在JavaScript中添加带整数字符串的奇怪行为？

Question

我只是发布了一些奇怪的东西（至少对我来说）发生在JavaScript中（不知道也许在其他语言中）。我有这个例子：

var a = "1"; 
var b = 1; 

console.log(a + b); // this one returns 11 as it suppose 
console.log(b + a); // this one returns also 11 as it suppose 
console.log(a + b + b); // this one returns 111? why not 12 ? 
console.log(b + b + a); // but this return 21 as it suppose 

有人可以向我解释为什么会发生这种情况？感谢先进的和抱歉，如果这是一个非常愚蠢的问题。

Answer 1

让我们打破你的代码：

var a = "1", // string 
    b = 1;  // number 

console.log(a + b); 
// a is a string and b is a number 
// read from left to right 
// string first, so when + goes to concatenate the two variables, you get 
// a + b.toString() == "1" + "1" == "11" 

console.log(b + a); 
// same thing as above because of the .toString() method 
// except the + operator sees number + string (versus string + number) 
// however, this still casts the number to a string, since that's how 
// strings and numbers are concatenated 

console.log(a + b + b) 
// same as above 
// read from left to right 
// string is encountered first, so everything after it is cast to a string 
// "1" + b.toString() + b.toString() == "1" + "2" + "2" == "122" 
// if you want it to be 12, you need to do 
// console.log(a + (b + b)) 

console.log(b + b + a) 
// the first + operator sees two integers, so they get added together first 
// now it's console.log(2 + a) 
// so (2).toString() + a == "2" + "1" == "21", if you follow the logic 
// from the second example 

Answer 2

a是一个字符串，所以当你对一个字符串进行数学运算时，它会简单地连接起来。让我们去通你：

console.log(a + b + b); // a + b = 11 - which is now a string + b = 111; 
console.log(b + b + a); // b + b = 2 + a string 1 = 21