快速返回乏型List对象或数值
User user= allData.rows.FindIndex(
delegate (User us)
{
return us.id == 1;
}
)
返回查找值的索引
List<int> list = new List<int>() { 1, 2, 3 };
int index = list.FindIndex(v1 => v1 == 2);
list[index] = 4;
posted @
2015-11-10 21:48
sonicit
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