如何将两个增量语句放在C ++'for'循环中？ [英] How to 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。

例如：

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

这是什么语法？

解决方案

一个常见的习惯是使用逗号运算符来计算两个操作数，并返回第二个操作数。因此：

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

但是它真的是逗号运算符吗？

现在写过了，一个注释者建议它在for语句中实际上是一些特殊的语法糖，而不是一个逗号运算符。我在GCC检查如下：

  int i = 0; 
 int a = 5; 
 int x = 0; 
 
 for（i; i <5; x = i ++，a ++）{
 printf（i =％da =％dx =％d \\\
，i，a，x ）; 
} 
  

我希望x能够获取a的原始值，已显示5,6,7 .. x。我得到的是这个

  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，如果我把表达式括起来强迫解析器真正看到一个逗号运算符，我得到

  int main（）{
 int i = 0; 
 int a = 5; 
 int x = 0; 
 
 for（i = 0; i <5; x =（i ++，a ++））{
 printf（i =％da =％dx =％d \\\
，i ，斧头）; 
} 
} 
 
i = 0 a = 5 x = 0 
i = 1 a = 6 x = 5 
i = 2 a = 7 x = 6 b 
 
 
最初，我认为这表明它不是一个逗号运算符，但事实证明，这只是一个优先问题 - 逗号运算符有最可能的优先级，因此表达式x = i ++，a ++被有效地解析为（x = i ++），a ++ 
 
 
 
感谢所有的意见，这是一个有趣的学习经验，我已经使用C多年！
 
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?
解决方案
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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