wtf
- Reference
- Development Environments
- Star Pattern 3 (2024.06.23)
- Star Pattern 2 (2024.06.08)
- Star Pattern (2024.06.07)
- Even Numbers (2024.06.06)
- A to Z (2024.06.05)
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Extension of Star Pattern 2 (2024.06.08) : From Triangle to Diamond
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Code and Result
Code : StarPattern3.bf
# Set constants ptr0 ++++ # Let ptr0 = 4 [ ptr0 > ptr1 +++++ +++++ < ptr0 - ] # Let ptr1 = (10 * 4) = 40 > [ ptr1 > ptr2 + > ptr3 + > ptr4 + <<< ptr1 - ] # Move ptr1 to ptr2 ptr3 ptr4 > ptr2 +++++ +++ # Let ptr2 = 48 ('0') > ptr3 ++ # Let ptr3 = 42 ('*') > ptr4 ----- --- # Let ptr4 = 32 (Space) > ptr5 +++++ +++++ # Let ptr5 = 10 (LF)
# Input a number > ptr6 , # Input ptr6 between '0' and '9' <<<< [ ptr2 >>>> ptr6 - <<<< ptr2 - ] # Convert ptr6 from char to int (ptr6 minus '0') >>>> [ ptr6 > ptr7 + > ptr8 + >>>> prt12 + <<<<< < ptr6 -] # Move ptr6 to ptr7 ptr8 ptr12 >>>> ptr10 + # Let ptr10 = 1
# Phase 1 : Increase stars and decrease spaces <<< [ ptr7 # ptr7: The loop times > [ ptr8 <<<< ptr4 . >>>>> ptr9 + < ptr8 - ] # ptr8 & ptr9 : The number of spaces >> [ ptr10 <<<<< << ptr3 . >>>>> >>> ptr11 + < ptr10 - ] # ptr10 & ptr11 : The number of stars > [ ptr11 < ptr10 + > ptr11 - ] < ptr10 ++ # Recover ptr10 from ptr11 and add 2 < [ ptr9 < ptr8 + > ptr9 - ] < ptr8 - # Recover ptr8 from ptr9 and subtract 1 <<< ptr5 . # Line replacement >> ptr7 - ]
# Phase 2 : Increase spaces and decrease stars > ptr8 ++ # Add 2 to ptr8 >> ptr10 ---- # Subtract 2 from ptr10 >> [ ptr12 <<<<< ptr7 + >>>>> ptr12 -] # Move ptr12 to ptr7 <<<<< ptr7 - # Subtract 1 from ptr7 [ ptr7 # ptr7: The loop times > [ ptr8 <<<< ptr4 . >>>>> ptr9 + < ptr8 - ] # ptr8 & ptr9: The number of spaces >> [ ptr10 <<<<< << ptr3 . >>>>> >>> ptr11 + < ptr10 - ] # ptr10 & ptr11: The number of stars > [ ptr11 < ptr10 + > ptr11 - ] < ptr10 -- # Recover ptr10 from ptr11 and suptract 2 < [ ptr9 < ptr8 + > ptr9 - ] < ptr8 + # Recover ptr8 from ptr9 and add 1 <<< ptr5 . # Line replacement >> ptr7 - ]
Result
9* *** ***** ******* ********* *********** ************* *************** ***************** *************** ************* *********** ********* ******* ***** *** *
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Acquiring new skills (slightly tricky)
- Possible to copy values
: Transfer values to two or more pointers while consuming the original pointer's value - Implementing loops with a general flow as
forusing an index variablei
- Possible to copy values
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Code and Result
Code : StarPattern2.bf
# Set constants < ptr0 ++++ # Let ptr0 = 4 [ ptr0 > ptr1 +++++ +++++ < ptr0 - ] # Let ptr1 = (10 * 4) = 40 > [ ptr1 > ptr2 + > ptr3 + > ptr4 + <<< ptr1 - ] # Move ptr1 to ptr2 ptr3 ptr4 > ptr2 +++++ +++ # Let ptr2 = 48 ('0') > ptr3 ++ # Let ptr3 = 42 ('*') > ptr4 ----- --- # Let ptr4 = 32 (Space) > ptr5 +++++ +++++ # Let ptr5 = 10 (LF)
# Input a number > ptr6 , # Input ptr6 between '0' and '9' <<<< [ ptr2 >>>> ptr6 - <<<< ptr2 - ] # Convert ptr6 from char to int (ptr6 minus '0') >>>> [ ptr6 > ptr7 + > ptr8 + << ptr6 -] # Move ptr6 to ptr7 and ptr8 >>>> ptr10 + # Let ptr10 = 1
# Print the pattern with general loop statement <<< [ ptr7 # ptr7: The loop times > [ ptr8 <<<< ptr4 . >>>>> ptr9 + < ptr8 - ] # ptr8 & ptr9: The number of spaces >> [ ptr10 <<<<< << ptr3 . >>>>> >>> ptr11 + < ptr10 - ] # ptr10 & ptr11: The number of stars > [ ptr11 < ptr10 + > ptr11 - ] < ptr10 ++ # Recover ptr10 from ptr11 and add 2 < [ ptr9 < ptr8 + > ptr9 - ] < ptr8 - # Recover ptr8 from ptr9 and subtract 1 <<< ptr5 . # Line replacement >> ptr7 - ]
Result
