我如何把两个增量语句放在C ++的'for循环中? [英] How do I put two increment statements in a C++ 'for' loop?
问题描述
我想在中为
-loop条件增加两个变量,而不是一个。
如:
for(int i = 0; i!= 5; ++ i和++ j)
do_something(i,j);
这是什么语法?
一个常见的用法是使用逗号运算符计算两个操作数,并返回第二个操作数。因此:
$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ do_something(I,J);
但它确实是一个逗号运算符吗?
<现在已经写了,一个评论者提出,它实际上是for语句中的一些特殊的语法糖,而不是一个逗号操作符。我在GCC中检查了如下:
int i = 0;
int a = 5;
int x = 0; (i; i <5; x = i ++,a ++){
printf(i =%da =%dx =%d \ n,i,a,x)的
);
}
我期待x拿起a的原始值,所以它应该已经显示5,6,7 ..为x。我得到的是这个
pre $ i = 0 a = 5 x = 0
$但是,如果我使用表达式强制解析器真的看到一个逗号运算符,那么我得到这个
i = 1 a = 6 x = 0
i = 2 a = 7 x = 1
i = 3 a = 8 x = 2
i = 4 a = 9 x = 3
<$> p $ p>
int main(){
int i = 0;
int a = 5;
int x = 0; (i = 0; i <5; x =(i ++,a ++)){
printf(i =%da =%dx =%d\\\
)的
,斧头);
i = 0 a = 5 x = 0
i = 1 a = 6 x = 5
i = 2 a = 7 x = 6
i = 3 a = 8 x = 7
i = 4 a = 9 x = 8
最初,我认为这表明它并不是一个逗号运算符,但事实证明,这只是一个优先问题 - 逗号运算符具有可能的最低优先级,所以表达式x = i ++,a ++被有效地解析为(x = i ++),a ++
感谢所有的评论,这是一个有趣的学习经历,我已经使用C多年了!
I would like to increment two variables in a for
-loop condition instead of one.
So something like:
for (int i = 0; i != 5; ++i and ++j)
do_something(i, j);
What is the syntax for this?
A common idiom is to use the comma operator which evaluates both operands, and returns the second operand. Thus:
for(int i = 0; i != 5; ++i,++j)
do_something(i,j);
But is it really a comma operator?
Now having wrote that, a commenter suggested it was actually some special syntactic sugar in the for statement, and not a comma operator at all. I checked that in GCC as follows:
int i=0;
int a=5;
int x=0;
for(i; i<5; x=i++,a++){
printf("i=%d a=%d x=%d\n",i,a,x);
}
I was expecting x to pick up the original value of a, so it should have displayed 5,6,7.. for x. What I got was this
i=0 a=5 x=0
i=1 a=6 x=0
i=2 a=7 x=1
i=3 a=8 x=2
i=4 a=9 x=3
However, if I bracketed the expression to force the parser into really seeing a comma operator, I get this
int main(){
int i=0;
int a=5;
int x=0;
for(i=0; i<5; x=(i++,a++)){
printf("i=%d a=%d x=%d\n",i,a,x);
}
}
i=0 a=5 x=0
i=1 a=6 x=5
i=2 a=7 x=6
i=3 a=8 x=7
i=4 a=9 x=8
Initially I thought that this showed it wasn't behaving as a comma operator at all, but as it turns out, this is simply a precedence issue - the comma operator has the lowest possible precedence, so the expression x=i++,a++ is effectively parsed as (x=i++),a++
Thanks for all the comments, it was an interesting learning experience, and I've been using C for many years!
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