Disable implicit conversion between typedefs - c++

Question

In C++11, is there a clean way to disable implicit conversion between typedefs, or do you have to do something nasty like wrap your int in a class and define and delete various operators?
typedef int Foo;
typedef int Bar;
Foo foo(1);
Bar bar(2);
bar = foo; // Implicit conversion!

Answer 1

HelloWorld explains why what you have cannot work. You'll need what's typically called a "strong" typedef to do what you want. An example implementation is BOOST_STRONG_TYPEDEF:
#include <boost/serialization/strong_typedef.hpp>
BOOST_STRONG_TYPEDEF(int, a)
void f(int x); // (1) function to handle simple integers
void f(a x); // (2) special function to handle integers of type a
int main(){
int x = 1;
a y;
y = x; // other operations permitted as a is converted as necessary
f(x); // chooses (1)
f(y); // chooses (2)
}
If we had done typedef int a;, then the code would be ambiguous.

Answer 2

The C++ standard says:
7.1.3 The typedef specifier
A name declared with the typedef specifier becomes a typedef-name. Within the scope of its declaration, a typedef-name is syntactically equivalent to a keyword and names the type associated with the identifier in
the way described in Clause 8. A typedef-name is thus a synonym for another type. A typedef-name does
not introduce a new type the way a class declaration (9.1) or enum declaration does
But e.g. class or struct introduce new types. In the following example uniqueUnused does actually nothing but is used to create a different type Value<int, 1> != Value<int, 2>. So maybe this is something you are looking for. Keep in mind there is no guarantee the compiler gets rid of the outer structure! The only guarantee this code gives you it's the same size as int
template<typename T, int uniqueUnused>
struct Value
{
Value() : _val({}) {}
Value(T val) : _val(val) { }
T _val;
operator T&() { return _val; }
// evaluate if you with or without refs for assignments
operator T() { return _val; }
};
using Foo = Value<int, 1>;
using Bar = Value<int, 2>;
static_assert(sizeof(Foo) == sizeof(int), "int must be of same size");
static_assert(sizeof(Bar) == sizeof(int), "int must be of same size");
If you want to create a new type based on a class you can simply go with this example (this doesn't work with scalar types since you can't inherit from ints):
class Foo : public Bar // introduces a new type called Foo
{
using Bar::Bar;
};

Answer 3

I wanted to do something similar to keep different indexes separated not only logically, but also enforced by the compiler. The solution I came up with is basically to just define structs with one element. In some ways it's more painful to use, but it works very well with my situation since I often don't need to deal with the actual value of the index for a lot of my code, just passing it around.
typedef struct{uint16_t i;} ExpressionIndex;
typedef struct{uint16_t i;} StatementIndex;
Now, trying to do
ExpressionIndex foo() {
StatementIndex i;
return i;
}
yields the error error: incompatible types when returning type ‘StatementIndex’ but ‘ExpressionIndex’ was expected
Converting between types is a bit painful, but that was the intent of my change.
ExpressionIndex exp = (ExpressionIndex){stmt.i};

Answer 4

It's not strict type-checking, but illegal conversions can made visible by using original, or Apps Hungarian Notation (H. N.). If you think H. N. means name-type-as-prefix, you're wrong (it's System H. N., and it's, hm, unnecessary naming overhead).
Using the (Apps) H. N., the variable prefix marks not the type (e.g. int), but the purpose, e.g. counter, length, seconds etc. So, when you add a counter to a variable contains elapsed time, you write cntSomethingCounter + secElapsedSinceLastSomething, and you can see that it smells. Compiler does not alerts, but it pokes your eyes.
Read more: http://www.joelonsoftware.com/articles/Wrong.html