I think to some extent it was probably an arbitrary decision based on typical use of switch. A switch can essentially be implemented in two ways (or in principle, a combination): for a small number of cases, or ones whose values are widely dispersed, a switch essentially becomes the equivalent of a series of ifs … Read more
My JVM does this pretty straightforward thing to the inner loop when you use longs: 0x00007fdd859dbb80: test %eax,0x5f7847a(%rip) /* fun JVM hack */ 0x00007fdd859dbb86: dec %r11 /* i– */ 0x00007fdd859dbb89: mov %r11,0x258(%r10) /* store i to memory */ 0x00007fdd859dbb90: test %r11,%r11 /* unnecessary test */ 0x00007fdd859dbb93: jge 0x00007fdd859dbb80 /* go back to the loop top … Read more
a) Can you explain to me the difference between int64_t and long (long int)? In my understanding, both are 64 bit integers. Is there any reason to choose one over the other? The former is a signed integer type with exactly 64 bits. The latter is a signed integer type with at least 32 bits. … Read more
JavaScript has a Number type which is a 64 bit floating point number*. If you’re looking to convert a string to a number, use either parseInt or parseFloat. If using parseInt, I’d recommend always passing the radix too. use the Unary + operator e.g. +”123456″ use the Number constructor e.g. var n = Number(“12343”) *there … Read more
Converting the output is too late; the calculation has already taken place in integer arithmetic. You need to convert the inputs to double: System.out.println((double)completed/(double)total); Note that you don’t actually need to convert both of the inputs. So long as one of them is double, the other will be implicitly converted. But I prefer to do … Read more
This code will work, but isn’t that pretty. It first divides all three values (it floors the values, so you ‘lose’ the remainder), and then divides the remainder: long n = x / 3 + y / 3 + z / 3 + ( x % 3 + y % 3 + z % 3 … Read more
Add a capital L to the end: long value = 9223372036854775807L; Otherwise, the compiler will try to parse the literal as an int, hence the error message
There are specific suffixes for long (e.g. 39832L), float (e.g. 2.4f) and double (e.g. -7.832d). If there is no suffix, and it is an integral type (e.g. 5623), it is assumed to be an int. If it is not an integral type (e.g. 3.14159), it is assumed to be a double. In all other cases … Read more
Java 8 has Math.multiplyExact, Math.addExact etc. for ints and long. These throw an unchecked ArithmeticException on overflow.
If you want to get the integer part of that number you must first convert it to a floating number then cast to long. long l1 = (long)Convert.ToDouble(“1100.25”); You can use Math class to round up the number as you like, or just truncate… Math.Round Math.Ceil