BIP: 441
Layer: Consensus (soft fork)
Title: Restoration of disabled script (Tapleaf 0xC2)
Authors: Rusty Russell <rusty@rustcorp.com.au>
Julian Moik <julianmoik@gmail.com>
Status: Draft
Type: Specification
Assigned: 2026-03-25
License: BSD-3-Clause
Discussion: https://groups.google.com/g/bitcoindev/c/GisTcPb8Jco/m/8znWcWwKAQAJ
Version: 0.2.1
Requires: 440
Introduction
Abstract
This BIP introduces a new tapleaf version (0xc2) which restores Bitcoin script to its pre-0.3.1 capability, relying on the Varops Budget in BIP440 to prevent the excessive computational time which caused CVE-2010-5137.
In particular, this BIP:
- Reenables disabled opcodes.
- Increases the maximum stack object size from 520 bytes to 4,000,000 bytes.
- Introduces a total stack byte limit of 8,000,000 bytes.
- Increases the maximum total number of stack objects from 1,000 to 32,768.
- Removes the 32-bit size restriction on numerical values.
- Treats all numerical values as unsigned.
All opcodes are described in exact (painstaking) byte-by-byte operations, so that their varops budget can be easily derived. Note that this level of detail is unnecessary to users of script, only being of interest to implementers.
Copyright
This document is licensed under the 3-clause BSD license.
Motivation
Since Bitcoin v0.3.1 (addressing CVE-2010-5137), Bitcoin's scripting capabilities have been significantly restricted to mitigate known vulnerabilities related to excessive computational time and memory usage. These early safeguards were necessary to prevent denial-of-service attacks and ensure the stability and reliability of the Bitcoin network.
Unfortunately, these restrictions removed much of the ability for users to control the exact spending conditions of their outputs, which has frustrated the long-held ideal of programmable money without third-party trust.
Execution of Tapscript 0xC2
If a taproot leaf has a version of 0xc2, execution of opcodes is as defined below. All opcodes not explicitly defined here are treated exactly as defined by BIP342.
Validation of a script fails if:
- It exceeds the remaining varops budget for the transaction.
- Any stack element exceeds 4,000,000 bytes.
- The total size of all stack (and altstack) elements exceeds 8,000,000 bytes.
- The number of stack elements (including altstack elements) exceeds 32,768.
Rationale
There needs to be some limit on memory usage, to avoid a memory-based denial of service.
Putting the entire transaction on the stack is a foreseeable use case, hence using the block size (4MB) as a limit makes sense. However, allowing 4MB stack elements is a significant increase in memory requirements, so a total limit of twice that many bytes (8MB) is introduced. Many stack operations require making at least one copy, so this allows such use.
Putting all outputs or inputs from the transaction on the stack as separate elements requires as much stack capacity as there are inputs or outputs. The smallest possible input is 41 bytes (allowing almost 24,390 inputs), and the smallest possible output is 9 bytes (allowing almost 111,111 outputs). However, empty outputs are rare and not economically interesting. Thus we consider smallest non-OP_RETURN standard output script, which is P2WPKH at 22 bytes, giving a minimum output size of 31 bytes, allowing 32,258 outputs in a maximally-sized transaction.
This makes 32,768 a reasonable upper limit for stack elements.
SUCCESS Opcodes
The following opcodes are renamed OP_SUCCESSx, and cause validation to immediately succeed:
- OP_1NEGATE = OP_SUCCESS79
- OP_NEGATE = OP_SUCCESS143
- OP_ABS = OP_SUCCESS144Anthony Towns suggested this could become an
opcode which normalized the value on the top of the stack by truncating any
trailing zeroes.</ref>
Rationale
Negative numbers are not natively supported in 0xC2 Tapscript. Arbitrary precision makes them difficult to manipulate and negative values are not used meaningfully in bitcoin transactions.
