代码的行为 [英] Behavior of the code
问题描述
大家好，
我正在练习一个C教程。
因为他们给了一个小代码并且询问了输出。
#include< stdio.h>
int main（无效）
{
int x = 0xFFFFFFF0;
signed char y;
y = x;
printf（"％x \ n"， y）;
返回0;
}
程序的输出是
fffffff0
他们给出的解释如下：
由于％x用于打印，printf语句将推广
y到整数，因此将打印fffffff0。
我对这个解释有疑问。
我的理解是
1）如果我们指定一个超过其最大尺寸的签名变量，它将溢出所以行为是未定义的。
2）在说明中它说％x将y提升为整数，但是我认为它是骗局的将其参数转换为十六进制。
我的理解错了吗？
请帮助。
问候，
Somenath
Hi All,
I was going through one of the exercise of one C tutorial .
In that they have given one small code and asked about the output.
#include <stdio.h>
int main(void)
{
int x = 0xFFFFFFF0;
signed char y;
y = x;
printf("%x\n", y);
return 0;
}
Output of the program is
fffffff0
The explanation they have given as bellow
Since %x is being used for printing, the printf statement will promote
y to an integer and hence fffffff0 will be printed.
I have doubt about the explanation.
My understanding is
1) if we assign a signed variable more than its max size it will
overflow so the behavior is undefined .
2) And in the explanation it says %x promote y to an integer but I
think it converts its argument to hexadecimal.
Is my understanding wrong?
Please help.
Regards,
Somenath
推荐答案
12月24日晚上8:57，somenath< somenath ... @ gmail.comwrote：
On Dec 24, 8:57 pm, somenath <somenath...@gmail.comwrote:
我正在练习一个C教程。
因为他们给了一个小代码并询问了输出。
#include< stdio.h>
int main（无效）
{
int x = 0xFFFFFFF0;
signed char y;
y = x;
printf（"％x \ n"，y）;
返回0;}
程序输出是
fffffff0
他们给出的解释如下：
由于％x用于打印，printf语句会将
$ b $提升为整数，因此将打印fffffff0 />
我对这个解释有疑问。
I was going through one of the exercise of one C tutorial .
In that they have given one small code and asked about the output.
#include <stdio.h>
int main(void)
{
int x = 0xFFFFFFF0;
signed char y;
y = x;
printf("%x\n", y);
return 0;}
Output of the program is
fffffff0
The explanation they have given as bellow
Since %x is being used for printing, the printf statement will promote
y to an integer and hence fffffff0 will be printed.
I have doubt about the explanation.
好。解释是错误的。
Good. The explanation is wrong.
我的理解是
1）如果我们分配的签名变量大于其最大大小它将
溢出，因此行为未定义。
My understanding is
1) if we assign a signed variable more than its max size it will
overflow so the behavior is undefined .
这并不是那么糟糕。如果0xfffffff0被解释为int
并且ints有一个32位的二进制补码表示，那么x是-16，
这是在signed char的范围内。对于其他实现
选项，例如64位整数，可能存在未定义的行为
正是你建议的原因。
这里有一些实现依赖：显然大小为
int，
但是（如果我正确地解释标准），是否
0xfffffff0
被解释为int或unsigned int常量。如果它是
解释
为unsigned int那么我认为你确实有未定义的行为
It''s not quite as bad as that. If 0xfffffff0 is interpreted as an int
and ints have a 32-bit two''s complement representation then x is -16,
which is within the range of signed char. For other implementation
choices, such as 64-bit ints, there may be undefined behaviour for
exactly the reason you suggest.
There is some implementation-dependence here: obviously the size of
int,
but also (if I am interpreting the standard correctly), whether
0xfffffff0
is interpreted as an int or unsigned int constant. If it is
interpreted
as unsigned int then I think you do have undefined behaviour
2）在解释中它说％x将y提升为整数，但是我认为它将其参数转换为十六进制。
2) And in the explanation it says %x promote y to an integer but I
think it converts its argument to hexadecimal.
是的。你是对的，解释是错误的。我们可以查看一个
简化版本的程序，它没有从int转换为令人困惑的
。
#include< stdio.h>
int main（无效）
{
signed char y = -16;
printf （％x \ n，y）;
返回0;
}
实际输出是实现指定的，但是在我的电脑上
这个
还打印
fffffff0
在这个程序中，因为y是可变长度参数的一部分
printf（）列表它将被默认参数
促销提升为int [1]。 printf（）接收int值为-16，
％x格式以十六进制格式请求。 ％x格式对于促销y到int没有任何结果。
-thomas
[1] I ''我不确定签名字符是否有可能具有不适合int的值
。我确定名单上有人。
Yes. You are right, the explanation is wrong. We can look at a
simplified version of the program that doesn''t have the confusing
conversion from int.
#include <stdio.h>
int main(void)
{
signed char y = -16;
printf("%x\n", y);
return 0;
}
The actual output is implementation-specified, but on my computer
this
also prints
fffffff0
In this program, since y is part of the variable-length argument
list of printf() it will be promoted[1] by the default argument
promotions to int. printf() receives an int value of -16, and the
%x format asks for this in hexadecimal. The %x format has nothing to
do with the promotion of y to int.
