将uint8_t数组转换为uint16_t

I have a uint8_t array that contains two elements:

我有一个包含两个元素的uint8_t数组：

uint8_t ui8[2]; // uint8_t array
ui8[0] = 70; // LSB
ui1[1] = 60; // MSB

I want to get a uint16_t number (not an array) from these two uin8_t values. I used this approach in order to get this result: uint16_t ui16 = 6070

我想从这两个uin8_t值得到一个uint16_t数字（不是一个数组）。我使用这种方法来得到这个结果：uint16_t ui16 = 6070

uint16_t ui16 = ui8[1] | (ui8[0] << 8);

But I got uint16_t ui16 = 15430;

但我得到了uint16_t ui16 = 15430;

Am using the wrong method to get what I need? Or is there something missing?

我用错误的方法得到我需要的东西？还是缺少一些东西？

3 个解决方案

#1

Maybe you meant to work with hexadecimal numbers :

也许你打算使用十六进制数字：

uint8_t ui8[2]; // uint8_t array
ui8[0] = 0x70; // LSB
ui1[1] = 0x60; // MSB

uint16_t ui16 = ui8[1] | (ui8[0] << 8);
printf("%x\n", ui16); // 7060

If you want to work with decimal number, when you need to multiply the "MSB" by 100 and add them. It's far more uncommon to work with decimal number for that.

如果要使用十进制数，则需要将“MSB”乘以100并添加它们。为此，使用十进制数更为常见。

uint8_t ui8[2]; // uint8_t array
ui8[0] = 70; // LSB
ui1[1] = 60; // MSB

uint16_t ui16 = ui8[1] + (ui8[0] * 100);
printf("%d\n", ui16); // 7060

Please not than in both case, "70" will be before the "60", because you're shifting the first element of the array (70). The 70 will be the MSB.

请不要在两种情况下，“70”将在“60”之前，因为你正在移动阵列的第一个元素（70）。 70将是MSB。

#2

you can also use union punning for it:

你也可以使用工会惩罚：

#include <stdio.h>
#include <stdint.h>

typedef union {
    uint8_t u8[2];
    uint16_t u16;
}data16;

int main() {
    data16 d16;

    d16.u8[0] = 0x60;
    d16.u8[1] = 0x70;

    printf("%hx\n", d16.u16);

    // it works in the opposite direction as well
    // lets try to store 7060 decimal in two bytes

    d16.u16 = 7060u;

    printf("storing %hu decimal in two bytes: LSB:0x%0hhx (%hhu decimal), MSB:0x%0hhx (%hhu decimal)\n", d16.u16, d16.u8[0], d16.u8[0], d16.u8[1], d16.u8[1]);

    return 0; }

#3

uint8_t ui8[2]; // uint8_t array
ui8[0] = 70; // LSB
ui1[1] = 60; // MSB

To copy the both values into a uint16, you can do as below:

要将这两个值复制到uint16，您可以执行以下操作：

uint16_t output = 0;
output |= (uint16_t)ui8[0] << 8;
output |= (uint16_t)ui8[1] << 0;

You can use the similar logic for writing the uint32_t/uint64_t from the array of uint8_t.

您可以使用类似的逻辑从uint8_t数组中写入uint32_t / uint64_t。

#1