java中,char类型能不能转换成int类型?能不能转换成string类型?能不能转换成double类型?


1. char类型可以转换成int类型。可以通过强制类型转换实现,例如:`(int) ch;`,其中ch是一个char类型的变量。 2. char类型可以转换成String类型。可以通过Character类的toString()方法或者String类的valueOf()方法实现,例如:`String.valueOf(ch);` 或者 `new String(ch);`。 3. char类型不能直接转换成double类型,但可以先转换成int类型,然后再转换成double类型。例如:`(double) (int) ch;`。 [2023-11-01 15:09:59 | AI小红书种草文 | 116点数解答]
相关提问
- dim caizhi1 as string dim caizhi2 as string dim caizhi as string dim dianzulv1 as double dim dianzulv2 as double dim midu1 as double dim midu2 as double dim zhijing as double dim jiejubi as double dim jiemianji as double dim zhouchang as double dim dianzusichangdu as double dim zhongliang as double dim quanshu as double dim jieju as double dim dianzulv as double dim midu as double dim zuzhi as double private sub command1_click() dim xlapp as object dim xlworkbook as object dim xlworkshee(32点数解答 | 2024-10-31 15:19:51)188
- // rod6.cpp : defines the entry point for the console application. // #include "stdafx.h" // consoleapplication2.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。 // #include <stdio.h> #include <math.h> #include <iostream> #include <conio.h> #include <cstdlib> using namespace std; double ro[10001]; double th[10001]; double e[10001]; double a[10001]; double b[10001]; double y[10001]; double x[10001]; double d[10001]; double d[10001]; double da[10001]; double db[10001]; double dc[10001]; double z[10001]; d(55点数解答 | 2024-08-16 15:22:27)201
- 采用c++语言,实现如下功能: 实现古典密码中的移位密码和仿射密码,具体实现如下接口: c++接口: <<<<<移位密码>>>>> bool is_valid_s(unsinged char k) { 判断k是否为合理的密钥 } int encrypt_s(unsigned char* p, unsigned char k) { 密钥合法则返回1,且密文覆盖明文: 密钥不合法则返回0. } int decrypt_s(unsigned char* c, unsigned char k) { 密钥合法则返回1,且明文覆盖密文: 密钥不合法则返回0. } <<<<<仿射密码>>>>> bool in_valid_a(unsinged char a, unsigned char b) { 判断a,b是否为合理的密钥 } int encrypt_a(unsigned char* p, unsigned char a, unsigned char b) { 密钥合法则返回1,且密文覆盖明文: 密钥不合法则返回0. } int decrypt_a(unsigned char* c, unsign(812点数解答 | 2024-12-18 16:02:36)224
- #define _crt_secure_no_warnings #include <iostream> #include <cstdlib> #include <cmath> #include <vector> #include <cstdio> using namespace std; // 你的代码... const int nx = 784, nb = 500, ny = 10;//输入层x,隐藏层b,输出层y节点数 const double eta = 0.1;//学习率 struct node { double val{}; double bias{}; vector<double> weight; } x[nx], b[nb], y[ny];//输入层x,隐藏层b,输出层y double g[ny], e[nb];//用于反向传播 double trainx[nx], trainy[ny];//训练数据 double sigmoid(double x) { return 1.0 / (1.0 + exp(-x)); } double(34点数解答 | 2024-09-06 08:43:52)254
- #define _crt_secure_no_warnings #include <iostream> #include <cstdlib> #include <cmath> #include <vector> #include <cstdio> using namespace std; // 你的代码... const int nx = 784, nb = 500, ny = 10;//输入层x,隐藏层b,输出层y节点数 const double eta = 0.1;//学习率 struct node { double val{}; double bias{}; vector<double> weight; } x[nx], b[nb], y[ny];//输入层x,隐藏层b,输出层y double g[ny], e[nb];//用于反向传播 double trainx[nx], trainy[ny];//训练数据 double sigmoid(double x) { return 1.0 / (1.0 + exp(-x)); } double(1309点数解答 | 2024-09-06 08:46:09)262
- #include <stdio.h> #include <math.h> #define pi 3.14159265358979323846 double meridianarclength(double b) { double a = 6378140.0; double e2 = 0.00669438002290; double m = a * (1 - e2) * (1 - e2 * sin(b) * sin(b)) * pow(1 - e2 * sin(b) * sin(b), 1.5); double n = a / sqrt(1 - e2 * sin(b) * sin(b)); return m + n * sin(b) * cos(b) * pow(tan(b),2) /2 + n * sin(b) * pow(cos(b), 3) * (5 - pow(tan(b), 2) + 9 * e2 * pow(cos(b), 2)) * pow(tan(b), 4) / 24; } double calculatebaselatitude(double x, double y)(565点数解答 | 2024-11-21 13:08:38)151
- 快速排序:# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"(710点数解答 | 2024-11-14 10:44:25)204
- 快速排序:# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"(951点数解答 | 2024-11-14 10:46:29)198
- 快速排序:# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"(472点数解答 | 2024-11-14 10:51:44)179
- c语言现在有n枚硬币,其中有一枚是假的,假的硬币比真币轻,但也只轻一点点,而**和真币的外观是一模一样的,从外观上无法辨别硬币的真假,请写程序找出**。 程序的要求: 必须使用递归调用函数实现 递归函数原型必须为:int findfakecoin(int *p, int low, int high) 程序中必须包含函数:int getsum(int *p, int start, int end) main函数已经写好,请编写程序剩余部分的代码并提交。 main函数如下: int main() { int coins[n]; int m; int index; int getsum(int *p, int start, int end); int findfakecoin(int *p, int low, int high); scanf("%d", &m); for (int i = 0; i < m; ++i(497点数解答 | 2024-12-02 22:03:42)184
- ```csharp using system; interface igetmoney { double callfee(int minutes); } class acard : igetmoney { private const double initialfee =100; private const double feeperminute = 0.1; private const int freeminutes = 3; private double balance; public acard() { balance = initialfee; } public double callfee(int minutes) { double fee = 0; if (minutes <= freeminutes) { fee = 0; } else { fee = (minutes - freeminutes) * feeperminute; } balance -= fee; return fee; } public double getbalance() { re(291点数解答 | 2024-05-27 16:22:31)236
- void __fastcall sub_80022ec(int a1, uint8_t *a2, uint8_t *a3, uint8_t *a4) { _byte *v4; // r4 unsigned __int8 *v5; // r4 int v6; // r4 int v7; // r4 int v8; // r4 int v9; // r4 int v10; // r4 int v11; // r4 int v12; // r4 int v13; // r4 int v14; // r4 int v15; // r4 int v16; // r4 int v17; // r4 int v18; // r4 int v19; // r4 int v20; // r5 int v21; // r6 int v22; // lr unsigned int v23; // r6(143点数解答 | 2024-11-01 18:27:18)182