Trade-off with const in legacy code

Today we argued an hour with a colleague regarding the following code:

void foo(T* t) {

The problem is that the function bar is a part of a legacy library which we cannot refactor right now. The signature of bar is void bar(T*). T is not const. But in reality bar never changes an object referenced by t. This is how it was implemented.

But foo is a part of a brand new API, and we want to make nice or clean. The contract of the function foo says that it doesn’t need to change its parameter.

I think the code should be like this:

void foo(const T* t) {

Why? The contact of foo doesn’t require the pointer t to be non-const. We must reflect this in the API by making t const. It doesn’t matter that for some reason a particular implementation of foo is based on the legacy bar function not having const in the argument but never changing it. Yes, we have to use the ugly const_cast but this bad code is nicely isolated inside foo only and doesn’t affect our nice and clean brand new API. Moreover, if we refactor foo at some point and get rid of legacy bar at all, the problem will disappear completely.

Here is a counterargument from my colleague: it may turn out that the function foo can have a bug and accidentally change t even it is declared as const. The solution is to simply keep the argument of foo non-const. In this case we don’t need that cast, we explicitly show to an end user of foo that she should expect its parameter to be const, and eventually we never violate the contact of the function foo.

Eventually we haven’t agreed. My flaw is that const doesn’t really protect from side effects coming from legacy bar and the argument of foo may be changed regardless being const. My friend’s flaw is that it is not easy to explain in the documentation how and why the argument of foo may be changed. Just because our particular implementation dictates this? Such approach spreads the drawback of the legacy code to our nice and shiny new code.


P.S. There is another esoteric approach — to create a temporary deep copy of T inside foo and pass it to bar by non-const pointer. Personally if I have to choose between quick but badly designed code and slow but nicely written code I usually go for the second one. Tomorrow we can buy another faster computer and the slow code will be faster, but that computer will make the bad code better.

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