K& R2运动问题 [英] K&R2 exercise question
问题描述
好的，我刚刚完成了一个C级课程，现在我正在通过K& R2（以及C99标准）工作
来*真正*学习C.
所以无论如何，我正在进行第一章的练习，它给了我一些奇怪的行为。这是我写的代码：
/ ******************************* ******************* ****************************
* K& R2练习1-9
*编写程序将其输入复制到其输出，替换字符串
空白
*，一张空白。
********************************* ***************** **************************** /
#include< stdio.h>
int main（无效）{
int c = 0，lastc = 0 ;（b = getchar（））！= EOF）{
if（c ==''''）{
if（lastc！=''''）{//如果c *是*空格而且lastc是
putchar（c）; // *不是* a空格，打印字符
lastc = c;
} //结束内部如果
} //结束外部如果
if（c！=''''）{//如果c不是空间不用担心它
putchar（c）;
lastc = c;
} //结束如果
} //结束时
re转0;
}
我遇到的问题是，当我运行程序时，我没有执行
执行，只是返回提示。但是，当我在它上面运行gdb
（即使我只是运行gdb中的程序）它可以工作，甚至可以实现我预期的b $ b。任何人都可以向我解释这个吗？
crr
2004年7月5日星期一17： 40:30 -0700，Chris Readle< c。****** @ cox.net>
在comp.lang.c中写道：
好的，我刚刚完成了一个C级课程，现在我正在通过K& R2（以及C99标准）来*真正*学习C.
所以无论如何，我正在进行第一章的练习，这给我带来了奇怪的行为。这是我写的代码：
/ *********************************** *************** ****************************
* K& R2练习1-9
*编写一个程序将其输入复制到其输出中，用一个空白替换空格
*
******* ******************************************* ******* ********************* /
#include< stdio.h>
int main（void）{
int c = 0，lastc = 0;
while（（c = getchar（））！= EOF）{
if（c == ''''）{
if（lastc！=''''）{//如果c *是*空格而lastc是
putchar（c）; // *不是*空格，打印字符
lastc = c;
} // end inner if
} // end outer if
if（c！=''''）{ //如果c不是空间我不担心它
我希望你知道上面的行可以替换为：
else {putchar（c）;
lastc = c;
} //结束如果
} //结束wh ile
返回0;
}
我遇到的问题是，当我运行程序时，我没有执行，只是返回提示。但是，当我在其上运行gdb
时（即使我只是运行gdb中的程序）它可以工作，甚至可以实现我的预期。有人可以向我解释这个吗？
crr
您的输入是否包含换行符？在C中，实现定义
是否输出到文本流的最后一行是否需要最终的
''\ n''。某些实现不会将
文本的最后一行输出到显示设备（如果它没有以换行符结尾）。
尝试添加：
putchar（''\ n''）;
....就在main（）返回的上方。
-
Jack Klein
主页： http://JK-Technology.Com
常见问题解答
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alt.comp.lang.learn.c-c ++
http：//www.contrib.andrew.cmu.edu/~a ... FAQ-acllc.html
Jack Klein写道：2004年7月5日星期一17:40:30 -0700，Chris Readle< c。****** @ cox.net>
在comp.lang.c中写道：
好的，我刚刚完成了一个C级课程，现在我正在通过K& R2（以及C99标准）来工作*真的*学习C.
所以无论如何，我正在第一章的练习中给我带来奇怪的行为。这是我写的代码：
/ *********************************** *************** ****************************
* K& R2练习1-9
*编写一个程序将其输入复制到其输出中，用一个空白替换空格
*
******* ***************************************** ********* ********************* /
#include< stdio.h>
int main（void）{
int c = 0，lastc = 0;
while（（c = getchar（））！= EOF）{
if（c == ''''）{
if（lastc！=''''）{//如果c *是*空格而lastc是
putchar（c）; // *不是*空格，打印字符
lastc = c;
} // end inner if
} // end outer if
if（c！=''''）{ //如果c不是空间我们不担心它
我希望你知道上面的行可以替换为：
else {
是的，我知道，但我正在努力追随这本书是的，因为它还没有包括其他的... bb> $ block $ =post_quotes> putchar（c）;
lastc = c;
} //结束如果
} //结束时
返回0;
}
我遇到的问题是，当我运行程序时，我没有执行
，只是返回提示。但是，当我在其上运行gdb
时（即使我只是运行gdb中的程序）它可以工作，甚至可以实现我的预期。任何人都可以向我解释这个吗？
crr
您的输入是否包含换行符？在C中，它是实现定义的
输出到文本流的最后一行是否需要最终的
''\ n''。如果没有以换行符结尾，某些实现不会将最后一行
文本输出到显示设备。
尝试添加：
putchar（'' \ n''）;
...正好在main（）的返回上方。
好吧，输入*会*包含换行符，但是当它有问题
之前描述的时候，我不能输入任何东西。基本上，我打电话给
程序，按回车键立即返回到提示符。
我已对此做了一些进一步的研究，我发现了如果
我启动了一个新的shell实例，它第一次运行时只是花花公子，
但后续运行会返回到提示而不允许任何输入。
我尝试在main（），
中的return语句之前添加换行符但是当我调用
程序。
crr
On Mon，05 Jul 2004 17:40:30 -0700， Chris Readle写道：
任何人都可以向我解释这个吗？
在猜测时，exec称为test。你在命令行中输入''test''来运行它。
尝试''测试'' - 不知何故我不知道我认为你正在运行你的程序
认为你正在运行。
Ok, I''ve just recently finished a beginning C class and now I''m working
through K&R2 (alongside the C99 standard) to *really* learn C.
