void outputString(const string &ss) {
cout << "outputString(const string& ) " + ss << endl;
}
void outputString(const string ss) {
cout << "outputString(const string ) " + ss << endl;
}
int main(void) {
//! outputString("ambigiousmethod");
const string constStr = "ambigiousmethod2";
//! outputString(constStr);
} ///:~
How to make distinct call?
EDIT: This piece of code could be compiled with g++ and MSVC.
thanks.
C++ does not allow you to overload functions where the only difference in the function signature is that one takes an object and another takes reference to an object. So something like:
void foo(int);
and
void foo(int&);
is not allowed.
You need to change the number and/or the type of the parameter.
In your case the function that accepts a reference, you can make it accept a pointer, if you want to allow the function to change its argument.
You could change the signature of one of the methods. It may not look pretty, however it is the simplest way.
So you could in principle have
void outputString(const string &ss, int notneeded) {
cout << "outputString(const string& ) " + ss << endl;
}
void outputString(const string ss) {
cout << "outputString(const string ) " + ss << endl;
}
and when you want to call the first function just call it with:
outputString("ambigiousmethod", 0);
which will result in a distinguishing call.
There is no other way (I'd love to be proven wrong on this one) since C++ does not allow overloading where passing (by value or by reference) is the only difference in signature.
Edit: as pointed out by bzabhi, you could also change the signature by changing the reference to a pointer. In the example you gave that would work, however you may have to change function code on some occasions.
According to your code, u need only
void outputString(const string &ss).
Because both methods cannot change the argument to the caller (because it's const reference or by-value passing).
Why do you need those 2 methods?
I recommend using the techinque of giving each and every function a unique name., i.e., do not use syntax overloading. I have been using it for years and I've only found advantages in it.
Related
Firstly please have a look at some simple codes that my questions derived from.
#include <iostream>
#include <string>
using namespace std;
string get_something()
{
cout << "output something";
return " and return something";
}
void print_something()
{
cout << "print something";
}
int main()
{
cout << get_something(); // will work
cout << print_something(); // will NOT work
return 0;
}
The only different thing I notice between get_something() and print_something() is that one is a return type and one isn't. As you can see I have added comments indicating that which one will work and not work.
However, I am really not sure what is happening behind the scene that makes it one possible and the other not possible.
I am not even sure how I should go about and search for this kind of question too.. so here I am asking a question.
Please enlighten me..
edit:
I am confused that how it is possible to do cout after cout..
both of the functions do that but one of them works and the other doesn't.
This seems to be a very common misunderstanding among beginners. Printing something via cout is not the same as returning a value from a function. Thats completely orthogonal things.
You can write:
std::string returned_value = get_something();
std::cout << returned_value;
But you cannot write:
??? returned_value = print_something();
std::cout << returned_value;
because print_something() does not return anything! void denotes the absence of a type. You cannot have an object of type void.
On the other hand, when you call a function, you can use the returned value (above), or you can ignore it, so this is correct code:
print_something(); // prints something
get_something(); // also print something and returned value is ignored
Note that the function get_something should get a better name, because it is not just "getting" a value. How about print_and_return_something()?
PS:
What I am really confused about is that, how is it possible to do a cout after a cout? Am I just missing what cout actually does?
Not sure If I understand, but I will try... std::cout is an object of type std::ostream. It has an operator<< that you can call, similar to calling methods of other objects. The following two are identical and just use different syntax:
std::cout.operator<<( "Hello World");
std::cout << "Hello World";
When you call print_something() then first the function is executed, then the return value is returned to the caller and execution continues with the caller. This:
std::cout << get_something();
is more or less the same as (well, its a crude simplification, but should be ok here):
// inside get_something
std::cout << "output something";
// return value
std::string result{"output something"};
// now execution continues in caller
std::cout << result;
Calling cout after cout is no different from calling some other function. Suppose you have a function print() that prints something then you can write
std::string print_and_return() {
std::string x{"Hello World"};
print(x);
return x;
}
The caller can do
std::string x = print_and_return(); // <- this already calls print()
print(x); // now we call it again
This is more or less the same as yours, just that I used some hypothetical print() instead of std::cout::operator<<.
