从早期的线程中神秘化的那种 [英] Sort of mystified from an earlier thread
问题描述
这取自以下内容：
http://groups.google.com/group/comp....3e9afae83d061c
我引用：
嗯，这也适用于char **，因为字符串文字的类型为
char * in c。但总体思路仍然存在。
令我恼火的是，尽管如此，在没有C的情况下违反const的情况是_trivial_
求助于所有这些。
const char foo [] =" mystring" ;;
char * constviol = strchr（foo，* foo）;
我不能得到的是'const char f [] =" mystring" ''定义
作为char，但原型定义如下：
char * strchr（const char * s，int c）;
当foo被取消引用时（即* foo），为什么编译器不会b / b $ b $抱怨''int'之间的区别和''char''？
提前致谢。
Chad
This was taken from the following:
http://groups.google.com/group/comp....3e9afae83d061c
And I quote:
"Well, that''s also ok for char**, since string literals are of type
char * in c. The general idea still stands, though.
The thing that irritates me is that despite all this, it''s _trivial_
to violate const in C without resorting to all this.
const char foo[] = "mystring";
char *constviol = strchr(foo,*foo); "
What I don''t get is that that ''const char f[]="mystring" '' is defined
as a char, but the prototype is defined as the following:
char *strchr(const char *s, int c);
When foo gets de-referenced (ie *foo), how come the compiler doesn''t
complain about the difference between ''int'' and ''char''?
Thanks in advance.
Chad
推荐答案
2005年10月29日星期六21:51:46 -0400，Chad< cd ***** @ gmail.com>写道：
On Sat, 29 Oct 2005 21:51:46 -0400, Chad <cd*****@gmail.com> wrote:
"嗯，这也适用于char **，因为字符串文字在c中的类型为
char *。然而，总体思路仍然存在。
令我恼火的是，尽管如此，在不诉诸所有这一切的情况下违反const中的const也是_trivial_
。
const char foo [] =" mystring" ;;
char * constviol = strchr（foo，* foo）;
我不能得到的是'const char f [] =" mystring" ''被定义为char，但原型定义如下：
char * strchr（const char * s，int c）;
当foo被取消引用时（即* foo），为什么编译器不会抱怨''int''和''char''之间的区别？
"Well, that''s also ok for char**, since string literals are of type
char * in c. The general idea still stands, though.
The thing that irritates me is that despite all this, it''s _trivial_
to violate const in C without resorting to all this.
const char foo[] = "mystring";
char *constviol = strchr(foo,*foo); "
What I don''t get is that that ''const char f[]="mystring" '' is defined
as a char, but the prototype is defined as the following:
char *strchr(const char *s, int c);
When foo gets de-referenced (ie *foo), how come the compiler doesn''t
complain about the difference between ''int'' and ''char''?
我实际上也在想这个。我知道char是一个
整数类型，但我仍然会认为int和char会带来某种警告或者什么不是。我不确定我的理解
这一切都是怎么回事。
- 北极
-
使用Opera革命性的电子邮件客户端： http://www.opera.com/mail/
Arctic Fidelity写道：
Arctic Fidelity wrote:
周六，29日2005年10月21:51:46 -0400，Chad< cd ***** @ gmail.com>写道：
On Sat, 29 Oct 2005 21:51:46 -0400, Chad <cd*****@gmail.com> wrote:
"嗯，这也适用于char **，因为字符串文字在c中的类型为
char *。然而，总体思路仍然存在。
令我恼火的是，尽管如此，在不诉诸所有这一切的情况下违反const中的const也是_trivial_
。
const char foo [] =" mystring" ;;
char * constviol = strchr（foo，* foo）;
我不能得到的是'const char f [] =" mystring" ''被定义为char，但原型定义如下：
char * strchr（const char * s，int c）;
当foo被取消引用时（即* foo），为什么编译器不会抱怨''int''和''char''之间的区别？
"Well, that''s also ok for char**, since string literals are of type
char * in c. The general idea still stands, though.
The thing that irritates me is that despite all this, it''s _trivial_
to violate const in C without resorting to all this.
const char foo[] = "mystring";
char *constviol = strchr(foo,*foo); "
What I don''t get is that that ''const char f[]="mystring" '' is defined
as a char, but the prototype is defined as the following:
char *strchr(const char *s, int c);
When foo gets de-referenced (ie *foo), how come the compiler doesn''t
complain about the difference between ''int'' and ''char''?
<我实际上也在想这个。我知道char是一个
整数类型，但我仍然会认为int和char会带来某种警告或什么不是。我不知道我是怎么理解
这一切是怎么回事。
- 北极
-
使用Opera'革命性的电子邮件客户端： http://www.opera.com/mail/
除了草率的措辞之外，这是我最好的猜测是什么。
on。
当我们去* foo时，我们从字符串中获取每个字符。
在每次传递的内部，我们将有一个存储''m'的变量，tnen
''y''等等。如果我们做了类似的事情，那将是一样的。
char internal_string =''m'';
然后char将自动转换为整数（在strchr
int c参数上）。这可能解释了即使我启用了强制警告标志，gnu编译器也没有抱怨
抱怨。
Outside of the sloppy wording, here is my best guess on what is going
on.
When we go *foo, we are getting each character from the string.
Internally at each pass, we would have a varibale storing ''m'', tnen
''y'', etc. This would be the same has we done something like
char internal_string = ''m'';
Then char would be automatically converted to integer (on the strchr
int c parameter). This might explain wny the gnu compiler didn''t
complain even when I hard warning flags enabled.
Chad写道：
这取自以下内容：
http://groups.google.com/group/comp....3e9afae83d061c
我引用：
嗯，这也适用于char **，因为字符串文字在c中的类型为
char *。然而，总体思路仍然存在。
令我恼火的是，尽管如此，在不诉诸所有这一切的情况下违反const中的const也是_trivial_
。
const char foo [] =" mystring" ;;
char * constviol = strchr（foo，* foo）;
我不能得到的是'const char f [] =" mystring" ''被定义为char，但原型定义如下：
char * strchr（const char * s，int c）;
当foo被取消引用时（即* foo），为什么编译器不会抱怨''int''和''char''之间的区别？
This was taken from the following:
http://groups.google.com/group/comp....3e9afae83d061c
And I quote:
"Well, that''s also ok for char**, since string literals are of type
char * in c. The general idea still stands, though.
The thing that irritates me is that despite all this, it''s _trivial_
to violate const in C without resorting to all this.
const char foo[] = "mystring";
char *constviol = strchr(foo,*foo); "
What I don''t get is that that ''const char f[]="mystring" '' is defined
as a char, but the prototype is defined as the following:
char *strchr(const char *s, int c);
When foo gets de-referenced (ie *foo), how come the compiler doesn''t
complain about the difference between ''int'' and ''char''?
因为将char转换为int没有问题，
除非（CHAR_MAX> INT_MAX）
看起来似乎没有在任何托管系统中都是如此。
-
pete
Because there''s no problem converting a char to an int,
unless (CHAR_MAX > INT_MAX)
which doesn''t seem to be the case in any hosted sysytems.
--
pete
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