What can be the reason for the problem? My method returns incorrect int values. When I give it hex value of AB or DC or something similar it returns int = 0 but when I give it a hex = 22 it returns me int = 22. (though int should be 34 in this case).
public int StatusBit(int Xx, int Rr) {
int Number;
int.TryParse(GetX(Xx,Rr), out Number);
return Number;
}
I tried to use Number = Convert.ToInt32(GetX(Xx,Rr)); but it gives same result but null instead of 0 for anything that includes letters.
Use Convert.ToInt32(string, int) instead. That way you can give a base the number should be interpreted in. E.g.
return Convert.ToInt32(GetX(Xx, Rr), 16);
(You also don't check the return value of TryParse which would give a hint that the parse failed.)
If you expect both decimal and hexadecimal numbers you need to branch according to how the number looks and use either base 10 or base 16. E.g. if your hexadeximal numbers always start with 0x you could use something along the following lines:
string temp = GetX(Xx, Rr);
return Convert.ToInt32(temp, temp.StartsWith("0x") ? 16 : 10);
But that would depend on how (if at all) you would distinguish the two. If everything is hexadecimal then there is no such need, of course.
Use NumberStyles.HexNumber:
using System;
using System.Globalization;
class Test
{
static void Main()
{
string text = "22";
int value;
int.TryParse(text, NumberStyles.HexNumber,
CultureInfo.InvariantCulture, out value);
Console.WriteLine(value); // Prints 34
}
}
Do you really want to silently return 0 if the value can't be parsed, by the way? If not, use the return value of int.TryParse to determine whether the parsing succeeded or not. (That's the reason it's returning 0 for "AB" in your original code.)
int.TryParse parses a base 10 integer.
Use Convert.ToUInt32(hex, 16) instead
here is my solution;
kTemp = int.Parse(xcc, System.Globalization.NumberStyles.HexNumber);
above kTemp is an integer, and xcc is a string.
xcc can be anything like; FE, 10BA, FE0912... that is to say; xcc is a string of hex characters in any length.
beware; I dont get the 0x prefix with my hex strings.
Related
Hello I am trying to convert String to Integer.
The code below shows a part from where I am trying to convert my string to integer.
if (other.gameObject.CompareTag("PickUp"))
{
if ( checkpointboolean == false)
{
string pickupName = other.ToString(); //other = Pickup4
//Remove first 6 letters thus remaining with '4'
string y = pickupName.Substring(6);
print(y); // 4 is being printed
int x = 0;
int.TryParse(y, out x);
print (x); // 0 is being printed
I also tried the below code and instead of '0' I am getting the following error:
if (other.gameObject.CompareTag("PickUp"))
{
if ( checkpointboolean == false)
{
//Get Object name ex: Pickup4
string pickupName = other.ToString();
//Remove first 6 letters thus remaining with '4'
string y = pickupName.Substring(6);
print(y);
int x = int.Parse(y);
FormatException: Input string was not in the correct format
System.Int32.Parse (System.String s)
int.TryParse returns a boolean, true if it succeeded and false if not. You need to wrap this in a if block and do something with that logic.
if(int.TryParse(y, out x))
print (x); // y was able to be converted to an int
else
// inform the caller that y was not numeric, your conversion to number failed
As far as why your number is not converted I could not say until you post what the string value is.
Your first attempt is the best for this case, that code works fine, The int.TryParse() giving 0 to the out parameter and returns false means the conversion is failed. The input string is not in the correct format/ it cannot be converted to an integer. You can check this by using the return value of the int.TryParse(). For this what you need to do is :-
if(int.TryParse(y, out x))
print (x); //
else
print ("Invalid input - Conversion failed");
First of all, ToString() usually uses for debug purpose so there's no guarantee that
other.ToString()
will return the expected "Pickup4" in the next version of the software. You, probably, want something like
int x = other.gameObject.SomeProperty;
int x = other.SomeOtherProperty;
int x = other.ComputePickUp();
...
If you, however, insist on .ToString() you'd rather not hardcode six letters (the reason is the same: debug information tends to change from version to version), but use regular expressions or something:
var match = Regex.Match(other.ToString(), "-?[0-9]+");
if (match.Success) {
int value;
if (int.TryParse(match.Value, out value))
print(value);
else
print(match.Value + " is not an integer value");
}
else
print("Unexpected value: " + other.ToString());
I'm trying to convert the string "127.0" to an integer.
