Overflow/underflow of []byte integer operations

General
,
1

i did read about byte operation here Opcodes - Algorand Developer Portal. so i think if need require large result number, i should use byte instead of uint64 (due to the value).
But i want to know what if the +,* failed on overflow or - operation failed on underflow for bytes (i see in uint operation it did)? what happended when they does not failed (the result will be unexpected)?

2

The byte integer operations b+ and b* do not overflow. The result will just be larger than 64 bytes.
b- fails on underflow.

See Transaction Execution Approval Language (TEAL) - Algorand Developer Portal for details.

1 Like
3

I still have some confuse when reading the doc.
Sorry if i’m not getting it but i really want to deeply understand about this.

  • with your answer The byte integer operations `b+` and `b*` do not overflow so what happend with the return value (if the result more than 64 bytes)?
  • The docs have this line there is an implicit length limit of 128 bytes on outputs, so does that mean the result of all operation (include b+,b*) must be less than 128 bytes, otherwise it will failed? or what happended with the return value?

Thanks again for your help.

4

The result is larger than 64 bytes.

It can never happen. When you add or multiply two 64-byte numbers, the result always fit in 128 bytes.

5

thanks for your answer. it help me a lot

just for clear, for example with this pyteal operation BytesDiv(BytesMul(a,b),c)
If a*b exceed 64 bytes, then the (a*b)/c will be failed (since left argument (which is a*b)) exceed 64 bytes)? the same this happend with b*,b+,b-?

6

and speaking to operation, i want to ask what exactly mulw , addw Opcodes - Algorand Developer Portal do?

can you give me some example, sorry for my bad understanding
Thanks

7

Yes, if a*b exceeds 64 bytes, the above operation will fail.
Note that it is “bytes” not “bits”. 64 bytes = 512 bits, which is very very large.

(Remember uint64 = 64 bits = 8 bytes.)

BytesMul in PyTeal compiles to b*.
BytesDiv in PyTeal compiles to b/.
And so on.

1 Like
8

These are used if you don’t want to use the b... operations.
Nowadays, I think b... are much easier.

1 Like
9

Thanks for your answer. i have another question that need to ask here for byte calculation here.
back to the scenario where i have BytesMul(a,b)
if a*b exceeds 64 bytes and i store the result in scratch slot or global/local slot (which defined as bytes), will it work?
If not, what does the result of a*b that exceeds 64 bytes can work? why doesnt it failed on overflow?

10

inputs of the b+, b*, … operations are limited to 64 bytes.
But outputs are not.
So b* output may be 128 bytes. There is never an overflow.

However, then you cannot use the 128-byte output in any of the b+, b*, … operations.

See The Algorand Virtual Machine (AVM) and TEAL. - Algorand Developer Portal

1 Like
11

thanks for your answer.
my question is can the 128-byte output be used to store in scratch slot or global variable in application (which type defined as bytes)?
If not, what does this 128-byte output can be used?

12

A 128-byte string can be stored in scratchpad or in the global/local storage with key "", the empty key (since you have 128 bytes for both the key and the value together).
You can also hash it, extract part of it, … or do more complex algorithms.

1 Like