How long is my number: Restricted Version

Question

Find the original challenge here

Challenge

Given an integer, Z in the range -2^31 < Z < 2^31, output the number of digits in that number (in base 10).

Rules

You must not use any string functions (in the case of overloading, you must not pass a string into functions which act as both string and integer functions). You are not allowed to store the number as a string.

All mathematical functions are allowed.

You may take input in any base, but the output must be the length of the number in base 10.

Do not count the minus sign for negative numbers. Number will never be a decimal.

Zero is effectively a leading zero, so it can have zero or one digit.

Examples

Input > Output

-45 > 2
1254 > 4
107638538 > 9
-20000 > 5
0 > 0 or 1
-18 > 2

Winning

Shortest code in bytes wins.

Answer 1

Mathematica, 13 bytes

IntegerLength

Well...

Answer 2

Python 2, 30 bytes

f=lambda x:x and-~f(abs(x)/10)

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Answer 3

JavaScript (ES6), 19 bytes

f=n=>n&&f(n/10|0)+1

console.log(f(-45))       // 2
console.log(f(1254))      // 4
console.log(f(107638538)) // 9
console.log(f(-20000))    // 5
console.log(f(0))         // 0
console.log(f(-18))       // 2

Answer 4

Japt, 5 3 bytes

ì l

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Answer 5

My answer from the other challenge still works:

Brachylog, 1 byte

l

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The l builtin is overloaded, but on integers, it takes the number of digits of the integer, ignoring sign.

Answer 6

Jelly, 3 2 bytes

1 byte saved thanks to Leaky Nun

DL

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Explanation

 L    Length of
D     Decimal expansion of input argument. Works for negative values too

Answer 7

APL (Dyalog Unicode), 7 bytes

⌈10⍟1+|

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Explanation:

⌈10⍟1+|
⌈        ⍝ round up the
 10⍟     ⍝ log10 of...
    1+   ⍝ incremented
      |  ⍝ absolute value of input

If format is allowed: 4 bytes - ≢⍕∘|

Answer 8

Alice, 16 bytes

/O
\I@/Hwa:].$Kq

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Explanation

/O
\I@/...

This is simply a framework for numerical input→mathematical processing→numerical output.

The rest of the code is the real algorithm:

Hwa:].$Kq
H            Compute absolute value
 w   .$K     While the result is not zero do:
  a:           divide the number by 10
    ]          move the tape head one cell forward
        q    Get the position of the tape head

Answer 9

Chaincode, 5 bytes

pqL_+

Note: This is exactly the same code as that from the other challenge

Explanation

pqL_+ print(
    +   succ(
   _      floor(
  L        log_10(
pq           abs(
               input())))))

Answer 10

dc, 1 byte

Z

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---

Not using a builtin, 18 bytes:

[d10/d0!=F]dsFxz1-

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Answer 11

R, 40 bytes

function(x)max(ceiling(log10(abs(x))),0)

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Answer 12

Pip, 14 bytes

LNABq/LNt//1+1

https://tio.run/##K8gs@P/fx8/RqVDfx69EX99Q2/D/f0MDczNjC1NjCwA

Explanation

LN              natural log of...      (change of base becasue this is the only log function they had)
  ABq           the absolute value of the input...
     /          divided by...
      LNt       the natural log of 10...   (change of base)
         //1    integer divided by 1 (no floor function)
            +1  added to 1

This was inspired by the Chaincode solution.

This could probably be optimized.

Answer 13

Java 8, 61 59 39 37 33 bytes

n->(int)Math.log10(n<0?-n:n+.5)+1

-4 bytes thanks to @MarkJeronimus.

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Explanation:

n->            // Method with integer as both parameter and return-type
  (int)        //  Convert the following double to an integer (truncating its decimals):
    Math.log10(//   The log_10 of:
      n<0?     //    If the input is negative:
          -n   //     Use its absolute value
         :     //    Else:
          n+.5)//     Add 0.5 to the input instead
  +1           //  And add 1 to the result at the end

Answer 14

S.I.L.O.S, 41 bytes

readIO
i|
lblb
i/10
a+1
if i b
printInt a

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Returns 1 for 0.