I tried this function:
int getInt(string numStr)
{
int result;
int.TryParse(numStr, out result);
return result;
}
But when I call it as int x = getInt("127.0"); then int.TryParse() sets result to 0.
When I rewrite the function like this:
int getInt(string numStr)
{
result=Convert.ToInt32(numStr);
return result;
}
the same getInt() call throws this exception:
Input string was not in a correct format.
The issue here is that "127.0" is not an integer, it's a floating point number. You will need to parse it using one of the other floating point types (i.e. double, float, Decimal, etc.).
You may want to consider either stripping off any values after the decimal point and attempting to parse it, or parsing it as another type and casting it as an integer :
int result = (int)Convert.ToDouble("1.270");
You could also take advantage of the Math.Truncate() function which would give you the integer portion of your value :
int result = (int)Math.Truncate(Convert.ToDouble("127.0"));
First off, you need to check the return value of int.TryParse(). If it returns false, then the string could not be converted.
Had you done that, you would see it returned false because 127.0 does not describe an integer value (it describes a floating point value).
Note that decimal.TryParse() would succeed here. You need to figure out if you need an integer or floating point value, and reject data that is incorrect.
An int cannot contain a decimal point; that makes it either a double, a float, or a decimal. Try to pull the number minus anything from the decimal point over to the right, like this:
int getInt(string numStr)
{
int result;
string[] splitup;
string number;
if (numstr.Contains('.'))
{
splitup = numstr.Split('.');
number = splitup[0];
int.TryParse(number, out result);
}
else
{
int.TryParse(numstr, out result);
}
return result;
}
Rion Williams is absolutely correct, IMHO.
Along with the fact that what you are parsing is not an integer, I'd personally use the TryParse method. Many of the .NET types have it, and it's quite a bit "safer" (it won't throw exceptions) than just parsing a string.
Example:
string stringValue = "127.0";
int intValue;
if(Int32.TryParse(stringValue, out intValue))
{
// return value
}
// handle the failure
If you don't like that, I'd wrap it in a try-catch...
This question already has answers here:
How to convert string to integer in C#
(12 answers)
Closed 8 years ago.
I need to convert a string to integer. My string can be of any type (float/int/string/special character).
For example:
If my string is "2.3", I need to convert to = 2
If my string is "anyCharacter", I need to convert to = 0
If my string is "2", I need to convert to = 2
I tried the following:
string a = "1.25";int b = Convert.ToInt32(a);
I got the error:
Input string was not in a correct format
How do I convert it?
Use Double.TryParse() and once you get the value from it, convert it to int using Convert.ToInt():
double parsedNum;
if (Double.TryParse(YourString, out parsedNum) {
newInt = Convert.ToInt32(num);
}
else {
newInt = 0;
}
Try to parse it as a floating point number, and convert to integer after that:
double num;
if (Double.TryParse(a, out num) {
b = (int)num;
} else {
b = 0;
}
This should help: treat any string as if it were a double, then Math.Floor() it to round it down to the nearest integer.
double theNum = 0;
string theString = "whatever"; // "2.3"; // "2";
if(double.TryParse(theString, out theNum) == false) theNum = 0;
//finally, cut the decimal part
int finalNum = (int)Math.Floor(theNum);
NOTE: the if might not be needed per-se, due to theNum initialization, but it's more readable this way.
I think Convert.ToInt32 is the wrong place to look for - I would use Integer.Tryparse and if TryParse evaluates to false, assign a 0 to the variable. Before the TryParse, you could simply delete any character after the dot, if you find it in the string.
Also, keep in mind that some languages use "," as a separator.
Try:
if (int.TryParse(string, out int)) {
variable = int.Parse(string);
}
As far as I know, there isn't any generic conversion, so you'd have to do a switch to find out the type of the variable and then use either of the following (for each type):
int.Parse(string)
or
int.TryParse(string, out int)
The second one will return a boolean which you can use to see if the conversion passed or failed.
Your best option would be to use double or decimal parsing as this won't remove any decimal places, unlike int.
bool Int32.TryParse(string, out int)
The boolean return value indicates if the conversion was successful or not.
