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33 is a simple esolang I created. You may have seen me use it in a few questions. You're not going to be writing a full interpreter. The interpreter you will be writing is for a simplified version of 33.

This simplified 33 has two numeric registers: the accumulator and the counter. The accumulator holds all the post-arithmetic values, the counter is where numbers in the program come from. Both registers are initialized to 0.

0-9 | Appends the digit to the counter a | Adds the counter to the accumulator m | Subtracts the counter from the accumulator x | Multiplies the accumulator by the counter d | Divides the accumulator by the counter r | Divides the accumulator by the counter, but stores the remainder in the accumulator z | Sets the counter to 0 c | Swaps the accumulator and counter ----- p | Outputs the current value of the accumulator as an ASCII character o | Outputs the current value of the accumulator as a formatted decimal number i | Outputs a newline character (0x0a) ----- P | Stores the ASCII value of the next character read in the accumulator O | Stores the next integer read into the accumulator ----- n | Skips the next instruction if the accumulator is not 0 N | Skips the next instruction if the accumulator is 0 g | Skips the next instruction if the accumulator is less than or equal to 0 G | Skips the next instruction if the accumulator is greater than 0 h | Skips the next instruction if the accumulator is greater than or equal to 0 H | Skips the next instruction if the accumulator is less than 0

Test cases

Program / Input -> Output 2o -> 0 25co -> 25 1a0aoi -> 11 (trailing newline) Oo / 42 -> 42 Op / 42 -> * 1cNoo -> 11 no -> 0 Ogo / 2 -> 2 Ogoi / -4 -> (newline) 50a -> (no output) On12co / 12 -> 2

Clarifications

  • The input to your interpreter will be a valid simplified 33 program.
  • The input may be given in any acceptable format.
  • The output may be given in any acceptable format.
  • When dividing, truncate any decimal places.
  • A trailing newline is acceptable, but it must be consistent; as in, if there's an i at the end of the program, it should have two trailing newlines.
  • The accumulator and counter must be able to hold at least a signed byte (-128 to 127)
  • You may assume that the p instruction will never be given when the accumulator has an invalid ASCII character in it.
  • You may assume the O instruction will never be given unless there is a valid integer string (such as 3, -54, +23) left in the input.
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  • \$\begingroup\$ How does O work? \$\endgroup\$ Commented Sep 10, 2019 at 10:57
  • \$\begingroup\$ @tsh Yes. I'll update that. \$\endgroup\$ Commented Sep 10, 2019 at 10:59
  • \$\begingroup\$ @EriktheOutgolfer It reads the integer in the same format that o outputs it \$\endgroup\$ Commented Sep 10, 2019 at 11:01
  • \$\begingroup\$ @TheOnlyMrCat If the remaining input is x-13, will O skip the x and read the -13? Or can we assume that there will always be an integer right where the O reads the input? \$\endgroup\$ Commented Sep 10, 2019 at 11:02
  • \$\begingroup\$ What is acceptable format for an integer input the interpreter must handle? Will +03 be valid input for integer? \$\endgroup\$ Commented Sep 10, 2019 at 11:21

3 Answers 3

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JavaScript (Node.js), 264 bytes

S=>I=>[...S].map(s=>K?K=0:1/s?C=+(C+s):eval("A*=C;[A,C]=[C,A];K=!A;;K=A;K=A>0;A%=C;K=A<=0;A=A/C|0;I=I.replace(/-?\\d+/,n=>(A=+n,''));K=A>=0;O+=A;K=A<0;A+=C;C=0;;[A]=B(I[0]),I=I.slice(1);O+=`\n`;O+=B([A]);A-=C".split`;`[B(s)[0]*73%378%22]),A=C=K=0,O='',B=Buffer)&&O

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Saved ~20 bytes, thanks to Arnauld's magic integer modulus.

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  • \$\begingroup\$ 253 bytes by converting the input to a buffer right away, re-using S as the result of the previous instruction and using a strict comparison with true. \$\endgroup\$ Commented Sep 11, 2019 at 7:22
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Lua 5.3, 342 bytes

Code is provided as a cmdline argument, input is provided via stdin.

H=string.char R=io.read A=0 C=0 i=1 s=...while i<=#s do b=s:byte(i)d=H(b)i=i+(({H=A<0,G=A>0,N=A==0,g=A<=0,h=A>=0,n=A~=0})[d]and 2or 1)io.write(({i='\n',o=A|0,p=H(A%255)})[d]or'')A,C=d=='O'and R'n'or d=='P'and R(1):byte()or d=='d'and A//C or d=='r'and A%C or({a=A+C,c=C,m=A-C,x=A*C})[d]or A,b>=48 and b<58 and b-48+C*10or({c=A,z=0})[d]or C end
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Julia 1.0, 355 bytes

p=print function f(s,a=0,c=0,i=1) for x=s z=findfirst(==(x),"amxdrzcpoiPOnNgGhH0123456789") y=z-9 i<1&&(i=1;continue) z<6 ? a=(+,-,*,÷,%)[z](a,c) : z<7 ? c=0 : z<8 ? ((a,c)=(c,a)) : z<9 ? p(Char(a)) : y<1 ? p(a) : y<2 ? p("\n") : y<3 ? a=0+read(stdin,UInt8) : y<4 ? a=parse(Int,readline()) : y<10 ? (!=,==,<=,>,>=,<)[y-3](a,0)&&(i=0) : c=10c+y-10 end end

Program provided as an argument, eg f("2o"), input taken from stdin.

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