Both your functions have a return type. It's just that one of them has a void return type.
The std::ostream class does not have an overload for << that takes a void type. This is sensible - what would be written to the stream in that case?
(cout is an instance of std::ostream that typically writes itself to the standard output which is normally the shell you're using to launch the program.)
Because print_something() has nothing to return, and cout want something to write to the console (the return value it is expecting). Therefore, it will give error.
get_something(), on the other hand, has something to return. So after executing it's rest of line (except return statement) it return the string, which gets printed by cout
get_something() returns something (what seems to be accepted by cout), so cout will receive the returned thing and will work.
On the other hand, print_something() returns nothing (because its return type is void), so cout cannot receive anything to print and won't work.
cout is a stream object.and we use << (insertion operator) to insert value like String,float,Int etc to it which will be displayed in output Screen.Since print_something() is not returning any value so nothing is inserted in stream ,That's why it is not working.
I recommend you to read about Streams in c++ ..
# include <iostream>
# include <string>
using std::string;
using std::cout;
using std::endl;
string func() {
string abc = "some string";
return abc;
}
void func1(string s) {
cout << "I got this: " << s << endl;
}
int main() {
func1(func());
}
This gives:
$ ./a.out
I got this: some string
How/why does this code work ? I wonder because abc went out of scope and got destroyed as soon as the call to func() completed. So a copy of abc cannot be/should not be available in variable s in function func1 Is this understanding correct ?
The return value is copied from the local variable, effectively creating a new string object.
However, RVO (returned value optimization) should eliminate this step.
Try single stepping your code in a debugger. You should see the std::string copy constructor called for the return line. Be sure to compile with debug enabled and optimizers off.
Your code is essentially asking:
"Call func1, and in order for func1 to work I have to receive a string which we can use by calling the copy constructor on that string. The parameter for func1 we want to come from the return value of func (which we know has to be a string since its explicitly defined".
abc goes out of scope only after the copy constructor is called on the return value of func() which passes the value of the string. In theory you could have written it passed by reference or constant reference:
void func1(string& s) {
cout << "I got this: " << s << endl;
}
Which allows func1 to directly access the string in memory through a pointer (and also change it, if your code was meant to.)
void func1(string const& s) {
cout << "I got this: " << s << endl;
}
Which provides a constant reference to the string from func(). This ensures that you get a pointer to the data and that you won't change its contents. Typically passing data by constant reference (const&) is desirable because it's very fast and keeps your code from accidentally changing data that it shouldn't.
You really only need to pass by value if you're going to manipulate the data once you pass it to the new function, saving you the resources of creating another new container to handle the manipulation:
void func1(string s) {
s += " some extra stuff to add to the end of the string"; //append some new data
cout << "I got this: " << s << endl;
}
swap(string1,string2) will swap two strings values easily while I use it in Main Function, But if I use it in another function and call it from main function it won't work!
this works:
int main()
{
string name1="A",name2="B";
swap(name1,name2);
}
but this one does not:
string name1="A",name2="B"; // Global Here
void exchange (string one,string two)
{
swap(one,two);
}
int main()
{
exchange(name1,name2);
}
where is the problem?
Pass the strings by reference instead of by value, otherwise exchange will modify local copies of one and two.
string name1="A", name2="B"; // Global Here
void exchange(string& one, string& two)
{
swap(one,two);
}
int main()
{
cout << name1 << "\n" << name2 << endl;
exchange(name1, name2);
cout << name1 << "\n" << name2 << endl;
}
Output:
A
B
B
A
Well, the values of the copies one and two actually are swapped. It won't affect the variables name1 and name2, of course. Assuming you want the values of these strings being swapped, you should pass the arguments by reference
void exchange(string& one, string& two) {
...
