I'm trying to learn to use namespaces declarations more definitive than not just say "using namespace std". I'm trying to format my data to 2 decimal places, and set the format to be fixed and not scientific. This is my main file:
#include <iostream>
#include <iomanip>
#include "SavingsAccount.h"
using std::cout;
using std::setprecision;
using std::ios_base;
int main()
{
SavingsAccount *saver1 = new SavingsAccount(2000.00);
SavingsAccount *saver2 = new SavingsAccount(3000.00);
SavingsAccount::modifyInterestRate(.03);
saver1->calculateMonthlyInterest();
saver2->calculateMonthlyInterest();
cout << ios_base::fixed << "saver1\n" << "monthlyInterestRate: " << saver1->getMonthlyInterest()
<< '\n' << "savingsBalance: " << saver1->getSavingsBalance() << '\n';
cout << "saver2\n" << "monthlyInterestRate: " << saver2->getMonthlyInterest()
<< '\n' << "savingsBalance: " << saver2->getSavingsBalance() << '\n';
}
On Visual Studio 2008, when I run my program, I get an output of "8192" before the data I want. Is there a reason for that?
Also, I don't think I am setting the fixed part or 2 decimal places correctly since I seem to get scientific notation once I added the setprecision(2). Thanks.
You want std::fixed (the other one just inserts its value into the stream, which is why you see 8192), and I don't see a call to std::setprecision in your code anywhere.
This'll fix it:
#include <iostream>
#include <iomanip>
using std::cout;
using std::setprecision;
using std::fixed;
int main()
{
cout << fixed << setprecision(2)
<< "saver1\n"
<< "monthlyInterestRate: " << 5.5 << '\n'
<< "savingsBalance: " << 10928.8383 << '\n';
cout << "saver2\n"
<< "monthlyInterestRate: " << 4.7 << '\n'
<< "savingsBalance: " << 22.44232 << '\n';
}
It might not be the answer you're looking for, but floating-point numbers are not suited to financial calculations because fractions like 1/100 cannot be represented exactly. You might be better off doing the formatting yourself. This can be encapsulated:
class money {
int cents;
public:
money( int in_cents ) : cents( in_cents ) {}
friend ostream &operator<< ( ostream &os, money const &rhs )
{ return os << '$' << m.cents / 100 << '.' << m.cents % 100; }
};
cout << money( 123 ) << endl; // prints $1.23
Better(?) yet, C++ has a facility called the monetary locale category which includes a money formatter which takes cents as an argument.
locale::global( locale("") );
use_facet< money_put<char> >( locale() ).put( cout, false, cout, ' ', 123 );
This should Do the Right thing internationally, printing the user's local currency and hiding the number of decimal places from your implementation. It even accepts fractions of a cent. Unfortunately, this does not seem to work on my system (Mac OS X), which has generally poor locale support. (Linux and Windows should fare better.)
cout << setiosflags(ios::fixed) << setprecision(2) << 1/3.;
ios_base::fixed is not manipulator it is a value (1 << 13) for the ios flag.
Related
I'm self learning C++ and for some reason "double" doesn't print more than 6 significant digits even after std::setprecision. Do I need to do something else? Most recent version of codeblocks if that helps. This is all the code:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
std::setprecision(9);
double A = 654321.987;
cout << A << endl;
return 0;
}
You need to feed the result of std::setprecision(9) to std::cout. Otherwise it has no way of knowing what output stream it applies to (and so it won't apply to anything).
std::cout << std::setprecision(9) << A << std::endl;
Or if you prefer you can do it separately:
std::cout << std::setprecision(9);
std::cout << A << std::endl;
How can I format my output in C++? In other words, what is the C++ equivalent to the use of printf like this:
printf("%05d", zipCode);
I know I could just use printf in C++, but I would prefer the output operator <<.
Would you just use the following?
std::cout << "ZIP code: " << sprintf("%05d", zipCode) << std::endl;
This will do the trick, at least for non-negative numbers(a) such as the ZIP codes(b) mentioned in your question.
#include <iostream>
#include <iomanip>
using namespace std;
cout << setw(5) << setfill('0') << zipCode << endl;
// or use this if you don't like 'using namespace std;'
std::cout << std::setw(5) << std::setfill('0') << zipCode << std::endl;
The most common IO manipulators that control padding are:
std::setw(width) sets the width of the field.
std::setfill(fillchar) sets the fill character.
std::setiosflags(align) sets the alignment, where align is ios::left or ios::right.
And just on your preference for using <<, I'd strongly suggest you look into the fmt library (see https://github.com/fmtlib/fmt). This has been a great addition to our toolkit for formatting stuff and is much nicer than massively length stream pipelines, allowing you to do things like:
cout << fmt::format("{:05d}", zipCode);
And it's currently being targeted by LEWG toward C++20 as well, meaning it will hopefully be a base part of the language at that point (or almost certainly later if it doesn't quite sneak in).
