[Solved] cout value without touching variable C++

Question

When you print the addresses of a and b, that forces the variables to actually have a memory address. Otherwise a value might be stored in a register, or not at all if never used.

It is a very common optimization for compilers to transform

const int a = 9;
cout << a;

into

cout << 9;

and never store a anywhere.

And anyway, there are no rules saying where and in what order the variables are to be stored in memory. That is totally up to the compiler. So your assumption that &a == &b + 1 (or &b - 1) is not valid.

Accepted Answer

When you print the addresses of a and b, that forces the variables to actually have a memory address. Otherwise a value might be stored in a register, or not at all if never used.

It is a very common optimization for compilers to transform

const int a = 9;
cout << a;

into

cout << 9;

and never store a anywhere.

And anyway, there are no rules saying where and in what order the variables are to be stored in memory. That is totally up to the compiler. So your assumption that &a == &b + 1 (or &b - 1) is not valid.