *)NULL;

Does the return of NULL or of an "unadorned 0" in a "return" statement
constitute an "assignment context", as that term is used in C FAQ 5.2?
In general, does the return of an arbitrary value in a "return"
statement constitute such an "assignment context"?
If it did, would that not be a problem if an unadorned 0 (not NULL)
was to be returned in a "return" statement with the intent that it be
interpreted as an integer 0 rather than as a (null) pointer?

Much thanks...

推荐答案

" LaEisem" < LA ***** @ aol.comnojunk>写了

....

"LaEisem" <la*****@aol.comnojunk> wrote
....

是否有必要在返回中进行强制转换。声明，一个NULL
（或unadorned 0，使用来自C FAQ 5.2的术语）
在该return中显示为待返回的值。陈述？
函数原型是否在范围内，对于这种返回函数的功能是否重要。声明出现了吗？

示例：如果返回下面的陈述出现在C函数中，
将是（char *）转换是否必要？
return（char *）NULL;

返回NULL或unadorned 0是否返回在返回中声明
构成一个赋值上下文，因为该术语在C FAQ 5.2中使用？
一般来说，return
语句中任意值的返回是否构成分配上下文？


尝试自己回答：如果函数返回怎么办，例如，

为double，你使用''return 42;''？会得到什么回报？为什么

会返回NULL（或0表示不同）？

如果确实如此，如果一个unadorned 0（不是NULL）<那就不会有问题
将以返回方式返回。语句是否被解释为整数0而不是（空）指针？

Is it ever necessary to "cast", in a "return" statement, a NULL
(or an "unadorned 0", to use terminology from C FAQ 5.2)
that appears as the to-be-returned value in that "return" statement?
Does it matter whether or not a function prototype is in scope
for the function in which such a "return" statement appears?

Example: If the "return" statement below appeared in a C function,
would the "(char *)" cast ever be necessary?
return (char *)NULL;

Does the return of NULL or of an "unadorned 0" in a "return" statement
constitute an "assignment context", as that term is used in C FAQ 5.2?
In general, does the return of an arbitrary value in a "return"
statement constitute such an "assignment context"?
Try answering this yourself: what if the function returned, for example,
a double and you used ''return 42;''? What would get returned? Why would
be returning NULL (or 0, for that matter) different?
If it did, would that not be a problem if an unadorned 0 (not NULL)
was to be returned in a "return" statement with the intent that it be
interpreted as an integer 0 rather than as a (null) pointer?

NULL在许多平台上#defined为0 。


Peter

NULL is on many platforms #defined as 0.

Peter

la * ****@aol.comn ojunk（LaEisem）写道：


< snip>

la*****@aol.comnojunk (LaEisem) wrote:

<snip>

是否有必要施放 ;在返回中;声明，一个NULL
（或unadorned 0，使用来自C FAQ 5.2的术语）
在该return中显示为待返回的值。陈述？
函数原型是否在范围内，对于这种返回函数的功能是否重要。声明出现了？


因为函数定义也是一个正确的函数原型，所以

无法定义函数返回没有原型的指针

范围。

示例：如果返回下面的陈述出现在C函数中，
将是（char *） cast是否必要？
return（char *）NULL;


如果函数被定义为返回一个指向char的指针，那么强制转换为
。否则演员没有意义。只需删除它。

返回NULL或unadorned 0是否返回在返回中声明
构成一个赋值上下文，因为该术语在C FAQ 5.2中使用？
一般来说，return
语句中任意值的返回是否构成分配上下文？


一般来说，是的。

如果确实如此，如果要返回一个未经修饰的0（非空）
，这是不是一个问题在返回中语句是否被解释为整数0而不是（空）指针？

Is it ever necessary to "cast", in a "return" statement, a NULL
(or an "unadorned 0", to use terminology from C FAQ 5.2)
that appears as the to-be-returned value in that "return" statement?
Does it matter whether or not a function prototype is in scope
for the function in which such a "return" statement appears?
As a function definition is also a proper function prototype there is
no way to define a function returning a pointer without a prototype in
scope.
Example: If the "return" statement below appeared in a C function,
would the "(char *)" cast ever be necessary?
return (char *)NULL;
If the function is defined to return a pointer to char the cast is
spurious. Otherwise the cast makes no sense. Just drop it.
Does the return of NULL or of an "unadorned 0" in a "return" statement
constitute an "assignment context", as that term is used in C FAQ 5.2?
In general, does the return of an arbitrary value in a "return"
statement constitute such an "assignment context"?
Generally, yes.
If it did, would that not be a problem if an unadorned 0 (not NULL)
was to be returned in a "return" statement with the intent that it be
interpreted as an integer 0 rather than as a (null) pointer?

任何解析为0的整数常量或者这样的常量转换为void *

用作正确的空指针常量。因此，在分配时，它将自动转换为有问题的指针类型。 NULL和0是指针上下文中可互换的
（除了

FAQ 5.2中提到的例外情况，但你已经知道了这一点。）


HTH

问候

-

Irrwahn

（ir ******* @ freenet .de）

Any integer constant resolving to 0 or such a constant casted to void*
serves as a proper null pointer constant. Thus it will automatically be
casted to the pointer type in question on assignment. NULL and 0 are
interchangeable in pointer contexts (with the exceptions mentioned in
FAQ 5.2, but you already know this).

HTH
Regards
--
Irrwahn
(ir*******@freenet.de)

Irrwahn Grausewitz写道：

Irrwahn Grausewitz wrote:

作为一个函数定义也是一个正确的函数原型有
无法定义一个函数在
范围内返回一个没有原型的指针。

As a function definition is also a proper function prototype there is
no way to define a function returning a pointer without a prototype in
scope.

并非所有的函数定义都是原型。以下是

定义，但不是原型：


int * f（）{return 0; }


Jeremy。

Not all function definitions are prototypes. The following is a
definition but not a prototype:

int *f() { return 0; }

Jeremy.

这篇关于是“演员”吗？在“返回”中有必要吗？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！