Calling payable function will directly transfer eth?

Question

It seems that calling a payable function will transfer eth directly even without explicitly calling transfer(). For example:

contract Storage {
    receive() external payable {}

    // Fallback function is called when msg.data is not empty
    fallback() external payable {}

    function getBalance() public view returns (uint) {
        return address(this).balance;
    }

    function getMoney() public payable returns (bool) {
        // payable(address(this)).transfer(msg.value);
        return true;
    }

}

Say I have an instance of Storage is at address x. If I call getMoney() with msg.value=1, the balance of x will increase by 1, both with and without "payable(address(this)).transfer(msg.value);". Is this expected?

Answer 1

Yes, it is correct. Each functions and addresses declared payable can receive ether. In this case if you declare getMoney() function in this way:

function getMoney() public payable returns (bool) {
   return true;
}

and you insert a value into msg.value box (in Remix IDE) the function will to transfer the amount at address of smart contract.

More information here.

IMPORTANT: Keep in mind payable is a modifier not a function!