So in my program I have some classes - Button, Window and WindowButton. Button consists only of text, Window - of a button and coordinates(x,y), and WindowButton consists of a Window.
In WindowButton, I have overloaded the << operator like this:
ostream& operator<<(ostream& out, WindowButton& ref)
{
ref.print();
return out;
}
Where the print function looks like:
void WindowButton::print()
{
theWindow->print();
}
And the window print function, in window class:
void Window::print()
{
char* buttonText = button->getText();
char* theText = new char[strlen(buttonText)+1];
strcpy(theText, buttonText);
cout << endl << "Window with coordinates (" << this->coord.x << "," << this->coord.y << ") , and button text \"" << theText << "\"" << endl;
}
In main:
WindowButton *test = new WindowButton();
cout << endl << test;
test->print();
The last line provides the right output, but the second line provides just a memory adress. What am I doing wrong? Everything should be working fine, because the test->print(); works fine.
You are passing a pointer to operator<< which expects a &.
cout << endl << *test;
You might also make it:
ostream& operator<<(ostream& out, const WindowButton& ref){
Which assumes print doesn't actually modify.
But, the bigger question is why are you using the cout ostream to trigger printing to theWindow -- these appear to be (though aren't) logically disconnected processes. You could pass the given stream into Window::print:
void Window::print(ostream& stream) {
and use that stream in place of cout. This avoids hard-coding cout into Window::print().
It's a pointer, so you'll need to dereference it for the operator to work:
cout << endl << *test;
This line
cout << endl << test;
prints a pointer to WindowButton, and there is an ostream& operator<< specialization for pointers, which prints the address. You can try de-referencing the pointer:
cout << endl << (*test);
As an aside, there is little point in overloading the ostream& operator<< in a way that eventually just prints to std::cout. The point of such an operator is that you can stream to any ostream, not just cout. You could fix this by modifying your print functions to take an ostream by reference, and modify it:
void WindowButton::print(std::ostream& out) const {
theWindow->print(out);
}
and
void Window::print(ostream& out) const {
// stuff
out << "Window with coordinates (" << this->coord.x << "," << this->coord.y << ") , and button text \"" << theText << "\"" << endl;
}
and finally
ostream& operator<<(ostream& out, const WindowButton& ref){
ref.print(out);
return out;
}
Related
I have some C++ console programs that display progress information on the last line of output, at regular intervals.
This progress line is cleared prior to writing the next real output (or updated progress information); this could be from a number of different places in the source, and I'm currently clearing the progress line on each one, e.g.:
cout << clearline << "Some real output" << endl;
...
cout << clearline << "Some other real output" << endl;
...
cout << clearline << setw(4) << ++icount << ") " << ... << endl;
...
cout << clearline << "Progress info number " << ++iprog << flush;
Here, 'clearline' is some (system dependent) string like "\r\33[2K" which clears the current last line.
I would prefer something cleaner, that localises source changes to the actual line that's going to be cleared, like simply:
cout << "Progress info number " << ++iprog << flush << defer_clearline;
where 'defer_clearline' causes the writing of 'clearline' to be deferred until just prior to the next cout output, wherever and whatever that happens to be. I then wouldn't need to use 'clearline' on all the other lines.
I thought it might be possible to do this if 'defer_clearline' is a manipulator, and/or using xalloc() and iword().
But I've not managed to get anything that works.
Is it possible to do this sort of thing, and if so how?
2020-12-30: edited to include missing 'flush's.
You can pretty easily setup an std::cout wrapper:
// Declare the empty struct clear_line and instantiate the object cls
struct clear_line { } cls;
class out {
private:
std::ostream &strm = std::cout;
bool is_next_clear = false;
public:
template <typename T>
out& operator<<(const T& obj) {
if(is_next_clear) {
strm << std::endl << std::endl << std::endl; // clear logic
is_next_clear = false;
}
strm << obj;
return *this;
}
out& operator<<(const clear_line& _) {
is_next_clear = true;
return *this;
}
};
This pretty simply stores an is_next_clear bool for whether or not the next regular output should be cleared. Then, in the general case (the templated operator<<()), we run your clear logic and flip the is_next_clear flag if applicable. Then just output as usual.
Then, the operator<<() is overloaded for the case of a clear_line object. So if one of those is sent, we know to flip the is_next_clear flag, but not actually output anything.
Here's an example use:
int main() {
out o;
o << "Some real output" << cls;
o << "Some other real output";
return 0;
}
Here it is in action: https://ideone.com/0Dzwlv
If you want to use endl, you'll need to add a special overload for it as this answer suggests: https://stackoverflow.com/a/1134467/2602718
// this is the type of std::cout
typedef std::basic_ostream<char, std::char_traits<char> > CoutType;
// this is the function signature of std::endl
typedef CoutType& (*StandardEndLine)(CoutType&);
// define an operator<< to take in std::endl
out& operator<<(StandardEndLine manip)
{
// call the function, but we cannot return its value
manip(strm);
return *this;
}
Live example: https://ideone.com/ACUMOo
I am trying to overload
<<
operator. For instance
cout << a << " " << b << " "; // I am not allowed to change this line
is given I have to print it in format
<literal_valueof_a><"\n>
<literal_valueof_b><"\n">
<"\n">
I tried to overload << operator giving string as argument but it is not working. So I guess literal
" "
is not a string. If it is not then what is it. And how to overload it?
