静态uint8_t数组的输入过程和类型 [英] Input process and type for static uint8_t array
问题描述
我目前正在尝试将整数变量转换为Arduino IDE中的静态uint8_t数组的值.
I am currently trying to convert an integer variable to the value of a static uint8_t array in the Arduino IDE.
我正在使用:
#include <U8x8lib.h>
我确实知道uint8_t的行为类似于字节类型.
And I do understand that uint8_t acts similarly to the byte type.
当前，数组具有一个设置值:
Currently, the array has a set value:
static uint8_t hello[] = "world";
从我的角度来看，世界"看起来像一个字符串，所以我认为我将从创建一个字符串变量开始:
From my perspective, "world" looks like a string, so I thought I would start with creating a string variable:
String world = "world";
static uint8_t hello[] = world;
这行不通，并给了我错误:
This did not work and gave me the error:
initializer fails to determine size of 'hello'
如果我也这样做，而是将"world"更改为如下所示的int ...
If I do the same, but instead change "world" to an int like below...
int world = 1;
static uint8_t hello[] = world;
我遇到了同样的错误:
initializer fails to determine size of 'hello'
我已成功通过以下过程将uint8_t数组转换为字符串:
I have been successful in converting the uint8_t array to a string through the following process:
static uint8_t hello[] = "world";
String helloconverted = String((const char*)hello);
我不理解以下内容:
-
uint8_t数组如何具有类似字符串的输入并可以正常工作，但是在涉及变量时却无法实现
How a uint8_t array can have a string-like input and work fine, but not when a variable is involved
如何创建字符串变量作为uint8_t数组的输入
How to create a string variable as the input for the uint8_t array
如何创建一个int变量作为uint8_t数组的输入
How to create a int variable as the input for the uint8_t array
在此先感谢您的帮助.
推荐答案
uint8_t数组如何具有类似字符串的输入并可以正常工作，但是 不是当涉及变量时
How a uint8_t array can have a string-like input and work fine, but not when a variable is involved
字符串字面量本质上是一个以null结尾的char数组.所以
String literal is essentially an array of chars with terminating null. So
static uint8_t hello[] = "world";
本质上是
static uint8_t hello[] = {'w','o','r','l','d','\0'};
这也是普通的数组副本初始化，并且所需的大小是从值中自动推导出的，这就是为什么可以使用[]而不是[size]
Which is also a normal array copy initialization, and the size needed is auto-deduced from the value, this is why you can use [] and not [size]
如何创建一个int变量作为uint8_t数组的输入
How to create a int variable as the input for the uint8_t array
由于int的大小在编译时是已知的，因此您可以创建大小为int的数组，然后使用memcpy逐字节地将int值复制到该数组中:
Since size of int is known at compile time you could create an array of size of int and copy int value to it byte by byte with memcpy:
int world = 1;
static uint8_t hello[sizeof(world)];
memcpy(hello, &world, sizeof(hello));
如何创建字符串变量作为uint8_t数组的输入
How to create a string variable as the input for the uint8_t array
您需要事先知道String的长度，以便创建一个足以容纳String值的数组:
You need to know the length of the String beforehand so you can create an array big enough to fit the String value:
String world = "Hello"; // 5 chars
static uint8_t hello[5];
world.toCharArray((char *)hello, sizeof(hello));
根据需要，您可能还需要处理终止空值.
Depending on what you need, you might want to also handle terminating null.
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