converting binary string into integer

[SYoon]

I have a binary number in string

a = "0b1001100"

How can I convert this into integer?

Thanks for any help.

[andrew cooke]

julia> parseint("0b1001"[3:], 2)

9

[Patrick O'Leary]

julia> a = "0b1001100"
"0b1001100"
julia> parse(a)
0x4c

[SYoon]

Great!

Thanks for both of you.

[andrew cooke]

this confused me for a moment; if it also confuses you it's because  julia parses non-decimal integers as unsigned, and prints unsigned integers as hex.

[Stefan Karpinski]

I would highly recommend that if you only need to parse integers you use parseint rather than parse. It will be faster and much safer.

[Patrick O'Leary]

On Tuesday, October 29, 2013 1:04:46 PM UTC-5, Stefan Karpinski wrote:

I would highly recommend that if you only need to parse integers you use parseint rather than parse. It will be faster and much safer.

Yes, that is generally more correct. You would need logic to handle prefixes, though, if they vary, unless that's been rolled into parseint()? I don't have an up-to-date Julia to play with atm (did this in the Forio web repl).

[Stefan Karpinski]

That's not a bad idea – when the base isn't given, we could infer it from the prefix.

[Stefan Karpinski]

This patch *kind* of makes this work, but it's not really good enough to commit to base: https://gist.github.com/StefanKarpinski/68a2b5f4e5d51a72c358.

[Stefan Karpinski]

Btw, my recent enhancements to parseint not only added overflow detection (modulo lack of 128-bit support in LLVM), but also added full support for base autodetection via 0b, 0o and 0x prefixes.

[John Myles White]

Nice!

— John

[Stefan Karpinski]

It should be noted that the base autodetection only happens when the base is not explicitly given. If an explicit base is given, the number may only contain digits in that base.

[Alberto Barradas]

Hi guys,

Now that `parseint()` got removed for version 0.5, Is `parse()` the only way to do this?

 How could I parse binary into a BigInt? More specifically, I want to see the integer number of the arecibo message. (73x23 so 1679 binary digits into a big int)

[Milan Bouchet-Valat]

parse() is meant to be called on Julia expressions.
Use parse(Int, ...) or parse(BigInt, ...) depending on your needs.



Regards

[David P. Sanders]

julia> s = join(rand(0:1, 64))

"1001011000111000000010011110001010010001111110100101001111100111"

julia> parse(Int64, s, 2)

------ OverflowError ------------------- Stacktrace (most recent call last)

julia> parse(BigInt, s, 2)   # the 2 at the end is the base (here, binary)

10824412573101347815 

[Alberto Barradas]

Thank you guys! This looks like a cleaner and more graceful solution!