C:将无符号字符数组转换为带符号的int(反之亦然)

问题描述：

我正在尝试将无符号的char数组缓冲区转换为带符号的int(反之亦然).

I'm trying to convert an unsigned char array buffer into a signed int (vice versa).

下面是一个演示代码:

int main(int argv, char* argc[])
{
    int original = 1054;
    unsigned int i = 1054;
    unsigned char c[4];
    int num;

    memcpy(c, (char*)&i, sizeof(int));

    //num  = *(int*) c;                          // method 1 get
    memcpy((char *)&num, c, sizeof(int));        // method 2 get
    printf("%d\n", num);

    return 0;
}

1)我应该使用哪种方法将unsigned char []转换为int?

方法1获取或方法2获取? (或任何建议)

method 1 get or method 2 get? (or any suggestion)

2)如何将int原件转换为未签名的char []?

我需要通过一个仅接受无符号char []的缓冲区发送该整数

I need to send this integer via a buffer that only accepts unsigned char[]

当前我正在做的是将int转换为unsigned int，然后转换为char []，例如:

Currently what i'm doing is converting the int to unsigned int then to char[], example :

int g = 1054;
unsigned char buf[4];
unsigned int n;
n = g;
memcpy(buf, (char*)&n, sizeof(int));

虽然可以正常工作，但我不确定它的正确方法还是安全?

Although it works fine but i'm not sure if its the correct way or is it safe?

PS.我正在尝试通过USB串行通信(在Raspberry Pi和Arduino之间)在2个设备之间发送数据

PS. I'm trying to send data between 2 devices via USB serial communication (between Raspberry Pi & Arduino)

答

您可以将int(或unsigned int)和unsigned char数组都存储为union.此方法称为键入punning ，并且自C99以来，它已完全通过标准进行了清理(这很常见尽早练习).假设sizeof(int) == 4:

You could store both int (or unsigned int) and unsigned char array as union. This method is called type punning and it is fully sanitized by standard since C99 (it was common practice earlier, though). Assuming that sizeof(int) == 4:

#include <stdio.h>

union device_buffer {
    int i;
    unsigned char c[4];
};

int main(int argv, char* argc[])
{
    int original = 1054;

    union device_buffer db;
    db.i = original;

    for (int i = 0; i < 4; i++) {
        printf("c[i] = 0x%x\n", db.c[i]);
    }
}

请注意，由于字节顺序(即字节顺序)而存储数组中的值.

Note that values in array are stored due to byte order, i.e. endianess.

C:将无符号字符数组转换为带符号的int(反之亦然)

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