为结构赋值 [英] Assigning values to a struct
问题描述
我正在从第三方移植一些代码，并且已经遇到以下问题： ba
a问题：
typedef struct {
unsigned char Red;
unsigned char Green;
unsigned char Blue;
} TBoxColour;
// -------------------------- -------------------------------------------------
无效DoSomething（无效）
{
TBoxColour Col;
Col =（TBoxColour）{ 10,20,30};
}
我的编译器（CodeGear C ++ Builder）抱怨错误：
不正确使用typedef''TBoxColour''
似乎完全合乎逻辑的是为结构赋值
成员，但这是合法的，还是原来的程序员只是在某个特定的编译器中利用一些怪癖（不知道哪个）？
谢谢
推荐答案
开启8月29日，上午12:59，Alf P. Steinbach， < al ... @ start.nowrote：
On Aug 29, 12:59 am, "Alf P. Steinbach" <al...@start.nowrote:
尝试
BoxColor const color = {10,20,30 };
干杯，& hth。，
- Alf
Try
BoxColor const color = { 10, 20, 30 };
Cheers, & hth.,
- Alf
你好Alf，
对不起，我应该说，这个TBoxColour的一个实例在整个代码中的100个位置分配了新的值。
我想是唯一的方式是：
Col.Red = 10;
Col.Green = 20;
Col.Blue = 30;
原始方法看起来更优雅，这只是丑陋，但更多
重要的是我用一个简单的<改变代码并不容易br />
搜索和替换。
Hi Alf,
Sorry, I should have said, a single instance of this TBoxColour gets
assigned new values in 100''s of locations throughout the code.
I suppose the only way is:
Col.Red = 10;
Col.Green = 20;
Col.Blue = 30;
The original method looks more elegant, This is just ugly, but more
importantly it''s not easy for me to change the code with a simple
search and replace.
Cliff写道：
Cliff wrote:
On 8月29日，上午12:59，Alf P. Steinbach， < al ... @ start.nowrote：
On Aug 29, 12:59 am, "Alf P. Steinbach" <al...@start.nowrote:
>尝试
BoxColor const color = {10,20,30};
干杯，& hth。，
- Alf
>Try
BoxColor const color = { 10, 20, 30 };
Cheers, & hth.,
- Alf
你好Alf，
对不起，我应该说，这个TBoxColour的单个实例在整个代码中的100个位置分配了新值。
我想唯一的方法是：
Col.Red = 10;
Col.Green = 20;
Col.Blue = 30;
原始方法看起来更优雅，这只是丑陋，但更多
重要的是我用一个简单的改变代码并不容易
搜索和替换。
Hi Alf,
Sorry, I should have said, a single instance of this TBoxColour gets
assigned new values in 100''s of locations throughout the code.
I suppose the only way is:
Col.Red = 10;
Col.Green = 20;
Col.Blue = 30;
The original method looks more elegant, This is just ugly, but more
importantly it''s not easy for me to change the code with a simple
search and replace.
你可以定义一个函数（一个伪构造函数）：
BoxColor createBoxColor（int a，int b，int c）{
BoxColor bc = {a，b，c};
返回bc;
}
然后在你需要指定的地方使用它：
Col = createBoxColor（10,20,30）;
（我明白我在没有阅读剩下的帖子的情况下插手，
对不起;如果我写的是虚假的，请原谅我并忽略它。
V
-
请在通过电子邮件回复时删除资金''A'
我没有回应top-发表回复，请不要问
You can define a function (a pseudo-constructor):
BoxColor createBoxColor(int a, int b, int c) {
BoxColor bc = { a, b, c };
return bc;
}
and then use it anywhere you need to assign:
Col = createBoxColor(10, 20, 30);
(I understand that I barge in without reading the rest of the thread,
sorry for that; if what I wrote is bogus, forgive me and ignore it)
V
--
Please remove capital ''A''s when replying by e-mail
I do not respond to top-posted replies, please don''t ask
Cliff< us ****** @ btconnect.comwrote in message .. 。
Cliff <us******@btconnect.comwrote in message...
8月29日上午12:59，Alf P. Steinbach < al ... @ start.nowrote：
On Aug 29, 12:59 am, "Alf P. Steinbach" <al...@start.nowrote:
尝试
BoxColor const color = {10,20,30};
干杯，& hth。，Alf
Try
BoxColor const color = { 10, 20, 30 };
Cheers, & hth., Alf
你好Alf，
对不起，我应该说，这个TBoxColour的一个实例得到
在整个代码中的100'位置分配了新值。
我想唯一的方法是：
Col.红色= 10;
Col.Green = 20;
Col.Blue = 30;
原始方法看起来更优雅，这只是丑陋的，但是更多
重要的是我用一个简单的
搜索和替换来改变代码并不容易。
Hi Alf,
Sorry, I should have said, a single instance of this TBoxColour gets
assigned new values in 100''s of locations throughout the code.
