Assigning memory variable to storage variable, what happens?

Question

I'm curious what happens if you try to do the next thing:

contract C {
    Struct S {
        uint a;
        uint b;
    }

    mapping(address => s) structs;

    function updateStructs(S sInstance) private {
        structs[msg.sender] = sInstance;
    }

    function addStruct (uint _a, uint _b) payable {
        S memory s = S({a: _a, b: _b});
        updateStructs(s);
    }
}

As I understood how variables work, the addStruct creates memory variable that has only function scope and will be deallocated once this transaction/message is processed. So what will remain in structs[msg.sender] after addStruct has been called? Will changing updateStructs(S sInstance) to updateStructs(S storage sInstance) change something?

Answer 1

When you store a memory variable into storage, then it will just copy the object into storage.

Using the storage keyword in updateStructs actually won't do anything. Structs and arrays in functions are by default storage variables, so the memory object would actually be copied to storage right when you call updateStructs without the storage keyword.

Edit: I stand corrected. @LibertyLocked is correct in their answer where they say you can't implicitly convert from memory to storage, so adding the storage keyword to the argument will make compilation fail.

Answer 2

In your example, if you'd change your code so that it looks like this

function updateStructs(S storage sInstance) private {
    structs[msg.sender] = sInstance;
}

function addStruct (uint _a, uint _b) payable {
    S memory s = S({a: _a, b: _b});
    updateStructs(s);
}

The code will not compile for the following reasons:

• updateStructs expects a struct instance from storage, but you are passing a struct instance in memory to it.
• There is no way for Solidity to figure out where to allocate and place the s in addStruct in storage.
• When you have S memory sInstance in your updateStruct, your s is copied into the stack frame of updateStructs. Also the s in addStruct is not deallocated until you return from addStruct (otherwise how would the function continue to run once it returns from updateStruct?).

To answer your question, in your example you must use memory instead of storage. In fact, it is implicitly memory in function arguments.