Arbitrary-length Values, Endianness, and Normalization of Results
The restoration of bit operations means that the little-endianness of stack values is once more exposed to the Script author, if they mix them with arithmetic operations. The restoration of arbitrary-length values additionally exposes the endianness to the implementation authors (who cannot simply load stack entries into registers), and requires explicit consideration when considering varops costs of operations.For example, removing trailing bytes from a stack element is almost free, whereas removing bytes from the front involves copying all remaining bytes.
Note that only arithmetic operations (those which treat operands as numbers) normalize their results: bit and byte operations do not.Such non-arithmetic operations can be used to operate on values such as preimages or (with introspection) parts of transactions, where truncation of zeros would be unexpected. One could argue that even arithmetic operators should not normalize, but that would be a gratuitous and surprising change. Note that "0 OP_ADD" can always be used to cheaply normalize the top stack element. Thus operations such as "0 OP_ADD" and "2 OP_MUL" will never result in a top stack entry with a trailing zero byte, but "0 OP_OR" and "1 OP_UPSHIFT" may.The original Bitcoin implementation had a similar operational split, but OP_LSHIFT and OP_RSHIFT did normalize, which was almost a requirement given that they also preserved the sign of the shifted operand
To be explicit, the following operations are defined as arithmetic and will normalize their results:
- OP_1ADD
- OP_1SUB
- OP_2MUL
- OP_2DIV
- OP_ADD
- OP_SUB
- OP_MUL
- OP_DIV
- OP_MOD
- OP_MIN
- OP_MAX
Non-Arithmetic Opcodes Dealing With Stack Numbers
The following opcodes are redefined in 0xC2 Tapscript to read numbers from the stack as arbitrary-length little-endian values (instead of CScriptNum):
- OP_CHECKLOCKTIMEVERIFY
- OP_CHECKSEQUENCEVERIFY
- OP_VERIFY
- OP_PICK
- OP_ROLL
- OP_IFDUP
- OP_CHECKSIGADD
These opcodes are redefined in 0xC2 Tapscript to write numbers to the stack as minimal-length little-endian values (instead of CScriptNum):
- OP_CHECKSIGADD
- OP_DEPTH
- OP_SIZE
In addition, the [[bip-0342.mediawiki#specification|BIP-342 success
requirement]] is modified to require a non-zero variable-length unsigned
integer value (not CastToBool()):
Previously:
4. (ii) If the execution results in anything but exactly one element on the stack which evaluates to true with `CastToBool()`, fail.
Now:
4. (ii) If the execution results in anything but exactly one element on the stack which contains one or more non-zero bytes, fail.
Enabled Opcodes
Fifteen opcodes that were removed in v0.3.1 are re-enabled in 0xC2 Tapscript.
If there are fewer than the required number of stack elements, these opcodes
fail validation. These are popped off the stack in right-to-left order,
i.e.
See BIP440 for the meaning of the annotations in the varops cost field.
Splice Opcodes
| Mnemonic | Opcode | Input Stack | Description | Definition | Varops Cost | Varops Reason |
|---|---|---|---|---|---|---|
| OP_CAT | 126 | Append B to A | (length(A) + length(B)) * 3 | COPYING | ||
| OP_SUBSTR | 127 | Extract bytes BEGIN through BEGIN+LEN of A | (length(LEN) + length(BEGIN)) * 2 + MIN(Value of LEN, MAX(length(A) - Value of BEGIN, 0)) * 3 | LENGTHCONV + COPYING | ||
| OP_LEFT | 128 | Extract the left OFFSET bytes of A | length(OFFSET) * 2 | LENGTHCONV | ||
| OP_RIGHT | 129 | Extract the right bytes of A, from OFFSET onwards | length(OFFSET) * 2 + value of OFFSET * 3 | LENGTHCONV + COPYING |
Rationale
OP_CAT may require a reallocation of A (hence, COPYING A) before appending B.
