So if I have a String
char string[4];
string = "A10";
How Can I get 10 as an Integer.
I tried getting 10 by itself using this but it didn't work.
char string[4];
char string2[2];
string = "A10";
string2[0] = string[1];
string2[1] = string[2];
I don't need to worry about the A, I know how to get that, I need to be able to get the 10 as an integer.
In java you can use the following code to convert string to integer,
int i=Integer.parseInt(s2);
for more information view this website,
https://www.javatpoint.com/java-string-to-int
I have a string coming from PC through serial to a microcontroller (Arduino), e.g.:
"HDD: 55 - CPU: 12.6 - Weather: Cloudy [...] $";
by this function I found:
String inputStringPC = "";
boolean stringCompletePC = false;
void serialEvent() {
while (Serial.available()) {
char inChar = (char)Serial.read();
inputStringPC += inChar;
if (inChar == '$') // end marker of the string
{
stringCompletePC = true;
}
}
}
I would like to extract the first number of it after the word HDD, CPU and also get the string after Weather (ie "cloudy"); my thinking is something like that:
int HDD = <function that does that>(Keyword HDD);
double CPU = <function that does that>(Keyword CPU);
char Weather[] = <function that does that>(Keyword Weather);
What is the right function to do that?
I looked into inputStringSerial.indexOf("HDD") but I am still a learner to properly understand what it does and don't know if theres a better function.
My approach yielded some syntax errors and confused me with the difference in usage between "String inputStringSerial" (class?) and "char inputStringSerial[]" (variable?). When I do 'string inputStringSerial = "";' PlatformIO complains that "string" is undefined. Any help to understand its usage here is greatly appreciated.
Thanks a bunch.
The String class provides member functions to search and copy the contents of the String. That class and all its member functions are documented in the Arduino Reference:
https://www.arduino.cc/reference/tr/language/variables/data-types/stringobject/
The other way a list of characters can be represented is a char array, confusingly also called a string or cstring. The functions to search and copy the contents of a char array are documented at
http://www.cplusplus.com/reference/cstring/
Here is a simple Sketch that copies and prints the value of the Weather field using a String object. Use this same pattern - with different head and terminator values - to copy the string values of the other fields.
Once you have the string values of HDD and CPU, you'll need to call functions to convert those string values into int and float values. See the String member functions toInt() and toFloat() at
https://www.arduino.cc/reference/en/language/variables/data-types/string/functions/toint/
or the char array functions atoi() and atof() at
http://www.cplusplus.com/reference/cstdlib/atoi/?kw=atoi
String inputStringPC = "HDD: 55 - CPU: 12.6 - Weather: Cloudy [...] $";
const char headWeather[] = "Weather: "; // the prefix of the weather value
const char dashTerminator[] = " -"; // one possible suffix of a value
const char dollarTerminator[] = " $"; // the other possible suffix of a value
void setup() {
int firstIndex; // index into inputStringPC of the first char of the value
int lastIndex; // index just past the last character of the value
Serial.begin(9600);
// find the Weather field and copy its string value.
// Use similar code to copy the values of the other fields.
// NOTE: This code contains no error checking for unexpected input values.
firstIndex = inputStringPC.indexOf(headWeather);
firstIndex += strlen(headWeather); // firstIndex is now the index of the char just past the head.
lastIndex = inputStringPC.indexOf(dollarTerminator, firstIndex);
String value = inputStringPC.substring(firstIndex, lastIndex);
Serial.print("Weather value = '");
Serial.print(value);
Serial.println("'");
}
void loop() {
// put your main code here, to run repeatedly:
}
When run on an Arduio Uno, this Sketch produces:
Weather value = 'Cloudy [...]'
I am trying to use windows form application picturebox method Load. My code to do so is the following:
string image = "image.jpg";
pictureBox2->Size = System::Drawing::Size(36, 40);
pictureBox2->Load(image);
Controls->Add( pictureBox2);
However, I am getting the following errors:
'void System::Windows::Forms::PictureBox::Load(System::String ^)' : cannot convert parameter 1 from 'std::basic_string<_Elem,_Traits,_Ax>' to 'System::String ^'
Any idea how can I convert string to String ^?
use simply :
String ^image = "image.png";
Load wait a managed string.
This should do it:
std::string image("image.png");
String^ MyString = gcnew String(image.c_str());
I'm trying to make as string out of multiple ints.
Let's say:
int year = 1995;
int month = 12;
int day = 18;
const char* date = ("%d-%d-%d", month, day, year);
I get:
error: invalid conversion from 'int' to 'const char*' [-fpermissive]
What's the best way to go about this?
