I have an int64 variable which contains a negative number and I wish to subtract it from an uint64 variable which contains a positive number:
var endTime uint64
now := time.Now().Unix()
endTime = uint64(now)
var interval int64
interval = -3600
endTime = endTime + uint64(interval)
The above code appears to work but I wonder if I can rely on this. I am surprised, being new to Go, that after casting a negative number to uint64 that it remains negative -- I had planned to subtract the now positive value (after casting) to get what I wanted.
Converting a signed number to an unsigned number will not remain negative, it can't, as the valid range of unsigned types doesn't include negative numbers. If you print uint(interval), you will certainly see a positive number printed.
What you experience is deterministic and you can rely on it (but it doesn't mean you should). It is the result of Go (and most other programming languages) storing signed integer types using the 2's completement representation.
What this means is that in case of negative numbers, using n bit, the value -x (where x is positive) is stored as the binary representation of the positive value 2^n - x. This has the advantage that numbers can be added bitwise, and the result will be correct regardless of whether they are negative or positive.
So when you have a signed negative number, it is basically stored in memory like if you would subtract its absolute value from 0. Which means that if you convert a negative, signed value to unsigned, and you add that to an unsigned value, the result will be correct because overflow will happen, in a useful way.
Converting a value of type int64 to uint64 does not change the memory layout, only the type. So what 8 bytes the int64 had, the converted uint64 will have those same 8 bytes. And as mentioned above, the representation stored in those 8 bytes is the bit pattern identical to the bit pattern of value 0 - abs(x). So the result of the conversion will be a number that you would get if you would subtract abs(x) from 0, in the unsigned world. Yes, this won't be negative (as the type is unsigned), instead it will be a "big" number, counting down from the max value of the uint64 type. But if you add an y number bigger than abs(x) to this "big" number, overflow will happen, and the result will be like y - abs(x).
See this simple example demonstrating what's happening (try it on the Go Playground):
a := uint8(100)
b := int8(-10)
fmt.Println(uint8(b)) // Prints 226 which is: 0 - 10 = 256 - 10
a = a + uint8(b)
fmt.Println(a) // Prints 90 which is: 100 + 226 = 326 = 90
// after overflow: 326 - 256 = 90
As mentioned above, you should not rely on this, as this may cause confusion. If you intend to work with negative numbers, then use signed types.
And if you work with a code base that already uses uint64 values, then do a subtraction instead of addition, using uint64 values:
interval := uint64(3600)
endTime -= interval
Also note that if you have time.Time values, you should take advantage of its Time.Add() method:
func (t Time) Add(d Duration) Time
You may specify a time.Duration to add to the time, which may be negative if you want to go back in time, like this:
t := time.Now()
t = t.Add(-3600 * time.Second)
time.Duration is more expressive: we see the value we specified above uses seconds explicitly.
Related
I've been reading this post on constants in Go, and I'm trying to understand how they are stored and used in memory. You can perform operations on very large constants in Go, and as long as the result fits in memory, you can coerce that result to a type. For example, this code prints 10, as you would expect:
const Huge = 1e1000
fmt.Println(Huge / 1e999)
How does this work under the hood? At some point, Go has to store 1e1000 and 1e999 in memory, in order to perform operations on them. So how are constants stored, and how does Go perform arithmetic on them?
Short summary (TL;DR) is at the end of the answer.
Untyped arbitrary-precision constants don't live at runtime, constants live only at compile time (during the compilation). That being said, Go does not have to represent constants with arbitrary precision at runtime, only when compiling your application.
Why? Because constants do not get compiled into the executable binaries. They don't have to be. Let's take your example:
const Huge = 1e1000
fmt.Println(Huge / 1e999)
There is a constant Huge in the source code (and will be in the package object), but it won't appear in your executable. Instead a function call to fmt.Println() will be recorded with a value passed to it, whose type will be float64. So in the executable only a float64 value being 10.0 will be recorded. There is no sign of any number being 1e1000 in the executable.
