是“演员”吗？在“返回”中有必要吗？

本文介绍了是“演员”吗？在“返回”中有必要吗？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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...

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" LaEisem" < LA ***** @ aolnojunk>写了 .... "LaEisem" <la*****@aolnojunk> 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 * ****@aoln ojunk（LaEisem）写道： < snip> la*****@aolnojunk (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 scopefor 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" statementconstitute 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 beinterpreted 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.