Why is the conversion constructor not called implicitly for i2?
class NumString
{
public:
NumString(const char* s)
{
}
int operator*( int i)
{
return 42;
}
};
int main(void)
{
int i1 = (NumString) "string" * 2; //OK
int i2 = "string" * 2; //ERROR
}
Because the compiler doesn't invoke user-defined conversions where there aren't any user-defined types involved.
The expression "string" * 2 involves just a const char * and an int, why should the compiler consider NumString in any way?
If it worked like you expect, how would the complier be expected to choose the correct conversion if more than one class had a suitable converting constructor?
Related
class A {
int x;
std::string s;
public:
A(std::string _s): x(0), s(_s) {
std::cout << "\tA(string)\n" ;
}
A(int _x): x(_x), s("") {
std::cout << "\tA(int)\n" ;
}
A(const A &other): x(other.x), s(other.s) {
std::cout << "\tA(A& other)\n" ;
}
};
int main() {
std::string str = "Hello";
A obj_1(str);
A obj_2 = str;
A obj_3(10);
A obj_4 = 10;
char ch = 'a';
A obj_5 = ch; // How is this working?
// A obj_6 = "Hello"; // And not this?
const char *ptr = "Hello";
// A obj_7 = ptr; // or this?
return 0;
}
On executing this code, the output is:
A(string)
A(string)
A(int)
A(int)
A(int)
As far as I understand, obj_1 and obj_3 are created directly by using the respective ctors. For obj_2 and obj_4, the compiler does an implicit conversion by calling their respective ctor (so only 1 implicit conversion is required in each case). However, for obj_5 the compiler first has to convert from char to int and then one more implicit conversion to call the int-ctor. But C++ standard allows only 1 implicit conversion. So what is happening here?
Also, in that case "Hello" should first get converted to a std::string then one more implicit conversion to call the string-ctor. But this doesn't work.
But C++ standard allows only 1 implicit conversion.
Thats not correct.
From cppreference:
Implicit conversion sequence consists of the following, in this order:
zero or one standard conversion sequence;
zero or one user-defined conversion;
zero or one standard conversion sequence.
From the language point of view, const char[N] -> std::string (or const char* to std::string) is a user-defined conversion. Hence, the commented out lines are errors. On the other hand,
A obj_5 = ch; // How is this working?
is fine, because there is only a single user-defined conversion involved.
'int' and 'double' conversion functions is 'explicit' and in this code why have I permit for use this conversion instead of error message?
If I delete all my conversion overload functions code occur 'conversion error'
class Person
{
public:
Person(string s = "", int age = 0) :Name(s), Age(age) {}
operator string() const { return Name; }
explicit operator int() const{ return 10; } // ! Explicit
explicit operator double()const { return 20; } //! Explicit
operator bool()const { if (Name != "") return true; return false; } // using this
};
int main(){
Person a;
int z = a;
std::cout << z << std::endl; // Why print "1"? Why uses bool conversion?
}
I this it is answer:
Because 'a' can not be convert to int or double it occur error, but because it has bool conversion function, which can convert to int and int to double, code use this function.
int z = a;
Looks innocuous, right?
Well, the above line calls the implicit bool-conversion-operator, because that's the only way you left it to get from Person to int, and that only needs one user-defined conversion (the maximum allowed).
This is the reason for the safe-bool-idiom before C++11, and the major reason for the general advice to mark conversion operators and single-argument ctors explicit unless they are not transformative nor lossy.
If you want to see it for yourself, fix your code and trace invocation of your operator bool.
I run into a strange c++ operator.
http://www.terralib.org/html/v410/classoracle_1_1occi_1_1_number.html#a0f2780081f0097af7530fe57a100b00d
class Number ｛
..
operator unsigned short () const;
};
I called this operator as:
a Number(..);
unsigned short b = a.operator unsigned short();
this works, but I can't understand how it works.
first, this operator don't have a return value.
seconds, a.operator unsigned short() is really strange to me. What is a better way to call this?
if I call :
unsigned short b = a; does the operator will get called? is there any c++ standard to say about this?
The function is a user defined conversion operator. More details can be found at http://en.cppreference.com/w/cpp/language/cast_operator.
