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type 'type' does not have an overloaded member 'operator ->'
Remarks
You need to define operator->() to use this pointer operation.
Examples
The following example generates C2819:
// C2819.cpp
// compile with: /c
class A {
public:
int i;
};
class B {};
void C(B j) {
j->i; // C2819
}
class D {
A* pA;
public:
A* operator->() {
return pA;
}
};
void F(D j) {
j->i;
}
C2819 can also occur when using C++ Stack Semantics for Reference Types. The following example generates C2819:
// C2819_b.cpp
// compile with: /clr
ref struct R {
void Test() {}
};
int main() {
R r;
r->Test(); // C2819
r.Test(); // OK
}