索引#define字符串 [英] Index a #define string
问题描述
我可以这样做吗:
#define MYSTRING" ABC"
..
..
..
char mychar = MYSTRING [0];
..
..
..
-Thanks,
s
Can I do this:
#define MYSTRING "ABC"
..
..
..
char mychar = MYSTRING[0];
..
..
..
-Thanks,
s
推荐答案
s写道:
我可以这样做:
#define MYSTRING" ABC"
。
。
。
char mychar = MYSTRING [0];
。
。
。
-Thanks,
s
Can I do this:
#define MYSTRING "ABC"
.
.
.
char mychar = MYSTRING[0];
.
.
.
-Thanks,
s
我建议使用:
const char * const MYSTRING =" ABC" ;;
AFAIK,你的#define字符串应该可以工作。
- Pete
I would suggest using:
const char* const MYSTRING = "ABC";
AFAIK, your #define string thing should work, though.
- Pete
" s" <哟********* @ home.with.yourself>在消息中写道
news:40 ************** @ home.with.yourself ...
"s" <yo*********@home.with.yourself> wrote in message
news:40**************@home.with.yourself...
我可以这样做:
#define MYSTRING" ABC"
char mychar = MYSTRING [0];
-Thanks,
s
Can I do this:
#define MYSTRING "ABC"
.
char mychar = MYSTRING[0];
-Thanks,
s
你好
用VC尝试它并且工作正常。
另外:char mychar = * MYSTRING工作正常。
-
Elias
Hello
Tried it with VC and it works fine.
Also: char mychar = *MYSTRING works fine.
--
Elias
星期五,2004年4月16日13:51:01 GMT在comp.lang.c ++中,Petec < x@x.x>
写道,
On Fri, 16 Apr 2004 13:51:01 GMT in comp.lang.c++, "Petec" <x@x.x>
wrote,
我建议使用:
const char * const MYSTRING =" ABC";
I would suggest using:
const char* const MYSTRING = "ABC";
为什么无偿指针?
const char mystring [] =" ABC" ;;
Why the gratuitous pointer?
const char mystring[] = "ABC";
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