Quote:
Originally Posted by Tolnaftate2004
The engine replaces inline function definitions with a function pointer (e.g., "function_1"). The return value of assignment should be the rvalue, so temp.f = function_1 returns that pointer. It's no mystery why it works. The mystery is why it doesn't work the way one would want it to. 
Probably has something to do with the way normal function are declared. |
Okay, I just tried it and it worked. That is a leap of logic I didn't make for the assignment operator to return the function pointer.
With this, you can also do:
PHP Code:
someOtherFunction(temp.f = function (temp.y) { return temp.y + 1; } );
But that is pretty nasty either way
.