编译器不允许将Char *的维度数组签名到另一个??? [英] compiler won't allow dimensioned array of Char* to be signed to another???
问题描述
我的编译器（m68k-gcc）不允许这样（简化示例;不是我的
真实代码），我不明白为什么不：
{
char * arrayCharPtr1 [] = {" a"，" ab"，NULL};
char * arrayCharPtr2 [] = {" A"，" AB"，NULL};
char * arrayCharPtrVar [] =
（（someBoolean）？arrayCharPtr1：arrayCharPtr2）; //< -error here！
StrCopy（someString，arrayCharPtr [0]）;
}
我也有尝试过=& arrayCharPtr1 [0]我认为应该工作，
;它没有。
我缺少什么？如何做到这一点？
My compiler (m68k-gcc) does not allow this (simplified example; not my
real code) and I don''t see why not:
{
char *arrayCharPtr1[] = {"a", "ab", NULL};
char *arrayCharPtr2[] = {"A", "AB", NULL};
char *arrayCharPtrVar[] =
((someBoolean) ? arrayCharPtr1 : arrayCharPtr2); // <-error here!
StrCopy(someString, arrayCharPtr[0]);
}
I have also tried "= &arrayCharPtr1[0]" which I assumed should work,
too; it doesn''t.
What am I missing? How can this be done?
推荐答案
Gregg Woodcock写道：
Gregg Woodcock wrote:
我的编译器（m68k-gcc）不允许这样做（简化示例;不是我的
真实代码），我不明白为什么不：
{* char * arrayCharPtr1 [] = {" a"，" ab"，NULL };
char * arrayCharPtr2 [] = {" A"，" AB"，NULL};
char * arrayCharPtrVar [] =
（（someBoolean）？arrayCharPtr1：arrayCharPtr2）; //< -error here！
StrCopy（someString，arrayCharPtr [0]）;
}
我还试过=& arrayCharPtr1 [0]我认为应该工作，
;它不是。
声明一个未知大小的数组只接受一个聚合
初始值设定项（即''{}'' - 封闭的初始化程序）。你提供的那个不是。
我缺少什么？怎么办呢？
My compiler (m68k-gcc) does not allow this (simplified example; not my
real code) and I don''t see why not:
{
char *arrayCharPtr1[] = {"a", "ab", NULL};
char *arrayCharPtr2[] = {"A", "AB", NULL};
char *arrayCharPtrVar[] =
((someBoolean) ? arrayCharPtr1 : arrayCharPtr2); // <-error here!
StrCopy(someString, arrayCharPtr[0]);
}
I have also tried "= &arrayCharPtr1[0]" which I assumed should work,
too; it doesn''t.
Declaration of an array of unknown size will only accept an aggregate
initializer (i.e. ''{}''-enclosed initializer). The one you provided is not.
What am I missing? How can this be done?
你究竟想做什么？
-
祝你好运，
Andrey Tarasevich
What exactly are you trying to do?
--
Best regards,
Andrey Tarasevich
" Gregg Woodcock" < WO ****** @ SonLightSoftware.com>在消息中写道
news：86 ************************** @ posting.google.c om ...
"Gregg Woodcock" <wo******@SonLightSoftware.com> wrote in message
news:86**************************@posting.google.c om...
我的编译器（m68k-gcc）不允许这个（简化示例;不是我的
真实代码），我不明白为什么不：
您可以在此代码中解释您的意图，我会尝试猜测
现在...
{
char * arrayCharPtr1 [ ] = {" a"，" ab"，NULL};
char * arrayCharPtr2 [] = {" A"，" AB"，NULL};
你应该在这里使用const char *，因为这些数组持有指向常量
字符串文字的指针。
char * arrayCharPtrVar [] =（/ someBoolean）？arrayCharPtr1：arrayCharPtr2）; //< -error在这里！
试试这个：char ** arrayCharPtrVar = someBoolean？ arrayCharPtr1：arrayCharPtr2;
如果按照我的建议使用const char ** ...
StrCopy（someString，arrayCharPtr [0]）;
StrCopy不是strcpy，它到底能做什么strcpy不能？
}
我也有尝试过=& arrayCharPtr1 [0]我认为应该工作，
;它不是。
问题在于类型本身：你应该声明一个指针，而不是一个
agregate。
我错过了什么？怎么办呢？
My compiler (m68k-gcc) does not allow this (simplified example; not my
real code) and I don''t see why not:
You could have explained what you intended in this code, I will try and guess
for now...
{
char *arrayCharPtr1[] = {"a", "ab", NULL};
char *arrayCharPtr2[] = {"A", "AB", NULL};
You should use const char * here because these arrays hold pointers to constant
string literals.
char *arrayCharPtrVar[] =
((someBoolean) ? arrayCharPtr1 : arrayCharPtr2); // <-error here!
Try this : char **arrayCharPtrVar = someBoolean ? arrayCharPtr1 : arrayCharPtr2;
Use const char **... if you followed my advice above.
StrCopy(someString, arrayCharPtr[0]);
StrCopy is not strcpy, what in hell can it do that strcpy can''t ?
}
I have also tried "= &arrayCharPtr1[0]" which I assumed should work,
too; it doesn''t.
The problem was with the type itself : you should declare a pointer, not an
agregate.
What am I missing? How can this be done?
就像我说的那样。
-
Chqrlie。
PS：你的命名约定也是幼稚的。 CharPtr没用，Var是变量名无意义的。
使用像lower_strings []和upper_strings []这样的名字......
Like I say.
--
Chqrlie.
PS: also your naming conventions are childish. CharPtr is useless, Var is
meaningless in variable names.
Use names like lower_strings[] and upper_strings[]...
Andrey Tarasevich< an ************** @ hotmail.com>在消息新闻中写道：< 10 ************* @ news.supernews.com> ...
Andrey Tarasevich <an**************@hotmail.com> wrote in message news:<10*************@news.supernews.com>...
Gregg Woodcock写道：
Gregg Woodcock wrote:
我的编译器（m68k-gcc）不允许这样（简化示例;不是我的
真实代码），我不明白为什么不：
{char /> char * arrayCharPtr1 [] = {" a"，" ab"，NULL};
char * arrayCharPtr2 [] = {" A"，" AB"，NULL};
char * arrayCharPtrVar [] =
（（someBoolean）？arrayCharPtr1：arrayCharPtr2）; //< -error here！
StrCopy（someString，arrayCharPtr [0]）;
}
我还试过=& arrayCharPtr1 [0]我认为应该工作，
;它不会。
My compiler (m68k-gcc) does not allow this (simplified example; not my
real code) and I don''t see why not:
{
char *arrayCharPtr1[] = {"a", "ab", NULL};
char *arrayCharPtr2[] = {"A", "AB", NULL};
char *arrayCharPtrVar[] =
((someBoolean) ? arrayCharPtr1 : arrayCharPtr2); // <-error here!
StrCopy(someString, arrayCharPtr[0]);
}
I have also tried "= &arrayCharPtr1[0]" which I assumed should work,
too; it doesn''t.
声明一个未知大小的数组只接受一个聚合的初始化器（即''{}'' - 封闭的初始化器）。你提供的那个不是。
Declaration of an array of unknown size will only accept an aggregate
initializer (i.e. ''{}''-enclosed initializer). The one you provided is not.
我错过了什么？怎么办呢？
What am I missing? How can this be done?
你究竟想做什么？
What exactly are you trying to do?
你做了一个相当不错的编译器模仿：非常准确，而且
完全没用。
我*知道*我不能做我想做的事情我想做的事
（编译器非常清楚）;这就是为什么我给出了
完整的上下文，显示了我想做的事情。我几乎不能更多
清楚。
我必须使用字符串数组。
我想分配一个或另一个数组到一个变量，这样我就可以用数组var [index]来索引数组中的任何字符串。
这很简单，我肯定不是不可能的。 ..
You do a pretty good compiler immitation: perfectly accurate and
totally worthless.
I *KNOW* I cannot do what I want to do the way I am trying to do it
(the compiler is quite clear about that); That''s why I gave the
complete context showing WHAT I want to do. I can hardly be more
clear.
I have to arrays of strings.
I want to assign one or the other array to a variable such that I can
index any string in the array with something like var[index].
It is very simple and I am quite sure not impossible...
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