为什么它在下面的函数中使用额外的行“sumOf()” - Swift?

时间:2015-10-26 00:57:58

标签: swift

以下 public static void main(String[] args) { // TODO code application logic here System.out.println("Please enter the size of the array: "); Scanner input1 = new Scanner(System.in); int numBits = input1.nextInt(); Scanner input2 = new Scanner(System.in); System.out.println("Please enter size of radix: "); int radixNum = input2.nextInt(); getBits(radixNum); ArrayList<Integer> sortIt = new ArrayList<Integer>(); radSort(sortIt, radixNum, numBits); System.out.println(sortIt.toString()); } private static int getBits(int radixNum) { double lognum = Math.log(10000); double logbase = Math.log(radixNum); int numBits = (int) Math.floor((lognum / logbase) + 1); System.out.println("Bits " + numBits); return numBits; } /** * Sorts the array in a radix sort where the radix is decided by the user * * @param sortIt * @param radixNum */ private static ArrayList radSort(ArrayList<Integer> sortIt, int radixNum, int numBits) { ArrayList<ArrayList<Integer>> bucketStorage = new ArrayList(); ArrayList<Integer> temp; temp = sortIt; int tempNumb; for (int i = 0; i < radixNum; i++) { ArrayList<Integer> bucket = new ArrayList<>(); bucketStorage.add(bucket); } for (int j = 0; j < numBits; j++) { while (!temp.isEmpty()) { tempNumb = temp.remove(0); bucketStorage.get(getBitNumber(tempNumb, j, radixNum)).add(tempNumb); } for (int m = 0; m < bucketStorage.size(); m++) { for (int n = 0; n < bucketStorage.get(m).size(); n++) { temp.add(bucketStorage.get(m).get(n)); } bucketStorage.get(m).clear(); } } return temp; } private static int getBitNumber(int a, int b, int R) { double expNum; int entry; expNum = Math.pow(R, b); entry = (int) Math.floor(a / expNum) % R; return entry; } }是对的。但我不明白为什么它使用额外的行public void function(int number,for(int b;b>number;b++){int param2, int param3,int param4,int param5,int param6}) { //start of body for(int b;b>number;b++) { object[b].setter(param2) object[b].setter(param3) object[b].setter(param4) object[b].setter(param5) object[b].setter(param6) } } ? “func”的含义是什么? 有人有任何线索吗?

sumOf ( )

2 个答案:

答案 0 :(得分:2)

sumOf()行证明你可以完全没有参数地调用该函数。

答案 1 :(得分:0)

sumOf的课程都是&#34;不必要的&#34;对于功能本身。它只显示了用法。

Int...是一个可变参数

  

可变参数接受零个或多个指定类型的值。   https://developer.apple.com/library/watchos/documentation/Swift/Conceptual/Swift_Programming_Language/Functions.html