在 Arduino 上将 int 或 String 转换为 char 数组

Question

I am getting an int value from one of the analog pins on my Arduino. How do I concatenate this to a String and then convert the String to a char[] ?

It was suggested that I try char msg[] = myString.getChars(); , but I am receiving a message that getChars does not exist.

Answer 1

1. To convert and append an integer, use operator += (or member function concat ):

    String stringOne = "A long integer: "; stringOne += 123456789;

2. To get the string as type char[] , use toCharArray() :

    char charBuf[50]; stringOne.toCharArray(charBuf, 50)

In the example, there is only space for 49 characters (presuming it is terminated by null). You may want to make the size dynamic.

Overhead

The cost of bringing in String (it is not included if not used anywhere in the sketch), is approximately 1212 bytes of program memory (flash) and 48 bytes RAM.

This was measured using Arduino IDE version 1.8.10 (2019-09-13) for an Arduino Leonardo sketch.

Answer 2

Just as a reference, below is an example of how to convert between String and char[] with a dynamic length -

// Define 
String str = "This is my string"; 

// Length (with one extra character for the null terminator)
int str_len = str.length() + 1; 

// Prepare the character array (the buffer) 
char char_array[str_len];

// Copy it over 
str.toCharArray(char_array, str_len);
 

Yes, this is painfully obtuse for something as simple as a type conversion, but somehow it's the easiest way.

Answer 3

You can convert it to char* if you don't need a modifiable string by using:

(char*) yourString.c_str();

This would be very useful when you want to publish a String variable via MQTT in arduino.

Answer 4

None of that stuff worked. Here's a much simpler way .. the label str is the pointer to what IS an array...

String str = String(yourNumber, DEC); // Obviously .. get your int or byte into the string

str = str + '\r' + '\n'; // Add the required carriage return, optional line feed

byte str_len = str.length();

// Get the length of the whole lot .. C will kindly
// place a null at the end of the string which makes
// it by default an array[].
// The [0] element is the highest digit... so we
// have a separate place counter for the array...

byte arrayPointer = 0;

while (str_len)
{
    // I was outputting the digits to the TX buffer

    if ((UCSR0A & (1<<UDRE0))) // Is the TX buffer empty?
    {
        UDR0 = str[arrayPointer];
        --str_len;
        ++arrayPointer;
    }
}

Answer 5

With all the answers here, I'm surprised no one has brought up using itoa already built in.

It inserts the string representation of the integer into the given pointer.

int a = 4625;
char cStr[5];       // number of digits + 1 for null terminator
itoa(a, cStr, 10);  // int value, pointer to string, base number

Or if you're unsure of the length of the string:

int b = 80085;
int len = String(b).length();
char cStr[len + 1];  // String.length() does not include the null terminator
itoa(b, cStr, 10);   // or you could use String(b).toCharArray(cStr, len);