LUA 5.4 - 如何将 64 位数字转换为十六进制

Question

I wanted to convert numbers greater than 64 bits, including up to 256 bits number from decimal to hex in lua.

Example:

num = 9223372036854775807
num = string.format("%x", num)
num = tostring(num)
print(num) -- output is 7fffffffffffffff

but if I already add a single number, it returns an error in the example below:

num = 9223372036854775808
num = string.format("%x", num)
num = tostring(num)
print(num) -- error lua54 - bad argument #2 to 'format' (number has no integer representation)

Does anyone have any ideas?

Answer 1

I wanted to convert numbers greater than 64 bits, including up to 256 bits number from decimal to hex in lua.

Well that's not possible without involving a big integer library such as this one . Lua 5.4 has two number types: 64-bit signed integers and 64-bit floats, which are both to limited to store arbitrary 256-bit integers.

The first num in your example, 9223372036854775807 , is just the upper limit of int64 bounds ( -2^63 to 2^63-1 , both inclusive). Adding 1 to this forces Lua to cast it into a float64, which can represent numbers way larger than that at the cost of precision. You're then left with an imprecise float which has no "integer representation" as Lua tells you.

You could trivially reimplement %x yourself, but that wouldn't help you extend the precision/size of floats & ints. You need to find another number representation and find or write a bigint library to go with it. Options are:

• String representation: Represent numbers as hex- or bytestrings (base 256).
• Table representation: Represent numbers as lists of numbers (base 2^x where x is < 64)