vector<const *T>, vector<*T>

Q:

 void f(vector<const T*>& p)
 {
 }
 int main()
 { 
  vector<T*> nonConstVec;
  f(nonConstVec);
 }

The following does not compile.The thing is that vector<T*> can not be converted to vector <const T*> , and that seems illogically to me , because there exists implicit conversion from T* to const T*. Why is this ?

vector<const T*> can not be converted to vector <T*> too, but that is expected because const T* can not be converted implicitly to T*.

A1:

I've added a few lines to your code. That's sufficient to make it clear why this is disallowed:

void f(vector<const T*>& p)
 {
    static const T ct;
    p.push_back(&ct); // adds a const T* to nonConstVec !
 }
 int main()
 { 
  vector<T*> nonConstVec;
  f(nonConstVec);
  nonConstVec.back()->nonConstFunction();
 }

A2:

It may be worth showing why it's a breach of const-correctness to perform the conversion you want:

#include <vector>
const int a = 1;

void addConst(std::vector<const int *> &v) {
    v.push_back(&a); // this is OK, adding a const int* to a vector of same
}

int main() {
    std::vector<int *> w;
    int b = 2;
    w.push_back(&b);  // this is OK, adding an int* to a vector of same
    *(w.back()) = 3;  // this is OK, assigning through an int*
    addConst(w);      // you want this to be OK, but it isn't...
    *(w.back()) = 3;  // ...because it would make this const-unsafe.
}

The problem is that vector<int*>.push_back takes a pointer-to-non-const (which I'll call a "non-const pointer" from now on). That means, it might modify the pointee of its parameter. Specifically in the case of vector, it might hand the pointer back out to someone else who modifies it. So you can't pass a const pointer to the push_back function of w, and the conversion you want is unsafe even if the template system supported it (which it doesn't). The purpose of const-safety is to stop you passing a const pointer to a function which takes a non-const pointer, and this is how it does its job. C++ requires you to specifically say if you want to do something unsafe, so the conversion certainly can't be implicit. In fact, because of how templates work, it's not possible at all (see later).

I think C++ could in principle preserve const-safety by allowing a conversion from vector<T*>& to const vector<const T*>&, just as int ** to const int *const * is safe. But that's because of the way vector is defined: it wouldn't necessarily be const-safe for other templates.

Likewise, it could in theory allow an explicit conversion. And in fact, it does allow an explicit conversion, but only for objects, not references ;-)

std::vector<const int*> x(w.begin(), w.end()); // conversion

The reason it can't do it for references is because the template system can't support it. Another example that would be broken if the conversion were allowed:

template<typename T> 
struct Foo {
    void Bar(T &);
};

template<>
struct Foo<const int *> {
    void Baz(int *);
};

Now, Foo<int*> doesn't have a Baz function. How on earth could a pointer or reference to Foo<int*> be converted to a pointer or reference to Foo<const int*>?

Foo<int *> f;
Foo<const int *> &g = f; // Not allowed, but suppose it was
int a;
g.Baz(&a); // Um. What happens? Calls Baz on the object f?

From:http://stackoverflow.com/questions/2102244/vector-and-const