是C宏隐式转换？ [英] Are C macros implicitly cast?
问题描述
我已经搜查SO，但还没有找到一个答案的具体问题。原谅我，如果它已经有了答案。
如果您具备以下条件：
的#define MACRO 40
您不要把它分配给一个变量，你在一个循环中使用它：
的for（int i = 0; I＆LT;万家乐，我++）{...
该perprocessor然后创建：
的for（int i = 0; I＆LT; 40;我++）{...
请问编译器则含蓄地将其转换为int，因为比较有型 INT I ？我看的#define 的变量的这个问题，类型，也有不少的答案下来埃德加博内特意味着有在该编译器选择如何看待宏观订单？
这个问题，<一个href=\"http://stackoverflow.com/questions/14603882/how-does-c-implicitly-cast-arguments-to-a-comparator-such-as\">How确实的C ++隐式转换参数给比较器例如＆lt ;? 时，也有人建议，但仅描述了铸造如何隐与两种类型的比较工作。由于宏并没有真正有型，我不知道这是否适用。
借助 preprocessor 扩展宏编译器甚至看到任何东西之前。我们可以看到，preprocessing号码不要将有一个类型的草案C99标准部分 6.4.8 的 preprocessing号的它说：
一个preprocessing数量没有类型或价值;它获得两个
转换成功后（如转换阶段7的一部分）到
浮点常量令牌或整型常量令牌。在C草案同一节++标准是
2.10。正如我们在çpreprocessor维基百科的文章宏扩展见发生在第4阶段。
在C语言术语和整型常量整型常量在C ++术语的转换涵盖了C99标准草案
6.4.4.1的整型常量的并在段落的 5 的它说见下表：
类型的整型常量是第一个对应列表
在其值可以被重新presented
八进制或十六进制
后缀十进制常数常数
-------------------------------------------------- -------------------------
没有INT INT
长整型无符号整型
久长整型长整型
unsigned long int类型
得到long long int
无符号长长整型
-------------------------------------------------- -------------------------
ü或U无符号整型无符号整型
unsigned long int类型unsigned long int类型
无符号长长整型无符号长长整型
-------------------------------------------------- -------------------------
I或L长整型长整型
得到long long int unsigned long int类型
得到long long int
无符号长长整型
-------------------------------------------------- -------------------------
无论ü或U unsigned long int类型unsigned long int类型
和L或L无符号长长整型无符号长长整型
-------------------------------------------------- -------------------------
LL或LL得到long long int得到long long int
unsigend得到long long int
-------------------------------------------------- -------------------------
无论ü或U unsigned long类型长整型无符号长长整型
和LL或LL
-------------------------------------------------- -------------------------表是从该回答的所述一个的修改版本。覆盖这在C草案的部分++标准是一款
2.14.2其中也有类似的表。因此，在您的例子
40没有后缀，是一个十进制常量和第一类它可以重新从表中该节psented $ P $是 INT 的在这一点上，我们现在使用的效果
40与＆LT结束;运营商。由于I和40均算术类型的话通常的算术转换将被执行，在这种情况下仍然会<青霉> INT 的。对于 C99 的这部分内容在第6.3.1.8和C ++部分的 5 的I've searched SO, but haven't found an answer to this specific questions. Forgive me if it's already been answered.
If you have the following:
#define MACRO 40You don't assign it to a variable you use it in a loop:
for(int i = 0; i < MACRO; i++) {...The perprocessor then creates:
for(int i = 0; i < 40; i++) {...Would the compiler then implicitly cast it to an int since the comparison is with type
int i? I've looked at this question Type of #define variables, and quite a few answers down Edgar Bonet implies that there is an order in which the compiler chooses how to treat the macro?This question, How does C++ implicitly cast arguments to a comparator such as <?, was also suggested, but only describes how implicit casting works with a comparison with two types. Since a macro doesn't really have a type I'm not sure if this applies.
解决方案The preprocessor expands macros before the compiler even sees anything. We can see that preprocessing numbers don't have a type by going to the draft C99 standard section
6.4.8Preprocessing numbers which says:A preprocessing number does not have type or a value; it acquires both after a successful conversion (as part of translation phase 7) to a floating constant token or an integer constant token.
The same section in the draft C++ standard is
2.10.As we can see in C preprocessor Wikipedia article macro expansion happens in phase 4.
The conversion of integer constants in C terminology and integer literals in C++ terminology is covered in the draft C99 standard section
6.4.4.1Integer constants and the following table in paragraph 5 which says:The type of an integer constant is the first of the corresponding list in which its value can be represented
Octal or Hexadecimal Suffix Decimal Constant Constant --------------------------------------------------------------------------- none int int long int unsigned int long long int long int unsigned long int long long int unsigned long long int --------------------------------------------------------------------------- u or U unsigned int unsigned int unsigned long int unsigned long int unsigned long long int unsigned long long int --------------------------------------------------------------------------- l or L long int long int long long int unsigned long int long long int unsigned long long int --------------------------------------------------------------------------- Both u or U unsigned long int unsigned long int and l or L unsigned long long int unsigned long long int --------------------------------------------------------------------------- ll or LL long long int long long int unsigend long long int --------------------------------------------------------------------------- Both u or U unsigned long long int unsigned long long int and ll or LL ---------------------------------------------------------------------------Table is a modified version of the one from this answer. The section that covers this in the draft C++ standard is section
2.14.2which also has a similar table.So in your example
40has no suffix and is a decimal constant and the first type it can be represented from that section of the table is int.At this point we now end up with the effects of using
40with the<operator. Sinceiand40are both arithmetic types then the usual arithmetic conversions will be performed, which in this case will still be int. For C99 this is covered in section6.3.1.8and C++ section 5.这篇关于是C宏隐式转换？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！
