How many bytes is it

Question

This time, we want render the count of bytes in a human readable way to user. Let's write a program (a full one or a function or other acceptable formats) to do this.

Input

An non-negative integer, in range \$0\leq n<2^{31}\$, which means n bytes.

Output

A string, the human friendly representation of n bytes.

Convention

• If n is less than 1000, add B after n and output; otherwise...
• Convert to a number less than 1000, plus one of these units: KiB, MiB, GiB; round the number to 3 figures
• For more details of conversion, check out these testcases

Testcases

0 -> 0B
1 -> 1B
42 -> 42B
999 -> 999B
1000 -> 0.98KiB
1024 -> 1.00KiB
2018 -> 1.97KiB
10086 -> 9.85KiB
100010 -> 97.7KiB
456789 -> 446KiB
20080705 -> 19.2MiB
954437177 -> 910MiB
1084587701 -> 1.01GiB
1207959551 -> 1.12GiB
2147483647 -> 2.00GiB

Rules

• This is code-golf, so shortest bytes win
• Standard loopholes forbidden
• You should output exactly the same to the testcase:
  • No white space or other symbols between number and units;
  • Use KiB, not KB, kib, or kb;
  • Leading or trailing white spaces are optional

Answer 1

JavaScript (ES6), 78 74 bytes

f=(n,i)=>n<999.5?i?n.toPrecision(n<1^3)+' KMG'[i]+'iB':n+'B':f(n/1024,-~i)

<input oninput=o.textContent=f(this.value)><pre id=o>

Edit: Saved 4 bytes thanks to @Arnauld.

Answer 2

Python3, 90 bytes

(76 70 bytes 3-most-significant-digits version)

h=lambda v,u=0:v>999.4and h(v/1024,u+2)or"%.*f%sB"%(u and(v<10)+(v<100),v,"KiMiGi"[u-2:u])

Version whose outputs are strictly adherent to test cases and author's suggestions.

list(map(lambda v: print("{} --> {}".format(v, h(v))), (0, 1, 42, 999, 1000, 124, 2018, 10086, 100010, 456789, 20080705, 954437177, 1084587701, 1207959551, 2147483647, 1023079)))

0 --> 0B
1 --> 1B
42 --> 42B
999 --> 999B
1000 --> 0.98KiB
1024 --> 1.00KiB
2018 --> 1.97KiB
10086 --> 9.85KiB
100010 --> 97.7KiB
456789 --> 446KiB
20080705 --> 19.2MiB
954437177 --> 910MiB
1084587701 --> 1.01GiB
1207959551 --> 1.12GiB
2147483647 --> 2.00GiB
1023079 --> 999KiB
1023488 --> 0.98MiB
1043333 --> 0.99MiB
1043334 --> 1.00MiB

Try it online!

Here, a version that produces the most 3 significant digits for the test cases.

h=lambda v,u=0:v<1e3and("%.3G%sB"%(v,"KiMiGi"[u-2:u]))or h(v/1024,u+2)

The original test cases:

list(map(lambda v: print("{} --> {}".format(v, h(v))), (0, 1, 42, 999, 1000, 124, 2018, 10086, 100010, 456789, 20080705, 954437177, 1084587701, 1207959551, 2147483647)))

0 --> 0B
1 --> 1B
42 --> 42B
999 --> 999B
1000 --> 0.977KiB
124 --> 124B
2018 --> 1.97KiB
10086 --> 9.85KiB
100010 --> 97.7KiB
456789 --> 446KiB
20080705 --> 19.2MiB
954437177 --> 910MiB
1084587701 --> 1.01GiB
1207959551 --> 1.12GiB
2147483647 --> 2GiB

Try it online!

Answer 3

C (gcc), 108 102 bytes

Works as Yimin Rong suggests.

f(float i){char*a="GiMiKi"+6;while(i>=999.5)i/=1024,a-=2;printf("%.*f%.2sB",*a&i<100?i<10?2:1:0,i,a);}

Try it online!

Unglofed

f(float i){
    char*a="GiMiKi"+6;
    while(i>=999.5)
        i/=1024,a-=2;
    printf("%.*f%.2sB",*a&i<100?i<10?2:1:0,i,a);
}

C (gcc), 108 101 99 bytes

Works as O.O.Balance suggests.

f(float i){char*a="GiMiKi"+6;while(i>999)i/=1024,a-=2;printf("%.*f%.2sB",*a&i<100?i<10?2:1:0,i,a);}

Try it online!

Unglofed

f(float i){
    char*a="GiMiKi"+6;
    while(i>999)
        i/=1024,a-=2;
    printf("%.*f%.2sB",*a&i<100?i<10?2:1:0,i,a);
}

Answer 4

Java 8, 130 112 108 bytes

n->n<1e3?n+"B":"".format("%.3G"+(n<1e9?n<1e6?"K":"M":"G")+"iB",n<1e9?n<1e6?n/1024.:n/1048576.:n/1073741824.)

Uses 3 significant figures, as per the original challenge. Try it online here.

Thanks to Jakob for golfing 4 bytes.

Ungolfed:

n -> // lambda 
n < 1e3 // special case: number of bytes is < 1000
    ? n + "B" // output the number of bytes
    : "".format("%.3G" // otherwise output 3 significant digits
    + (n < 1e9 // if it's less than 1 billion bytes
        ? n < 1e6 // and it's less than a million bytes
            ? "K" // output kibibytes
            : "M" // else output mebibytes
        : "G") // else output gibibytes
        + "iB", // common suffix
        n < 1e9 // if it's less than 1 billion bytes
            ? n < 1e6 // if it's less than 1 million bytes
                ? n/1024.     // convert to kibibytes
                : n/1048576.  // else convert to mebibytes
            : n/1073741824. ) // else convert to gibibytes

Answer 5

APL (Dyalog Classic), 60 bytes

{'BKMGi'[i,4 0/⍨×i],⍨v⍕⍨(3-≢⍕⌊v←⍵÷i⊃p)××i←⍵+.≥1e3×p←1024*⍳4}

Try it online!

Answer 6

Rust, 275 bytes

fn h(n:u32)->String{let i=[(30,"Gi"),(20,"Mi"),(10,"Ki"),(0,"")].iter().map(|x| (n as f32/2f32.powi(x.0),x.1)).filter(|y| y.0<999.5).last().unwrap();format!("{}{}B",if n<1000{format!("{}",n)}else{format!("{:.4}",format!("{:.2}",i.0)).trim_right_matches(".").to_string()},i.1)}

The basic idea is borrowed from the other answers, but formatting the string to have the proper number of significant digits, not have trailing decimal points, and not have numbers like 423.78 takes over half of the code... and it still rounds wrong (19.1 should be 19.2)

Answer 7

Stax, 34 bytes

ì▌╚ƒ░¿n↕─.êN┴▀m▌@àΔ*¡I⌠'}¿Äüz♥∞└6]

Run and debug it

Writing this, I ran into a bug in stax's log10 operation. Log10 of 1000 yields 2.9999...., which is particularly unhelpful for this challenge. I ended up using string length instead. Unpacked, ungolfed, and commented, it looks like this.

Vk<             is input less than 1000?
{               start block one
  yp'BP         print input followed by "B"
}               end block one
{               start block two
  x             push input as number
  y%v3/X        push (len(input) - 1) / 3, and store in the X register
  A*|2          multiply by 10, then get calculate two-power
  :_2j          floating point division, rounded to 2 places as a string
  p             print without trailing newline
  "GKM"x@.iB+P  use x register to index into "GKM" string, then concat "ib" and print
}               end block two
?               if-else based on initial condition

Run this one

Answer 8

Python 3, 87 81 80 bytes

def f(n):
 for u in'','Ki','Mi','Gi':
  if n<1e3:return"%.3G%sB"%(n,u)
  n/=1024

Try it online!

This does not work, I will need to think about how to fix it.

^{-6 bytes from Jo King}

^{-1 bytes from Stewie Griffin}

Answer 9

C, 154 165 191 189 bytes (full program)

Big thanks to O.O.Balance and l4m2 for suggestions and providing this TIO and this TIO!

double atof(),x;main(i,v)char**v;{x=atof(*++v);if(x<1e3)printf("%sB",*v);else{for(;x>=999.5;++i)x/=1024;printf("%.*f%ciB",2-(int)log10(x),x,"KMG"[i-3]);}}

Save as doit.c, compile using gcc -o doit doit.c -lm, and run:

> ./doit 1000
0.98KiB

Test cases:

> for i in 0 1 42 999 1000 1024 2018 10086 100010 456789 1023079 1023488 20080705 954437177 1084587701 1207959551 2147483647 ; do ./doit $i ; echo; done
0B
1B
42B
999B
0.98KiB
1.00KiB
1.97KiB
9.85KiB
97.7KiB
446KiB
999KiB
0.98MiB
19.2MiB
910MiB
1.01GiB
1.12GiB
2.00GiB

Answer 10

C (gcc), 135 bytes

e;double atof(),x;main(i,v)int**v;{for(x=atof(*++v);x>=999.5;e++)x/=1024;printf(e?"%.*f%ciB":"%.*fB",e?2-(int)log10(x):0,x,".KMG"[e]);}

Try it online!

Quite change from Yimin Rong's solution