StackOverflow2013

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2013-04-11T08:40:04.700
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You got an error because the conversion is illegal. The reason it is illegal is simple: it breaks <code>const</code> without requiring a <code>const_cast</code>. As an example of why it is forbidden, imagine that <code>foo</code> was: <pre><code>void foo( char const*& str ) { str = "abc"; } int main() { char* h; foo( h ); *h = '1'; } </code></pre> If you're not going to modify <code>str</code> in <code>foo</code>, pass by value. Pass by reference will works if you have <code>char const* const&</code>, but there's no reason to use it here. It works because the additional const means that you can bind a temporary to it (as in the case of <code>foo( "hello" )</code>, where the argument is a temporary resulting from the conversion of <code>char const[6]</code> to <code>char const*</code>, and <code>foo( h )</code> works, because the implicit const conversions will work (in C++, but not in C!), as long as you add const everywhere (and not just at one level). Also, your code also uses a deprecated conversion to initialize <code>h</code>. You should get a warning here. And <code>void main</code> is an error, and shouldn't compile. <h2>Edit:</h2> Just to be clear, there's no problem with: <pre><code>void f( char const*& str ); </code></pre> But you can only call it with an lvalue of type <code>char const*</code>; anything else will either result in an illegal implicit const conversion, or try to initialize a non-const reference with an rvalue, which is illegal.
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1. POcan't convert from char * to const char *&
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1. USJames Kanze
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1. COCan you briefly explain what `char const* const&` does? Why do you need the "const" twice?
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2. CO@akristmann There are two things which can be immutable: the pointer itself, and the data it points to. `const` applies to whatever is to the left of it, so in `char const* const&`, the first const means that the pointed to `char` are constant, and the second that the pointer is constant.
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