字符常量 [英] Character Constants
问题描述
由于字符常量是一个int值，用单引号表示为字符
，所以它被视为1字节整数现在看看
以下代码段。
#include< stdio.h>
int main（无效）
{
char a =''Abbc'';
printf（"％c"，a）;
返回0;
}
请解释如何解释初始化，编译时
给出警告多字符字符常量。但运行没有
错误也可能是单引号中的最后一个字符是
存储在一个。
我在TC2.0中编译时收到错误消息
错误：函数main中字符常量太长
Akhil写道：
char a =''Abbc'';
请解释如何解释初始化，
初始化被解释为
编译器想要的任何方式。 />
C标准没有定义它。
-
pete
Akhil写道：
由于字符常量是一个以单引号表示为字符的int值，因此它被视为1字节整数现在看起来在
以下片段。
＃include< stdio.h>
int main（无效）
{/ a> char a =''Abbc'' ;
printf（"％c"，a）;
返回0;
}
请解释在编译时如何解释初始化它给出了一个警告多字符字符常量。但运行没有
错误也可能是单引号中的最后一个字符存储在a。
有关初始化的说明，请参阅你的编译器文档，
标准表示它的实现是定义的。
至于警告，你的编译器很好，让你知道
你可能正在做一些你不想要的事情。我还告诉我
隐式转换已经溢出。
-
BR，弗拉基米尔
我是工人的朋友，我宁愿做他的朋友
而不是一个。
- Clarence Darrow
Since a character constant is an int value represented as a character
in single quotes,so it is treated as a 1 byte integer now look at the
following snippet.
#include<stdio.h>
int main(void)
{
char a=''Abbc'';
printf("%c",a);
return 0;
}
Please explain how the initialization is interpreted,on compilation it
gives a warning "multi-character character constant" but runs without
error also possibly the last character within the single quotes is
being stored in a.
I got an error message when compile it in TC2.0
Error: Character constant too long in function main
Akhil wrote:
char a=''Abbc''; Please explain how the initialization is interpreted,
The initialization is interpreted
any way that the compiler wants to.
The C standard doesn''t define it.
--
pete
Akhil wrote:
Since a character constant is an int value represented as a character
in single quotes,so it is treated as a 1 byte integer now look at the
following snippet.
#include<stdio.h>
int main(void)
{
char a=''Abbc'';
printf("%c",a);
return 0;
}
Please explain how the initialization is interpreted,on compilation it
gives a warning "multi-character character constant" but runs without
error also possibly the last character within the single quotes is
being stored in a.
For explanation of initialisation, refer to your compiler documentation,
as Standard says it''s implementation defined.
As for the warning, your compiler is just being nice, letting you know
that you may be doing something you don''t want. Mine also tells me that
implicit conversion has overflown.
--
BR, Vladimir
I am a friend of the working man, and I would rather be his friend
than be one.
-- Clarence Darrow
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