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Why is the following allowed to be compiled in C++?
#include<iostream> using namespace std; class mytest { public: operator int() { return 10; } operator const int() { return 5; } }; int main() { mytest mt; //int x = mt; //ERROR ambigious //const int x = mt; //ERROR ambigious } Why does it make sense to allow different versions (based on constness) of the conversion operator to be compiled when their use always results in ambiguity?
Can someone clarify what I am missing here?
789
So in your question, "int const *" means that the int is constant, while "int * const" would mean that the pointer is constant. If someone decides to put it at the very front (eg: "const int *"), as a special exception in that case it applies to the thing after it.
const int * And int const * are the same. const int * const And int const * const are the same. If you ever face confusion in reading such symbols, remember the Spiral rule: Start from the name of the variable and move clockwise to the next pointer or type.
int *const is a constant pointer to integer This means that the variable being declared is a constant pointer pointing to an integer.
The const keyword specifies that a variable's value is constant and tells the compiler to prevent the programmer from modifying it.
For conversion they're ambiguous; but you might call them explicitly. e.g.
int x = mt.operator int(); const int x = mt.operator const int(); If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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