C语言 声明没有声明任何东西：警告？
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declaration does not declare anything : warning?
提问by kevin gomes
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
int main()
{
struct emp
{
struct address
{
int a;
};
struct address a1;
};
}
This code shows a warning:-
此代码显示警告：-
warning : declaration does not declare anything (enabled by default)
警告：声明没有声明任何东西（默认启用）
Where as the following code shows no warning
以下代码显示没有警告的地方
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
int main()
{
struct emp
{
struct address
{
int a;
}a1;
};
}
Why 'warning' is displayed in the first code only?
为什么“警告”仅显示在第一个代码中？
采纳答案by juan.facorro
The reason why the compiler is showing the warning is because it doesn't see a name for the variable of type addressyou defined for the empstruct, even though you dodeclare something using addresson the next line, but I guess the compiler is not smart enough to figure that out.
编译器显示警告的原因是因为它没有看到address您为emp结构定义的类型变量的名称，即使您确实address在下一行声明了一些内容，但我猜编译器不够聪明弄清楚这一点。
As you showed, this produces a warning:
正如你所展示的，这会产生一个警告：
struct emp {
struct address {}; // This statement doesn't declare any variable for the emp struct.
struct address a1;
};
But not this:
但不是这个：
struct emp {
struct address {} a1; // This statement defines the address struct and the a1 variable.
};
Or this:
或这个：
struct address {};
struct emp {
struct address a1; //the only statement declare a variable of type struct address
};
The struct emp {}doesn't show any warnings since this statement is not inside a struct defintion block. If you did put it inside one of those then the compiler will show a warning for that as well. The following will show two warnings:
在struct emp {}不显示任何警告，因为这种说法是不是结构确定指标块内。如果您确实将其放入其中之一，那么编译器也会为此显示警告。以下将显示两个警告：
struct emp {
struct phone {};
struct name {};
};
回答by rullof
The syntax of a structure definition is:
结构定义的语法是：
struct identifier {
type member_name;
// ...
};
If you add an identifier just after the closing curly brace, you're declaring a variable with that defined struct.
如果您在右花括号之后添加一个标识符，那么您就是在声明一个具有该定义结构的变量。
In your first example the compiler consider the addressstruct as member type. it's like if you writes:
在您的第一个示例中，编译器将address结构视为成员类型。就像你这样写：
struct identifier {
type ; // No member name is specified
type a1;
// ...
}
But in the second example you specified the member name:
但在第二个示例中，您指定了成员名称：
struct identifier {
type a1; // Member name specified
// ...
}
And here is an example of the warning: http://ideone.com/KrnYiE.
这是警告的示例：http: //ideone.com/KrnYiE。
回答by Antti Haapala
The reason the warning is displayed is that the first excerpt is not proper C - it has a constraint violation that a standards-compliant C compiler mustproduce a diagnostisc message for. It violates the C11 6.7.2.1p2:
显示警告的原因是第一个摘录不是正确的 C - 它违反了约束，符合标准的 C 编译器必须为其生成诊断消息。它违反了C11 6.7.2.1p2：
Constraints
- A struct-declarationthat does not declare an anonymous structure or anonymous union shall contain a struct-declarator-list.
约束
- 一个结构声明不声明匿名结构或匿名联合应包含结构说明符列表。
Meaning that it is OK to write
意思是可以写
struct foo {
struct {
int a;
};
};
since the inner structdeclares an anonymousstructure, i.e. it is not named.
因为内部struct声明了一个匿名结构，即它没有命名。
But in your example the struct addresshas a name - address- and therefore it musthave a declarator list after the closing brace - declarator list being for example a1as in your example, or more complex foo, *bar, **baz[23][45].
但是在您的示例中，它struct address有一个名称 -address因此它必须在右大括号之后有一个声明符列表 - 例如a1在您的示例中，或者更复杂的声明符列表foo, *bar, **baz[23][45]。
