javascript javascript中++和+=1的区别
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Difference between ++ and +=1 in javascript
提问by RobinL
Could somebody explain why the following functions give different results. The first does not seem to work, but the second does. I'm puzzled because I thought +=1 and ++ did the same thing.
有人可以解释为什么以下函数会给出不同的结果。第一个似乎不起作用，但第二个起作用。我很困惑，因为我认为 +=1 和 ++ 做了同样的事情。
(I'm not intending to actually use this code, it's just to demonstrate the difference).
（我不打算实际使用此代码，只是为了演示差异）。
/*function 1*/
function incrementIfZero1(base,element) {
if (element == 0) {
return base++;
}
else
{
return base;
}
};
/*function 2*/
function incrementIfZero2(base,element) {
if (element == 0) {
return base+=1;
}
else
{
return base;
}
};
incrementIfZero1(1,0) /* -> 1*/
incrementIfZero2(1,0) /* -> 2*/
Any help is very much appreciated.
很感谢任何形式的帮助。
Thanks,
谢谢，
Robin
罗宾
[Edit:]
[编辑：]
Thank you for your replies, it makes sense now. I had also tried the following statement, which resulted in the same thing as function 1:
谢谢你的回复，现在说得通了。我还尝试了以下语句，结果与函数 1 相同：
return (base++)
I'm now surprised that this doesn't give the same result as function 2 - I would have expected the brackets to 'force' it to be evaluated before returning. Any idea why this is not the case?
我现在很惊讶这不会给出与函数 2 相同的结果 - 我原以为括号会“强制”在返回之前对其进行评估。知道为什么不是这样吗？
采纳答案by Stephan
when you return base++it returns the value of base just before it gets incremented. You want to do ++baseto make sure the increment happens first then it gets returned
当你return base++在它增加之前返回 base 的值时。你想++base确保增量首先发生然后它被返回
otherwise ++is the same as +=1
否则++与+=1
[edit] in response to your edit, i tried wrapping random statements in parentheses and most mathematical operators respond as expected, this incrementation seems to be exempt, likely because the syntax of pre-incrementation vs post-incrementation is highly intentional and the statement itself returnsa specific value regardless of whether or not you wrap it in parentheses
[编辑] 为了响应您的编辑，我尝试将随机语句括在括号中，并且大多数数学运算符都按预期响应，这种增量似乎可以免除，可能是因为预增量与后增量的语法是高度有意的，并且语句本身无论您是否将其括在括号中，都将返回一个特定值
回答by sonoftunk
You are returning base++. This is the postfix increment, thus, the increment is handled after the return. For this specific case, return ++base;would be correct
你回来了base++。这是后缀增量，因此，在返回后处理增量。对于这种特定情况，return ++base;是正确的
回答by bfavaretto
return (base++)I'm now surprised that this doesn't give the same result as function 2 - I would have expected the brackets to 'force' it to be evaluated before returning. Any idea why this is not the case?
return (base++)我现在很惊讶这不会给出与函数 2 相同的结果 - 我原以为括号会“强制”在返回之前对其进行评估。知道为什么不是这样吗？
The increment operation isevaluated, regardless of the brackets. The catch here is that expressions in JavaScript always result in a value, which is then returned. For example, when you do var x = 2 + 2, 2 + 2is evaluated, then "returned" to the =operator, which then assigns that value to x. The ++postfix operator (as in base++) works differently: it does increment the variable preceding it, but the expression returns the value before incrementation. The prefix increment operator, on the other hand, works as you want, ++basereturns the value after incrementing it.
该增值业务的评估，无论括号。这里的问题是 JavaScript 中的表达式总是产生一个值，然后返回该值。例如，当您执行var x = 2 + 2,2 + 2被评估，然后“返回”给=运算符，然后将该值分配给x。所述++后缀运算符（如base++）的工作方式不同：它确实增加其之前的变量，但递增之前表达式返回的值。另一方面，前缀增量运算符按您的需要工作，++base在增加它后返回值。
回答by Dan455
return base++;
Is post-increment. It increments the value held by base, and then returns the original value before the increment happened.
是后增量。它递增 base 持有的值，然后返回递增发生前的原始值。
return base += 1;
Adds one to base, and then returns base. This is a pre-increment. This can also be written as:
给基数加一，然后返回基数。这是一个预增量。这也可以写成：
return ++base;
回答by Dany Caissy
Just to clarify a little.
只是为了澄清一点。
"variable += 1" is the equivalent of "variable = variable + 1". It is NOT the equivalent of variable++!
“变量+= 1”相当于“变量=变量+ 1”。它不等同于变量++！
Something like variable++ can be used in-place, the former is an assignation.
像 variable++ 这样的东西可以就地使用，前者是赋值。
回答by Danny
Warning! There's a difference when two different types, a string and a number, are mutated using ++and +=:
警告！使用++and 对两种不同类型（字符串和数字）进行变异时会有所不同+=：
When countis a string, count += 1seems to do type conversion and converts the 2nd number to a string, whereas count++takes the string countargument, converts to a number, and increments by 1.
当count是字符串时，count += 1似乎进行类型转换并将第二个数字转换为字符串，而count++采用字符串count参数，转换为数字，并以 1 递增。
let count = process.argv[2]
while (count < 5) {
console.log('Inside of the loop. Count is', count)
count += 1 // NOT likely what you want! Now, try it with count++
}
console.log('Outside of loop. Count is', count)
Another example:
另一个例子：
> let num = '2'
undefined
> num++
2
> num
3
// VS using +=
> let num2 = '2'
undefined
> num2 += 1
'21'
> num2
'21'
回答by Danny
be carefull :
小心点 ：
x = 1; y = (x++ * 10 + 1);
x = 1; y = (x++ * 10 + 1);
equal to :
等于 ：
y = 1 * 10 + 1 = 11
y = 1 * 10 + 1 = 11
indeed :
的确 ：
x = 1; y = (++x * 10 + 1);
x = 1; y = (++x * 10 + 1);
equal to :
等于 ：
y = 2 * 10 + 1 = 21;
y = 2 * 10 + 1 = 21；
to finish :
完成 ：
++x ;
++x;
or
或者
x++ ;
x++ ;
there are not difference !
没有区别！
回答by Usual Suspect
In The first function what happens is:
在第一个函数中发生的是：
return base++;// at first base value is returned and then incremented.
return base++;// 首先返回基值，然后递增。
In the 2nd Function :
在第二个功能中：
return base+=1;//here the incrementation is done first and then the variable is returned.
return base+=1;//这里首先完成增量，然后返回变量。
shorthand assignment has higher priority i.e. it finishes it's job in that line itself
速记分配具有更高的优先级，即它在该行本身完成它的工作
