Question

我有一些包含嵌套if语句的代码：

if(numberOfNeighbors == 1){

                //go through comparison again, add Pixel(i,j) to current linked list -> complist[numberOfComponents]
                    //  break out of large check ??

                    if(ji.getPixelColor(i, j) == (ji.getPixelColor(i-1,j-1))){ //compare to top left
                        complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break; 
                    }
                    if(ji.getPixelColor(i, j) == (ji.getPixelColor(i,j-1))){ // compare to top
                        complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break;
                    }

                    if(ji.getPixelColor(i, j) == (ji.getPixelColor(i+1,j-1))){ // compare to top right
                        complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break;
                    }
                    if(ji.getPixelColor(i, j) == (ji.getPixelColor(i-1,j))){ // compare to left
                        complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break;
                    }
} // end of if(numberOfNeighbors == 1)

基本上我想做的事情，无论多么低效，都是比较一下4次，但如果事实证明它是匹配的，那么就要突破4个嵌套if语句的集合，以及外部if语句。

这会有用吗？或者它是否会突破嵌套，如果它当前处于并继续到下一个直到它通过所有4？

Answer 1

重要提示：break语句用于循环而不是分支

我理解了您的问题，但使用break语句来消除for，while，do while等循环。当满足条件并执行if分支内的语句时，您可以退出if语句。如果您不想在满足第一个if时检查其他条件，则必须使用if else个分支，而不是使用4个if语句。这两个链接可能很有用

见下面的例子

if(condition) {
    if(condition) { //if this evaluates to true, logic1 is executed
        logic1;
    }
    else if(condition) { //if the above condition fails, but this condition satisfies then logic 2 is executed
        logic2;
    }
    else { //if the above 2 conditions fail, you can execute logic3
        logic3;
    }
}

Answer 2

您在寻找else吗？

if(numberOfNeighbors == 1){

            //go through comparison again, add Pixel(i,j) to current linked list -> complist[numberOfComponents]
                //  break out of large check ??

                if(ji.getPixelColor(i, j) == (ji.getPixelColor(i-1,j-1))){ //compare to top left
                    complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break; 
                }
                else if(ji.getPixelColor(i, j) == (ji.getPixelColor(i,j-1))){ // compare to top
                    complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break;
                }

                else if(ji.getPixelColor(i, j) == (ji.getPixelColor(i+1,j-1))){ // compare to top right
                    complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break;
                }
                else if(ji.getPixelColor(i, j) == (ji.getPixelColor(i-1,j))){ // compare to left
                    complist[numberOfComponents].addFirst(new Pixel(i,j,numberOfComponents)); break;
                }
} // end of if(numberOfNeighbors == 1)

Answer 3

中断;不适用于if语句，只适用于循环，切换等。

您可以这样做以避免嵌套：

if(mainCondition)
{
  if(condition1)
    goto LabelContinue;

  bool condition2 = logic...;

  if(condition2)
    goto LabelContinue;

  //Code
}
LabelContinue:
//Other code.

突破嵌套if

3 个答案: