Discussion between hrkrshnn and Nika Kurashvili

« first day (2 days earlier) last day (15 days later) »

hrkrshnn

10:04

Hi again. So the problem is that, in general the copy memory -> memory is the abi encoding of the memory object. In general, it may not be the same.

In this case, the encoding is the same, though.

gitter.im/ethereum/solidity-dev?at=6013f97b4f7d1b68e519419c relevant.

and gitter.im/ethereum/solidity-dev?at=6017cfb04f7d1b68e522e3a9

@NikaKurashvili tagging just in case, you don't get notified otherwise.

2 hours later…

Nika Kurashvili

11:41

still there ?

@hrkrshnn

hrkrshnn

11:53

Yep

Nika Kurashvili

I didn't quite understand something

even if it does abi encoding, the space will still be required in memory, the same size or more, because in the encoding, we also need to store sizes and all that if it's dynamic

makes sense ?

@hrkrshnn

hrkrshnn

But the abi encoding can be different. Which is why you need a copy.

In your example, you don't need a copy.

Nika Kurashvili

1. but it still copies

doesn't it ?

hrkrshnn

Yes. This is the job of the optimizer. It currently can't handle this case.

That is, to remove the unnecessary copy.

Nika Kurashvili

why does it store abi encoding at all ?

so it copied from calldata to memory let's say [99,101] array

now as you're saying it puts abi.encode([99,101]) in the memory again before calling the function, right ?

hrkrshnn

12:02

Yeah.

Nika Kurashvili

I just don't get why .

hrkrshnn

You need to store it before you do the external call.

Nika Kurashvili

That's the thing! why ? :D

hrkrshnn

How else would you do it?

Nika Kurashvili

you already have it in memory

use it !!!

hrkrshnn

12:04

For one, it's not in the memory after you do the external call.

Nika Kurashvili

when you do the external call, it goes as the calldata for the callee function

so what you're saying is the following

it abi encodes the data

puts it in memory again

and before using call opcode, it takes this newly data (abi encoded) in memory and passes as calldata

am I right ?

hrkrshnn

```
contract C {
    function f() public returns(uint r) {
        assembly {
            r := mload(10000)
        }
    }
}

contract D {
    C c = new C();

    function g() public returns(uint old, uint intme, uint aft, uint Cmemory) {
        assembly {
            old := mload(10000)
            mstore(10000, 200)
            intme := mload(10000)
        }

        Cmemory = c.f();

        assembly {
            aft := mload(10000)
        }
    }
}
```

Nika Kurashvili

I will take a look at it, but could you agree to this ?

so what you're saying is the following
it abi encodes the data
puts it in memory again
and before using call opcode, it takes this newly data (abi encoded) in memory and passes as calldata
am I right ?

hrkrshnn

In this example, r would return 0.

Yeah. That sounds right.

Nika Kurashvili

i think

that abi encoded data will always be bigger than the actual data