警告：初始化程序周围缺少大括号 [英] warning: missing braces around initializer
问题描述
你能告诉hw写这段代码是为了不接受来自主题的警告：
#include< ; iostream>
使用命名空间std;
使用命名空间__gnu_cxx;
struct entry {
union {
int a;
};
union {
int b;
};
};
static struct entry en [] = {{12,13}};
int main（）{
（void）en;
};
我在linux上编译它如下命令：
g ++ -Wall -g -O0 -o / tmp / p2 p2.cpp
问候，
Pawel
Pawel写道：
可能是：
你好小组成员
你能告诉hw写这段代码是为了不接受来自主题的警告：
#include< iostream>
使用命名空间std;
使用命名空间__gnu_cxx;
struct entry {
union u_a {
< blockquote class =post_quotes>
int a;
};
union u_a {
int b;
};
};
static struct entry en [] = {{12,13}};
int main（）{
（void）en;
};
我使用以下命令在linux上编译它：
g ++ -Wall -g -O0 -o / tmp / p2 p2.cpp
问候，
Pawel
Pawel写道：
你好组成员
你能否告诉hw写这段代码为了不接受来自主题的警告：
struct entry {
union {
int a;
};
union {
int b;
};
};
static struct entry en [] = {{12,13}};
要静态初始化类型为entry的对象，您必须静态地使用
初始化其每个字段。由于每个字段都是
聚合类型，编译器会警告您没有使用大括号
来标记这些字段的初始值设定项。在这种情况下它没关系，因为
每个字段只需要一个初始化程序。所以
第一个联盟初始化为12，第二个联盟初始化为13.如果你更改
那些联合体结构（当然是添加名称）初始化
仍然可以以相同的方式工作。但是如果你在第一个字段中添加一个字段，那么12和13将用于初始化它的两个字段，并且第二个结构将初始化为0。 。
所以，为了摆脱警告，在初始化程序周围加上大括号为
每个汇总成员：
static struct entry en [] = {{{12}，{13}}};
-
- Pete
标准C ++库扩展：教程和
参考的作者。有关本书的更多信息，请参阅
www.petebecker.com/tr1book。
Pawel写道：
你好组成员
你能告诉hw编写这段代码是为了不收到主题的警告：
#include< iostream>
使用命名空间std;
使用命名空间__gnu_cxx;
struct entry {
union {
int a;
};
union {
int b;
}; < br $>
};
static struct entry en [] = {{12,13}};
静态条目en [] = {{{12}，{13}}};
int main（）{
（void）en;
};
删除最后一个分号。请注意，最后一个语句没有效果。
这是一件好事，因为：
void n;是非法的
和en实际衰减到一个指针。所以你可以争辩说：
int main（）
{
void * pvoid = static_cast< void *>（& en）;
std :: cout<< pvoid = << pvoid<< std :: endl;
}
>
我使用以下命令在linux上编译它：
g ++ -Wall -g -O0 -o / tmp / p2 p2.cpp
问候，
Pawel
Hallo group members
Could You tell hw to write this code in order not to get warning from subject:
#include <iostream>
using namespace std;
using namespace __gnu_cxx;
struct entry {
union {
int a;
};
union {
int b;
};
};
static struct entry en[] = { { 12, 13 } };
int main() {
(void)en;
};
I compile it on linux with following command:
g++ -Wall -g -O0 -o /tmp/p2 p2.cpp
regards,
Pawel
Pawel wrote:
May be:Hallo group members
Could You tell hw to write this code in order not to get warning from subject:
#include <iostream>
using namespace std;
using namespace __gnu_cxx;
struct entry {union u_a{
int a;
};union u_a{
int b;
};
};
static struct entry en[] = { { 12, 13 } };
int main() {
(void)en;
};
I compile it on linux with following command:
g++ -Wall -g -O0 -o /tmp/p2 p2.cpp
regards,
Pawel
Pawel wrote:Hallo group members
Could You tell hw to write this code in order not to get warning from subject:
struct entry {
union {
int a;
};
union {
int b;
};
};
static struct entry en[] = { { 12, 13 } };To statically initialize an object of type entry you have to statically
initialize each of its fields. Since each of those fields is in turn an
aggregate type, the compiler is warning you that you haven''t used braces
to mark the initializers for those fields. In this case it''s okay, since
there''s only one initializer needed for each of those fields. So the
first union is initialzied with 12 and the second with 13. If you change
those unions to structs (adding names, of course) the initialization
would still work the same way. But if you then added a field to the
first one, the 12 and 13 would be used to initialize its two fields, and
the second struct would be initialized to 0.
So, to get rid of the warning, put braces around the initializer for
each of the aggregate members of entry:
static struct entry en[] = { { {12}, {13} } };
--
-- Pete
Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." For more information about this book, see
www.petebecker.com/tr1book.
Pawel wrote:Hallo group members
Could You tell hw to write this code in order not to get warning from subject:
#include <iostream>
using namespace std;
using namespace __gnu_cxx;
struct entry {
union {
int a;
};
union {
int b;
};
};
static struct entry en[] = { { 12, 13 } };static entry en[] = { { {12},{13} } };
int main() {
(void)en;
};remove that last semicolon. Note that the last statement has no effect.
Which is a good thing since:
void n; is illegal
and en actually decays to a pointer. So you could argue:
int main()
{
void* pvoid = static_cast<void*>(&en);
std::cout << "pvoid = " << pvoid << std::endl;
}
>
I compile it on linux with following command:
g++ -Wall -g -O0 -o /tmp/p2 p2.cpp
regards,
Pawel
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