一些源代码需要帮助 [英] Help wanted on some source codes
问题描述
1。=========================================== <无线电通信/>
/ * ac * /
int x;
int y;
void main（）< br $>
{
f（）;
printf（"％x％x \ n"，x，y）;
}
/ * bc * /
double x;
void f（）
{
x = -0.0;
}
2. ============ ===============================
/ * ac * /
int x = 1;
int y;
void main（）
{
f（）;
printf（"％x％x \ n"，x，y）;
}
/ * bc * /
双倍x;
void f（）
{
x = -0.0;
}
3。======================= ====================
/ * ac * /
int x;
int y = 1;
void main（）
{
f（）;
printf（"％x％x \ n"，x，y）;
}
/ * bc * /
double x;
void f（）
{
x = -0.0;
}
4。=================================== ========
/ * ac * /
int x = 1;
int y = 1;
void main（）
{
f（）;
printf（"％x％x \\ \\ n"，x，y）;
}
/ * bc * /
double x;
void f（）
{
x = -0.0;
}
5. ===========================================
/ * ac * /
int x = 1;
int y = 1;
void main（）
{
f（）;
printf（"％x％x \ n"，x，y）;
}
/ * bc * /
double x;
无效f（）
{
x = -0;
}
备注======== ===============================
上述程序从1到5都构成
两个单独的文件，ac& bc，可以在gcc下编译
。
例如
/ * Linux * /
$ gcc - o测试ac bc
$ ./test
/ * Windows * /
C：\ gcc -o test.exe ac bc
C：\ test
问题============================= ==========
我想知道程序产生的奇怪结果
。请帮我。我想了解
为什么结果是这样的。
顺便说一句，上面的源代码是旧的K& R风格
并且将由gcc警告，你可以将它们更改为
新的ISO / ANSI样式:-)
在此先感谢&最好的问候！
Becker
30-Nov-2005
1.===========================================
/* a.c */
int x;
int y;
void main()
{
f();
printf("%x %x\n", x, y);
}
/* b.c */
double x;
void f()
{
x = -0.0;
}
2.===========================================
/* a.c */
int x = 1;
int y;
void main()
{
f();
printf("%x %x\n", x, y);
}
/* b.c */
double x;
void f()
{
x = -0.0;
}
3.===========================================
/* a.c */
int x;
int y = 1;
void main()
{
f();
printf("%x %x\n", x, y);
}
/* b.c */
double x;
void f()
{
x = -0.0;
}
4.===========================================
/* a.c */
int x = 1;
int y = 1;
void main()
{
f();
printf("%x %x\n", x, y);
}
/* b.c */
double x;
void f()
{
x = -0.0;
}
5.===========================================
/* a.c */
int x = 1;
int y = 1;
void main()
{
f();
printf("%x %x\n", x, y);
}
/* b.c */
double x;
void f()
{
x = -0;
}
Remark=======================================
The above programs from 1 to 5 are all constituted
of two separate files, a.c & b.c, and can be
compiled under gcc.
e.g.
/* Linux */
$ gcc -o test a.c b.c
$ ./test
/* Windows */
C:\gcc -o test.exe a.c b.c
C:\test
Question=======================================
I''m wondering about the strange results generated
by the programs. Please help me. I want to understand
why the results are like that.
By the way, the source codes above are in old K&R style
and will be warned by the gcc, you can change them to
new ISO/ANSI style :-)
Thanks in advance & best regards!
Becker
30-Nov-2005
推荐答案
gcc -o test ac bc
gcc -o test a.c b.c
./ test
/ * Windows * /
C ：\ gcc -o test.exe ac bc
C：\ test
问题================= ======================
我想知道产生的奇怪结果
by the程式。请帮我。我想了解
为什么结果是这样的。
顺便说一句，上面的源代码是旧的K& R风格
并且将由gcc警告，你可以将它们更改为
新的ISO / ANSI样式:-)
在此先感谢&最好的问候！
Becker
30-Nov-2005
./test
/* Windows */
C:\gcc -o test.exe a.c b.c
C:\test
Question=======================================
I''m wondering about the strange results generated
by the programs. Please help me. I want to understand
why the results are like that.
By the way, the source codes above are in old K&R style
and will be warned by the gcc, you can change them to
new ISO/ANSI style :-)
Thanks in advance & best regards!
Becker
30-Nov-2005
文章< 11 ********************** @ z14g2000cwz.googlegroups .com> ;,
Becker< we ***** **@devhr.com>写道：
In article <11**********************@z14g2000cwz.googlegroups .com>,
Becker <we*******@devhr.com> wrote:
1. ====================================== =====
/ * ac * /
int x;
int y;
void main（）
{
f （）;
printf（"％x％x \ n"，x，y）;
}
％x需要一个unsigned int，而不是一个int。
/ * bc * /
双x;
void f（）
{
x = -0.0;
}
你永远不会初始化y，并且你将x初始化为错误的
类型。 double可能与int的大小相同，也可能不同;它不是很长，但这并不确定。如果它更长
那么你 - 最终可能会覆盖y，但x和
$ b $的相对顺序不是由标准确定的，而double可能大于两个
一起注入所以你可能会在随机的内存部分涂鸦。
如果您碰巧使用IEEE754浮点数，则-0.0是
被认为不同于0.0。即使在不是的系统上，
浮点表示为0并不一定都是二进制的
0'（尽管 - 浮点数标准要求所有二进制0'的
必须被视为浮点0.）
您的程序2到4只是完全相同主题的变体
各种不同的值初始化内存可能或者可能不会被潦草地写下来。
5。==== =======================================
void f（）
{
x = -0;
}
1.===========================================
/* a.c */
int x;
int y;
void main()
{
f();
printf("%x %x\n", x, y);
}
%x expects an unsigned int, not an int.
/* b.c */
double x;
void f()
{
x = -0.0;
}
You never initialize y at all, and you initialize x to the wrong
type. double might or might not be the same size as int; it is not
uncommon for it to be longer, but that is not certain. If it is longer
then you -might- end up overwritting y, but the relative order of x and
y are not fixed by the standard, and double might be larger than two
ints together so you might be scribbling on a random portion of memory.
If you happen to be using IEEE754 floating point, then -0.0 is
considered to be different than 0.0 . Even on systems that it is not,
the floating point representation of 0 is not certain to be all binary
0''s (though -some- floating point standards require that all-binary 0''s
must be treated as floating point 0.)
Your programs 2 thru 4 are just variations on exactly the same themes
with various different values initializing the memory that might or might
not be scribbled over.
5.=========================================== void f()
{
x = -0;
}
-0是一个整数常量，它与除了
相当罕见的有符号幅度的机器（
标准允许这样做。） - 0作为整数因此与0作为整数相同。 />
然后将该整数强制转换为您声明为x的双精度值
在此文件中。在一些不同于浮动的系统上
指向-0.0。
-
重要的是要记住，当它到来时对于法律，电脑
从不复制，只有人类复制。计算机给出了
命令，而非许可。只有人才能获得许可。
- Brad Templeton
-0 is an integer constant, and it is the same as 0 on all but the
rather-uncommon signed-magnitude machines (which are allowed for by
the standard.) -0 as an integer is thus the same as 0 as an integer.
That integer is then cast to the double that you have declared x to
be in this file. On some systems that is different than the floating
point -0.0 .
--
"It is important to remember that when it comes to law, computers
never make copies, only human beings make copies. Computers are given
commands, not permission. Only people can be given permission."
-- Brad Templeton
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