问题描述

我想在 for 循环条件中增加两个变量而不是一个.

I would like to increment two variables in a for-loop condition instead of one.

比如:

for (int i = 0; i != 5; ++i and ++j) 
    do_something(i, j);

这是什么语法?

推荐答案

一个常见的习惯用法是使用 逗号运算符 计算两个操作数，并返回第二个操作数.因此:

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);

但它真的是逗号运算符吗?

写完之后，一位评论者认为它实际上是 for 语句中的一些特殊语法糖，根本不是逗号运算符.我在 GCC 中进行了如下检查:

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
",i,a,x);
}

我期望 x 获取 a 的原始值，因此它应该为 x 显示 5,6,7...我得到的是这个

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

但是，如果我将表达式括起来以强制解析器真正看到逗号运算符，我会得到这个

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
",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

最初我认为这表明它根本不像逗号运算符，但事实证明，这只是一个优先级问题 - 逗号运算符具有 最低可能的优先级，所以表达式 x=i++,a++ 被有效地解析为 (x=i++),a++

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++

感谢所有的评论，这是一次有趣的学习经历，我已经使用 C 多年了！

Thanks for all the comments, it was an interesting learning experience, and I've been using C for many years!

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