= *(last-2)) { return(*(unsigned char *)(first-2) - *(unsigned char *)(last
=0; int First=0; int mid=0; while (k<last) { First =0; //3.剩余大于等于两个子序列的元素个数 while (First<=last
= *(last-2)) { return(*(unsigned char *)(first-2) - *(unsigned char *)(last
得到数组b中的第三个元素3;'$.d.a'得到的就是"inner_a";'$.d.b[2]'得到的就是"inner_b";'$.b[1 to 2]'返回[2, 3];'$.b[last]'返回5;'$.b[last