Assuming you want lower case letters: var chr = String.fromCharCode(97 + n); // where n is 0, 1, 2 … 97 is the ASCII code for lower case ‘a’. If you want uppercase letters, replace 97 with 65 (uppercase ‘A’). Note that if n > 25, you will get out of the range of letters.
Use sprintf(): int someInt = 368; char str[12]; sprintf(str, “%d”, someInt); All numbers that are representable by int will fit in a 12-char-array without overflow, unless your compiler is somehow using more than 32-bits for int. When using numbers with greater bitsize, e.g. long with most 64-bit compilers, you need to increase the array size—at … Read more
Use the modulo (%) operator to check if there’s a remainder when dividing by 2: if (x % 2) { /* x is odd */ } A few people have criticized my answer above stating that using x & 1 is “faster” or “more efficient”. I do not believe this to be the case. Out … Read more
Actually, valueOf uses parseInt internally. The difference is parseInt returns an int primitive while valueOf returns an Integer object. Consider from the Integer.class source: public static int parseInt(String s) throws NumberFormatException { return parseInt(s, 10); } public static Integer valueOf(String s, int radix) throws NumberFormatException { return Integer.valueOf(parseInt(s, radix)); } public static Integer valueOf(String s) … Read more
When you compile a number literal in Java and assign it to a Integer (capital I) the compiler emits: Integer b2 =Integer.valueOf(127) This line of code is also generated when you use autoboxing. valueOf is implemented such that certain numbers are “pooled”, and it returns the same instance for values smaller than 128. From the … Read more
“Integer” is an arbitrary precision type: it will hold any number no matter how big, up to the limit of your machine’s memory…. This means you never have arithmetic overflows. On the other hand it also means your arithmetic is relatively slow. Lisp users may recognise the “bignum” type here. “Int” is the more common … Read more
See the Integer class and the static parseInt() method: http://developer.android.com/reference/java/lang/Integer.html Integer.parseInt(et.getText().toString()); You will need to catch NumberFormatException though in case of problems whilst parsing, so: int myNum = 0; try { myNum = Integer.parseInt(et.getText().toString()); } catch(NumberFormatException nfe) { System.out.println(“Could not parse ” + nfe); }
System.out.println(NumberFormat.getNumberInstance(Locale.US).format(35634646)); Output: 35,634,646
This is what list comprehensions are for: numbers = [ int(x) for x in numbers ]
using Java NIO’s ByteBuffer is very simple: byte[] bytes = ByteBuffer.allocate(4).putInt(1695609641).array(); for (byte b : bytes) { System.out.format(“0x%x “, b); } output: 0x65 0x10 0xf3 0x29