Question

I wanted to know how the following works @ compiler level.

int const iVal = 5;  (int&)iVal = 10;  

A bit of m/c or compiler level answer would be great full.

Thanks in advance.

Answer 1

It is undefined behavior.

In the first line you define a constant integer. Henceforth, in your program, the compiler is permitted to just substitute iVal with the value 5. It may load it from memory instead, but probably won’t, because that would bring no benefit.

The second line writes to the memory location that your compiler tells you contains the number 5. However, this is not guaranteed to have any effect, as you’ve already told the compiler that the value won’t change.

For example, the following will define an array of 5 elements, and print an undefined value (or it can do anything it wants! it’s undefined)

int const iVal = 5; (int&)iVal = 10; char arr[iVal]; cout << iVal; 

The generated assembly might look something like:

sub ESP, 9      ; allocate mem for arr and iVal. hardcoded 5+sizeof(int) bytes                 ; (iVal isn't _required_ to have space allocated to it) mov $iVal, 10   ; the compiler might do this, assuming that you know what                 ; you're doing. But then again, it might not. push $cout push 5 call $operator_ltlt__ostream_int add ESP, 9