I suppose the only way is:
Col.Red = 10;
Col.Green = 20;
Col.Blue = 30;
The original method looks more elegant, This is just ugly, but more
importantly it''s not easy for me to change the code with a simple
search and replace.
嗯，因为你不会说，也许这会把它拖出你：
struct BoxColor {
unsigned char红色;
unsigned char绿色;
unsigned char蓝色;
BoxColor（）：红色（0），绿色（0），蓝色（ 0）{}
BoxColor（unsigned char R，unsigned char G，
unsigned char B）：红色（R），绿色（G），蓝色（B）{ }
void颜色（unsigned char R，unsigned char G，
unsigned char B）{
红色= R;绿= G;蓝色= B;
返回;
}
};
//如果你不能更改此结构，请参阅BoxColor2
typedef struct {
unsigned char Red;
unsigned char Green;
unsigned char蓝色;
} TBoxColour;
struct BoxColor2：TBoxColour {
BoxColor2（）{Red = 0;绿色= 0;蓝色= 0; } $ / $
BoxColor2（unsigned char R，unsigned char G，
unsigned char B）{
Red = R;
Green = G;
Blue = B;
}
};
int main（）{
BoxColor bc1（10,20,30）;
cout<<" bc1 Red ="<< int（bc1.Red）<< ;"绿色="
<< int（bc1.Green）<<"蓝色="<<< int（bc1.Blue）<< std :: endl;
BoxColor bc2;
cout<<" bc2 Red = QUOT;<< INT（bc2.Red）LT;<"绿色="
<< int（bc2.Green）<<"蓝色="<<< int（bc2.Blue）<< std :: endl;
bc2.Color（40,50,60）;
cout<<" bc2 Red ="<< int（bc2.Red）<<""绿色="
<< int（bc2.Green）<<" Blue ="<< int（bc2.Blue）<< std :: endl;
BoxColor2 BC2（10,20,30）;
TBoxColour TBC;
TBC = BC2;
cout<<"TBC Red ="<< int（TBC.Red）<< ;"绿色="
<< int（TBC.Green）<<" Blue ="<< int（TBC.Blue）<< std :: endl;
TBC = BoxColor2（30,40,50）;
cout<<"" TBC Red ="<< int（TBC.Red）<<<"绿色="
<< int（TBC.Green）<<"蓝色="<<< int（TBC.Blue）<< std :: endl;
// ----------------- -------------------
返回0;
} // main（）
那，或者使用Victor的例子。
-
Bob R
POVrookie
Well, since you won''t say, maybe this will drag it out of you:
struct BoxColor{
unsigned char Red;
unsigned char Green;
unsigned char Blue;
BoxColor() : Red(0),Green(0),Blue(0){}
BoxColor( unsigned char R, unsigned char G,
unsigned char B ) : Red(R), Green(G), Blue(B){}
void Color( unsigned char R, unsigned char G,
unsigned char B ){
Red = R; Green = G; Blue = B;
return;
}
};
// if you cannot change this struct, see BoxColor2
typedef struct{
unsigned char Red;
unsigned char Green;
unsigned char Blue;
}TBoxColour;
struct BoxColor2 : TBoxColour{
BoxColor2(){ Red = 0; Green = 0; Blue = 0; }
BoxColor2( unsigned char R, unsigned char G,
unsigned char B ){
Red = R;
Green = G;
Blue = B;
}
};
int main(){
BoxColor bc1( 10, 20, 30 );
cout<<"bc1 Red="<<int(bc1.Red)<<" Green="
<<int(bc1.Green)<<" Blue="<<int(bc1.Blue)<<std::endl;
BoxColor bc2;
cout<<"bc2 Red="<<int(bc2.Red)<<" Green="
<<int(bc2.Green)<<" Blue="<<int(bc2.Blue)<<std::endl;
bc2.Color( 40, 50, 60 );
cout<<"bc2 Red="<<int(bc2.Red)<<" Green="
<<int(bc2.Green)<<" Blue="<<int(bc2.Blue)<<std::endl;
BoxColor2 BC2( 10, 20, 30 );
TBoxColour TBC;
TBC = BC2;
cout<<"TBC Red="<<int(TBC.Red)<<" Green="
<<int(TBC.Green)<<" Blue="<<int(TBC.Blue)<<std::endl;
TBC = BoxColor2( 30, 40, 50 );
cout<<"TBC Red="<<int(TBC.Red)<<" Green="
<<int(TBC.Green)<<" Blue="<<int(TBC.Blue)<<std::endl;
// ------------------------------------
return 0;
} // main()
That, or use Victor''s example.
--
Bob R
POVrookie
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