This float64 type is derived from the default type of the untyped constant Huge. 1e1000 is a floating-point literal. To verify it:
const Huge = 1e1000
x := Huge / 1e999
fmt.Printf("%T", x) // Prints float64
Back to the arbitrary precision:
Spec: Constants:
Numeric constants represent exact values of arbitrary precision and do not overflow.
So constants represent exact values of arbitrary precision. As we saw, there is no need to represent constants with arbitrary precision at runtime, but the compiler still has to do something at compile time. And it does!
Obviously "infinite" precision cannot be dealt with. But there is no need, as the source code itself is not "infinite" (size of the source is finite). Still, it's not practical to allow truly arbitrary precision. So the spec gives some freedom to compilers regarding to this:
Implementation restriction: Although numeric constants have arbitrary precision in the language, a compiler may implement them using an internal representation with limited precision. That said, every implementation must:
Represent integer constants with at least 256 bits.
Represent floating-point constants, including the parts of a complex constant, with a mantissa of at least 256 bits and a signed exponent of at least 32 bits.
Give an error if unable to represent an integer constant precisely.
Give an error if unable to represent a floating-point or complex constant due to overflow.
Round to the nearest representable constant if unable to represent a floating-point or complex constant due to limits on precision.
These requirements apply both to literal constants and to the result of evaluating constant expressions.
However, also note that when all the above said, the standard package provides you the means to still represent and work with values (constants) with "arbitrary" precision, see package go/constant. You may look into its source to get an idea how it's implemented.
Implementation is in go/constant/value.go. Types representing such values:
// A Value represents the value of a Go constant.
type Value interface {
// Kind returns the value kind.
Kind() Kind
// String returns a short, human-readable form of the value.
// For numeric values, the result may be an approximation;
// for String values the result may be a shortened string.
// Use ExactString for a string representing a value exactly.
String() string
// ExactString returns an exact, printable form of the value.
ExactString() string
// Prevent external implementations.
implementsValue()
}
type (
unknownVal struct{}
boolVal bool
stringVal string
int64Val int64 // Int values representable as an int64
intVal struct{ val *big.Int } // Int values not representable as an int64
ratVal struct{ val *big.Rat } // Float values representable as a fraction
floatVal struct{ val *big.Float } // Float values not representable as a fraction
complexVal struct{ re, im Value }
)
As you can see, the math/big package is used to represent untyped arbitrary precision values. big.Int is for example (from math/big/int.go):
// An Int represents a signed multi-precision integer.
// The zero value for an Int represents the value 0.
type Int struct {
neg bool // sign
abs nat // absolute value of the integer
}
Where nat is (from math/big/nat.go):
// An unsigned integer x of the form
//
// x = x[n-1]*_B^(n-1) + x[n-2]*_B^(n-2) + ... + x[1]*_B + x[0]
//
// with 0 <= x[i] < _B and 0 <= i < n is stored in a slice of length n,
// with the digits x[i] as the slice elements.
//
// A number is normalized if the slice contains no leading 0 digits.
// During arithmetic operations, denormalized values may occur but are
// always normalized before returning the final result. The normalized
// representation of 0 is the empty or nil slice (length = 0).
//
type nat []Word
And finally Word is (from math/big/arith.go)
// A Word represents a single digit of a multi-precision unsigned integer.
type Word uintptr
Summary
At runtime: predefined types provide limited precision, but you can "mimic" arbitrary precision with certain packages, such as math/big and go/constant. At compile time: constants seemingly provide arbitrary precision, but in reality a compiler may not live up to this (doesn't have to); but still the spec provides minimal precision for constants that all compiler must support, e.g. integer constants must be represented with at least 256 bits which is 32 bytes (compared to int64 which is "only" 8 bytes).
When an executable binary is created, results of constant expressions (with arbitrary precision) have to be converted and represented with values of finite precision types – which may not be possible and thus may result in compile-time errors. Note that only results –not intermediate operands– have to be converted to finite precision, constant operations are carried out with arbitrary precision.
How this arbitrary or enhanced precision is implemented is not defined by the spec, math/big for example stores "digits" of the number in a slice (where digits is not a digit of the base 10 representation, but "digit" is an uintptr which is like base 4294967295 representation on 32-bit architectures, and even bigger on 64-bit architectures).
Go constants are not allocated to memory. They are used in context by the compiler. The blog post you refer to gives the example of Pi:
Pi = 3.14159265358979323846264338327950288419716939937510582097494459
If you assign Pi to a float32 it will lose precision to fit, but if you assign it to a float64, it will lose less precision, but the compiler will determine what type to use.
Here is the contract repo. https://github.com/Learn-NEAR/NCD.L1.sample--lottery
I don't understand the play method here
https://github.com/Learn-NEAR/NCD.L1.sample--lottery/blob/2bd11bc1092004409e32b75736f78adee821f35b/src/lottery/assembly/lottery.ts#L11-L16
play(): bool {
const rng = new RNG<u32>(1, u32.MAX_VALUE);
const roll = rng.next();
logging.log("roll: " + roll.toString());
return roll <= <u32>(<f64>u32.MAX_VALUE * this.chance);
}
I don't understand the winning process but I'm sure it is hidden inside this method. So can someone explain how this play method works in detail?
To understand the winning process we should take a look at the play method in the lottery.ts file in the contract.
https://github.com/Learn-NEAR/NCD.L1.sample--lottery/blob/2bd11bc1092004409e32b75736f78adee821f35b/src/lottery/assembly/lottery.ts#L11-L16
play(): bool {
const rng = new RNG<u32>(1, u32.MAX_VALUE);
const roll = rng.next();
logging.log("roll: " + roll.toString());
return roll <= <u32>(<f64>u32.MAX_VALUE * this.chance);
}
There are a couple of things we should know about before we read this code.
bool
u32
f64
RNG<32>
bool means that our play method should only return true or false.
u32 is a 32-bit unsigned integer. It is a positive integer stored using 32 bits.
u8 has a max value of 255. u16 has a max value of 65535. u32 has a max value of 4294967295. u64 has a max value of 18446744073709551615. So, these unsigned integers can't be negative values.
f64 is a number that has a decimal place. This type can represent a wide range of decimal numbers, like 3.5, 27, -113.75, 0.0078125, 34359738368, 0, -1. So unlike integer types (such as i32), floating-point types can represent non-integer numbers, too.
RNG stands for Random Number Generator. It basically gives you a random number in the range of u32. And it takes two parameters that define the range of your method. In that case, the range is between 1 and u32.MAX_VALUE. In other words, it is 1 and 4294967296.
The next line creates a variable called roll and assigned it to the value of rng.next().
So, what does next() do? Think rng as a big machine which only has one big red button on it. When you hit that big red button, it gives you a number that this machine is capable of producing. Meaning, every time you hit that button, it gives you a number between 1 and u32.MAX_VALUE
The third line is just about logging the roll into the console. You should see something like that in your console roll: 3845432649
The last line looks confusing at the beginning but let's take a look piece by piece.
Here, u32.MAX_VALUE * this.chance we multiply this max value with a variable called chance which we defined as 0.2 in the Lottery class.
Then, we put <f64> at the beginning of this calculation because the result always will be a floating number due to 0.2.
Then, we put <32> at the beginning of all to convert that floating number into unsigned integer because we need to compare it with the roll which is an unsigned integer. You can't compare floating numbers with unsigned integers.
Finally, if the roll less than or equals to <u32>(<f64>u32.MAX_VALUE * this.chance) this, player wins.
var x uint64 = 257
var y int = 257
fmt.Println("rv1 is ", byte(x)) // ok
fmt.Println("rv2 is ", byte(y)) // ok
fmt.Println("rv3 is ", byte(257)) // constant 257 overflows byte
fmt.Println("rv4 is ", byte(int(257))) // constant 257 overflows byte
It is strange.
All of them are converting int to byte,so all of them should be error.
But case 1,2 is ok!
How could be that?
Variable numeric values can be converted to smaller types, with the normal loss of the high bits.
The compiler refuses to do this for constant values (that is clearly always an error). This is required by the spec (emphasize mine):
Every implementation must:
Represent integer constants with at least 256 bits.
Represent floating-point constants, including the parts of a complex constant, > with a mantissa of at least 256 bits and a signed binary exponent of at least 16 bits.
Give an error if unable to represent an integer constant precisely.
Give an error if unable to represent a floating-point or complex constant due to overflow.
Round to the nearest representable constant if unable to represent a floating-point or complex constant due to limits on precision.
These requirements apply both to literal constants and to the result of evaluating constant expressions.
Consequently, if you change var x and var y to const x and const y, you get an error for all four cases.
I've been reading this post on constants in Go, and I'm trying to understand how they are stored and used in memory. You can perform operations on very large constants in Go, and as long as the result fits in memory, you can coerce that result to a type. For example, this code prints 10, as you would expect:
const Huge = 1e1000
fmt.Println(Huge / 1e999)
How does this work under the hood? At some point, Go has to store 1e1000 and 1e999 in memory, in order to perform operations on them. So how are constants stored, and how does Go perform arithmetic on them?
Short summary (TL;DR) is at the end of the answer.
Untyped arbitrary-precision constants don't live at runtime, constants live only at compile time (during the compilation). That being said, Go does not have to represent constants with arbitrary precision at runtime, only when compiling your application.
Why? Because constants do not get compiled into the executable binaries. They don't have to be. Let's take your example:
const Huge = 1e1000
fmt.Println(Huge / 1e999)
There is a constant Huge in the source code (and will be in the package object), but it won't appear in your executable. Instead a function call to fmt.Println() will be recorded with a value passed to it, whose type will be float64. So in the executable only a float64 value being 10.0 will be recorded. There is no sign of any number being 1e1000 in the executable.
This float64 type is derived from the default type of the untyped constant Huge. 1e1000 is a floating-point literal. To verify it:
const Huge = 1e1000
x := Huge / 1e999
fmt.Printf("%T", x) // Prints float64
Back to the arbitrary precision:
Spec: Constants:
Numeric constants represent exact values of arbitrary precision and do not overflow.
So constants represent exact values of arbitrary precision. As we saw, there is no need to represent constants with arbitrary precision at runtime, but the compiler still has to do something at compile time. And it does!
Obviously "infinite" precision cannot be dealt with. But there is no need, as the source code itself is not "infinite" (size of the source is finite). Still, it's not practical to allow truly arbitrary precision. So the spec gives some freedom to compilers regarding to this:
Implementation restriction: Although numeric constants have arbitrary precision in the language, a compiler may implement them using an internal representation with limited precision. That said, every implementation must:
Represent integer constants with at least 256 bits.
Represent floating-point constants, including the parts of a complex constant, with a mantissa of at least 256 bits and a signed exponent of at least 32 bits.
Give an error if unable to represent an integer constant precisely.
Give an error if unable to represent a floating-point or complex constant due to overflow.
Round to the nearest representable constant if unable to represent a floating-point or complex constant due to limits on precision.
These requirements apply both to literal constants and to the result of evaluating constant expressions.
However, also note that when all the above said, the standard package provides you the means to still represent and work with values (constants) with "arbitrary" precision, see package go/constant. You may look into its source to get an idea how it's implemented.
Implementation is in go/constant/value.go. Types representing such values:
// A Value represents the value of a Go constant.
type Value interface {
// Kind returns the value kind.
Kind() Kind
// String returns a short, human-readable form of the value.