OP_SUBSTR may have to copy LEN bytes, but also needs to read its two numeric operands. LEN is limited to the length of the operand minus BEGIN.
OP_LEFT only needs to read its OFFSET operand (truncation is free), whereas OP_RIGHT must copy the bytes, which depends on the OFFSET value.
Bit Operation Opcodes
| Mnemonic | Opcode | Input Stack | Description | Definition | Varops Cost | Varops Reason |
|---|---|---|---|---|---|---|
| OP_INVERT | 131 | Bitwise invert A | length(A) * 4 | OTHER | ||
| OP_AND | 132 | Binary AND of A and B | (length(A) + length(B)) * 2 | OTHER + ZEROING | ||
| OP_OR | 133 | Binary OR of A and B | MIN(length(A), length(B)) * 4 | OTHER | ||
| OP_XOR | 134 | Binary exclusive-OR of A and B | MIN(length(A), length(B)) * 4 | OTHER |
Rationale
OP_AND, OP_OR and OP_XOR are assumed to fold the results into the longer of the two operands. This is an OTHER operation (i.e. cost is 4 per byte), but OP_AND needs to do this until one operand is exhausted, and then zero the rest (ZEROING, cost 2 per byte). OP_OR and OP_XOR can stop processing the operands as soon as the shorter operand is exhausted.
Bitshift Opcodes
Note that these are raw bitshifts, unlike the sign-preserving arithmetic shifts in Bitcoin v0.3.0, and as such they also do not truncate trailing zeroes from results: they are renamed OP_UPSHIFT (née OP_LSHIFT) and OP_DOWNSHIFT (née OP_RSHIFT).
| Mnemonic | Opcode | Input Stack | Description | Definition | Varops Cost | Varops Reason |
|---|---|---|---|---|---|---|
| OP_UPSHIFT | 152 | Move bits of A right by BITS (numerically increase) | length(BITS) * 2 + (Value of BITS) / 8 * 2 + length(A) * 3. If BITS % 8 != 0, add length(A) * 4 | LENGTHCONV + ZEROING + COPYING. If BITS % 8 != 0, + OTHER. | ||
| OP_DOWNSHIFT | 153 | Move bits of A left by BITS (numerically decrease) | length(BITS) * 2 + MAX((length(A) - (Value of BITS) / 8), 0) * 3 | LENGTHCONV + COPYING |
Rationale
DOWNSHIFT needs to read the value of the second operand BITS. It then needs to move the remainder of A (the part after offset BITS/8 bytes). In practice this should be implemented in word-size chunks, not bit-by-bit!
UPSHIFT also needs to read BITS. In general, it may need to reallocate (copying A and zeroing out remaining words). If not moving an exact number of bytes (BITS % 8 != 0), another pass is needed to perform the bitshift.
OP_UPSHIFT can produce huge results, and so must be checked for limits prior to evaluation. It is also carefully defined to avoid reallocating twice (reallocating to prepend bytes, then again to append a single byte) which has the practical advantage of being able to share the same downward bitshift routine as OP_DOWNSHIFT.
Multiply and Divide Opcodes
| Mnemonic | Opcode | Input Stack | Description | Definition | Varops Cost | Varops Reason |
|---|---|---|---|---|---|---|
| OP_2MUL | 141 | Multiply A by 2 | length(A) * 7 | OTHER + COPYING | ||
| OP_2DIV | 142 | Divide A by 2 | length(A) * 4 | OTHER | ||
| OP_MUL | 149 | Multiply A by B | (length(A) + length(B)) * 3 + (length(A) + 7) / 8 * length(B) * 27 (BEWARE OVERFLOW) | See Appendix | ||
| OP_DIV | 150 | Divide A by (non-zero) B | length(A) * 18 + length(B) * 4 + length(A)^2 * 2 / 3 (BEWARE OVERFLOW) | See Appendix | ||
| OP_MOD | 151 | Replace A with remainder when A divided by (non-zero) B | length(A) * 18 + length(B) * 4 + length(A)^2 * 2 / 3 (BEWARE OVERFLOW) | See Appendix |
Rationale
These opcodes can be computationally intensive, which is why the varops budget must be checked before operations. OP_2MUL and OP_2DIV are far simpler, equivalent to OP_UPSHIFT and OP_DOWNSHIFT by 1 bit, except truncating the most-significant zero bytes.