You said,
this operator don't have a return value. seconds,
The return values of the user define conversion operators is the explicit type. In your case, the return type is unsigned short.
You asked:
What is a better way to call this?
You could do an explicit cast to invoke the function.
Number n;
unsigned short s = (unsigned short)v;
It is also called when an conversion is required by the compiler.
void foo(unsigned short s) {}
Number n;
foo(n); // Number::operator unsigned short() is called to cast
// n to an unsigned short.
You asked:
if I call : unsigned short b = a; does the operator will get called? is there any c++ standard to say about this?
Yes. The user defined operator function gets called.
Here's the relevant sections from the C++ Draft Standard (N3337):
12.3.2 Conversion functions
1 A member function of a class X having no parameters with a name of the form
...
[ Example:
struct X {
operator int();
};
void f(X a) {
int i = int(a);
i = (int)a;
i = a;
}
In all three cases the value assigned will be converted by X::operator int(). — end example ]
This is the conversion operator. A conversion function typically has the general form
operator type() const;
where type represents a type. It means objects of type Number can be converted to short int.
The conversion operator have no explicitly stated return type and no parameters, because the return type is exactly the type in the signature.
It's a conversion function, called to convert your type into a specific other type under various conditions, and it's covered in ISO C++11 12.3.2 Conversion functions.
In your case, it's called when the Number instance needs to be converted into an unsigned short.
By providing conversion operators, you can take full control over what happens during the conversion process, including such evil as the following:
#include <iostream>
struct X {
int val;
X(int v) { val = v; };
operator int() { return val + 1; }; // pure evil
friend std::ostream& operator<< (std::ostream&, X&);
};
std::ostream& operator<< (std::ostream &out, X &x) {
out << x.val;
return out;
}
int main (void) {
X xyzzy (42);;
std::cout << xyzzy << '\n';
std::cout << (int)xyzzy << '\n';
return 0;
}
which will output the value when you use the instance directly, but output something totally different when you cast it.
Now granted, that's rather evil and not a really good use case but you can use this for things such as rounding floats rather than truncating them, when converting to an integer:
#include <iostream>
struct X {
double val;
X(double v) { val = v; };
operator int() { return (int)(val + 0.5); };
friend std::ostream& operator<< (std::ostream&, X&);
};
std::ostream& operator<< (std::ostream &out, X &x) {
out << x.val;
return out;
}
#define E 2.718281828456
int main (void) {
X xyzzy (E);
double plugh = E;
std::cout << plugh << " -> " << (int)plugh << '\n';
std::cout << xyzzy << " -> " << (int)xyzzy << '\n';
return 0;
}
The output of that code is:
2.71828 -> 2
2.71828 -> 3
As a supplement for the first answer, when you use keyword explicit, note the difference, using explicit would force the programmer to assert his intention to convert using a cast:
class Number {
private:
int num;
public:
explicit Number(int number) : num(number) {} // constructor
explicit operator unsigned short () const { // conversion operator
return num;
}
};
int main() {
Number classTypeNumber(10);
// unsigned short convertToUshortNumber = classTypeNumber; // error
// implicit conversion is not allowed.
// now you should explicit convert the instance first.
// typedef unsigned short int __u_short in types.h file.
unsigned short convertToUshortNumber = static_cast<__u_short>(classTypeNumber);
cout << convertToUshortNumber;
}
I know that there is no legal overload based on return type in C++; i.e.
you cannot do something like:
int operator ++ getOwner();
char operator ++ getOwner();
However, I stumbled upon the following:
https://stackoverflow.com/a/9569120/1376317
class Proxy
{
My const* myOwner;
public:
Proxy( My const* owner ) : myOwner( owner ) {}
operator int() const
{
return myOwner->getInt();
}
operator char() const
{
return myOwner->getChar();
}
};
My question is how does operator overload work in this configuration. How do you call this in your main.cpp to get this kind of overloading. How does the compiler deduce , how to call the right overload?
My question is how does operator overload work in this configuration.
These operators provide implicit conversions. That means that this class can be used in many contexts where an int or char is expected, and will use these operators to provide the expected value.
How do you call this in your main.cpp to get this kind of overloading.
