Power64: Assume r3 = a, r4 = b, zero-extendedĪdd r3, r3, r4 64-bit addition: r3 = r3 + r4 Insll a1, #0, a1 a1 = a1 zero-extended to 64-bit valueĪddq a0, a1, v0 64-bit addition: v0 = a0 + a1 Insll a0, #0, a0 a0 = a0 zero-extended to 64-bit value Alpha AXP: Assume a0 = a, a1 = b, both in canonical form (shift + zero-extend in one instruction) Ubfx x0, x0, 1, 32 Extract bits 1 through 32 from result Uxtw x0, w0 x0 = w0 zero-extended to 64-bit valueĪdd x0, w1, uxtw 64-bit addition: x0 = x0 + (uint32_t)w1 AArch64 (ARM 64-bit): Assume w0 = a, w1 = b, upper 32 bits unknown Mov edx, edx rdx = edx zero-extended to 64-bit valueĪdd rax, rdx 64-bit addition: rax = rax + rdx Mov eax, ecx rax = ecx zero-extended to 64-bit value (I follow the processor’s natural calling convention for what is in the upper 32 bits of 64-bit registers.) // x86-64: Assume ecx = a, edx = b, upper 32 bits unknown The results would look something like this for processor with native 64-bit registers. 'unsigned long long' is a 64-bit type. Suppose 'unsigned' is a 32-bit type and If your compiler supports integers larger than the size of an unsigned, say because unsigned is a 32-bit value but the native register size is 64-bit, or because the compiler supports multiword arithmetic, then you can cast to the larger data type: unsigned average(unsigned a, unsigned b) This will be too low if the original addition contained a carry from bit 0 to bit 1, which happens if bit 0 is set in both of the terms, so we detect that case and make the necessary adjustment.Īnd then there’s the technique in the style known as SWAR, which stands for “SIMD within a register”. The trick here is to pre-divide the values before adding. patent for which expired in 2016: unsigned average(unsigned a, unsigned b) There’s another algorithm that doesn’t depend on knowing which value is larger, the U.S. If you know which number is the larger number (which is often the case), then you can calculate the width and halve it: unsigned average(unsigned low, unsigned high) However, this gives the wrong answer in the face of integer overflow: For example, if unsigned integers are 32 bits wide, then it says that average(0x80000000U, 0x80000000U) is zero. So what happens if we attempt to assign a too-small value to a variable of type double? Let’s look at an example: for(int i = 1073 i <= 1076 i++) else if(pare(+0.Finding the average of two unsigned integers, rounding toward zero, sounds easy: unsigned average(unsigned a, unsigned b) That means the smallest positive value a double can have is Math.pow(2, -1074), which is equal to 4.9e-324.Īs a consequence, the precision of a double in Java does not support values between 0 and 4.9e-324, or between -4.9e-324 and 0 for negative values. The minimum exponent for the binary representation of a double is given as -1074. The chapter about Types, Values, and Variables in the Java SE language specification describes how floating-point types are represented. See the Java documentation for a list of all these methods.įurthermore, there are exact conversion methods, which throw an exception if there is an overflow during the conversion to another data type.įor the conversion from a long to an int: In addition to addExact(), the Math package in Java 8 provides corresponding exact methods for all arithmetic operations. The static method addExact() performs a normal addition, but throws an exception if the operation results in an overflow or underflow: 2147483646Įxception in thread 'main' : integer overflowĪt (Math.java:790)Īt (OverUnderflow.java:115)

Let’s look at an example first: int value = Integer.MAX_VALUE-1 There are situations where we don’t want to allow larger values, nor do we want an overflow to occur, and we want to throw an exception instead.Īs of Java 8, we can use the methods for exact arithmetic operations.

Fl studio integer overflow