c - 如何将混合类型数组转换为 char 数组并返回？

Question

我必须通过 SPI 将混合类型数组( union )从一个 uController(源)传输到另一个 uController(目标)，因此我必须在传输之前将整个传输 block 放入 uint8_t 数组，然后在将 uC 目标返回到混合类型 union 数组。

我尝试了以下方法，但它不起作用。 (简体)

int i;
int len=4;

union dummy{
    float f32;
    uint32_t u32;
    uint8_t u8[4];
};

union dummy inst1[len]; //First instance on source uC
union dummy inst2[len]; //Second instance on target uC


inst1[0].u8[0]=73;
inst1[1].f32=17.5;
//... and so on

printf("Source: %d, %f\n",inst1[0].u8[0],inst1[1].f32);

//Prepare SPI uint8_t array
uint8_t spi_arr[4*len];

*spi_arr=*inst1; //Wrong. What to do here?

//SPI uint8_t array arrived on target. Convert it back to union type

*inst2=*spi_arr; //Wrong. What to do here?

printf("Target: %d, %f\n",inst2[0].u8[0],inst2[1].f32);

它给我错误:

error: incompatible types when assigning to type 'uint8_t {aka unsigned char}' from type 'union dummy'

我想要的只是按原样来回复制数组的内容。如何做到这一点？

Answer 1

这是一个示例程序，展示了如何使用指针访问两种不同类型的数据:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

const int len = 4;

union dummy{
    float f32;
    uint32_t u32;
    uint8_t u8[4];
};

int main(void) {
    union dummy inst1[len]; //First instance on source uC
    union dummy *inst2; //Second instance on target uC
    uint8_t *spi_arr;

    for (int index = 0; index < len; index++) {
        for (int inner_index = 0; inner_index < 4; inner_index++) {
            inst1[index].u8[inner_index] = '1' + inner_index;
        }
    }

    spi_arr = (uint8_t *) inst1;

    for (int index = 0; index < sizeof(inst1); index++) {
        printf("%c", spi_arr[index]);
    }
    printf("\n\n");

    inst2 = (union dummy *) spi_arr;

    for (int index = 0; index < len; index++) {
        for (int inner_index = 0; inner_index < 4; inner_index++) {
            printf("%c", inst2[0].u8[inner_index]);
        }
        printf("\n");
    }

    return 0;
}

输出

1234123412341234

1234
1234
1234
1234