Formally, you cannot do this. A char containing a character value has a different bit configuration than an int. Attempting to convert between the two is non-portable.
However, in the ASCII character set, the character '0' has the value 48, or 0x30. If you subtract 48 from the char you will get the int value, but only if the char was one of the digits '0' through '9'.
It is better to use the library routines to convert, such as atoi and sscanf, because they will give you predictable, portable results.
You can't convert the data type of any variable.
Yes: unsigned char CircLeft (unsigned char value) { if (value&0x80) return (value<<1) + 1; else return (value<<1); } unsigned char CircRight (unsigned char value) { if (value&0x01) return (value>>1) + 0x80; else return (value>>1); }
Character.toString('c')
Assuming - char mychar ; and int myint have been properly declared, myint = (int) mychar; // converts This is a feature of Java= to change types, put the type you want to convert into in parenthes before the variable that stores the converted value. ^Will convert your char into an int, but it will be the ascii value. For example, mychar = 3; myint = (int) mychar; //returns 51 To make an accurate one that will not return an error, you can use a try-catch statement Example char mychar = '3'; try { int myint = (int) mychar; }catch(NumberFormatException e) { //Do whatever you want with it! }
The main difference is that int values are treated as being integers whereas char values are treated as being character codes. Thus if you output a char with value 65 you will get the symbol 'A', but if you output an int with the value 65 you will get the value 65 instead. In order to output the symbol 'A' you would have to cast the int to a char.
In C a character already is its ASCII value: char c= 'A'; printf ("%c is %d (0x%x hexa)\n", c, c, c);
That depends on what language you're using. In PHP for example, it would be like this: $c = chr($i); In C, it would be: char c = (char)i; in BASIC, you'd use: LET C$ = CHR$(I)
char
const char *p means the char value pointed by 'p' is constant we can't change anyway but the address(location) of 'p' can change. char const *p means the char value pointed by 'p' can change but the location of p can't be change it is constant.
It depends on what you mean by "convert an int to char". If you simply want to cast the value in the int, you can cast it using Java's typecast notation: int i = 97; // 97 is 'a' in ASCII char c = (char) i; // c is now 'a' If you mean transforming the integer 1 into the character '1', you can do it like this: if (i >= 0 && i <= 9) { char c = Character.forDigit(i, 10); .... }
The statement char ch = 'z'; would store the character 'z' in the variable ch. This means that the variable ch would hold the value 'z'.
In order to print a character using its ASCII value, you need to first assign it to a char value like this: char c = (char) 65; In this example, we are casting the int 65 to a char, which converts it to an 'A', since 65 is the ASCII value for the capital letter 'a'. Next, you can print it out if you want: System.out.println(c); That's pretty much all there is to it!