// For numeric values, the result may be an approximation;
// for String values the result may be a shortened string.
// Use ExactString for a string representing a value exactly.
String() string
// ExactString returns an exact, printable form of the value.
ExactString() string
// Prevent external implementations.
implementsValue()
}
type (
unknownVal struct{}
boolVal bool
stringVal string
int64Val int64 // Int values representable as an int64
intVal struct{ val *big.Int } // Int values not representable as an int64
ratVal struct{ val *big.Rat } // Float values representable as a fraction
floatVal struct{ val *big.Float } // Float values not representable as a fraction
complexVal struct{ re, im Value }
)
As you can see, the math/big package is used to represent untyped arbitrary precision values. big.Int is for example (from math/big/int.go):
// An Int represents a signed multi-precision integer.
// The zero value for an Int represents the value 0.
type Int struct {
neg bool // sign
abs nat // absolute value of the integer
}
Where nat is (from math/big/nat.go):
// An unsigned integer x of the form
//
// x = x[n-1]*_B^(n-1) + x[n-2]*_B^(n-2) + ... + x[1]*_B + x[0]
//
// with 0 <= x[i] < _B and 0 <= i < n is stored in a slice of length n,
// with the digits x[i] as the slice elements.
//
// A number is normalized if the slice contains no leading 0 digits.
// During arithmetic operations, denormalized values may occur but are
// always normalized before returning the final result. The normalized
// representation of 0 is the empty or nil slice (length = 0).
//
type nat []Word
And finally Word is (from math/big/arith.go)
// A Word represents a single digit of a multi-precision unsigned integer.
type Word uintptr
Summary
At runtime: predefined types provide limited precision, but you can "mimic" arbitrary precision with certain packages, such as math/big and go/constant. At compile time: constants seemingly provide arbitrary precision, but in reality a compiler may not live up to this (doesn't have to); but still the spec provides minimal precision for constants that all compiler must support, e.g. integer constants must be represented with at least 256 bits which is 32 bytes (compared to int64 which is "only" 8 bytes).
When an executable binary is created, results of constant expressions (with arbitrary precision) have to be converted and represented with values of finite precision types – which may not be possible and thus may result in compile-time errors. Note that only results –not intermediate operands– have to be converted to finite precision, constant operations are carried out with arbitrary precision.
How this arbitrary or enhanced precision is implemented is not defined by the spec, math/big for example stores "digits" of the number in a slice (where digits is not a digit of the base 10 representation, but "digit" is an uintptr which is like base 4294967295 representation on 32-bit architectures, and even bigger on 64-bit architectures).
Go constants are not allocated to memory. They are used in context by the compiler. The blog post you refer to gives the example of Pi:
Pi = 3.14159265358979323846264338327950288419716939937510582097494459
If you assign Pi to a float32 it will lose precision to fit, but if you assign it to a float64, it will lose less precision, but the compiler will determine what type to use.
I'm using the hash function murmur2 which returns me an uint64.
I want then to store it in PostgreSQL, which only support BIGINT (signed 64 bits).
As I'm not interested in the number itself, but just the binary value (as I use it as an id for detecting uniqueness (my set of values being of ~1000 values, a 64bit hash is enough for me) I would like to convert it into int64 by "just" changing the type.
How does one do that in a way that pleases the compiler?
You can simply use a type conversion:
i := uint64(0xffffffffffffffff)
i2 := int64(i)
fmt.Println(i, i2)
Output:
18446744073709551615 -1
Converting uint64 to int64 always succeeds: it doesn't change the memory representation just the type. What may confuse you is if you try to convert an untyped integer constant value to int64:
i3 := int64(0xffffffffffffffff) // Compile time error!
This is a compile time error as the constant value 0xffffffffffffffff (which is represented with arbitrary precision) does not fit into int64 because the max value that fits into int64 is 0x7fffffffffffffff:
constant 18446744073709551615 overflows int64