一个突然出现的问题 [英] A popping question
问题描述
有人可以帮我理解K& R II练习的一个方面;
（第76页）
给定功能：
双流行（无效）{
....如果存在则从堆栈返回双倍
}
这是如何工作的？
/ *增加2双* /
案例''+''：
push（pop（）+ pop（））;
我原以为你不得不这么做类似
double firstpop，secondpop;
firstpop = pop（）;
secondpop = pop（） ;
push（firstpop + secondpop）;
即如果没有分配，双打去哪里。它显然在某个地方发生了，它确实有效，只是我看不出来。
提前致谢。
Could someone help me understand one aspect of an exercise in K&R II;
(page 76)
Given function:
double pop( void){
....return a double from a stack if present
}
how does this work?
/* adds 2 doubles*/
case ''+'' :
push ( pop() + pop()) ;
I would have thought you would have to do something like
double firstpop, secondpop;
firstpop=pop();
secondpop=pop();
push ( firstpop+secondpop);
ie where do the doubles go if they are not assigned. It obviously goes
somewhere and it does work, it''s just that I do not see how.
Thanks in advance.
推荐答案
" mdh" < md ** @ comcast.netwrites：
"mdh" <md**@comcast.netwrites:
这是如何工作的？
how does this work?
....
....
push（pop（）+ pop（））;
我原本以为你必须要做一些像
double firstpop，secondpop;
firstpop = pop（）;
secondpop = pop（）;
push（firstpop + secondpop）;
即在哪里如果他们没有被分配，他们会去。它显然在某个地方发生了b $ b，它确实有效，只是我不知道如何。
push ( pop() + pop()) ;
I would have thought you would have to do something like
double firstpop, secondpop;
firstpop=pop();
secondpop=pop();
push ( firstpop+secondpop);
ie where do the doubles go if they are not assigned. It obviously goes
somewhere and it does work, it''s just that I do not see how.
没关系。编译器的工作就是为表达式中使用的
temporaries预留空间。程序员没有必要担心这个问题。
-
程序员有权不了解许多细节您的代码
仍然会做出合理的更改。
--Kernighan和Plauger，_软件工具_
It doesn''t matter. It''s the compiler''s job to reserve space for
temporaries used in expressions. The programmer doesn''t have to
worry about it.
--
"Programmers have the right to be ignorant of many details of your code
and still make reasonable changes."
--Kernighan and Plauger, _Software Tools_
" mdh" < md ** @ comcast.netwrites：
"mdh" <md**@comcast.netwrites:
即如果没有分配，双打去哪里。它显然在某个地方发生了b $ b，它确实有效，只是我不知道如何。
ie where do the doubles go if they are not assigned. It obviously goes
somewhere and it does work, it''s just that I do not see how.
Ben Pfaff写道：
Ben Pfaff wrote:
没关系。编译器的工作就是为表达式中使用的
temporaries预留空间。程序员不必担心这个问题。
It doesn''t matter. It''s the compiler''s job to reserve space for
temporaries used in expressions. The programmer doesn''t have to
worry about it.
因此，假设pop是公平的在某种意义上，它取而代之的是它的价值，并且任何赋值都是additonal。但是没有必要执行pop？
So is it fair to assume that "pop" is in a sense replaced by it''s
value, and any assignment is "additonal" but not necessary to the
execution of "pop"?
" mdh" < md ** @ comcast.netwrites：
"mdh" <md**@comcast.netwrites:
>" mdh" < md ** @ comcast.netwrites：
>"mdh" <md**@comcast.netwrites:
即如果没有分配，双打去哪里。它显然在某个地方发生了b $ b，它确实有效，只是我不知道如何。
ie where do the doubles go if they are not assigned. It obviously goes
somewhere and it does work, it''s just that I do not see how.
Ben Pfaff写道：
Ben Pfaff wrote:
>这没关系。为表达式中使用的临时表预留空间是编译器的工作。程序员不必担心它。
>It doesn''t matter. It''s the compiler''s job to reserve space for
temporaries used in expressions. The programmer doesn''t have to
worry about it.
因此，假设pop是公平的在某种意义上，它取而代之的是它的价值，并且任何赋值都是additonal。但是没有必要执行pop？
So is it fair to assume that "pop" is in a sense replaced by it''s
value, and any assignment is "additonal" but not necessary to the
execution of "pop"?
您是否知道函数的结果必须立即分配给变量，或者否则会丢失？
这就是我解释你的话的方式。这个想法是错误的。给定一个适当的函数foo
和变量x：
foo（）;允许在C中允许以下所有
; / *抛弃返回值。 * /
x = foo（）; / *指定返回值。 * /
x = foo（）+ 1; / *指定返回值加1. * /
x = foo（）+ foo（）; / *调用foo两次，分配总和。 * /
-
int main（void）{char p [] =" ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz。\
\ n" ，* q =" kl BIcNBFr.NKEzjwCIxNJC" ;; int i = sizeof p / 2; char * strchr（）; int putchar（\
）; while（* q）{i + = strchr （p，* q ++） - p; if（i> =（int）sizeof p）i- = sizeof p-1; putchar（p [i] \
）;} return 0; }
Do you have an idea that the result of a function must be
immediately assigned to a variable, or that otherwise it is lost?
That is how I interpret your words. This idea is wrong. All of
the following are allowed in C, given an appropriate function foo
and variable x:
foo(); /* Throws away return value. */
x = foo(); /* Assigns return value. */
x = foo() + 1; /* Assigns return value plus 1. */
x = foo() + foo(); /* Calls foo twice, assigns the sum. */
--
int main(void){char p[]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz.\
\n",*q="kl BIcNBFr.NKEzjwCIxNJC";int i=sizeof p/2;char *strchr();int putchar(\
);while(*q){i+=strchr(p,*q++)-p;if(i>=(int)sizeof p)i-=sizeof p-1;putchar(p[i]\
);}return 0;}
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