Asked  7 Months ago    Answers:  5   Viewed   43 times

I have a char in c#:

char foo = '2';

Now I want to get the 2 into an int. I find that Convert.ToInt32 returns the actual decimal value of the char and not the number 2. The following will work:

int bar = Convert.ToInt32(new string(foo, 1));

int.parse only works on strings as well.

Is there no native function in C# to go from a char to int without making it a string? I know this is trivial but it just seems odd that there's nothing native to directly make the conversion.

 Answers

32

Interesting answers but the docs say differently:

Use the GetNumericValue methods to convert a Char object that represents a number to a numeric value type. Use Parse and TryParse to convert a character in a string into a Char object. Use ToString to convert a Char object to a String object.

http://msdn.microsoft.com/en-us/library/system.char.aspx

Tuesday, June 1, 2021
 
Xatoo
answered 7 Months ago
77

The value of a char can be 0-255, where the different characters are mapped to one of these values. The numeric digits are also stored in order '0' through '9', but they're also not typically stored as the first ten char values. That is, the character '0' doesn't have an ASCII value of 0. The char value of 0 is almost always the null character.

Without knowing anything else about ASCII, it's pretty straightforward how subtracting a '0' character from any other numeric character will result in the char value of the original character.

So, it's simple math:

'0' - '0' = 0  // Char value of character 0 minus char value of character 0
// In ASCII, that is equivalent to this:
48  -  48 = 0 // '0' has a value of 48 on ASCII chart

So, similarly, I can do integer math with any of the char numberics...

(('3' - '0') + ('5' - '0') - ('2' - '0')) + '0') = '6'

The difference between 3, 5, or 2 and 0 on the ASCII chart is exactly equal to the face value we typically think of when we see that numeric digit. Subtracting the char '0' from each, adding them together, and then adding a '0' back at the end will give us the char value that represent the char that would be the result of doing that simple math.

The code snippet above emulates 3 + 5 - 2, but in ASCII, it's actually doing this:

((51 - 48) + (53 - 48) - (50 - 48)) + 48) = 54

Because on the ASCII chart:

0 = 48
2 = 50
3 = 51
5 = 53
6 = 54
Wednesday, June 9, 2021
 
cbcp
answered 6 Months ago
96

In Swift 2.0, toInt(), etc., have been replaced with initializers. (In this case, Int(someString).)

Because not all strings can be converted to ints, this initializer is failable, which means it returns an optional int (Int?) instead of just an Int. The best thing to do is unwrap this optional using if let.

I'm not sure exactly what you're going for, but this code works in Swift 2, and accomplishes what I think you're trying to do:

let unsolved = "123abc"

var fileLines = [Int]()

for i in unsolved.characters {
    let someString = String(i)
    if let someInt = Int(someString) {
        fileLines += [someInt]
    }
    print(i)
}

Or, for a Swiftier solution:

let unsolved = "123abc"

let fileLines = unsolved.characters.filter({ Int(String($0)) != nil }).map({ Int(String($0))! })

// fileLines = [1, 2, 3]

You can shorten this more with flatMap:

let fileLines = unsolved.characters.flatMap { Int(String($0)) }

flatMap returns "an Array containing the non-nil results of mapping transform over self"… so when Int(String($0)) is nil, the result is discarded.

Wednesday, June 30, 2021
 
Freddie
answered 5 Months ago
59

The conversion from char (a 2-byte type) to int (a 4-byte type) is implicit in Java, because this is a widening conversion -- all of the possible values you can store in a char you can also store in an int. The reverse conversion is not implicit because it is a narrowing conversion -- it can lose information (the upper two bytes of the int are discarded). You must always explicitly cast in such scenarios, as a way of telling the compiler "yes, I know this may lose information, but I still want to do it."

Thursday, July 29, 2021
 
tdous
answered 4 Months ago
19

chars are automatically promoted to integers in C expressions

Yes, they are. C99 section 6.3.1.8, Usual arithmetic conversions:

Many operators that expect operands of arithmetic type cause conversions and yield result types in a similar way. The purpose is to determine a common real type for the operands and result. For the specified operands, each operand is converted, without change of type domain, to a type whose corresponding real type is the common real type. Unless explicitly stated otherwise, the common real type is also the corresponding real type of the result, whose type domain is the type domain of the operands if they are the same, and complex otherwise. This pattern is called the usual arithmetic conversions:

  • First, if the corresponding real type of either operand is long double, the other operand is converted, without change of type domain, to a type whose corresponding real type is long double.
  • Otherwise, if the corresponding real type of either operand is double, the other operand is converted, without change of type domain, to a type whose corresponding real type is double.
  • Otherwise, if the corresponding real type of either operand is float, the other operand is converted, without change of type domain, to a type whose corresponding real type is float.62)
  • Otherwise, the integer promotions are performed on both operands. Then the following rules are applied to the promoted operands:
    • If both operands have the same type, then no further conversion is needed.
    • Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank.
    • Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.
    • Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.
    • Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

Integer promotions are described on Section 6.3.1.1.2:

The following may be used in an expression wherever an int or unsigned int may be used:

  • An object or expression with an integer type whose integer conversion rank is less than or equal to the rank of int and unsigned int.
  • A bit-field of type _Bool, int, signed int, or unsigned int

If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions. All other types are unchanges by the integer promotions.

The rank of a char is less than or equal to that of an int, so char is included in here.

(As a footnote, it is mentioned that integer promotions are only applied as part of the usual arithmetic conversions, to certain argument expressions, to the operands of the unary +, - and ~, and to both operands of the shift operators).

As mentioned in the comments, integer promotion is also performed on function-call arguments.

Saturday, November 13, 2021
 
Arbalest
answered 3 Weeks ago
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