C++ int with output manipulators - c++

Question

I am wrecking my brain trying to figure this. Please bear with me as I'm only in my forth week into learning C++.
The code below works fine when having all the code just in the main()
int main()
{
cout << setfill('0') << setw(1) << hundreds << setw(1) << tens
<< setw(1) << units << "\n";
system("PAUSE");
return 0;
}
However I've been asked to make it work so that the code is split, but I can't seem to do it. My attempt is below. If someone could send me on the right track that would be great. Please bare in mind that we haven't learnt anything to complex so it shouldn't be hard.
int main()
{
cout << "The encrypted number is: " << recomposedEncryptedNumber() << "\n";
system("PAUSE");
return (0);
}
int recomposedEncryptedNumber()
{
return setfill('0') << setw(1) << hundreds << setw(1) << tens << setw(1) << units;
}

Answer 1

the expression cout << stuff has cout as return value so you need to do the same in your function:
std::ostream& recomposedEncryptedNumber()
{
return cout << setfill('0') << setw(1) << hundreds << setw(1) << tens << setw(1) << units;
}

Answer 2

Your initial code combines a lot on a single line. I'm breaking it apart for explanation:
int main()
{
cout // cout is "console output" -- i.e. how you print to the screen
<< setfill('0') // This is the operation "cout << setfill('0')"
// which creates a modifier "setfill", passes it to cout,
// and now you've "taught" cout to use '0' as the fill
// character.
// Importantly, "cout << setfill('0')" returns cout...
<< setw(1) // This is the operation "cout << setw(1)".
// It's applying the "<< setw(1)" to the return of the
// previous command. This is also a special modifier,
// and again, this returns 'cout'.
<< hundreds // Now you're finally printing something, because this
// is doing "cout << hundreds"
<< setw(1)
<< tens
<< setw(1)
<< units
<< "\n";
system("PAUSE");
return 0;
}
Importantly, every operation in that long line is cout << {something}, and the result of the << operation is the cout again -- which is why you can chain the statements together.
Knowing that, you should now be able to see why this fails:
int recomposedEncryptedNumber()
{
return
setfill('0') // This gives you the setfill modifier.
// -- but that's _not_ cout.
<< setw(1) // "setfill << setw"?? Not what you want at all.
// You want "cout << setfill" and "cout << setw".
<< hundreds << setw(1) << tens << setw(1) << units;
}
You have many options on how to proceed. You could do something like this:
int main() {
cout << "The encrypted number is: ";
printRecomposedEncryptedNumber();
cout << "\n";
}
These are three separate statements. Now printRecomposedEncryptedNumber can print your details to cout directly, and return void (i.e. not return anything).
If you want to do it inline:
int main() {
cout << "The encrypted number is: " << recomposedEncryptedNumber() << "\n";
}
Then your recomposedEncryptedNumber must return something that can be given to cout. Since you're trying to do some special formatting, you don't want to use int -- cout will just display the int the way it wants to. So you should use a string. Then your function can do whatever formatting it wants, and return the properly formatted string:
string recomposedEncryptedNumber(int hundreds, int tens, int units)
{
std::ostringstream msg; // 'o' is for output
// 'string' is because we're building a string
// 'stream' is because this uses streaming methods
// i.e. 'ostringstream' lets us build a string,
// by using streaming mechanisms.
// Same stream machinations you were doing before, just with 'msg'
// instead of 'cout'
msg << setfill('0') << setw(1) << hundreds << setw(1) << tens << setw(1) << units;
return msg.str(); // return the string we've built.
}
Lastly, in order for your function to access hundreds, tens, units, the function must be given those variables (i.e. the variables must be passed into the function as parameters).
Welcome to C++!

Answer 3

Did you declare your prototype before main? for example. I do not have more time to explain but here is how you would do it
#include <iostream>
#include <iomanip>
using namespace std;
void recomposedEncryptedNumber(int hun, int tens, int units);
int main()
{
int hun = 1;
int tens = 1;
int units = 1;
cout << "The encrypted number is: " << "\n";
recomposedEncryptedNumber(hun,tens,units);
system("PAUSE");
return (0);
}
void recomposedEncryptedNumber(int hun, int tens, int units)
{
cout << setfill('0') << setw(1) << hun << setw(1) << tens << setw(1) << units;
}`

Answer 4

cout print on the standard output, you're trying to mix 2 types and that's why it doesn't work.
Try it :
std::string recomposedEncryptedNumber()
{
std::string str;
concat(str,setfill('0'));
concat(str,setw(1));
...
return str;
}

Answer 5

Well,
thats how I would implement what I think you need:
#include <iostream>
#include <iomanip>
using namespace std;
class EncryptedNumber
{
public:
EncryptedNumber( unsigned h, unsigned t, unsigned u )
: hundreds( h ),
tens( t ),
units( u )
{}
protected:
unsigned hundreds,
tens,
units;
friend std::ostream & operator << ( std::ostream& outs, const EncryptedNumber& en );
};
std::ostream & operator << ( std::ostream& outs, const EncryptedNumber& en )
{
outs << setfill( '0' ) << setw( 1 ) << en.hundreds
<< setw( 1 ) << en.tens
<< setw( 1 ) << en.units;
return outs;
}
int main()
{
EncryptedNumber en( 1, 2, 3 );
cout << "The encrypted number is: " << en << "\n";
system("PAUSE");
return (0);
}
Of course it is a quick solution and I haven't created getter/setter methods, didn't split the concept from implementantion and nothing useful inside EncryptedNumber.
The point is that code will look far more elegant and mantainable this way.
Please tell me if you want a more detailed explanation.