const-correctness

First, why would you ever want to have the ability to change data in a class that's declared const? This gets at the heart of what constness means, and there are two ways of thinking about it. One idea is that of "bitwise constness", which basically means that a const class should have exactly the same representation in memory at all times. Unfortunately (or fortunately), this is not the paradigm used by the C++ standard; instead, C++ uses "conceptual constness". Conceptual constness refers to the idea that the output of the const class should always be the same. This means that the underlying data might change as long as the fundamental behavior remains the same. (In essence, the "concept" is constant, but the representation may vary.)

http://www.cprogramming.com/tutorial/const_correctness.html

Whenever you see

foo class Something {} bar;

It's the same as

class Something {};foo Something bar;

Or in the case of

const class test{int a;};

The same as

class test {  int a;};const test;

const test; by itself is basically a pointless, but legal C++ statement, in the same way const int; by itself is a pointless but legal C++ statement.

Const-correctness has no runtime overhead as all checking is done at compile-time.

Quote:

Original post by scjohnno Const-correctness has no runtime overhead as all checking is done at compile-time.

Yeah.Thats what I could think of.That in the worst situation it will have compile-time overhead as it will try to ensure that const-correctness has been maintained.Thnks all you guys

const can also enhance performance, allowing the compiler to avoid the creation of temporary variables that (if const'ed) would never vary. Ie., As far as I know:

vector3d v = negate(v);

where

const vector3d negate(const vector3d& par)
{
return vector3d(-par.x, -par.y, -par.z);
}

in this case no temporary variable will be created when the function returns.
This is what they told me, so I hope that it is correct :-)

Quote:

Original post by cignox1 const can also enhance performance, allowing the compiler to avoid the creation of temporary variables that (if const'ed) would never vary. Ie., As far as I know: vector3d v = negate(v); where const vector3d negate(const vector3d& par) { return vector3d(-par.x, -par.y, -par.z); } in this case no temporary variable will be created when the function returns. This is what they told me, so I hope that it is correct :-)

That optimization can be applied even if the return type is non-const.

See: http://www.gotw.ca/gotw/081.htm

Specifically, this is called the Named Return Value Optimization in MSVC.

Quote:

Original post by RDragon1 Quote: Original post by cignox1 const can also enhance performance, allowing the compiler to avoid the creation of temporary variables that (if const'ed) would never vary. Ie., As far as I know: vector3d v = negate(v); where const vector3d negate(const vector3d& par) { return vector3d(-par.x, -par.y, -par.z); } in this case no temporary variable will be created when the function returns. This is what they told me, so I hope that it is correct :-) That optimization can be applied even if the return type is non-const. See: http://www.gotw.ca/gotw/081.htm

Oh, ok, thank you for this precisation :-)