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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;
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原文地址:http://www.cnblogs.com/sonicit/p/4954449.html