I dont mean a variable in a class but a default value for the class as a whole.
struct Scalar {
unsigned int value;
void toThePowerOf(int power);
// etc.
}
I'd like to be able to do something like
Scaler foo;
foo.value = 2;
foo.toThePowerOf(2);
std::cout << foo << std::endl; // Outputs 4!
开发者_运维技巧
Is that possible in C++?
No, classes do not have a value in the way you're thinking.
What you probably mean to do is to overload the << operator:
ostream& operator<<(ostream& output, const Scalar& val)
{
output << val.value;
return output;
}
I meant a default value for the class, so if you called that object by just its name
foo
it would returnfoo.value
by default.
It is actually possible to define an implicit conversion from Scalar
to int
:
struct Scalar
{
unsigned int value;
operator int() const
{
return value;
}
};
int main()
{
Scalar foo = {2};
std::cout << foo << std::endl;
}
But implicit conversions are generally frowned upon in the C++ community, because it can make code quite hard to read. (I guess this is why noone mentioned conversion operators yet.)
No, you cannot but you can overload operator << for your class and ostream to get the desired effect
std::ostream& operator << (std::ostream& out, const Scaler& foo)
{
return out << foo.value;
}
Your code will now work and produce the desires result
Yes. it is possible. Just initialize all the values in the constructor of the class. Use class instead of struct.
Use a default value in the ctor. Make the ctor explicit if you don't want implicit conversions.
struct Scalar {
unsigned int value;
Scalar(int value=0) : value (value) {}
void toThePowerOf(int power) {
// naive implementation just for show
int new_value = 1;
assert(power >= 0); // or other error checking
for (; power > 0; --power) {
new_value *= value;
}
value = new_value;
}
friend std::ostream& operator<<(std::ostream &out, Scalar const &x) {
out << x.value;
return out;
}
};
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