So anyway, I''m working on an exercise in chapter one which give me
strange behavior. Here is the code I''ve written:
/************************************************** ****************************
* K&R2 Exercise 1-9
* Write a program to copy its input to its output, replacing strings
of blanks
* with a single blank.
************************************************** ****************************/
#include <stdio.h>
int main(void) {
int c= 0, lastc= 0;
while((c= getchar()) != EOF) {
if (c == '' '') {
if (lastc != '' '') {//if c *is* a space and lastc is
putchar(c);//*not* a space, print the char
lastc= c;
}//end inner if
}//end outer if
if (c != '' '') {//if c isn''t a space don''t worry about it
putchar(c);
lastc =c;
}//end if
}//end while
return 0;
}
The problem I''m having is that when I run the program, I get no
execution, just returned to a prompt. However, when I run gdb on it
(even if I just "run" the program within gdb) it works, and even does
what I expected it to. Can anyone explain this one to me?
crr
On Mon, 05 Jul 2004 17:40:30 -0700, Chris Readle <c.******@cox.net>
wrote in comp.lang.c:
Ok, I''ve just recently finished a beginning C class and now I''m working
through K&R2 (alongside the C99 standard) to *really* learn C.
So anyway, I''m working on an exercise in chapter one which give me
strange behavior. Here is the code I''ve written:
/************************************************** ****************************
* K&R2 Exercise 1-9
* Write a program to copy its input to its output, replacing strings
of blanks
* with a single blank.
************************************************** ****************************/
#include <stdio.h>
int main(void) {
int c= 0, lastc= 0;
while((c= getchar()) != EOF) {
if (c == '' '') {
if (lastc != '' '') {//if c *is* a space and lastc is
putchar(c);//*not* a space, print the char
lastc= c;
}//end inner if
}//end outer if
if (c != '' '') {//if c isn''t a space don''t worry about it
I hope you know that the line above could be replaced with:
else { putchar(c);
lastc =c;
}//end if
}//end while
return 0;
}
The problem I''m having is that when I run the program, I get no
execution, just returned to a prompt. However, when I run gdb on it
(even if I just "run" the program within gdb) it works, and even does
what I expected it to. Can anyone explain this one to me?
crr
Does your input contain a newline? In C it is implementation defined
whether the final line of output to a text stream requires a final
''\n'' or not. Some implementations will not output the final line of
text to the display device if it does not end in a newline.
Try adding:
putchar(''\n'');
....just above the return from main().
--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html
Jack Klein wrote:On Mon, 05 Jul 2004 17:40:30 -0700, Chris Readle <c.******@cox.net>
wrote in comp.lang.c:Ok, I''ve just recently finished a beginning C class and now I''m working
through K&R2 (alongside the C99 standard) to *really* learn C.
So anyway, I''m working on an exercise in chapter one which give me
strange behavior. Here is the code I''ve written:
/************************************************** ****************************
* K&R2 Exercise 1-9
* Write a program to copy its input to its output, replacing strings
of blanks
* with a single blank.
************************************************ ******************************/
#include <stdio.h>
int main(void) {
int c= 0, lastc= 0;
while((c= getchar()) != EOF) {
if (c == '' '') {
if (lastc != '' '') {//if c *is* a space and lastc is
putchar(c);//*not* a space, print the char
lastc= c;
}//end inner if
}//end outer if
if (c != '' '') {//if c isn''t a space don''t worry about it
I hope you know that the line above could be replaced with:
else {
Yes, I knew that, but I''m trying to follow the book closely and since it
hasn''t covered else yet...putchar(c);
lastc =c;
}//end if
}//end while
return 0;
}
The problem I''m having is that when I run the program, I get no
execution, just returned to a prompt. However, when I run gdb on it
(even if I just "run" the program within gdb) it works, and even does
what I expected it to. Can anyone explain this one to me?
crr
Does your input contain a newline? In C it is implementation defined
whether the final line of output to a text stream requires a final
''\n'' or not. Some implementations will not output the final line of
text to the display device if it does not end in a newline.
Try adding:
putchar(''\n'');
...just above the return from main().
Well, the input *would* contain a newline, but when it has the problem
previously described, I don''t get to input anything. Basically, I call
the program, press enter and am immediately returned to the prompt.
I have done some further research on this, and I have discovered that if
I start a new instance of the shell, it runs just dandy the first time,
but subsequent runs return to the prompt without allowing any input.
I tried adding the newline just before the return statement in main(),
but all that does is put an additional newline on the screen when I call
the program.
crr
On Mon, 05 Jul 2004 17:40:30 -0700, Chris Readle wrote:
Can anyone explain this one to me?
At a guess- the exec is called test. You''re running it by typing ''test'' in
the command line.
Try ''which test'' - somehow I don''t think you''re running the program you
think you''re running.
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