What happens in CALL when gas is set to 0?

Question

Reading the documentation on the CALL opcode :

"call contract at address a with input mem[in..(in+insize)) providing g gas and v wei and output area mem[out..(out+outsize)) returning 0 on error (eg. out of gas) and 1 on success"

Where CALL is

call(g, a, v, in, insize, out, outsize)

If I test a contract which uses the transfer function and then check what happens in assembly, I see that the stack does indeed represent the right address a at the first position of the stack and the right value v at the second position of the stack. However, the 0th postion of the stack is set to 0. This would mean 0 gas is sent with this call. In transfer, 2300 gas has to be sent, but this apparently does not happen.

How does this work?

EDIT Contract code

pragma solidity ^0.4.24;
contract Transfer{
function test() public payable {msg.sender.transfer(msg.value);}
}

Answer 1

That is highly unlikely... With Solidity 0.4.25, compiling address.call() would give you:

...
gas
call

As you can see, the last parameter pushed on the stack is the total amount of available amount of gas after the execution of the GAS opcode itself.

However, in your example, you are compiling address.transfer(), which, per the Solidity docs, forwards a standard gas stipend of 2300.

  0x8fc
  ...
  dup9  // duplicates 0x8fc=2300
  call