The language specification is where null is defined, and it says There is also a special null type, the type of the expression null, which has no name. Because the null type has no name, it is impossible to declare a variable of the null type or to cast to the null type. The null … Read more
They are completely different types see standard: 3.9.1 Fundamental types [basic.fundamental] 1 Objects declared as characters char) shall be large enough to store any member of the implementation’s basic character set. If a character from this set is stored in a character object, the integral value of that character object is equal to the value … Read more
For strings you can just write “abc” * 3, which works via StringOps and uses a StringBuffer behind the scenes. For characters I think your solution is pretty reasonable, although char.toString * n is arguably clearer. Do you have any reason to suspect the List.fill version isn’t efficient enough for your needs? You could write … Read more
It’s a multi-character literal. An ordinary character literal that contains more than one c-char is a multicharacter literal . A multicharacter literal has type int and implementation-defined value. Also from 6.4.4.4/10 in C11 specs An integer character constant has type int. The value of an integer character constant containing a single character that maps to … Read more
A char can be assigned to an int without a cast because that is a widening conversion. To do the reverse, an int to a char requires a cast because it is a narrowing conversion. See also JLS. Chapter 5. Conversions and Promotions.
Unlike C, a string and an array of char are different. A string in C# can be viewed as an array of char but you should consider them different, therefore the ‘==’ comparison isn’t appropriate. One easy way to see this is with the following simple expression if (“a” == ‘a’) { /* do something … Read more
Because in C character constants, such as ‘a’ have the type int. There’s a C FAQ about this suject: Perhaps surprisingly, character constants in C are of type int, so sizeof(‘a’) is sizeof(int) (though this is another area where C++ differs).
In Java, char is a numeric type. When you add 1 to a char, you get to the next unicode code point. In case of ‘A’, the next code point is ‘B’: char x=’A’; x+=1; System.out.println(x); Note that you cannot use x=x+1 because it causes an implicit narrowing conversion. You need to use either x++ … Read more
Here’s the way I’d do it (minimal number of variables needed): for (i=0; s[i]; s[i]==’.’ ? i++ : *s++);
You’re getting that because it’s adding the ASCII value of the char. You must convert it to an int first.