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Home/ Questions/Q 1079205
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T21:50:06+00:00 2026-05-16T21:50:06+00:00

I want to enforce explicit conversion between structs kind of like native types: int

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I want to enforce explicit conversion between structs kind of like native types:

int i1;
i1 = some_float; // this generates a warning
i1 = int(some_float): // this is OK
int i3 = some_float; // this generates a warning

I thought to use an assignment operator and copy constructor to do the same thing, but the behavior is different:

Struct s1;
s1 = other_struct; // this calls the assignment operator which generates my warning
s1 = Struct(other_struct) // this calls the copy constructor to generate a new Struct and then passes that new instance to s1's assignment operator
Struct s3 = other_struct; // this calls the COPY CONSTRUCTOR and succeeds with no warning

Are there any tricks to get that third case Struct s3 = other_struct; construct s3 with the default constructor and then call the assignment operator?

This all compiles and runs as it should. The default behavior of C++ is to call the copy constructor instead of the assignment operator when you create a new instance and call the copy constructor at once, (i.e. MyStruct s = other_struct;becomes MyStruct s(other_struct); not MyStruct s; s = other_struct;. I’m just wondering if there are any tricks to get around that.

EDIT: The “explicit” keyword is just what I needed!

class foo {
    foo(const foo& f) { ... }
    explicit foo(const bar& b) { ... }
    foo& operator =(const foo& f) { ... }
};

foo f;
bar b;
foo f2 = f; // this works
foo f3 = b; // this doesn't, thanks to the explicit keyword!
foo f4 = foo(b); // this works - you're forced to do an "explicit conversion"
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T21:50:07+00:00Added an answer on May 16, 2026 at 9:50 pm

    You can get around this if you overload the type cast operator for other_struct, and edit the original structure accordingly. That said, it’s extremely messy and there generally isn’t a good reason to do so.


    #include <iostream>
    
    using namespace std;
    
    struct bar;
    
    struct foo {
        explicit foo() {
            cout << "In foo default constructor." << endl;
        }
    
        explicit foo(bar const &) {
            cout << "In foo 'bar' contructor." << endl;
        }
    
        foo(foo const &) {
            cout << "In foo constructor." << endl;
        }
    
        foo const & operator=(bar const &) {
            cout << "In foo = operator." << endl;
            return *this;
        }
    };
    
    struct bar {
        operator foo() {
            cout << "In bar cast overload." << endl;
            foo x;
            x = *this;
            return x;
        }
    };
    
    int main() {
        bar b;
        foo f = b;
        return 0;
    }
    

    Outputs:

    In bar cast overload.
    In foo default constructor.
    In foo = operator.
    In foo constructor.
    In foo constructor.
    
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