The detailed rationale for these costs can be found in Appendix A.
Limited Hashing Opcodes
OP_RIPEMD160 and OP_SHA1 are now defined to FAIL validation if their operands exceed 520 bytes.1
Extended Opcodes
The opcodes OP_ADD, OP_SUB, OP_1ADD and OP_1SUB are redefined in 0xC2 Tapscript to operate on variable-length unsigned integers. These always produce minimal values (no trailing zero bytes).
| Mnemonic | Opcode | Input Stack | Description | Definition | Varops Cost | Varops Reason |
|---|---|---|---|---|---|---|
| OP_ADD | 147 | Add A and B | MAX(length(A), length(B)) * 9 | ARITH + COPYING | ||
| OP_1ADD | 139 | Add one to A | MAX(1, length(A)) * 9 | ARITH + COPYING | ||
| OP_SUB | 148 | Subtract B from A where B is <= A | MAX(length(A), length(B)) * 6 | ARITH | ||
| OP_1SUB | 140 | Subtract 1 from (non-zero) A | MAX(1, length(A)) * 6 | ARITH |
Rationale
Note that the basic cost for ADD is six times the maximum operand length (ARITH), but then considers the case where a reallocation and copy needs to occur to append the final carry byte (COPYING, which costs 3 units per byte).
Subtraction is cheaper because underflow does not occur: that is a validation failure, as mathematicians agree the result would not be natural.
Misc Operators
The following opcodes have costs below:
| Opcode | Varops Budget Cost | Varops Reason |
|---|---|---|
| OP_CHECKLOCKTIMEVERIFY | Length of operand * 2 | LENGTHCONV |
| OP_CHECKSEQUENCEVERIFY | Length of operand * 2 | LENGTHCONV |
| OP_CHECKSIGADD | MAX(1, length(number operand)) * 9 + 500,000 | ARITH + COPYING + SIGCHECK |
| OP_CHECKSIG | 500,000 | SIGCHECK |
| OP_CHECKSIGVERIFY | 500,000 | SIGCHECK |
Rationale
OP_CHECKSIGADD does an OP_1ADD on success, so we use the same cost as that. For simplicity, this is charged whether the OP_CHECKSIGADD succeeds or not.
Other Operators
The varops costs of the following opcodes are defined in BIP440:
- OP_VERIFY
- OP_NOT
- OP_0NOTEQUAL
- OP_EQUAL
- OP_EQUALVERIFY
- OP_2DUP
- OP_3DUP
- OP_2OVER
- OP_IFDUP
- OP_DUP
- OP_OVER
- OP_PICK
- OP_TUCK
- OP_ROLL
- OP_BOOLOR
- OP_NUMEQUAL
- OP_NUMEQUALVERIFY
- OP_NUMNOTEQUAL
- OP_LESSTHAN
- OP_GREATERTHAN
- OP_LESSTHANOREQUAL
- OP_GREATERTHANOREQUAL
- OP_MIN
- OP_MAX
- OP_WITHIN
- OP_SHA256
- OP_HASH160
- OP_HASH256
Any opcodes not mentioned in this document or the preceding list have a cost of 0 (they do not operate on variable-length stack objects).
Backwards compatibility
This BIP defines a previously unused (and thus, always-successful) tapscript version, for backwards compatibility.