}
void exchange (string& one,string& two)
{
swap(one,two);
}
This should work. The amperstand (&) means that you are passing the arguments by reference, and that the function is allowed to modify the initial strings that were passed as parameters. If you do not use &, the strings will be passed by value and you will only modify copies of the actual strings.
From what I understand, this should work:
const char* x = "x";
std::cout << x << std::endl;
Passing x into this function:
void myClass::passAsVoid(void* v) {
std::cout << (const char*)v << std::endl;
}
The first example prints "x";
The second example prints "\350\224A";
I want to learn what's going on, and the correct approach to do this!
The actual code:
float delay = 1;
std::string txt = "random filler text that is not lorum ipsum";
for (int i = 0; i < txt.length(); ++i) {
const char* x = "x";
std::cout << x << "code1" << std::endl;
CCSequence* seq = CCSequence::create(CCDelayTime::create(i*delay),
CCCallFuncND::create( this, callfuncND_selector(OverWorldView::setString), (void*)x ),
NULL);
this->runAction(seq);
}
Callback function:
void OverWorldView::setString(void* x) {
std::cout << (const char*)x << "code2" << std::endl;
label1->setString( (const char*)x );
}
I'm using cocos2dx 2.1.4
It looks like you are not using the API correctly.
The documentation says you need to use a SEL_CallFuncND type callback, which receives two arguments, not one.
For everyone's convenience, callfuncND_selector is a macro that hides a static_cast, or probably even a C-style cast (could not find other API versions online) which lets you use just about anything as a callback without getting any compilation errors. Pure joy.
This won't even compile because you're passing a const pointer to a function which accepts a non-const pointer. If you can't change the type of the callback then instead of passing a string literal, pass a character array.
char buffer[100];
callback(buffer); //buffer decays to char* which is implicitly converted to void*
If you can change the type of the callback though then modify the parameter to be const void*.
I would like to be able to do:
foo(stringstream()<<"number = " << 500);
EDIT: single line solution is crucial since this is for logging purposes. These will be all around the code.
inside foo will print the string to screen or something of the sort.
now since stringstream's operator<< returns ostream&, foo's signature must be:
foo(ostream& o);
but how can I convert ostream& to string? (or char*).
Different approaches to achieving this use case are welcome as well.
The obvious solution is to use dynamic_cast in foo. But the given
code still won't work. (Your example will compile, but it won't do what
you think it should.) The expression std::ostringstream() is a
temporary, you can't initialize a non-const reference with a temporary,
and the first argument of std::operator<<( std::ostream&, char const*)
is a non-const reference. (You can call a member function on a
temporary. Like std::ostream::operator<<( void const* ). So the code
will compile, but it won't do what you expect.
You can work around this problem, using something like:
foo( std::ostringstream().flush() << "number = " << 500 );
std::ostream::flush() returns a non-const reference, so there are no
further problems. And on a freshly created stream, it is a no-op.
Still, I think you'll agree that it isn't the most elegant or intuitive
solution.
What I usually do in such cases is create a wrapper class, which
contains it's own std::ostringstream, and provides a templated
member operator<< which forwards to the contained
std::ostringstream. Your function foo would take a const
reference to this—or what I offen do is have the destructor call
foo directly, so that the client code doesn't even have to worry about
it; it does something like:
log() << "number = " << 500;
The function log() returns an instance of the wrapper class (but see
below), and the (final) destructor of this class calls your function
foo.
There is one slight problem with this. The return value may be copied,
and destructed immediately after the copy. Which will wreck havoc with
what I just explained; in fact, since std::ostringstream isn't
copyable, it won't even compile. The solution here is to put all of the
actual logic, including the instance of std::ostringstream and the
destructor logic calling foo in a separate implementation class, have
the public wrapper have a boost::shared_ptr to it, and forward. Or
just reimplement a bit of the shared pointer logic in your class:
class LogWrapper
{
std::ostringstream* collector;
int* useCount;
public:
LogWrapper()
: collector(new std::ostringstream)
, useCount(new int(1))
{
}
~LogWrapper()
{
-- *useCount;
if ( *useCount == 0 ) {
foo( collector->str() );
delete collector;
delete useCount;
}
}
template<typename T>
LogWrapper& operator<<( T const& value )
{
(*collector) << value;
return *this;
}
};
Note that it's easy to extend this to support optional logging; just
provide a constructor for the LogWrapper which sets collector to
NULL, and test for this in the operator<<.
