&安培;&安培;和||运营商优先执行 [英] && and || Operator precedence enforcement
问题描述
嗨伙计，
这个问题之前可能已经被问到并得到了解答。但是，抱歉
我无法从档案中找到一个。
我输入了这个程序并用gcc编译了3.3.2
main（）{
int i = -3，j = 2，k = 0，m;
m = ++ i || ++ j&& ++ k;
printf（" \ n％d％d％d％d \ n"，i，j，k，m）;
}
并得到输出
-2 2 0 1
我认为答案应该是
-3 3 1 1
作为++ j和++ k首先进行评估&&运算符应用于
获得结果为TRUE（3&& 1）。因此，++我不会被评估为
作为||的一部分是真的。
ANSI C规范明确指出&&优先于||。
有人可以解释为什么这个奇怪的-2 2 0 1程序输出
获得？
干杯
Vivek
Hi Folks,
This question may well have been asked and answered before. But, sorry
that I couldn''t find one from the archives.
I typed up this program and compiled it with gcc 3.3.2
main() {
int i = -3,j= 2,k=0,m;
m = ++i || ++j && ++k;
printf("\n%d %d %d %d\n",i,j,k,m);
}
and got the output to be
-2 2 0 1
I presumed that the answer should have well been
-3 3 1 1
as ++j and ++k would first be evaluated and && operator applied to
obtain the result to be TRUE (3 && 1). Hence, ++i wouldn''t be evaluated
as one part of || goes TRUE.
The ANSI C specification clearly states that && has precedence over ||.
Could someone explain why this strange "-2 2 0 1" output for the program
is obtained?
Cheers
Vivek
推荐答案
Vivek N写道：
Vivek N wrote:
嗨伙计，
这个问题可能已经被问及以前的答案。但是，抱歉
我无法从档案中找到一个。
我输入了这个程序并用gcc编译了3.3.2
main（）{
int i = -3，j = 2，k = 0，m;
m = ++ i || ++ j&& ++ k;
printf（" \ n％d％d％d％d \ n"，i，j，k，m）;
}
并得到输出是
-2 2 0 1
我认为答案应该是很好的，因为++ j和首先评估++ k并且&&运算符应用于
获得结果为TRUE（3&& 1）。
因此，++我不会被评估为||的一部分是真的。
ANSI C规范明确指出&&
优先于||。
（a || b&& c），与（a ||（b&& c））相同。
有人可以解释为什么这个奇怪的-2 2 0 1
节目输出？
Hi Folks,
This question may well have been asked and answered before. But, sorry
that I couldn''t find one from the archives.
I typed up this program and compiled it with gcc 3.3.2
main() {
int i = -3,j= 2,k=0,m;
m = ++i || ++j && ++k;
printf("\n%d %d %d %d\n",i,j,k,m);
}
and got the output to be
-2 2 0 1
I presumed that the answer should have well been
-3 3 1 1
as ++j and ++k would first be evaluated and && operator applied to
obtain the result to be TRUE (3 && 1).
Hence, ++i wouldn''t be evaluated
as one part of || goes TRUE.
The ANSI C specification clearly states that &&
has precedence over ||.
(a || b && c), is the same as (a || (b && c)).
Could someone explain why this strange "-2 2 0 1"
output for the program is obtained?
j和k根本没有评估。
||的左操作数在右边之前进行评估。
||的右操作数只有当左操作数等于零时才有条件地评估。
你的左操作数（++ i），不等于零。
你的右操作数（++ j&& ++ k）未被评估。
-
pete
j and k are not evaluated at all.
The left operand of || is evaluted before the right.
The right operand of || is evaluated conditionally only if
the left operand is equal to zero.
Your left operand (++i), is not equal to zero.
Your right operand (++j && ++k) is not evaluated.
--
pete
Vivek N< ke *********** @ yahoo.com>这样说：
Vivek N <ke***********@yahoo.com> spoke thus:
main（）{
允许函数的返回类型默认为int是非法的
C99，否则可疑。
int i = -3，j = 2，k = 0，m;
m = ++ i || ++ j&& ++ k;
printf（" \ n％d％d％d％d \ n，i，j，k，m）;
哪里是< stdio.h>？打开gcc'的警告（-Wall -pedantic -ansi）。
有人可以解释为什么这个奇怪的-2 2 0 1获得该程序的输出
？
main() {
Allowing the return type of a function to default to int is illegal in
C99 and dubious otherwise.
int i = -3,j= 2,k=0,m; m = ++i || ++j && ++k;
printf("\n%d %d %d %d\n",i,j,k,m);
Where is <stdio.h>? Turn up gcc''s warnings (-Wall -pedantic -ansi).
Could someone explain why this strange "-2 2 0 1" output for the program
is obtained?
由于短路评估，一点都不奇怪。
&&优先于||只表示对m的赋值是
相当于
m = ++ i || （++ j& ++ k）;
++我首先计算，如果它是非零（真），那么整个
$ b无论是什么（++ j&& ++ k），$ b语句都保证为真。
在C中，这意味着（++ j&& ++ k ）不会被评估。那是什么短路评估是什么？这就是为什么只有我在你发布的代码中增加了
。
-
Christopher Benson-Manica |我*应该*知道我在说什么 - 如果我
ataru（at）cyberspace.org |不，我需要知道。火焰欢迎。
Not strange at all, thanks to short-circuit evaluation. The fact that
&& has precedence over || only means that the assignment to m is
equivalent to
m=++i || (++j && ++k);
++i is computed first, and if it is nonzero (true), then the entire
statement is guaranteed to be true regardless of what (++j && ++k) is.
In C, this means that (++j && ++k) will not be evaluated. That is
what short-circuit evaluation is, and that is why only i is
incremented in the code you posted.
--
Christopher Benson-Manica | I *should* know what I''m talking about - if I
ataru(at)cyberspace.org | don''t, I need to know. Flames welcome.
ke ****** *****@yahoo.com （Vivek N）写道：
ke***********@yahoo.com (Vivek N) wrote:
作为++ j和++ k将首先被评估和&&运算符应用于
获得结果为TRUE（3&& 1）。因此，++我不会被评估为||的一部分是真的。
ANSI C规范明确指出&&优先于||。
as ++j and ++k would first be evaluated and && operator applied to
obtain the result to be TRUE (3 && 1). Hence, ++i wouldn''t be evaluated
as one part of || goes TRUE.
The ANSI C specification clearly states that && has precedence over ||.
ISO C规范没有说明这样的事情，因为优先级是是什么？b $ b不明确。
它是什么_does_状态是这样的：
*首先，所有逻辑运算符都是从左到右计算的，用
进行短路评估。这意味着它可以这样做：
if（ptr！= NULL&& * ptr！=''A''）...
并确保在被解除引用之前测试指针。
*逻辑和（&&）运算符比逻辑或
（||）。也就是说，x || y&& z表示x || （y&& z），而不是（x || y）&& ž。然而，
评价顺序保持不变。
Richard
The ISO C specification states no such thing, since "precedence" is
ambiguous.
What it _does_ state is this:
* To begin with, all logical operators are evaluated left-to-right, with
short-circuit evaluation. This means that it''s possible to do this:
if (ptr!=NULL && *ptr!=''A'') ...
and be certain that your pointer is tested before being dereferenced.
* The logical and (&&) operator binds more closely than the logical or
(||). That is, x || y && z means x || (y && z), not (x || y) && z. The
order of evaluation remains the same, however.
Richard
这篇关于&安培;&安培;和||运营商优先执行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！
