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Vistascode check :i expect IndexPlus is 2 but it seems it becom 1+1=11,why?
description :
i expect the indexPlus 2 but it comes 11，what is wrong? and why? see it in line 28 .
code:
var inputArr = [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0]
console.log('before', inputArr)
var arr = []
//先遍历inputArr确定0的位置
for (key in inputArr) {
if (inputArr[key] === 0) {
arr.push(key)
}
}
//将0移到数组最后面，通过位置交换实现
for (value of arr) {
for (i = value; i < inputArr.length - 1; i++) {
swap(i, i + 1)
}
}
function swap(index, indexPlus) {
var temp = inputArr[index]
inputArr[index] = inputArr[indexPlus]
inputArr[indexPlus] = temp
}
console.log('after', inputArr)
·
Juan Pablo Isaza
2 Respuestas
Responde la pregunta0
When you do for (key in inputArr), the keys are strings, not integers. So later when you do swap(i, i + 1), i is a string, and i + 1 does string concatenation, not integer addition.
Change the first loop to loop over indexes, not keys.
var inputArr = [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0]
console.log('before', inputArr)
var arr = []
//先遍历inputArr确定0的位置
for (let key = 0; key < inputArr.length; key++) {
if (inputArr[key] === 0) {
arr.push(key)
}
}
//将0移到数组最后面，通过位置交换实现
for (value of arr) {
for (i = value; i < inputArr.length - 1; i++) {
swap(i, i + 1)
}
}
function swap(index, indexPlus) {
var temp = inputArr[index]
inputArr[index] = inputArr[indexPlus]
inputArr[indexPlus] = temp
}
console.log('after', inputArr)
·
Juan Pablo Isaza
Denunciar
0
When you use for ( in ), in every single loop, you will get a key(position) as a string. So when you take it and + 1 you will get result as a string.
·
Juan Pablo Isaza
Denunciar
Responde la pregunta