Answer 15

Lua, 40 bytes

Port from my python answer

print(math.log10(math.abs(10*...)+1)//1)

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Answer 16

C, 27 bytes

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f(n){return n?1+f(n/10):0;}

C (gcc), 22 bytes

f(n){n=n?1+f(n/10):0;}

Using math, 29 bytes

f(n){return 1+log10(abs(n));}

Answer 17

ARM Thumb-2 (no div instruction, no libgcc), 30 bytes

Raw machine code:

2800 d00b bfb8 4240 2201 2100 3101 380a
dafc 3201 0008 280a daf7 0010 4770     

Uncommented assembly:

        .syntax unified
        .globl count_digits
        .thumb
        .thumb_func
count_digits:
        cmp     r0, #0
        beq     .Lret
        it      lt
        neglt   r0, r0
        movs    r2, #1
.Lcountloop:
        movs    r1, #0
.Ldivloop:
        adds    r1, #1
        subs    r0, #10
        bge     .Ldivloop
.Ldivloop_end:
        adds    r2, #1
        movs    r0, r1
        cmp     r0, #10
        bge     .Lcountloop
.Lcountloop_end:
        movs    r0, r2
.Lret:
        bx      lr

Returns 0 if 0.

Explanation

C function signature:

int32_t count_digits(int32_t val);

First, we compare val to zero.

If it is zero, we return zero. If it is less than zero, we negate it.

count_digits:
        cmp     r0, #0
        beq     .Lret
        it      lt
        neglt   r0, r0

Set up our digit counter for the outer loop.

        movs    r2, #1

Now, a naïve subtraction based division loop

        movs    r1, #0
.Ldivloop:
        adds    r1, #1
        subs    r0, #10
        bge     .Ldivloop

Increment the digits counter, then loop to the outer loop if we are still more than 10.

.Ldivloop_end:
        adds    r2, #1
        movs    r0, r1
        cmp     r0, #10
        bge     .Lcountloop

Move the result into the return register and return.

.Lcountloop_end:
        movs    r0, r2
.Lret:
        bx      lr

Answer 18

Julia, 7 bytes

ndigits

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Answer 19

PowerShell Core, 46 bytes

param($a)for($s=0;$a){$s+=!!($a=[int]$a/10)}$s

Implementation Details

param($a)              # Defines the input parameter
for($s=0;$a){          # Initialise the result to 0 and iterates while the parameter is not 0 
$s+=!!($a=[int]$a/10)  # Divides the parameter by 10 and increment the result variable
                       # by one if the result is not 0
}$s                    # Returns the result

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Answer 20

Excel, 26 bytes

=INT(LOG(A1^2+(A1=0))/2)+1

Log of the number squared / 2 is 1 byte shorter than ABS

Answer 21

K (ngn/k), 10 9 7 bytes

-3 bytes from @ngn (see Am I an insignificant array?)

#10\#!:

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• #!: get the absolute value of the input (literally, take the count of the range of each value; e.g. #!-2 -> #-2 -1 -> 2)
• 10\ "digit-ize" the input
• # take the count

A solution using $ instead of 10\ save two bytes, but may be invalid given the question's rules.

Answer 22

TI-Basic, 9 8 bytes

int(log(1+10abs(Ans

Takes input in Ans.

-1 byte thanks to MarcMush.

Answer 23

05AB1E, 6 bytes

Ä>T.nî

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Ä      # Absolute value
 >     # Increment
  T.n  # Log base 10
     î # Round up

Answer 24

Jelly, 5 bytes

A‘l⁵Ċ

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Answer 25

C#, 49 56 bytes

namespace System.Math{n=>n==0?1:Floor(Log10(Abs(n))+1);}

Answer 26

Python 2, 48 bytes

-3 thanks to ovs -1 thanks to pizzapants

lambda x:math.log10(abs(10*x)+1)//1
import math

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Answer 27

Ruby, 27 bytes

f=->x{x==0?0:1+f[x.abs/10]}

As a test:

tests = [[-45 , 2],
         [1254 , 4],
         [107638538 , 9],
         [-20000 , 5],
         [0 , 0 ],
         [-18 , 2]]

tests.each do |i, o|
  p f.call(i) == o
end

It outputs:

true
true
true
true
true
true

Answer 28

@yBASIC, 53 bytes.

_#=@_>.@_
_%=_%/(_#*_#+!.)_=_+!.GOTO(@_)+"_"*!_%@__?_

Input should be stored in variable _%

Answer 29

Rust, 41 bytes

fn f(n:i32)->u8{n!=0&&return f(n/10)+1;0}

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Answer 30

Pyth, 10 bytes

.xhs.l.aQT

Test suite

Explanation:

.xhs.l.aQT  # Code
.xhs.l.aQTQ # With implicit variables
            # Print (implicit):
    .l   T  #   the log base 10 of:
      .aQ   #    the absolute value of the input
   s        #   floored
  h         #   plus 1
.x        Q #  unless it throws an error, in which case the input

Python 3 translation:

import math
Q=eval(input())
try:
    print(int(math.log(abs(Q),10))+1)
except:
    print(Q)