javascript 计数器++与计数器=计数器+1；
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counter ++ vs counter = counter+1;
提问by user3562812
var counter = 0;
var increment = function(){
return counter++;
// return counter=counter+1;
}
console.log(increment());
Why does counter ++;returns 0 but counter = counter +1;returns 1?
为什么计数器++；返回 0 但计数器 = 计数器 +1；返回 1?
回答by Amit Joki
The post-fix increment operator returns the current valuefor evaluation and thenincrements it, so the value change is noticeable when it is referenced for the second time.
post-fix increment 运算符返回当前值进行评估，然后将其递增，因此第二次引用时值的变化很明显。
So when the returnstatement is executed, the value has not yetbeen incremented. You can cross-check by doing the following:
所以，当return执行语句时，值没有尚未被递增。您可以通过执行以下操作进行交叉检查：
console.log(counter++); // 0
console.log(counter); // 1
return counter; // obviously 1
Try the pre-fix operator which increments and thenreturns the incrementedvalue to be evaluated.
尝试使用前缀运算符递增，然后返回要评估的递增值。
return ++counter; // 1
回答by Alex McMillan
This is called prefix (++x) vs postfix (x++)and the only difference is really the order of operations:
这称为前缀 ( ++x) 与后缀 ( x++)，唯一的区别实际上是操作顺序：
counter;
evaluates to a value. (0)
评估为一个值。(0)
counter++;
evaluates to a value (0), performs a calculation (0 -> 1) and modifies a variable (counter-> 1).
计算为值 (0)，执行计算 (0 -> 1) 并修改变量 ( counter-> 1)。
++counter;
performs a calculation (0 + 1), modifies a variable (counter-> 1) and evaluates to a value (1).
执行计算 (0 + 1)，修改变量 ( counter-> 1) 并计算出值 (1)。
回答by Mehedi Hasan
var counter = 0;
无功计数器 = 0;
var increment = function(){
// now counter is 0 and after return it increment it's value to 1
// return 0 then 0+1=1;
return counter++;
// calculate like math, counter = 0+1 then counter = 1, now return 1;
// return counter=counter+1;
}
console.log(increment());
In first scenario,
在第一个场景中，
return counter++;
This statement is postfix, and evaluates like
这个语句是后缀，并且评估如下
return 0; and then 0+1=1
In second scenario,
在第二种情况下，
return counter=counter+1;
Calculate like math,
像数学一样计算，
return counter = 0+1 then,
return counter = 1,
return 1;