(a) If you do need to handle negative numbers, you can use std::internal as follows:
cout << internal << setw(5) << setfill('0') << zipCode << endl;
This places the fill character between the sign and the magnitude.
(b) This ("all ZIP codes are non-negative") is an assumption on my part but a reasonably safe one, I'd warrant :-)
Use the setw and setfill calls:
std::cout << std::setw(5) << std::setfill('0') << zipCode << std::endl;
In C++20 you'll be able to do:
std::cout << std::format("{:05}", zipCode);
In the meantime you can use the {fmt} library, std::format is based on.
Disclaimer: I'm the author of {fmt} and C++20 std::format.
cout << setw(4) << setfill('0') << n << endl;
from:
http://www.fredosaurus.com/notes-cpp/io/omanipulators.html
or,
char t[32];
sprintf_s(t, "%05d", 1);
will output 00001 as the OP already wanted to do
Simple answer but it works!
ostream &operator<<(ostream &os, const Clock &c){
// format the output - if single digit, then needs to be padded with a 0
int hours = c.getHour();
// if hour is 1 digit, then pad with a 0, otherwise just print the hour
(hours < 10) ? os << '0' << hours : os << hours;
return os; // return the stream
}
I'm using a ternary operator but it can be translated into an if/else statement as follows
if(c.hour < 10){
os << '0' << hours;
}
else{
os << hours;
}
I have a problem where I need to print some double value using basic C++ output system (such as iostream or sstream) and this value somehow gets truncated a lot.
For example:
double a = 283752.24234;
std::cout << 283752.24234 << std::endl;
std::cout << a << std::endl;
Both of the outputs will be 283752
Why is this happening and what can I do to get complete output of any double value?
Why is this happening ...
That's because the default precision is 6 and thus only the foremost 6 digits are rendered 283752.
... and what can I do to get complete output of any double value?
First option is to use a bigger precision value using the std::setprecision() I/O manipulator:
#include <iostream>
#include <iomanip>
int main(){
double a = 283752.24234;
std::cout << std::setprecision(12) << 283752.24234 << std::endl;
std::cout << std::setprecision(12) << a << std::endl;
}
See Live Demo
Second option is to use the std::fixed I/O manipulator to show the values after the decimal point:
#include <iostream>
#include <iomanip>
int main(){
double a = 283752.24234;
std::cout << std::fixed << 283752.24234 << std::endl;
std::cout << std::fixed << a << std::endl;
}
See Live Demo
Output:
283752.242340
283752.242340
I want to just hard code these values into a table. when I try to use 2D arrays, I run into the problem of dealing with characters and integers. When I do a struct I have this so far but it doesn't divide the information up in columns, and I'm not sure how to format it that way. (I only did 3 rows to start off with, if I get them working, the rest will just be the same)
#include <iostream>
#include <string>
#include <iomanip>
using namespace std;
typedef struct table
{
std::string game;
int year;
float rating;
int num_voters;
}t;
void processTab(t*);
int main()
{
t tabl[2] = {0,0};
int i;
processTab(tabl);
for(i=0; i<2; i++)
{
std::cout << "Game: " << setw(20) << tabl[i].game;
std::cout << "\tYear: " << setw(4) << tabl[i].year;
std::cout << "\tRating: " << fixed << setprecision(2) << tabl[i].rating;
std::cout << "\tVoters: " << setw(6) << tabl[i].num_voters;
}
system("pause");
return 0;
}
void processTab(t*tab)
{
(tab[0].game, "twilight struggles");
tab[0].year = 2005;
tab[0].rating = 8.226;
tab[0].num_voters = 10690;
(tab[1].game, "Agricloa");
tab[1].year = 2007;
tab[1].rating = 8.17;
tab[1].num_voters = 23738;
(tab[2].game, "Puerto Rico");
tab[2].year = 2002;
tab[2].rating = 8.163;
tab[2].num_voters = 27433;
}
Table Data:
Game (0) Year (1) Rating (2) Num Voters (3)
Twilight Struggle 2005 8.226 10690
Agricola 2007 8.17 23738
Puerto Rico 2002 8.163 27433
Through the Ages 2006 8.153 8137
Power Grid 2004 8.02 21655
Le Havre 2008 7.972 9258
Eclipse 2011 7.968 3194
Brass 2007 7.911 5814
Dominion: Intrigue 2009 7.895 10889
Caylus 2005 7.878 13878
What I think you are looking for is <iomanip>
#include <iomanip>
std::cout << "Game: " << setw(20) << tabl[i].game;
std::cout << "\tYear: " << setw(4) << tabl[i].year;
std::cout << "\tRating: " << fixed << setprecision(3) << tabl[i].rating;
std::cout << "\tVoters: " << setw(6) << tabl[i].num_voters;
std::cout << std::end;
Notes:
setw adds padding when writing out stuff, so it will always be at least a certain width
setprecision specifies how many decimal places to display
fixed makes floating point never use scientific notation
Your game member is a letter, and you're attemptying to assign it a string. Don't use strcpy in C++, use the std::string class instead.