Reference Implementation
Work in progress:
https://github.com/jmoik/bitcoin/tree/gsr
Changelog
- 0.2.1: 2023-03-27: fix OP_MUL cost to round length(B) up
- 0.2.0: 2025-02-21: change costs to match those in varops budget
- 0.1.0: 2025-09-27: first public posting
Thanks
This BIP would not exist without the thoughtful contributions of coders who considered all the facets carefully and thoroughly, and also my inspirational wife Alex and my kids who have been tirelessly supportive of my esoteric-seeming endeavors such as this!
In alphabetical order:
- Anthony Towns
- Brandon Black (aka Reardencode)
- John Light
- Jonas Nick
- Mark "Murch" Erhardt
- Rijndael (aka rot13maxi)
- Steven Roose
- FIXME: your name here!
Appendix A: Cost Model Calculations for Multiply and Divide
Multiplication and division require multiple passes over the operands, meaning a cost proportional to the square of the lengths involved, and the word size used for that iteration makes a difference. We assume 8 bytes (64 bits) at a time are evaluated, and the ability to multiply two 64-bit numbers and receive a 128-bit result, and divide a 128-bit number by a 64-bit number to receive a 128-bit quotient and remainder. This is true on modern 64-bit CPUs (sometimes using multiple instructions).
Multiplication Cost
For multiplication, the steps break down like so:
- Allocate and zero the result: cost = (length(A) + length(B)) * 2 (ZEROING)
- For each word in A:
- Multiply by each word in B, into a scratch vector: cost = 6 * ((length(B) + 7) / 8) * 8 (ARITH)
- Sum scratch vector at the word offset into the result: cost = 6 * ((length(B) + 7) / 8) * 8 (ARITH)
We increase the length of B here to the next word boundary, using "((length(B) + 7) / 8) * 8", as the multiplication below makes the difference of that from the simple "length(B)" significant.
Note: we do not assume Karatsuba, Toom-Cook or other optimizations.
The theoretical cost is: (length(A) + length(B)) * 2 + (length(A) + 7) / 8 * ((length(B) + 7) / 8) * 8 * 12.
However, benchmarking reveals that the inner loop overhead (branch misprediction, cache effects on small elements) is undercosted by the theoretical model. A 2.25× multiplier on the quadratic term accounts for this, giving a cost of: (length(A) + length(B)) * 3 + (length(A) + 7) / 8 * ((length(B) + 7) / 8) * 8 * 27.
This is slightly asymmetric: in practice an implementation usually finds that CPU pipelining means choosing B as the larger operand is optimal.
Division Cost
For division, the steps break down like so:
-
Bit shift both operands to set top bit of B (OP_UPSHIFT, without overflow for B): cost = length(A) * 6 + length(B) * 4
-
Trim trailing bytes. This costs according to the number of byte removed, but since that is subtractive on future costs, we ignore it.
-
If B is longer, the answer is 0 already. So assume A is longer from now on (or equal length).
-
Compare: cost = length(A) * 2 (COMPARING)
-
Subtract: cost = length(A) * 6 (ARITH)
-
for (length(A) - NormalizedLength(B)) in words:
- Multiply word by B -> scratch: cost = NormalizedLength(B) * 6 (ARITH)
- Subtract scratch from A: cost = length(A) * 6 (ARITH)
- Add B into A (no overflow): cost = length(A) * 6 (ARITH)
- Shrink A by 1 word.
-
OP_MOD: shift A down, trim trailing zeroes: cost = length(A) * 4
-
OP_DIV: trim trailing zeros: cost = length(A) * 4
Note that the loop at step 6 shrinks A every time, so the average cost of each iteration is (NormalizedLength(B) * 6 + length(A) * 12) / 2. The cost of step 6 is:
(length(A) - NormalizedLength(B)) / 8 * (NormalizedLength(B) * 6 + length(A) * 12) / 2
The worst case is when NormalizedLength(B) is 0: length(A) * length(A) * 2 / 3.
The cost for all the steps is: length(A) * 18 + length(B) * 4 + length(A) * length(A) * 2 / 3.