First, convert your parameters to strings with std::to_string().
string m = std::to_string(month);
string d = std::to_string(day);
string y = std::to_string(year);
Then, concatenate them:
string datestr = m + d + y;
Finally, convert that string into a char const*, using c_str(), which converts a string int to a C-like string.
char const* date = date.c_str();
I would like to parse strings like 1 or 32.23 into integers and doubles. How can I do this with Dart?
You can parse a string into an integer with int.parse(). For example:
var myInt = int.parse('12345');
assert(myInt is int);
print(myInt); // 12345
Note that int.parse() accepts 0x prefixed strings. Otherwise the input is treated as base-10.
You can parse a string into a double with double.parse(). For example:
var myDouble = double.parse('123.45');
assert(myDouble is double);
print(myDouble); // 123.45
parse() will throw FormatException if it cannot parse the input.
In Dart 2 int.tryParse is available.
It returns null for invalid inputs instead of throwing. You can use it like this:
int val = int.tryParse(text) ?? defaultValue;
Convert String to Int
var myInt = int.parse('12345');
assert(myInt is int);
print(myInt); // 12345
print(myInt.runtimeType);
Convert String to Double
var myDouble = double.parse('123.45');
assert(myInt is double);
print(myDouble); // 123.45
print(myDouble.runtimeType);
Example in DartPad
As per dart 2.6
The optional onError parameter of int.parse is deprecated. Therefore, you should use int.tryParse instead.
Note:
The same applies to double.parse. Therefore, use double.tryParse instead.
/**
* ...
*
* The [onError] parameter is deprecated and will be removed.
* Instead of `int.parse(string, onError: (string) => ...)`,
* you should use `int.tryParse(string) ?? (...)`.
*
* ...
*/
external static int parse(String source, {int radix, #deprecated int onError(String source)});
The difference is that int.tryParse returns null if the source string is invalid.
/**
* Parse [source] as a, possibly signed, integer literal and return its value.
*
* Like [parse] except that this function returns `null` where a
* similar call to [parse] would throw a [FormatException],
* and the [source] must still not be `null`.
*/
external static int tryParse(String source, {int radix});
So, in your case it should look like:
// Valid source value
int parsedValue1 = int.tryParse('12345');
print(parsedValue1); // 12345
// Error handling
int parsedValue2 = int.tryParse('');
if (parsedValue2 == null) {
print(parsedValue2); // null
//
// handle the error here ...
//
}
void main(){
var x = "4";
int number = int.parse(x);//STRING to INT
var y = "4.6";
double doubleNum = double.parse(y);//STRING to DOUBLE
var z = 55;
String myStr = z.toString();//INT to STRING
}
int.parse() and double.parse() can throw an error when it couldn't parse the String
Above solutions will not work for String like:
String str = '123 km';
So, the answer in a single line, that works in every situation for me will be:
int r = int.tryParse(str.replaceAll(RegExp(r'[^0-9]'), '')) ?? defaultValue;
or
int? r = int.tryParse(str.replaceAll(RegExp(r'[^0-9]'), ''));
But be warned that it will not work for the below kind of string
String problemString = 'I am a fraction 123.45';
String moreProblem = '20 and 30 is friend';
If you want to extract double which will work in every kind then use:
double d = double.tryParse(str.replaceAll(RegExp(r'[^0-9\.]'), '')) ?? defaultValue;
or
double? d = double.tryParse(str.replaceAll(RegExp(r'[^0-9\.]'), ''));
This will work for problemString but not for moreProblem.
you can parse string with int.parse('your string value');.
Example:- int num = int.parse('110011'); print(num); // prints 110011 ;
If you don't know whether your type is string or int you can do like this:
int parseInt(dynamic s){
if(s.runtimeType==String) return int.parse(s);
return s as int;
}
For double:
double parseDouble(dynamic s){
if(s.runtimeType==String) return double.parse(s);
return s as double;
}
Therefore you can do parseInt('1') or parseInt(1)
void main(){
String myString ='111';
int data = int.parse(myString);
print(data);
}
String age = stdin.readLineSync()!; // first take the input from user in string form
int.parse(age); // then parse it to integer that's it
You can do this for easy conversion like this
Example Code Here
void main() {
var myInt = int.parse('12345');
var number = myInt.toInt();
print(number); // 12345
print(number.runtimeType); // int
var myDouble = double.parse('123.45');
var double_int = myDouble.toDouble();
print(double_int); // 123.45
print(double_int.runtimeType);
}