7* *** ***** ******* ********* *********** *************
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Only can input a number as a character between 0 and 9
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Learned nested
[]brackets and the behavior of<>within them -
Code and Result
Code : StarPattern.bf
< ptr0 +++++ [ > ptr1 +++++ +++++ < ptr0 - ] > ptr1 -- # Let ptr1 = (10 * 5) minus 2 = 48('0') < ptr0 ++++ [ >> ptr2 +++++ +++++ << ptr0 - ] >> ptr2 ++ # Let ptr2 = (10 * 4) plus 2 = 42('*') > ptr3 +++++ +++++ # Let ptr3 = 10 (LF)
> prt4 , # Input ptr4 between '0' and '9' <<< [ >>> ptr4 - <<< ptr1 - ] # Convert ptr4 from char to int (ptr4 minus '0')
>>> ptr4 [ << ptr2 . > ptr3 . > ptr4 - [ << ptr2 .. > ptr3 . > ptr4 - [ << ptr2 ... > ptr3 . > ptr4 - [ << ptr2 .... > ptr3 . > ptr4 - [ << ptr2 ..... > ptr3 . > ptr4 - [ << ptr2 ..... . > ptr3 . > ptr4 - [ << ptr2 ..... .. > ptr3 . > ptr4 - [ << ptr2 ..... ... > ptr3 . > ptr4 - [ << ptr2 ..... .... > ptr3 . > ptr4 - ]]]]] ]]]]
Result
5* ** *** **** *****
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Limitations of Using Loops to Approximate Conditional Statements
- Loops could only approximate conditional behavior, failing to fully replicate the functionality of standard conditional statements.
- Reusing pointer values that had been set to 0 within [] brackets was not possible when checking conditional requirements.
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Code and Result
Code : EvenNumbers.bf
# 1) Add 2 from 48('0') 4 times at ptr0 ptr0 +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++ # Let ptr0 = 48('0') ++. ++. ++. ++. # Print 2 4 6 8
# 2) Use Loop statement ptr0 [-] # Reset ptr0 as 0 +++++ +++++ . [-] # Let ptr0 = 10 (LF) & Line replacement & Reset ptr0 ptr0 +++++ # Let ptr0 = 5 [ > ptr1 +++++ +++++ < ptr0 -] > ptr1 -- # Let ptr1 = (10 * 5) minus 2 = 48('0') < ptr0 ++++ # Let ptr0 = 4 [ > ptr1 ++ . < ptr0 - ] # Print 2 4 6 8 with Loop statement
# 3) Use Conditional statement ptr0 [-] # Reset ptr0 as 0 > prt1 [-] # Reset ptr1 as 0 < ptr0 +++++ +++++ . [-] # Let ptr0 = 10 (LF) & Line replacement & Reset ptr0 ptr0 +++++ # Let ptr0 = 5 [ > ptr1 +++++ +++++ < ptr0 -] > ptr1 -- # Let ptr1 = (10 * 5) minus 2 = 48('0') < ptr0 +++++ +++ # Let ptr0 = 8 [ > ptr1 + < ptr0 - # If odd then not print > ptr1 + . < ptr0 - # If even then print ] # Unable to use new ptr to judge if odd or even because it can't be reused in loop statement
Result
2468 2468 2468
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My initial Brainf*** practice; Print the whole Alphabet
- I think I've become ready to deal with Assembly!
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Code and Result
Code : Alphabet.bf
# 1) Add 1 from 65('A') 25 times at ptr0 ptr0 +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ . # Let ptr0 = 65 and print 'A' +.+.+.+.+.+.+.+.+.+.+.+.+.+.+.+.+.+.+.+.+.+.+.+.+. # Increment and print ptr0 25 times
# 2) Use square brackets statements ptr0 [-] # Reset ptr0 as 0 ptr0 +++++ +++++ . [-] # Let ptr0 = 10 (LF) & Line replacement & Reset ptr0 ptr0 +++++ +++++ +++ # Let ptr0 = 13 [ > ptr1 +++++ < ptr0 - ] > ptr1 . # Increment ptr1 by 5 13 times to get 65 ('A') & print < ptr0 +++++ +++++ +++++ +++++ +++++ # Let ptr0 = 25 [ > ptr1 + . < ptr0 - ] # Loop to increment 25 times & print ptr1 from 66 ('B') to 90 ('Z')
Result
ABCDEFGHIJKLMNOPQRSTUVWXYZ ABCDEFGHIJKLMNOPQRSTUVWXYZ