EDITED:
One other thing occurs to me: you'll probably want to check whether the
destructor is being called as a result of an exception, and not call
foo in that case. Logically, I'd hope that the only exception you
might get is std::bad_alloc, but there will always be a user who
writes something like:
log() << a + b;
where the + is a user defined overload which throws.
I would suggest you to use this utility struct:
struct stringbuilder
{
std::stringstream ss;
template<typename T>
stringbuilder & operator << (const T &data)
{
ss << data;
return *this;
}
operator std::string() { return ss.str(); }
};
And use it as:
void f(const std::string & s );
int main()
{
char const *const pc = "hello";
f(stringbuilder() << '{' << pc << '}' );
//this is my most favorite line
std::string s = stringbuilder() << 25 << " is greater than " << 5 ;
}
Demo (with few more example) : http://ideone.com/J995r
More on my blog : Create string on the fly just in one line
You could use a proxy object for this; this is a bit of framework, but if you want to use this notation in a lot of places then it may be worth it:
#include <iostream>
#include <sstream>
static void foo( std::string const &s )
{
std::cout << s << std::endl;
}
struct StreamProxy
{
std::stringstream stream;
operator std::string() { return stream.str(); }
};
template <typename T>
StreamProxy &operator<<( StreamProxy &s, T v )
{
s.stream << v;
return s;
}
static StreamProxy make_stream()
{
return StreamProxy();
}
int main()
{
foo( make_stream() << "number = " << 500 );
}
This program prints
number = 500
The idea is to have a little wrapper class which can be implicitely converted into a std::string. The << operator is simply forwarded to the contained std::stringstream. The make_stream() function is strictly speaking not necessary (you could also say StreamProxy(), but I thought it looks a bit nicer.
A couple of options other than the nice proxy solution just presented by Frerich Raabe:
Define a static string stream variable in the header that defines the logging function and use the comma operator in your invocation of the logging function so that this variable is passed rather than the ostream& returned by the stream insertion operator. You can use a logging macro to hide this ugliness. The problem with this solution is that it is a bit on the ugly side, but this is a commonly used approach to logging.
Don't use C++ I/O. Use a varargs C-style solution instead. Pass a format string as the first argument, with the remaining arguments being targets for that format string. A problem with this solution is that even if your compiler is smart enough to ensure that printf and its cousins are safe, the compiler probably won't know that this new function is a part of the printf family. Nonetheless, this is also a commonly used approach.
If you don't mind using macros functions, you can make the logging function accept const string&, and use the following macro
#define build_string(expr) \
(static_cast<ostringstream*>(&(ostringstream().flush() << expr))->str())
And suppose you foo has signature void foo(const string&), you only need the one-liner
foo(build_string("number = " << 500))
This was inspired by James Kanze's answer about static_cast and stringstream.flush. Without the .flush() the above method fails with unexpected output.
Please note that this method should not leak memory, as temporary values, whether in the pointer form or not, are still allocated on the stack and hence destroyed upon return.
Since you're converting to string anyways, why not
void foo(const std::string& s)
{
std::cout << "foo: " << s << std::endl;
}
...
std::stringstream ss;
ss << "number = " << 500;
foo(ss.str());
This is not possible. As the name ostream implies, it is used for output, for writing to it. You could change the parameter to stringstream&. This class has the method str() which returns a std::string for your use.
EDIT I did not read the issue with operator << returning ostream&. So I guess you cannot simply write your statements within the functions argument list but have to write it before.
You can create a small wrapper around std::ostringstream that will convert back to std::string on use, and have the function take a std::string const &. The first approach to this solution can be found in this answer to a different question.
On top of that, you can add support for manipulators (std::hex) if needed.