我收到的端口号是2个字节(最低有效字节在前),我想将其转换为整数以便进行操作。我已经编写了以下代码:
char buf[2]; //Where the received bytes are
char port[2];
port[0]=buf[1];
port[1]=buf[0];
int number=0;
number = (*((int *)port));
然而,有些问题,因为我没有获得正确的端口号。有任何想法吗?
我收到的端口号是2个字节(最低有效字节在前),我想将其转换为整数以便进行操作。我已经编写了以下代码:
char buf[2]; //Where the received bytes are
char port[2];
port[0]=buf[1];
port[1]=buf[0];
int number=0;
number = (*((int *)port));
然而,有些问题,因为我没有获得正确的端口号。有任何想法吗?
我会收到一个由2个字节构成的端口号(从低位字节开始)
然后您可以进行以下操作:
int number = buf[0] | buf[1] << 8;
char
是否被视为有符号数。 - Joachim Isaksson如果将buf
改为unsigned char buf[2];
,您可以将其简化为:
number = (buf[1] << 8) + buf[0];
我很感谢您已经得到了合理的答复。然而,另一种技术是在您的代码中定义一个宏,例如:
// bytes_to_int_example.cpp
// Output: port = 514
// I am assuming that the bytes the bytes need to be treated as 0-255 and combined MSB -> LSB
// This creates a macro in your code that does the conversion and can be tweaked as necessary
#define bytes_to_u16(MSB,LSB) (((unsigned int) ((unsigned char) MSB)) & 255)<<8 | (((unsigned char) LSB)&255)
// Note: #define statements do not typically have semi-colons
#include <stdio.h>
int main()
{
char buf[2];
// Fill buf with example numbers
buf[0]=2; // (Least significant byte)
buf[1]=2; // (Most significant byte)
// If endian is other way around swap bytes!
unsigned int port=bytes_to_u16(buf[1],buf[0]);
printf("port = %u \n",port);
return 0;
}
最不显著字节: int number = (uint8_t)buf[0] | (uint8_t)buf[1] << 8;
最显著字节: int number = (uint8_t)buf[1] << 8 | (uint8_t)buf[0];
char buf[2]; //Where the received bytes are
int number;
number = *((int*)&buf[0]);
&buf[0]
取得 buf 中第一个字节的地址。
(int*)
将其转换为整数指针。
最左边的 *
从该内存地址读取整数。
如果需要交换字节序:
char buf[2]; //Where the received bytes are
int number;
*((char*)&number) = buf[1];
*((char*)&number+1) = buf[0];