首先我们来看看 (int) 强制类型转换: #include <stdio.h> #include <stdlib.h> int main(){ int a = (int)(-1.9); int b = (int)(1.9); printf("%d\n%d\n", a, b); return 0; } int 是将数值绝对值的小数部分去掉,再取相同的正负号.所以依据输入浮点数的正负性.能够得到四舍五入函数: int func(float x) { if(x &g
从网上得知 /etc/resolv.conf中的DNS配置是从/etc/resolvconf/resolv.conf.d/head中加载而来,所以每回改resolv.conf都会失效,在此文件里面已经有说明 打开/etc/resolv.conf内容如下:cat /etc/resolv.conf # Dynamic resolv.conf(5) file for glibc resolver(3) generated by resolvconf(8)# DO NOT EDIT THIS FILE
Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array. For example, Given nums = [0, 1, 3] return 2. Note: Your algorithm should run in linear runtime complexity. Could you implement it u
写出将字符串中的数字转换为整型的方法,如:"as31d2v"->312,并写出相应的单元测试,输入超过int范围时提示不合法输入. public struct ConvertResult { public ConvertState State; public int Number; } public enum ConvertState { // 输入不合法 InValid = , // 输入合法 Valid = } public class StringHelper { publ