Try something like this:
public int ForceToInt(string input)
{
int value; //Default is zero
int.TryParse(str, out value);
return value;
}
This will do the trick. However I don't recommend taking this approach. It is better to control your input whereever you get it.
I am really stumped on this one. In C# there is a hexadecimal constants representation format as below :
int a = 0xAF2323F5;
is there a binary constants representation format?
Nope, no binary literals in C#. You can of course parse a string in binary format using Convert.ToInt32, but I don't think that would be a great solution.
int bin = Convert.ToInt32( "1010", 2 );
As of C#7 you can represent a binary literal value in code:
private static void BinaryLiteralsFeature()
{
var employeeNumber = 0b00100010; //binary equivalent of whole number 34. Underlying data type defaults to System.Int32
Console.WriteLine(employeeNumber); //prints 34 on console.
long empNumberWithLongBackingType = 0b00100010; //here backing data type is long (System.Int64)
Console.WriteLine(empNumberWithLongBackingType); //prints 34 on console.
int employeeNumber_WithCapitalPrefix = 0B00100010; //0b and 0B prefixes are equivalent.
Console.WriteLine(employeeNumber_WithCapitalPrefix); //prints 34 on console.
}
Further information can be found here.
You could use an extension method:
public static int ToBinary(this string binary)
{
return Convert.ToInt32( binary, 2 );
}
However, whether this is wise I'll leave up to you (given the fact it will operate on any string).
Since Visual Studio 2017, binary literals like 0b00001 are supported.
I have a Double which could have a value from around 0.000001 to 1,000,000,000.000
I wish to format this number as a string but conditionally depending on its size. So if it's very small I want to format it with something like:
String.Format("{0:.000000000}", number);
if it's not that small, say 0.001 then I want to use something like
String.Format("{0:.00000}", number);
and if it's over, say 1,000 then format it as:
String.Format("{0:.0}", number);
Is there a clever way to construct this format string based on the size of the value I'm going to format?
Use Math.Log10 of the absolute value of the double to figure out how many 0's you need either left (if positive) or right (if negative) of the decimal place. Choose the format string based on this value. You'll need handle zero values separately.
string s;
double epislon = 0.0000001; // or however near zero you want to consider as zero
if (Math.Abs(value) < epislon) {
int digits = Math.Log10( Math.Abs( value ));
// if (digits >= 0) ++digits; // if you care about the exact number
if (digits < -5) {
s = string.Format( "{0:0.000000000}", value );
}
else if (digits < 0) {
s = string.Format( "{0:0.00000})", value );
}
else {
s = string.Format( "{0:#,###,###,##0.000}", value );
}
}
else {
s = "0";
}
Or construct it dynamically based on the number of digits.
Use the # character for optional positions in the string:
string.Format("{0:#,###,##0.000}", number);
I don't think you can control the number of decimal places like that as the precision of the double will likely mess things up.
To encapsulate the logic of deciding how many decimal places to output you could look at creating a custom formatter.
The first two String.Format in your question can be solved by automatically removing trailing zeros:
String.Format("{0:#,##0.########}", number);
And the last one you could solve by calling Math.Round(number,1) for values over 1000 and then use the same String.Format.
Something like:
String.Format("{0:#,##0.########}", number<1000 ? number : Math.Round(number,1));
Following up on OwenP's (and by "extension" tvanfosson):
If it's common enough, and you're on C# 3.0, I'd turn it into an extension method on the double:
class MyExtensions
{
public static string ToFormmatedString(this double d)
{
// Take d and implement tvanfosson's code
}
}
Now anywhere you have a double you can do:
double d = 1.005343;
string d_formatted = d.ToFormattedString();
If it were me, I'd write a custom wrapper class and put tvanfosson's code into its ToString method. That way you could still work with the double value, but you'd get the right string representation in just about all cases. It'd look something like this:
class FormattedDouble
{
public double Value { get; set; }
protected overrides void ToString()
{
// tvanfosson's code to produce the right string
}
}
Maybe it might be better to make it a struct, but I doubt it would make a big difference. You could use the class like this:
var myDouble = new FormattedDouble();
myDouble.Value = Math.Pi;
Console.WriteLine(myDouble);