程序输出？ [英] program output?
问题描述
我对C进行了测试。
有一个关于程序输出类型的客观问题。
以下是该计划：
main（）
{
char ch;
int i = 2;
for（ch = 0; ch <128; ch ++）
i + = 2;
printf（"％d" ，i）;
}
i写的答案256将打印出来。
但我们的导师告诉我它将是无限循环。
是吗，我不确定答案。
问候，
Bhavik
Hi,
I took a test on C.
there was an objective question for program output type.
following is the program:
main()
{
char ch;
int i =2;
for (ch=0;ch<128;ch++)
i+=2;
printf("%d",i);
}
i wrote the answer 256 will be printed.
But our instructor told me that it will be infinite loop.
Is that right, I am not sure of the answer.
Regards,
Bhavik
推荐答案
2004年5月24日04:20:00 -0700， bh ********** @ yahoo.com （bnp）写道：
On 24 May 2004 04:20:00 -0700, bh**********@yahoo.com (bnp) wrote:
我对C进行了测试。
对程序输出类型有一个客观的问题。
以下是程序：
main（）
{
char ch;
int i = 2;
for（ch = 0; ch <128; ch ++）
i + = 2;
打印f（％d，i）;
}
我写的答案256将被打印。
但是我们的导师告诉我它将是无限循环。
是那是对的，我不确定答案。
有点棘手。
首先，默认的签名是char类型是
平台依赖。如果它是8位和已签名，这是典型的基于PC的平台上的默认值，那么这确实会进入无限循环。
例如，8位字符的值范围是-128到+127。它b / b
总是小于+128。
如果chars的默认值是无符号的，那么程序输出是258 [I
没想出来，我只是跑了;-)]
-leor
问候，
Bhavik
Hi,
I took a test on C.
there was an objective question for program output type.
following is the program:
main()
{
char ch;
int i =2;
for (ch=0;ch<128;ch++)
i+=2;
printf("%d",i);
}
i wrote the answer 256 will be printed.
But our instructor told me that it will be infinite loop.
Is that right, I am not sure of the answer.
Kinda tricky.
First of all, the default "signedness" of the char type is
platform-dependent. If it is 8 bits and "signed", as is the default on
typical PC-based platforms, then this does indeed go into an infinite loop.
In that case, the range of values for an 8-bit char is -128 to +127. It
will always be less than +128.
If the default for chars is unsigned, then the program output is 258 [I
didn''t figure it out, I just ran it ;-) ]
-leor
Regards,
Bhavik
-
Leor Zolman --- BD软件--- www.bdsoft.com
C / C ++，Java，Perl和Unix的现场培训
C ++用户：下载BD Software的免费STL错误消息解密器：
www.bdsoft.com/tools/stlfilt.html
2004年5月24日星期一04:20:00 -0700，bnp写道：
On Mon, 24 May 2004 04:20:00 -0700, bnp wrote:
我对C进行了测试。
有一个关于程序输出类型的客观问题。
以下是程序：
main（）
int main（）或int main（void）是首选。
{
正确的缩进是一件好事（tm），即使对于诸如
这样的微不足道的例子也是如此。
char ch;
int i = 2;
for（ch = 0; ch< 128; ch ++）
i + = 2;
printf（" ;％d"，i）;
}
我写的答案256将被打印。
但我们的导师告诉我它将是无限循环。
是吗，是的，我不确定答案。
Hi,
I took a test on C.
there was an objective question for program output type.
following is the program:
main()
int main () or int main (void) is preferred.
{
Proper indentation is a Good Thing(tm), even for trivial examples such as
this.
char ch;
int i =2;
for (ch=0;ch<128;ch++)
i+=2;
printf("%d",i);
}
i wrote the answer 256 will be printed.
But our instructor told me that it will be infinite loop.
Is that right, I am not sure of the answer.
你的导师可能是对的。这一切都取决于CHAR_MAX，如果char签名，可能是
为127。如果是这种情况，ch将永远不会是> = 128，
并且循环永远不会结束。如果CHAR_MAX大于127（因为它是
无符号，或者因为char有超过8个值位），那么它可以打印
258（并且将用于确定你是否打印换行符或刷新标准符号。
-
int main（void）{int putchar（int），i = 0; unsigned long t = 500555079，n [] = {t
，159418370,88921539,286883974,80500161,0};而（n [i]）putchar（*（！（t& 1） >
）+！（t ||！++ i）+"＃\ n"）），（t&& 1+（t>> = i-〜-i））|| （t = n [i] ^ n [i-1]）;返回0;}
Your instructor may have been right. It all depends on CHAR_MAX, which may
be 127 if char is signed. If this is the case, ch will never be >= 128,
and the loop will never end. If CHAR_MAX is greater than 127 (because it''s
unsigned, or because char has more than 8 value bits), then it may print
258 (and will for sure if you print a newline or flush stdout).
--
int main(void){int putchar(int),i=0;unsigned long t=500555079,n[]={t
,159418370,88921539,286883974,80500161,0};while(n[i])putchar(*(!(t&1
)+!(t||!++i)+"# \n")),(t&&1+(t>>=i-~-i))||(t=n[i]^n[i-1]);return 0;}
星期一， 2004年5月24日，bnp写道：
On Mon, 24 May 2004, bnp wrote:
我对C进行了测试。
有一个关于程序输出类型的客观问题。
以下是程序：
好的风格要求包含< stdio.h> （见下面有关
''printf''的错误。）
main（）
_int_ main（）需要C99 ;
C89和C99中_int_ main（_void_）首选。 （其中_this_表示强调，而不是文字下划线。）
{
char ch;
缺少缩进是一件坏事，应该避免，尤其是在Usenet和学校测试中使用
。此外，其他地方。
int i = 2;
for（ch = 0; ch< 128; ch ++）
非严格 - 一致性：程序行为取决于
普通字符的签名。
可能的未定义行为：如果char已签名且有8位，则
有符号整数溢出将发生。
i + = 2;
printf（"％d"，i）;
可能的链接器错误：''printf''未定义。
未定义的行为：variadic函数''printf''调用没有
范围内的原型。
约束违规（这是正确的吗？）：程序输出结束
而不终止换行''\ n''。
}
我写了答案256将被打印。
但我们的导师告诉我它将是无限循环。
是吗，我不确定答案。
I took a test on C.
there was an objective question for program output type.
following is the program:
Good style dictates the inclusion of <stdio.h> (see errors regarding
''printf'' below).
main()
_int_ main() required for C99; _int_ main(_void_) preferred in both
C89 and C99. (Where _this_ indicates emphasis, not literal underscores.)
{
char ch;
Lack of indentation is a bad thing and should be avoided, especially
on Usenet and on school tests. Also, everywhere else.
int i =2;
for (ch=0;ch<128;ch++)
Non-strict-conformance: program behavior depends on signedness of
plain char.
Possible undefined behavior: If char is signed and has eight bits,
signed integer overflow will occur.
i+=2;
printf("%d",i);
Possible linker error: ''printf'' undefined.
Undefined behavior: variadic function ''printf'' called without
prototype in scope.
Constraint violation (is that correct?): program output ends
without terminating newline ''\n''.
}
i wrote the answer 256 will be printed.
But our instructor told me that it will be infinite loop.
Is that right, I am not sure of the answer.
是的，你们两个都是正确的。只有程序错了。
-Arthur
Yes, you are both "right." Only the program is wrong.
-Arthur
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