Division by powers of 2 is faster with unsigned int, because it can be optimized into a single shift instruction. With signed int, it usually requires more machine instructions, because division rounds towards zero, but shifting to the right rounds down. Example: int foo(int x, unsigned y) { x /= 8; y /= 8; return … Read more
Assuming: You have 2’s-complement representations in mind; and, By (unsigned long) you mean unsigned 32-bit integer, then you just need to add 2**32 (or 1 << 32) to the negative value. For example, apply this to -1: >>> -1 -1 >>> _ + 2**32 4294967295L >>> bin(_) ‘0b11111111111111111111111111111111’ Assumption #1 means you want -1 to … Read more
Starting Java 8, there is support for unsigned long (unsigned 64 bits). The way you can use it is: Long l1 = Long.parseUnsignedLong(“17916881237904312345”); To print it, you can not simply print l1, but you have to first: String l1Str = Long.toUnsignedString(l1) Then System.out.println(l1Str);
%lu is the correct format for unsigned long. Sounds like there are other issues at play here, such as memory corruption or an uninitialized variable. Perhaps show us a larger picture?
document.write( (1 << 31) +”<br/>”); The << operator is defined as working on signed 32-bit integers (converted from the native Number storage of double-precision float). So 1<<31 must result in a negative number. The only JavaScript operator that works using unsigned 32-bit integers is >>>. You can exploit this to convert a signed-integer-in-Number you’ve been … Read more
Well, even in Java 8, long and int are still signed, only some methods treat them as if they were unsigned. If you want to write unsigned long literal like that, you can do static long values = Long.parseUnsignedLong(“18446744073709551615”); public static void main(String[] args) { System.out.println(values); // -1 System.out.println(Long.toUnsignedString(values)); // 18446744073709551615 }
There’s no dedicated “character type” in C language. char is an integer type, same (in that regard) as int, short and other integer types. char just happens to be the smallest integer type. So, just like any other integer type, it can be signed or unsigned. It is true that (as the name suggests) char … Read more
Printing %d will read the integer as a signed decimal number, regardless of its defined type. To print unsigned numbers, use %u. This happens because of C’s way to handle variable arguments. The compiler just pulls values from the stack (typed as void* and pointing to the call stack) and printf has to figure out … Read more
Because the int value is promoted to an unsigned int. specifically 0xFFFFFFFC on a 32-bit machine, which as an unsigned int is 4294967292, considerably larger than 10 C99 6.3.1.1-p2 If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; … Read more