Here are a few examples of implicit conversions:
Proxy p = whatever();
int i = p; // convert to int
char c = p; // convert to char
long l = p; // ERROR: ambiguous
void f(int);
f(p); // convert to int
void g(int);
void g(char);
g(p); // ERROR: ambiguous
You can also request explicit conversions using the usual cast notations:
long l = static_cast<int>(p); // convert to int, then to long
g((char)p); // convert to char
How does the compiler deduce , how to call the right overload?
Whenever there's a type mismatch, the compiler looks for a conversion sequence. The rules are quite complicated, but basically the sequence can include at most one user-defined conversion (using either an operator like this, or a converting construction), as well as standard conversions such as int to long.
This is sometimes called the Return Type Resolver idiom or "overload on return type". The conversion operator to call is selected thanks to the context of use where an implicit conversion is needed (for example based on the type of an object to initialize or assign to). For example:
#include <stdio.h>
class RtR {
public:
operator int() const
{
puts("operator int()");
return 42;
}
operator double() const
{
puts("operator double()");
return 3.14;
}
};
void f(int) {}
int main()
{
RtR x;
int i = x; // or int i = RtR();
double d = x;
f(x);
}
output:
operator int()
operator double()
operator int()
See it live.
In 13.1 Overloadable declarations:
Function declarations that differ only in the return type cannot be
overloaded. [ Note: ... It does not apply to sets of functions
fabricated as a result of name lookup (e.g., because of using-directives) or overload resolution (e.g.,
for operator functions) ... ]
Hence this is valid:
struct X {
// Conversion function:
operator int () { return 1; }
operator double () { return 2; }
};
In addition (not relating directly to the question):
struct Y
{
// Operator (Function call):
int operator () (int) { return 1; }
double operator () (double) { return 2; }
// Operator (Subscripting):
int operator [] (int) { return 1; }
double operator [] (double) { return 2; }
// Operator (Shift):
int operator << (int) { return 1; }
double operator << (double) { return 2; }
// and more ...
};
The above code is for operator type conversions and provide an implicit way to cast the Proxy type to int and char.
The compiler "knows" based on the context of the conversion calls, for instance:
Proxy p;
// p_int will be set equal to p.my_owner->getInt()
int p_int = p;
// p_char will be set equal to p.my_owner->getChar()
char p_char = p;
The fact that it's a proxy is irrelevant; the same thing works for any class. Those are conversion operators, and the compiler does select the right version based on how it's used in the calling code.
struct S {
operator int() const { return 1; }
operator double() const { return 2.0; }
};
int main() {
S s;
int i = s;
double d = s;
std::cout << i << ' ' << d << '\n';
return 0;
}
My question is regarding data type conversion in c++. Does c++ provides a implicit conversion for built-in data types ( int, float) to user defined data types?
In the following example, I am trying to add a double ( t4 = t3 + 1.0) with test object type and its working fine using + operator so is the double implicitly converted to test type object?
class test {
double d;
int m;
public:
test() {
d=0;
m=0;
}
test(double n) {
d=n;
}
const test operator+(const test& t) {
test temp;
temp.d = d+ t.d;
return temp;
}
};
int main() {
test t1(1.2);
test t2(2.5);
test t3, t4;
t3= t1+ t2;
t4 = t3 + 1.0;
return 0;
}
The constructor test(double n) declares an implicit conversion from double to test. In general, any constructor that is callable with exactly one argument (that includes constructors that can take more arguments, but have default-values for those), can be used as an implicit conversion, unless it is marked explicit:
struct Foo {
Foo(int x); // implicit conversion from int to Foo
explicit Foo(char c); // marked explicit - no implicit conversion
Foo(std::string a, double pi = 3.14159); // can be used as implicit
// conversion due to default
// argument
};
Edit: t4 = 1.0 + t3 does not work, because you have overloaded your operator+ as a member-function, and thus it is only considered if the first operand is of the type test - implicit conversion are not tried in this case. To make this work, make your operator a free function:
test operator+(const test& lhs, const test& rhs);
Yes, as long as your constructor test(double) is not explicit
// implicite convertible
test(double n)
{
d=n;
}
// not implicite convertible
explicit test(double n)
{
d=n;
}