增量实施 [英] Increment implementation
问题描述
大家好，
这是C中的错误（或未定义）实现吗？
/ * start * /
#include< stdio.h>
#define min（a，b）（a< b？a：b）
int main（无效）
{
int i，j，k;
i = 1;
j = 2;
k = min（i ++，j）; / *是这个exp未定义因为我正在使用
增量运算符* /
printf（" i =％d j =％d k = ％d \ n"，i，j，k）;
返回0;
}
/ *结束* /
感谢n Regds，
Nishu
" Nishu" < na ********** @ gmail.comwrites：
大家好，
是在C中这个错误的（或未定义的）实现？
/ * start * /
#include< stdio.h>
#define min（a，b）（a< b？a：b）
宏应该在所有大写中，并且它们的参数应该是完全的
括号。
int main（void）
{
int i ，j，k;
i = 1;
j = 2;
k = min（i ++，j）; / *是这个exp未定义因为我正在使用
增量运算符* /
是的。这就是为什么你需要ALL CAPS警告告诉你使用宏你是
，你应该小心。
printf（i =％d j =％d k =％d \ n，i，j，k）;
返回0;
}
/ *结束* /
感谢n Regds，
不要使用愚蠢的缩写。它们的阻碍远远超过它们的帮助。
Nishu
-
Andrew Poelstra< http://www.wpsoftware.net/projects>
要通过电子邮件与我联系，请在上述域名中使用apoelstra。
" ;是否需要插入电缆的两端？ -Anon。
Andrew Poelstra写道：
" Nishu" < na ********** @ gmail.comwrites：
/ * start * /
#include< stdio.h>
#define min（a，b）（a< b？a：b）
宏应该在所有大写中，并且他们的参数应该是完全的
括号。
为什么宏应该是全部大写？我怀疑*应该*是否由
标准指定。
谢谢，要指出，最好使用括号。
#define min（（a）<（b））（（（a）<（b））？（a）:( b））
>
int main（void）
{
int i，j，k;
i = 1;
j = 2;
k = min（i ++，j）; / *是这个exp未定义因为我正在使用
增量运算符* /
是的。这就是为什么你需要ALL CAPS警告，告诉你使用宏你是b
$ b，你应该小心。
你的意思是这是未定义的？ k =（i ++< j）？ （i ++）:( j）;请
告诉我原因。不是我应该从右到左继续
表达式增加运算符吗？
>
printf（" i =％d j =％d k =％d \ n"，i，j，k）;
返回0;
}
/ *结束* /
感谢n Regds，
不要使用愚蠢的缩写。他们的阻力远大于他们的帮助。
好的。谢谢你&&此致，
Nishu
Andrew Poelstra写道：
" Nishu" < na ********** @ gmail.comwrites：
大家好，
是在C中这个错误的（或未定义的）实现？
/ * start * /
#include< stdio.h>
#define min（a，b）（a< b？a：b）
宏应该在所有大写
这是风格问题。
和他们的论点应该完整
括号内容。
同意。
int main（void）
{
int i，j，k;
i = 1;
j = 2;
k = min（i ++，j）; / *是这个exp未定义因为我正在使用
增量运算符* /
是的。
不，不是。这扩展为（为了清晰起见而增加了空格）：
k =（i ++< j？i ++：j）;
这是完全合法的，即使后增量表达式也是如此由于条件运算符引入了所需的序列点，所以两次计算了
。结果是OP意图
是另一回事。
Robert Gamble
Hi All,
Is this wrong(or undefined) implementation in C?
/* start */
#include<stdio.h>
#define min(a,b) (a<b? a:b)
int main(void)
{
int i,j,k;
i = 1;
j = 2;
k = min(i++, j); /* is this exp undefined because I''m using
increment operator */
printf("i = %d j = %d k = %d\n", i,j,k);
return 0;
}
/* end */
Thank n Regds,
Nishu
"Nishu" <na**********@gmail.comwrites:
Hi All,
Is this wrong(or undefined) implementation in C?
/* start */
#include<stdio.h>
#define min(a,b) (a<b? a:b)
Macros should be in ALL CAPS, and their arguments should be fully
parenthesized.
int main(void)
{
int i,j,k;
i = 1;
j = 2;
k = min(i++, j); /* is this exp undefined because I''m using
increment operator */
Yes. That''s why you need the ALL CAPS warning to tell you that you''re
using a macro, and you should be careful.
printf("i = %d j = %d k = %d\n", i,j,k);
return 0;
}
/* end */
Thank n Regds,Don''t use silly abbreviations. They hinder far more than they help.
Nishu
--
Andrew Poelstra <http://www.wpsoftware.net/projects>
To reach me by email, use `apoelstra'' at the above domain.
"Do BOTH ends of the cable need to be plugged in?" -Anon.
Andrew Poelstra wrote:"Nishu" <na**********@gmail.comwrites:
/* start */
#include<stdio.h>
#define min(a,b) (a<b? a:b)
Macros should be in ALL CAPS, and their arguments should be fully
parenthesized.Why Macros should be ALL CAPS? I doubt if *should* is specified by the
standard.
Thanks, to point out, it is better to use parentheses.
#define min((a) < (b)) ( ((a) < (b))? (a) : (b))
>int main(void)
{
int i,j,k;
i = 1;
j = 2;
k = min(i++, j); /* is this exp undefined because I''m using
increment operator */
Yes. That''s why you need the ALL CAPS warning to tell you that you''re
using a macro, and you should be careful.you mean that this is undefined? k = (i++ < j)? (i++) : (j) ; Please
tell me the reason too. Isnt that I should proceed from right to left
of the expression in case of increment operators?
>printf("i = %d j = %d k = %d\n", i,j,k);
return 0;
}
/* end */
Thank n Regds,
Don''t use silly abbreviations. They hinder far more than they help.
Ok. Thank you && Regards,
Nishu
Andrew Poelstra wrote:"Nishu" <na**********@gmail.comwrites:
Hi All,
Is this wrong(or undefined) implementation in C?
/* start */
#include<stdio.h>
#define min(a,b) (a<b? a:b)
Macros should be in ALL CAPSThat''s a matter of style.
and their arguments should be fully
parenthesized.Agreed.
int main(void)
{
int i,j,k;
i = 1;
j = 2;
k = min(i++, j); /* is this exp undefined because I''m using
increment operator */
Yes.No, it isn''t. This expands to (spaces added for clarity):
k = ( i++ < j ? i++ : j );
which is completely legal, even if the post-increment expression is
evaluated twice since the required sequence points are introduced by
the conditional operator. Whether the result is what the OP intended
is a different matter.
Robert Gamble
这篇关于增量实施的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