Kindly help;
Full code
//Begin Program
// Begin -> Non - Editable
#include <iostream>
#include <string>
using namespace std;
// End -> Non -Editable
//---------------------------------------------------------------------
// Begin -> Editable (I have written )
ostream& operator << (ostream& os, const string& str) {
string s = " ";
if(str == " ") {
os << '\n';
}
else {
for(int i = 0; i < str.length(); ++i)
os << str[i];
}
return os;
}
// End -> Editable
//--------------------------------------------------------------------------
// Begin -> No-Editable
int main() {
int a, b;
double s, t;
string mr, ms;
cin >> a >> b >> s >> t ;
cin >> mr >> ms ;
cout << a << " " << b << " " ;
cout << s << " " << t << " " ;
cout << mr << " " << ms ;
return 0;
}
// End -> Non-Editable
//End Program
Inputs and outputs
Input
30 20 5.6 2.3 hello world
Output
30
20
5.6
2.3
hello
world
" " is a string-literal of length one, and thus has type const char[2]. std::string is not related.
Theoretically, you could thus overload it as:
auto& operator<<(std::ostream& os, const char (&s)[2]) {
return os << (*s == ' ' && !s[1] ? +"\n" : +s);
}
While that trumps all the other overloads, now things get really hairy. The problem is that some_ostream << " " is likely not uncommon, even in templates, and now no longer resolves to calling the standard function. Those templates now have a different definition in the affected translation-units than in non-affected ones, thus violating the one-definition-rule.
What you should do, is not try to apply a global solution to a very local problem:
Preferably, modify your code currently streaming the space-character.
Alternatively, write your own stream-buffer which translates it as you wish, into newline.
Sure this is possible, as I have tested. It should be portable since you are specifying an override of a templated function operator<<() included from <iostream>. The " " string in your code is not a std::string, but rather a C-style string (i.e. a const char *). The following definition works correctly:
ostream& operator << (ostream& os, const char *str) {
if(strcmp(str, " ") == 0) {
os << '\n';
} else {
// Call the standard library implementation
operator<< < std::char_traits<char> > (os, str);
}
return os;
}
Note that the space after std::char_traits<char> is necessary only if you are pre-c++11.
Edit 1
I agree with Deduplicator that this is a potentially dangerous solution as it may cause undesirable consequences elsewhere in the code base. If it is needed only in the current file, you could make it a static function (by putting it within an unnamed namespace). Perhaps if you shared more about the specifics of your problem, we could come up with a cleaner solution for you.
You might want to go with a user defined literal, e.g.
struct NewLine {};
std::ostream& operator << (std::ostream& os, NewLine)
{
return os << "\n";
}
NewLine operator ""_nl(const char*, std::size_t) // "nl" for newline
{
return {};
}
This can be used as follows.
int main(int, char **)
{
std::cout << 42 << ""_nl << "43" << ""_nl;
return 0;
}
Note three things here:
You can pass any string literal followed by the literal identifier, ""_nl does the same thing as " "_nl or "hello, world"_nl. You can change this by adjusting the function returning the NewLine object.
This solution is more of an awkward and confusing hack. The only real use case I can imagine is pertaining the option to easily change the behavior at a later point in time.
When doing something non-standard, it's best to make that obvious and explicit - here, the user defined literal indeed shines, because << ""_nl is more likely to catch readers' attention than << " ".
I'm trying to convert a string streamed phone lookup program into file streamed.. I'm missing something, but I'm stuck.. what members can I use in the ofstream process to get this working?
ofstream& process (ofstream &os, vector<PersonInfo> people)
{
// for each entry in people
for (vector<PersonInfo>::const_iterator entry = people.begin();
entry != people.end(); ++entry) {
ofstream formatted, badNums; // objects created on each loop
// for each number
for (vector<string>::const_iterator nums = entry->phones.begin();
nums != entry->phones.end(); ++nums) {
if (!valid(*nums)) {
badNums << " " << *nums; // string in badNums
} else
// ``writes'' to formatted's string
formatted << " " << format(*nums);
}
if (badNums.empty()) // there were no bad numbers
os << entry->name << " " // print the name
<< formatted.str() << endl; // and reformatted numbers
else // otherwise, print the name and bad numbers
cerr << "input error: " << entry->name
<< " invalid number(s) " << badNums.str() << endl;
}
return os;
}
First, you don't want an ofstream, except at the point you're opening
the file (creating the instance). The output stream interface is
defined by std::ostream; std::ofstream derives from this, as does
std::ostringstream (output can become an std::string), and in most
applications, a couple of others written by the local programmers. In
your case (if I've understood the problem correctly), what you want is:
std::ostream& process( std::ostream& os,
std::vector<PersonInfo> const& people )
// Note the use of a const reference above. No point
// in copying the entire vector if you're not going to
// modify it.
