是否定义了行为 [英] Is the behaviour defined
问题描述
我的同事认为
后续计划的行为是定义的，但我有点担心。
#include< stdio.h>
#include< string.h>
int main（）
{
char * c;
c =& c;
strcpy（c，" abc"）;
put（& c）;
retun 0;
}
程序打印相同值abc在多个平台上，我甚至用b / b
尝试使用多个编译器。我可以看出它试图写入指向地址的
，所以可能它写的最大值是3个字节+
''\'''。但即使我尝试复制更多的4个字节，它也会打印整个
字符串而不会发生任何崩溃。
更改了行>> strcpy（c，" abcdefg"）;
现在我知道这个行为是未定义的（写入超过4个字节）
作为指针的大小是4个字节（在我测试的机器上）。
任何人都可以发表评论这是否符合规范并且行为是保证。
保证。
谢谢
~
Hi,
A collegue of mine is of the opinion that the behaviour of the
following program is defined,but I am a little apprehensive.
#include<stdio.h>
#include<string.h>
int main()
{
char *c;
c = &c;
strcpy(c,"abc");
puts(&c);
retun 0;
}
The program prints the same value "abc" on multiple platforms and I even
tried it with multiple compilers.I can make out that its trying to write
to the pointer address and so probably the max it can write is 3 bytes +
''\0''.But even if I try to copy more that 4 bytes it prints the whole
string without any crashes.
Changed line >> strcpy(c,"abcdefg");
Now I know that this behaviour is undefined(writing more than 4 bytes)
as the sizeof the pointer is 4 bytes(on the machine I tested on).
Can anyone comment if this is compliant code and is the behaviour
guaranteed.
Thanks
~
推荐答案
> int main（）
> int main()
{
char * c;
c =& c;
strcpy（c，" abc"）;
puts （& c）;
重新调整0;
}
{
char *c;
c = &c;
strcpy(c,"abc");
puts(&c);
retun 0;
}
应该返回0;
返回的咒语是在上面的测试程序中有错误。我没有在我编译的测试程序中有b $ b，但是在编写这封邮件的时候添加了它，担心会受到重创用C语言
纯粹主义者;）。
It should be return 0;
The spell of the return is wrong in the test program above.I did not
have it in my test program which I compiled , but added it while
composing this mail , of the fear of getting battered by the C language
purists ;).
grid在01/10/05写道：
grid wrote on 01/10/05 :
一位同事我认为以下
程序的行为是定义的，
至少，它是依赖于实现的。但是将指针中任何值放入
的结果可能会有一个未定义的结果（陷阱
表示）。有可能存在不确定的行为。
但我有点担心。
你可以！
#include< stdio.h>
#include< string.h>
int main（）
{*> char * c;
c =& c;
类型不兼容。 （c是char *，& c是char **）
strcpy（c，" abc"）;
''sizeof char * == sizeof" abc"''表达式的结果是
依赖于实现。
你需要一个4字节的空间用于abc。
例如，在x86的实模式（16位）模型中，指针是
16位宽，这意味着这台机器上有2个字节。对于abc来说太小了
放（& c）;
重新调整0;
}
A collegue of mine is of the opinion that the behaviour of the following
program is defined,
At minimum, it''s implementation-dependent. But the result of putting
any value in a pointer may have an undefined result (trap
representation). There is a possibility of undefined behaviour.
but I am a little apprehensive.
You can!
#include<stdio.h>
#include<string.h>
int main()
{
char *c;
c = &c;
Types are incompatible. (c is a char * and &c is a char**)
strcpy(c,"abc");
The result of the ''sizeof char* == sizeof "abc"'' expression is
implementation-dependent.
You need a space of 4 bytes for "abc".
For example, on a x86 in real mode (16-bit) model small, pointers are
16-bit wide, which means 2 bytes on this machine. Too small for "abc"
puts(&c);
retun 0;
}
-
Emmanuel
C-FAQ： http://www.eskimo.com/~scs/C-faq/faq.html
C库： http://www.dinkumware.com/refxc.html
..sig正在修理
--
Emmanuel
The C-FAQ: http://www.eskimo.com/~scs/C-faq/faq.html
The C-library: http://www.dinkumware.com/refxc.html
..sig under repair
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新闻：QK ************ @ news.oracle.com ...
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news:QK************@news.oracle.com...
我的同事属于认为
以下程序的行为已被定义，但我有点担心。
＃include< stdio.h>
#include< string.h>
int main（）
{*> char * c;
c =& c;
strcpy（c，" abc"）;
puts（& c）;
retun 0;
}
A collegue of mine is of the opinion that the behaviour of the
following program is defined,but I am a little apprehensive.
#include<stdio.h>
#include<string.h>
int main()
{
char *c;
c = &c;
strcpy(c,"abc");
puts(&c);
retun 0;
}
以上代码已损坏。你试图将4个字符/字节
（1 + strlen（" abc"）== 4）存储到变量c中，但你不知道是否c
（指向char的指针）足以容纳4个字符/字节。在某些平台上
它可能是2甚至1，因此你有可能覆盖别的东西。
在建议代码之前先告诉你的同事更好地学习C比如
以上。
Alex
The above code is broken. You attempt to store 4 chars/bytes
(1+strlen("abc")==4) to the variable c, but you don''t know whether c
(pointer to a char) is big enough to hold 4 chars/bytes. On some platforms
it can be 2 or even 1, hence you''re risking to overwrite something else.
Tell your collegue first to learn C better before advising a code such as
the above.
Alex
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