static_cast 的奇怪用法
Strange usage of static_cast
我在我们的生产环境中遇到了以下代码构造(但是大大简化了)。
#include <iostream>
typedef struct
{
char entry[10];
}
inn_struct;
typedef struct
{
inn_struct directory;
}
out_struct;
struct test
{
static const int
ENTRY_LENGTH = (sizeof((static_cast<out_struct*>(0))->directory.entry)
/ sizeof((static_cast<out_struct*>(0))->directory.entry[0]));
};
int main()
{
test a;
std::cout << test::ENTRY_LENGTH;
}
现在不考虑它明显混淆的性质,因为它只是确定数组长度的旧 C 方法......我真的被 0
值的 static_cast
困扰了. ... 这个代码可以接受吗?你能否在你的回复中附上一些 C++ 标准的段落,告诉我(如果)为什么这段代码没问题?
是的,这段代码完全可以接受。参见§5.3.3/1(强调我的)。
The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is an unevaluated operand (Clause 5), or a paranthesized type-id.
表达式未被计算,因此看起来取消引用空指针没有问题。
另请注意,在 C++11 中,您无需跳过那个圈套,只需直接使用 sizeof 引用 class 成员即可,感谢 §5/8(强调我的) :
In some contexts, unevaluated operands appear (5.2.8, 5.3.3, 5.3.7, 7.1.6.2). An unevaluated operand is not evaluated. An unevaluated operand is considered a full-expression. [ Note: In an unevaluated operand, a non-static class member may be named (5.1) and naming of objects or functions does not, by itself, require that a definition be provided (3.2). — end note ]
和§5.1.1/13:
An id-expression that denotes a non-static data member or non-static member function of a class can only be used:
..
if that id-expression denotes a non-static data member and it appears in an unevaluated context [Example:
struct S {
int m;
};
int i = sizeof(S::m); // OK
int j = sizeof(S::m + 42); // OK
- end example]
我在我们的生产环境中遇到了以下代码构造(但是大大简化了)。
#include <iostream>
typedef struct
{
char entry[10];
}
inn_struct;
typedef struct
{
inn_struct directory;
}
out_struct;
struct test
{
static const int
ENTRY_LENGTH = (sizeof((static_cast<out_struct*>(0))->directory.entry)
/ sizeof((static_cast<out_struct*>(0))->directory.entry[0]));
};
int main()
{
test a;
std::cout << test::ENTRY_LENGTH;
}
现在不考虑它明显混淆的性质,因为它只是确定数组长度的旧 C 方法......我真的被 0
值的 static_cast
困扰了. ... 这个代码可以接受吗?你能否在你的回复中附上一些 C++ 标准的段落,告诉我(如果)为什么这段代码没问题?
是的,这段代码完全可以接受。参见§5.3.3/1(强调我的)。
The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is an unevaluated operand (Clause 5), or a paranthesized type-id.
表达式未被计算,因此看起来取消引用空指针没有问题。
另请注意,在 C++11 中,您无需跳过那个圈套,只需直接使用 sizeof 引用 class 成员即可,感谢 §5/8(强调我的) :
In some contexts, unevaluated operands appear (5.2.8, 5.3.3, 5.3.7, 7.1.6.2). An unevaluated operand is not evaluated. An unevaluated operand is considered a full-expression. [ Note: In an unevaluated operand, a non-static class member may be named (5.1) and naming of objects or functions does not, by itself, require that a definition be provided (3.2). — end note ]
和§5.1.1/13:
An id-expression that denotes a non-static data member or non-static member function of a class can only be used:
..
if that id-expression denotes a non-static data member and it appears in an unevaluated context [Example:
struct S {
int m;
};
int i = sizeof(S::m); // OK
int j = sizeof(S::m + 42); // OK
- end example]