How unions are used to avoid type coversion

Question

I am reading The C++ Programming Language by Bjarne Stroustrup. It states on unions:

"Unions are occasionally used deliberately to avoid type conversion. One might, for example, use a Fudge to find the representation of the pointer 0:

union Fudge{
    int i;
    int* p;
};

int main()
{
    Fudge foo;
    foo.p= 0;
    cout<< "the integer value of the pointer 0 is" << foo.i<< ´\n´;
} "

I get that this code fragment gives the representaion of pointer 0. But how is it avoiding type-conversion as mentioned in the text?

This is not valid C++ code. And didn't we just have this? https://stackoverflow.com/questions/48843760/conversion-of-int-to-int-using-union posted by you — , Feb 17 '18 at 18:42
std::variant might be of assistance, although it might add some overhead — JVApen, Feb 17 '18 at 18:43
Keep in mind: that book is more than thirty years old. Stroustrup today, in his C++ “Guidelines,” recommends using `reinterpret_cast` instead of this kind of type-punning through a `union`, because both are undefined behavior, but at least the cast makes it explicit. However, in the intervening years, the C (but not C++) language standard made type-punning with `unions` legal. The difficulty with doing it in C++ is that the union member might not be trivially-copyable, and the constructor would not have been called. — Davislor, Feb 17 '18 at 18:56

Some programmer dude · Accepted Answer · 2018-02-17T18:48:40.893

1

It's called type punning and is explicitly allowed using unions in C but not in C++.

It works (in C) because all members of a union occupy the same memory, so there's really no conversion. You just interpret the bits of the memory differently.

edited Feb 17 '18 at 18:48

answered Feb 17 '18 at 18:42

Some programmer dude

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How unions are used to avoid type coversion

1 Answers1