C# 计算偏移量的更简洁的方法是什么?
此代码:C# 计算偏移量的更简洁的方法是什么?,c#,arrays,algorithm,refactoring,C#,Arrays,Algorithm,Refactoring,此代码: private int GetColumnToPopulate(int columnBase, int offset) { int duckbillColNum = 0; switch (columnBase) { case 0: duckbillColNum = 1; break; case 1: duckbillColNum = 5; br
private int GetColumnToPopulate(int columnBase, int offset) {
int duckbillColNum = 0;
switch (columnBase) {
case 0:
duckbillColNum = 1;
break;
case 1:
duckbillColNum = 5;
break;
case 2:
duckbillColNum = 9;
break;
case 3:
duckbillColNum = 13;
break;
case 4:
duckbillColNum = 17;
break;
case 5:
duckbillColNum = 21;
break;
case 6:
duckbillColNum = 25;
break;
case 7:
duckbillColNum = 29;
break;
}
duckbillColNum += offset;
return duckbillColNum;
}
…很容易摸索,但冗长笨拙。我想精简一下。我的伪代码想法是:
int[] colBases
{
0..7
}
int[] offsets
{
1,5,9,13,17,21,25,29
}
int idx = colBases.IndexOf(columnBase);
duckbillColNum = offsets[idx];
…或者是一个多维整数数组?无论如何,“IndexOf”业务当然不符合[ile,ute]。在我看来,您应该用以下表达式替换开关
duckbillColNum=columnBase*4+1代码>不是吗?这只是一个简单的数学模式:
private int GetColumnToPopulate(int columnBase, int offset)
{ return 1 + (columnBase * 4) + offset; }
那么:
private int GetColumnToPopulate(int columnBase, int offset) {
return (columnBase * 4) + 1 + offset;
}
考虑使用字典
,您可以这样定义它:
Dictionary<int, int> offsets = new Dictionary<int, int>()
{
{ 0, 1 },
{ 1, 5 },
{ 2, 9 },
{ 3, 13 },
{ 4, 17 },
{ 5, 21 },
{ 6, 25 },
{ 7, 29 },
}
摆脱Colbase,一切都好你试过什么?你试过你的两个想法吗?这似乎不是一个经过充分研究的问题。它给我的印象是columnBase*4+1
。。。当然,如果你觉得足够的话。
return offsets[columnBase];