sizeof()什么时候返回0

[ibingow]

在看webkit，看到JavaScriptCore/wtf/OwnPtrCommon.h有这么一段

    template <typename T> inline void deleteOwnedPtr(T* ptr)
    {
        typedef char known[sizeof(T) ? 1 : -1];
        if (sizeof(known))
            delete ptr;
    }

很奇怪，sizeof()还会返回0？什么时候会发生？

[fishion]

。。还真不知道。。类都有1字节的占位的

[muyi66]

没见识过...

[风雪夜归人]

未知。。。

[wjb_yd]

typedef char ch[0]; typedef char ch[-1] 在VS2008下均编译不通过。

[Jinhao]

简单地说，没有，但是编译器都有扩展。其中有两个扩展。

int buf[0];
sizeof(buf); //0

另一个是
当sizeof 非法时，也就是说产生了编译错误，如同lz给出的代码。
试试这样
void * p = 0;
deleteOwnedPtr(p); //编译错误 void 不能用于sizeof

[Coder_Y_Jao]

申明一个全局变量,
然后extern

比如：extern int a[];
sizeof(a)

这个有些编译器会给警告，有些编译器编译不过，，不过在删掉sizeof那句话后在调试中取看sizeof(a)的话都是0

[evencoming]

[Quote=引用 5 楼 的回复:]
简单地说，没有，但是编译器都有扩展。其中有两个扩展。

int buf[0];
sizeof(buf); //0

另一个是
当sizeof 非法时，也就是说产生了编译错误，如同lz给出的代码。
试试这样
void * p = 0;
deleteOwnedPtr(p); //编译错误 void 不能用于sizeof
[/Quote]
路过学习。

[csnd_life]

真还不清楚，想学学

[vcf_reader]

出错时，可以

[logiciel]

http://stackoverflow.com/questions/2632021/can-sizeof-return-0-zero

Is it possible for the sizeof operator to ever return 0 (zero) in C or C++? If it is possible, is it correct from a standards point of view?

Answer:
In C++ an empty class or struct has a sizeof at least 1 by definition. From the C++ standard, 9/3 "Classes": "Complete objects and member subobjects of class type shall have nonzero size."

In C an empty struct is not permitted, except by extension (or a flaw in the compiler).

This is a consequence of the grammar (which requires that there be something inside the braces) along with this sentence from 6.7.2.1/7 "Structure and union specifiers": "If the struct-declaration-list contains no named members, the behavior is undefined".

If a zero-sized structure is permitted, then it's a language extension (or a flaw in the compiler). For example, in GCC the extension is documented in "Structures with No Members", which says:


GCC permits a C structure to have no members:
struct empty {
};

The structure will have size zero. In C++, empty structures are part of the language. G++ treats empty structures as if they had a single member of type char.

[sducn1]

哪位老大帮我顶一下我的帖子啊
http://topic.csdn.net/u/20120324/10/a128ed09-e604-4fb7-86f4-305c320566f0.html

[Kaile]

可能是想编译期检查一些错误

[ibingow]

试了下下标为0果然可以，至少gcc编译通过。以前看到过一些静态诊断用的就是数组大小不能为0，看来也不正确。
[Quote=引用 5 楼 的回复:]

简单地说，没有，但是编译器都有扩展。其中有两个扩展。

int buf[0];
sizeof(buf); //0

另一个是
当sizeof 非法时，也就是说产生了编译错误，如同lz给出的代码。
试试这样
void * p = 0;
deleteOwnedPtr(p); //编译错误 void 不能用于sizeof
[/Quote]

[RabbitLBJ]

#define sizeof(X) 0

[pengzhixi]

In a declarationT D where D has the form
D1 [constant-expressionopt]
and the type of the identifier in the declarationT D1 is “derived-declarator-type-list T,” then the type of the
identifier of D is an array type.

If the constant-expression
(5.19) is present, it shall be an integral constant expression and its value shall be greater than zero.

所以支持数组元素个数为0的是没有可移植性的

[ibingow]

有些编译器支持，所以为了周全，webkit梨还是这么写了，应该这样

[Quote=引用 15 楼 的回复:]

In a declarationT D where D has the form
D1 [constant-expressionopt]
and the type of the identifier in the declarationT D1 is “derived-declarator-type-list T,” then the type of the
identifier of D……
[/Quote]

[tinyhare]

我打了个代码sizeof返回零了，求解。 代码： long long int a=3; printf("%ld,%d\n",a,sizeof(a)); 编译警告： warning: format ‘%ld’ expects type ‘long int’, but argument 2 has type ‘long long int’ 运行结果： 3,0 如果将第一个%ld按警告改为%lld则无警告，运行结果就是3,8为什么呢？ ubuntu gcc编译器， gcc --version gcc (Ubuntu/Linaro 4.4.4-14ubuntu5.1) 4.4.5