问：如何设置用于if条件的变量答案

【问题标题】：Q: How to set variable which use for if condition问：如何设置用于if条件的变量
【发布时间】：2018-12-03 00:36:05
【问题描述】：

我的代码如下所示并且 if 或 while 的条件太长了，所以我想使用变量作为条件我该怎么做？

我的示例代码是：

aTempArray.add(sInput[iStringOnPosition].toString())
//                    bCheckNextisDigit=true
                    while (iStringOnPosition + 1 < sInput.length &&
                        (sInput[iStringOnPosition + 1]).isDigit() || sInput[iStringOnPosition + 1] == '.') { 
                        aTempArray[aTempArray.size - 1] = aTempArray[aTempArray.size - 1] +
                                sInput[iStringOnPosition + 1]
                        iStringOnPosition++
                    }

和

if (sInput[iStringOnPosition] == ' ') {
                while (iStringOnPosition < sInput.length && sInput[iStringOnPosition] == ' ') {
                    iStringOnPosition++
                }
            } else if (sInput[iStringOnPosition] == '-' && (iStringOnPosition == 0 //checking value for negativity

                        || sInput[iStringOnPosition - 1] == '-'
                        || sInput[iStringOnPosition - 1] == '+'
                        || sInput[iStringOnPosition - 1] == '*'
                        || sInput[iStringOnPosition - 1] == '/')
            ) {
                bNegativePresent = true
                iStringOnPosition++

            } else if ((sInput[iStringOnPosition] == '-'             //checking for repeating operators
                        || sInput[iStringOnPosition] == '+'
                        || sInput[iStringOnPosition] == '*'
                        || sInput[iStringOnPosition] == '/') && (iStringOnPosition == 0
                        || sInput[iStringOnPosition - 1] == '-'
                        || sInput[iStringOnPosition - 1] == '+'
                        || sInput[iStringOnPosition - 1] == '*'
                        || sInput[iStringOnPosition - 1] == '/')
            ) {
                bError = true
                return "Error: two or more operators in a row"
            }

我想这样写

aTempArray.add(sInput[iStringOnPosition].toString())
var bCheckNextisDigit = iStringOnPosition + 1 < sInput.length &&
                            (sInput[iStringOnPosition + 1]).isDigit() || sInput[iStringOnPosition + 1] == '.'
                        bCheckNextisDigit=true
                        while (bCheckNextisDigit) { 
                            aTempArray[aTempArray.size - 1] = aTempArray[aTempArray.size - 1] +
                                    sInput[iStringOnPosition + 1]
                            iStringOnPosition++
                        }

和

if (first_condition) {
                        iStringOnPosition++
                    }
                } else if (second_condition) {
                    bNegativePresent = true
                    iStringOnPosition++

                } else if (last_condition) {
                    bError = true
                    return "Error: two or more operators in a row"
                }

请帮忙解决这个问题

【问题讨论】：

标签： java android android-studio intellij-idea kotlin

【解决方案1】：

如果需要，您可以将它们全部保存为布尔值。

像这样：

first_condition = sInput[iStringOnPosition] == ' '
second_condition = sInput[iStringOnPosition] == '-' && (iStringOnPosition == 0 || sInput[iStringOnPosition - 1] == '-' || sInput[iStringOnPosition - 1] == '+' || sInput[iStringOnPosition - 1] == '*' || sInput[iStringOnPosition - 1] == '/')

或者

second_condition1 = iStringOnPosition == 0
second_condition2 = sInput[iStringOnPosition - 1] == '-'
second_condition3 = sInput[iStringOnPosition - 1] == '+'
second_condition4 = sInput[iStringOnPosition - 1] == '*'
second_condition5 = sInput[iStringOnPosition - 1] == '/'

second_conditionTotal = first_condition && (second_condition1 || second_condition2 || second_condition3 || second_condition4 || second_condition5)

【讨论】：

【解决方案2】：

我这样重写怎么样？

var blank_cond: Boolean = false
var blank_loop_cond: Boolean = false
var negertive_First_cond: Boolean = false
var negertive_sec_cond1: Boolean = false
var negertive_sec_cond2: Boolean = false
var negertive_sec_cond3: Boolean = false
var negertive_sec_cond4: Boolean = false
var negertive_sec_cond5: Boolean = false
var negertive_sec_condToT: Boolean = false


    //check Blooean blank
    blank_cond = sInput[iStringOnPosition] == ' '
    //check Blooean negertive
    negertive_First_cond = sInput[iStringOnPosition] == '-'
    negertive_sec_cond1 = iStringOnPosition == 0
    negertive_sec_cond2 = sInput[iStringOnPosition - 1] == '-'
    negertive_sec_cond3 = sInput[iStringOnPosition - 1] == '+'
    negertive_sec_cond4 = sInput[iStringOnPosition - 1] == '*'
    negertive_sec_cond5 = sInput[iStringOnPosition - 1] == '/'
    negertive_sec_condToT = negertive_First_cond &&
            (negertive_sec_cond1 || negertive_sec_cond2 || negertive_sec_cond3 || negertive_sec_cond4 || negertive_sec_cond5)
    bCheckNegative = negertive_sec_condToT
    when (true) {
        bCheckBlank -> {
            blank_loop_cond = true
            while (blank_loop_cond) {
                blank_loop_cond = iStringOnPosition < sInput.length && sInput[iStringOnPosition] == ' '
                iStringOnPosition++
            }
        }
        bCheckNegative -> {
            bNegativePresent = true
            iStringOnPosition++
        }
    }

做一些我必须解决的事情？

【讨论】：