询问一段代码的合法性

starwing83

这个事儿嘛。其实是有两个方面的考虑的：

6.2.5.19 The void type comprises an empty set of values; it is an incomplete type that cannot be completed.
6.5.3.2.4 The unary * operator denotes indirection. If the operand points to a function, the result is
a function designator; if it points to an object, the result is an lvalue designating the
object. If the operand has type ‘‘pointer to type’’, the result has type ‘‘type’’. If an
invalid value has been assigned to the pointer, the behavior of the unary * operator is
undefined.84)

所以，其实*p这个操作本身是合法的。它的类型为void，问题是，void被定义成了一个不完全类型，然而：

6.5.6.2 For addition, either both operands shall have arithmetic type, or one operand shall be a
pointer to an object type and the other shall have integer type. (Incrementing is
equivalent to adding 1.)

因此，实际上真正的错误是类型不匹配的错误：+运算符需要两个arithmetic type，或者至少一个pointer一个Integer，你却给了一个void，显然就error了。

赋值的情况与之类似。

starwing83

另外，玩几个游戏吧：

1. 下面的代码能否编译通过？为什么？

void *p = ∑
int *pp = &*p;

2. 下面的代码能否编译通过？为什么？

int a[] = {1, 2, 3};
printf("%p\n", &2[p]);

slackware12

回复 12# starwing83


    先谢谢了.
我得先好好研读研读你那些回复。

聪聪知不道

另外，玩几个游戏吧：

另外，玩几个游戏吧：

1. 下面的代码能否编译通过？为什么？

void *p = ∑
int *pp = &*p;

2. 下面的代码能否编译通过？为什么？

int a[] = {1, 2, 3};
printf("%p\n", &2[p]);

starwing83 发表于 2010-06-19 14:51

第一个不能编译吧：*p是void类型，void类型不能做lvalue，而&要求一个lvaue。
第二个可以编译，相当于printf("%p\n",&p[2]);

tcwm1

int* 型指针+1时 地址移动4个字节
char* 型指针+1时 地址移动两个字节
那void*+1时 楼主认为应该移动几个字节？

聪聪知不道

又进一步的考虑了一下，对一个void *的指针做解除引用操作，得到一个void类型的东东，然后按照标准
6.3.2.2 void
1 The (nonexistent) value of a void expression (an expression that has type void) shall not
be used in any way, and implicit or explicit conversions (except to void) shall not be
applied to such an expression. If an expression of any other type is evaluated as a void
expression, its value or designator is discarded. (A void expression is evaluated for its
side effects.)
标准上说The value of void expression shall not be used in any way.
那我对他做取地址操作会发生啥呢？
不过我编译只给了一个警告，说我对void*指针做了解除引用操作。而不是一个对void类型做&操作的错误。
还得研究研究。

小翼给答案啊~

聪聪知不道

加上这个貌似就能解释了。

6.5.3.2 Address and indirection operators
Constraints
1 The operand of the unary & operator shall be either a function designator, the result of a
[] or unary * operator, or an lvalue that designates an object that is not a bit-field and is
not declared with the register storage-class specifier.
2 The operand of the unary * operator shall have pointer type.
Semantics
3 The unary & operator returns the address of its operand. If the operand has type ‘‘type’’,
the result has type ‘‘pointer to type’’. If the operand is the result of a unary * operator,
neither that operator nor the & operator is evaluated and the result is as if both were
omitted, except that the constraints on the operators still apply and the result is not an
lvalue.

小翼快来公布答案

peidright

回复 2# starwing83


    从来没看过c99标准， 弱弱的问一下， 这个东西，在什么场合比较重要。。

ehust_cu

回复 1# slackware12

这个跟标准没有关系吧，语法错误，p指针指向的数据类型未知，怎么进行取值操作呢？

starwing83

答案啊，编译一下不就知道了吗？~~~