-thomas
[1] I''m not sure whether it is possible for signed char to have values
that don''t fit in an int. I''m sure that someone on the list is.
Thomas Lumley写道：
Thomas Lumley wrote:
12月24日，8：57 pm，somenath< somenath ... @ gmail.comwrote：
On Dec 24, 8:57 pm, somenath <somenath...@gmail.comwrote:
>我正在进行一个C教程的练习。
因为他们给出了一个小代码并询问了输出。
#include< stdio.h>
int main（无效）
{
int x = 0xFFFFFFF0;
签名char y;
y = x;
printf（"％x \ n"，y）;
返回0;}
该程序的输出是
fffffff0
他们给出的解释如下：
因为％x被用于打印时，printf语句会将y提升为整数，因此会打印出fffffff0。
我对这个解释有疑问。
>I was going through one of the exercise of one C tutorial .
In that they have given one small code and asked about the output.
#include <stdio.h>
int main(void)
{
int x = 0xFFFFFFF0;
signed char y;
y = x;
printf("%x\n", y);
return 0;}
Output of the program is
fffffff0
The explanation they have given as bellow
Since %x is being used for printing, the printf statement will
promote y to an integer and hence fffffff0 will be printed.
I have doubt about the explanation.
好。解释是错误的。
Good. The explanation is wrong.
>我的理解是
1）如果我们分配的签名变量大于其最大大小它
将溢出，因此行为未定义。
>My understanding is
1) if we assign a signed variable more than its max size it
will overflow so the behavior is undefined .
这并不是那么糟糕。如果0xfffffff0被解释为int
并且int有32位二进制补码表示[...]
It''s not quite as bad as that. If 0xfffffff0 is interpreted as an int
and ints have a 32-bit two''s complement representation [ ... ]
C标准明确地为这种限制辩解。 int既不需要
为32位也不以二进制形式表示。它需要
保持在-32767到32767范围内的值。它可以是二进制补码，
补码或符号和大小等等。它没有明确说明，但我们可能假定int至少有16个值位，
因为C需要二进制表示。
The C standard explicitly excuses such restrictions. int neither needs
to be 32 bits nor be represented in twos-complement form. It needs to
hold values in the range -32767 to 32767. It could be twos-complement,
ones-complement or sign-and-magnitude or something else. It is not
explicitly stated but we may presume at least 16 value bits for an int,
as C needs binary representation.
12月25日，10：57 * am，santosh< santosh .... @ gmail.comwrote：
On Dec 25, 10:57*am, santosh <santosh....@gmail.comwrote:
Thomas Lumley写道：
Thomas Lumley wrote:
12月24日晚上8:57，somenath< somenath ... @ gmail.comwrote：
On Dec 24, 8:57 pm, somenath <somenath...@gmail.comwrote:
我正在进行一个C教程的练习。
因为他们给了一个小代码并询问了输出。
I was going through one of the exercise of one C tutorial .
In that they have given one small code and asked about the output.
#include< stdio.h>
int main （无效）
{
* * * * int x = 0xFFFFFFF0;
* * * * signed char y;
* * * * y = x;
* * * * printf（"％x \ n"，y）;
* * * * return 0;}
#include <stdio.h>
int main(void)
{
* * * * int x = 0xFFFFFFF0;
* * * * signed char y;
* * * * y = x;
* * * * printf("%x\n", y);
* * * * return 0;}
*程序的输出是
Output of the *program is
fffffff0
fffffff0
他们给出的解释如下：
The explanation they have given as bellow
由于％x正在用于打印，printf语句将
将y提升为整数，因此将打印fffffff0 。
Since %x is being used for printing, the printf statement will
promote y to an integer and hence fffffff0 will be printed.
我对这个解释有疑问。
I have doubt about the explanation.
好。 *解释错了。
Good. * The explanation is wrong.
我的理解是
1）* * *如果我们分配一个签名变量超过它的最大尺寸*它
将溢出，因此行为未定义。
My understanding is
1) * * *if we assign a signed variable more than its max size *it
will overflow so the behavior is undefined .
这并不是那么糟糕。 *如果0xfffffff0被解释为int
且ints有32位二进制补码表示[...]
It''s not quite as bad as that. *If 0xfffffff0 is interpreted as an int
and ints have a 32-bit two''s complement representation [ ... ]
C标准明确地为这些限制辩解。 int既不需要
为32位也不以二进制形式表示。它需要
保持在-32767到32767范围内的值。它可以是二进制补码，
补码或符号和大小等等。这不是明确说明的b $ b，但是我们可能假设一个int至少有16个值位，
The C standard explicitly excuses such restrictions. int neither needs
to be 32 bits nor be represented in twos-complement form. It needs to
hold values in the range -32767 to 32767. It could be twos-complement,
ones-complement or sign-and-magnitude or something else. It is not
explicitly stated but we may presume at least 16 value bits for an int,
更清楚一点。 />
因此，如果我将128分配给已签名的字符，则无需担保它将
转换为-128
例如
签名字符x = 0x80;
x并不总是包含-128？
只有2'的补码表示才有可能吗？
To get little bit clearer .
So if i assigned 128 to signed char it is not guranted that it will
be converted to -128
For example
signed char x = 0x80;
x will not always contains -128 ?
Is it only possible in case of 2''s complement representation ?
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