{
for ( std::vector<PersonInfo>::const_iterator entry = people.begin();
entry != people.end();
++ entry ) {
std::ostringstream formatted;
std::ostringstream badNums;
// ...
if ( badNums.str().empty() ) {
os << ... << formatted.str() << std::endl;
} else {
os << ... << badNums.str() << std::endl;
}
}
return os;
}
Note the different types: std::ostream formats output, independently
of the destination type. std::ofstream derives from it, and provides
a file as destination. std::ostringstream derives from it, and
provides a std::string as destination type. And the std::ostream
takes a std::streambuf* as argument, and you provide the destination
type.
You never associate a file with ostream, so the compiler doesn't know what to do with the data you write into it.
ofstream& process (ofstream &os, vector<PersonInfo> people)
{
os.open("Data.txt"); //open file to be used
if(!os.is_open())
std::cerr << "Error opening file!\n";
//rest of code goes here
}
EDIT: after reading through your program again, i noticed you're using ofstream wrong. Ofstream is for opening and writing FILES.The program has a lot of syntax and logical errors i would read up on it more here.
It looks like you don't need to use ofstreams for the internal parts of this function. In fact you don't need to use streams at all, a std::string would do:
ofstream& process (ofstream &os, vector<PersonInfo> people)
{
// for each entry in people
for (vector<PersonInfo>::const_iterator entry = people.begin();
entry != people.end(); ++entry) {
string formatted, badNums; // objects created on each loop
// for each number
for (vector<string>::const_iterator nums = entry->phones.begin();
nums != entry->phones.end(); ++nums) {
if (!valid(*nums)) {
badNums += " " + *nums; // string in badNums
} else
// ``writes'' to formatted's string
formatted += " " + format(*nums);
}
if (badNums.empty()) // there were no bad numbers
os << entry->name << " " // print the name
<< formatted << endl; // and reformatted numbers
else // otherwise, print the name and bad numbers
cerr << "input error: " << entry->name
<< " invalid number(s) " << badNums << endl;
}
return os;
}
When you're overloading the << operator for a class (pretend this is defined as a friend in SomeClass), why do you both take a reference to the ostream and return that ostream?
ostream& operator<<(ostream& s, const SomeClass& c) {
//whatever
return s;
}
What benefit can returning the ostream be when it was already directly modifiable by reference? This seems redundant to me - though I'm sure it's not :)
It allows to "chain" output together. As in :
std::cout << someObj << someValue;
This is equivalent to something like :
operator<<(operator<<(std::cout, someObj), someValue);
This is not redundant, but useful for chaining calls. It's easier to see with functions like std::string::append, so I'll start with that:
std::string mystring("first");
mystring.append(" second");
mystring.append(" third");
can be rewritten as:
std::string mystring("first").append(" second").append(" third");
This is possible because .append() returns a reference to the string it modified, so we can keep adding .append(...) to the end. The code correlating to what you are doing is changing from this:
std::cout << "first";
std::cout << " second";
std::cout << " third";
into this. Since operator<< returns the stream, we can also chain these!
std::cout << "first" << " second" << " third";
see the similarity, and usefulness?
So that you can write chained-invocation of operator<< as:
stream << s1 << s2 << s3 ;
If you don't return ostream&, then you cannot write it more than once.
You can think of that either as:
operator<<(operator<<(operator<<(stream, s1), s2), s3);
Or as,
((stream << s1) << s2) << s3 ;
First (stream << s1) returns stream& on which you again invoke << and it becomes stream << s2 which returns stream& on which you again invoke << and it becomes stream << s3.
Ah, this is so linked output, like
cout << "this " << "is " << "a pen" << endl;
will still work.
How would someone do that?
for example I do like:
std::cout << "something";
then it should print the time before "something"
Make your own stream for that :) This should work:
class TimedStream {
public:
template<typename T>
TimedStream& operator<<(const T& t) {
std::cout << getSomeFormattedTimeAsString() << t << std::endl;
return *this;
}
};
TimedStream timed_cout;
void func() {
timed_cout << "123";
}
You'd be able to use this class for every type for which std::cout << obj can be done, so no further work is needed.
But please note that the time will be written before every <<, so you cannot chain them easily. Another solution with explicit timestamp is:
class TimestampDummy {} timestamp;
ostream& operator<<(ostream& o, TimestampDummy& t) {
o << yourFancyFormattedTimestamp();
}
void func() {
cout << timestamp << "123 " << 456 << endl;
}
You could use a simple function that prints the timestamp and then returns the stream for further printing:
std::ostream& tcout() {
// Todo: get a timestamp in the desired format
return std::cout << timestamp << ": ";
}
You would then call this function instead of using std::cout directly, whenever you want a timestamp inserted:
tcout() << "Hello" << std::endl;
ostream& printTimeWithString(ostream& out, const string& value)
{
out << currentTime() << ' ' << value << std::endl;
return out;
}
Generate current time using your favourite Boost.DateTime output format.
This looks like homework. You want something in the line of:
std::cout << time << "something";
Find a way the retrieve the time on your system, using a system call.
Then you'll have to implement a << operator for your system-dependent time class/struct.