#include <string>
struct table
{
std::string game;
int year;
double rating;
int num_voters;
};
Avoid using namespace std;, when you get to complex code with many namespaces, those few letters are a small price to pay for avoiding confusion.
Avoid endl: it flushes buffers which is slow. If you just want a newline, use '\n'.
Also, you can use the new initialization syntax to initialize your list:
std::vector<table> tbal = {
{"twilight struggles ", 2005, 8.226, 10690},
{"Agricola ", 2007, 8.17 , 23738},
{"Puerto Rico ", 2002, 8.163, 27433}
};
Short answer: use std::vector, not a raw array.
The following is as close to your original code as I could make it, introducing a minimum number of new features for you:
#include <assert.h> // assert
#include <iostream> // std::cout, std::endl
#include <stddef.h> // ptrdiff_t
#include <string> // std::string
#include <utility> // std::begin, std::end
#include <vector> // std::vector
using namespace std;
typedef ptrdiff_t Size;
template< class Container >
Size countOf( Container& c ) { return end( c ) - begin( c ); }
struct Game
{
string game;
int year;
double rating;
int num_voters;
};
void initialize( vector<Game>& games )
{
assert( countOf( games ) == 0 );
games.resize( 3 );
games[0].game = "twilight struggles";
games[0].year = 2005;
games[0].rating = 8.226;
games[0].num_voters = 10690;
games[1].game = "Agricloa";
games[1].year = 2007;
games[1].rating = 8.17;
games[1].num_voters = 23738;
games[2].game = "Puerto Rico";
games[2].year = 2002;
games[2].rating = 8.163;
games[2].num_voters = 27433;
}
int main()
{
vector<Game> games;
initialize( games );
for( int i = 0; i < countOf( games ); ++i )
{
cout << i << endl;
cout <<"\tGame: " << games[i].game << endl;
cout<<"\tYear: " << games[i].year << endl;
cout<<"\trating: " << games[i].rating << endl;
cout<<"\tnum voters: " << games[i].num_voters << endl;
}
}
There are ways to just declare the data more directly, including brace initializers; check out your C++ textbook.
First you need to define your table (bad name for a struct, by the way) correctly. You're trying to use game to store a string, but have defined it as only a single char. You probably want to change that to a std::string instead.
Then you probably want to do your formatting in an operator<< overloaded to take a reference to table as the type. #MooingDuck has already covered the formatting itself quite well, so it's mostly a matter of how you package that:
std::ostream &operator<<(std::ostream &os, table const &t) {
os << "Game: " << setw(20) << t.game;
os << "\tYear: " << setw(4) << t.year;
os << "\tRating: " << fixed << setprecision(2) << t.rating;
return os << "\tVoters: " << setw(6) << t.num_voters;
}
Along with that, you almost certainly want to change tabl from an array to std::vector<table>:
std::vector<tabl> tabl;
Then processing the table becomes:
std::copy(tabl.begin(), tabl.end(), std::ostream_iterator<table>(std::cout, "\n"));
One other minor detail: you seem to have two entirely different/separate functions, both named processTab. You probably want to rename at least one of those. Just glancing at them, I'd probably call one initializeTable or something on that order.
I would like to print floating point numbers in a format like this.
01234.56
I use the following function to round a float to n digits.
double round(double val, int precision)
{
std::stringstream s;
s << std::setprecision(precision) << std::setiosflags(std::ios_base::fixed) << val;
s >> val;
return val;
}
And the following line to output any number with leading zeroes to a fixed length.
setfill ('0') << setw(5) << number
But when I try to combine the two, I end up with non-consistent length, like in the following examples:
8.2 => 008.2
4.57 => 04.56
What I would like to have is an output like this:
8.2 => 08.20
4.57 => 04.57
Can you show me a function like string myround (double d, intpart n, floatpart f) which returns:
myround (1234.5, 5, 2) => 01234.50
myround (1234.569, 5, 2) => 01234.57
I imageine it has been asked before, but I couldn't find it using the internal search engine.
the fixed manipulator will do the trick I think, my example below outputs what you wanted:
#include <iostream>
#include <iomanip>
using namespace std;
int main(void) {
cout << setprecision(2) << setfill ('0') << setw(5) << fixed << 8.2 << endl;
cout << setprecision(2) << setfill ('0') << setw(5) << fixed << 4.57 << endl;
}
just put it into your stringstream and it should work. However, the fixed information will be lost when the string is converted to a double value again. I think it is no good idea to round a float using a string method.