将一下字节数组转为long byte[] bytes = { 0x6f, 0x12, 0x83, 0xc0, 0xca, 0x23, 0x63, 0x40 };
时间: 2024-10-28 20:08:15 浏览: 14
将给定的字节数组转换为`long`类型的整数,可以按照大端字节序(Big Endian),也就是从数组的第一个元素开始,逐字节向右组合。对于这个例子:
```java
byte[] bytes = { 0x6f, 0x12, 0x83, 0xc0, 0xca, 0x23, 0x63, 0x40 };
long longValue = ((long)bytes[0] << 56) |
((long)bytes[1] << 48) |
((long)bytes[2] << 40) |
((long)bytes[3] << 32) |
((long)bytes[4] << 24) |
((long)bytes[5] << 16) |
((long)bytes[6] << 8) |
((long)bytes[7]);
```
上面的代码首先左移第一个字节56位(因为`long`有64位,第一位是符号位),然后依次左移后续每个字节的位数。这样就得到了对应的`long`值。
相关问题
C# 将一下字节数组转为double byte[] bytes = { 0x6f, 0x12, 0x83, 0xc0, 0xca, 0x23, 0x63, 0x40 }; 转为经度
在C#中,将字节数组转换为双精度浮点数(double类型)并解析成地理坐标(如经度)通常需要一些额外步骤,因为字节序列不一定直接对应经纬度值。这个特定的字节数组看起来像是某种二进制数据表示,比如网络字节序(Big Endian),它可能是由某些编码算法生成的。
假设字节顺序是正确的,并且这些字节代表了经纬度的一部分(例如,前几个字节表示经度部分),你可以这样做:
```csharp
byte[] bytes = { 0x6f, 0x12, 0x83, 0xc0, 0xca, 0x23, 0x63, 0x40 };
double longitudeValue;
if (BitConverter.IsLittleEndian)
{
// 如果是小端模式,需要先反序
Array.Reverse(bytes);
}
longitudeValue = BitConverter.ToDouble(bytes, 0); // 双精度浮点数,从第一个字节开始
// 地理坐标计算可能涉及到进一步的处理和校验
double degrees = longitudeValue;
double minutes = degrees * 60 % 60; // 分钟
double seconds = minutes * 60; // 秒
// 实际上,你需要一个具体的坐标系统(比如WGS84)来确定经度
// double longitude = degrees + minutes / 60 + seconds / 3600;
```
但是请注意,上述代码仅仅是示例,并非实际的地理坐标解析过程。对于真实的地理位置数据,还需要考虑偏移量、转换公式或其他相关信息。
#include "reg51.h" #include "intrins.h" #define FOSC 11059200UL #define BRT (65536 - FOSC / 115200 / 4) sfr AUXR = 0x8e; sfr T2H = 0xd6; sfr T2L = 0xd7; sfr S2CON = 0x9a; sfr S2BUF = 0x9b; sfr IE2 = 0xaf; sfr P0M1 = 0x93; sfr P0M0 = 0x94; sfr P1M1 = 0x91; sfr P1M0 = 0x92; sfr P2M1 = 0x95; sfr P2M0 = 0x96; sfr P3M1 = 0xb1; sfr P3M0 = 0xb2; sfr P4M1 = 0xb3; sfr P4M0 = 0xb4; sfr P5M1 = 0xc9; sfr P5M0 = 0xca; bit busy; char wptr; char rptr; char buffer[16]; void Uart2Isr() interrupt 8 { if (S2CON & 0x02) { S2CON &= ~0x02; busy = 0; } if (S2CON & 0x01) { S2CON &= ~0x01; buffer[wptr++] = S2BUF; wptr &= 0x0f; } } void Uart2Init() { S2CON = 0x10; T2L = BRT; T2H = BRT >> 8; AUXR = 0x14; wptr = 0x00; rptr = 0x00; busy = 0; } void Uart2Send(char dat) { while (busy); busy = 1; S2BUF = dat; } void Uart2SendStr(char *p) { while (*p) { Uart2Send(*p++); } } void main() { P0M0 = 0x00; P0M1 = 0x00; P1M0 = 0x00; P1M1 = 0x00; P2M0 = 0x00; P2M1 = 0x00; P3M0 = 0x00; P3M1 = 0x00; P4M0 = 0x00; P4M1 = 0x00; P5M0 = 0x00; P5M1 = 0x00; Uart2Init(); IE2 = 0x01; EA = 1; Uart2SendStr("Uart Test !\r\n"); while (1) { if (rptr != wptr) { Uart2Send(buffer[rptr++]); rptr &= 0x0f; } } } 修改此程序使发送者发送字母o接收者连接的板子中端口P2.5灯开始亮其余灯灭
修改后的程序如下:
```c
#include "reg51.h"
#include "intrins.h"
#define FOSC 11059200UL
#define BRT (65536 - FOSC / 115200 / 4)
sfr AUXR = 0x8e;
sfr T2H = 0xd6;
sfr T2L = 0xd7;
sfr S2CON = 0x9a;
sfr S2BUF = 0x9b;
sfr IE2 = 0xaf;
sfr P0M1 = 0x93;
sfr P0M0 = 0x94;
sfr P1M1 = 0x91;
sfr P1M0 = 0x92;
sfr P2M1 = 0x95;
sfr P2M0 = 0x96;
sfr P3M1 = 0xb1;
sfr P3M0 = 0xb2;
sfr P4M1 = 0xb3;
sfr P4M0 = 0xb4;
sfr P5M1 = 0xc9;
sfr P5M0 = 0xca;
bit busy;
char wptr;
char rptr;
char buffer[16];
void Uart2Isr() interrupt 8 {
if (S2CON & 0x02) {
S2CON &= ~0x02;
busy = 0;
}
if (S2CON & 0x01) {
S2CON &= ~0x01;
buffer[wptr++] = S2BUF;
wptr &= 0x0f;
}
}
void Uart2Init() {
S2CON = 0x10;
T2L = BRT;
T2H = BRT >> 8;
AUXR = 0x14;
wptr = 0x00;
rptr = 0x00;
busy = 0;
}
void Uart2Send(char dat) {
while (busy);
busy = 1;
S2BUF = dat;
}
void Uart2SendStr(char *p) {
while (*p) {
Uart2Send(*p++);
}
}
void main() {
P0M0 = 0x00;
P0M1 = 0x00;
P1M0 = 0x00;
P1M1 = 0x00;
P2M0 = 0x00;
P2M1 = 0x00;
P3M0 = 0x00;
P3M1 = 0x00;
P4M0 = 0x00;
P4M1 = 0x00;
P5M0 = 0x00;
P5M1 = 0x00;
Uart2Init();
IE2 = 0x01;
EA = 1;
Uart2SendStr("Uart Test !\r\n");
while (1) {
if (rptr != wptr) {
char received = buffer[rptr++];
rptr &= 0x0f;
if (received == 'o') {
P2 = 0x20;
} else {
P2 = 0x00;
}
Uart2Send(received);
}
}
}
```
其中,修改的部分为:
```c
if (received == 'o') {
P2 = 0x20;
} else {
P2 = 0x00;
}
```
该部分代码会检测接收到的字符是否为字母'o',如果是,则点亮P2.5灯(二进制为0010 0000),否则将P2口置为0(灭灯)。
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