I recently started programming in c++ and I've bumped into a small problem. If I want my output to be structured (let's say that every line starts with a name and then a number) in a way that the names are written normally to the screen (every first letter of every name starts at the beginning of each new line) and I want the numbers that follow to be lined up in a column, how would I do this? I want the programs output to look like this:
Gary 0
LongName 0
VerylongName 0
I want my program to print something in the way above, but with different lengths of names (and the '0' in this case, lined up in a column).
Try the following: if you know the maximum length of all the names you intend to print (e.g. 20), then use the C++ i/o manipulators to set the width of the output (and left-justification). This will force the output to take up max characters.
Code snippet:
#include <iostream>
#include <iomanip>
...
// for each entry
std::cout << std::setw(20) << std::left << "Gary" << 10 << "\n";
...
std::cout << std::flush;
Here's some more information...
I'm shooting in the dark here since you haven't really included much information... HOWEVER one way you can do this is to make sure that you create the columns with padding around the name - and not worry about the numbers. Formatted output is one case where C has an advantage over C++ (IMHO). In C++ you can also do this with something like this:
cout << setw(15) << name << number << "\n";
Bonus points if you figure out ahead of time the maximum length of the name you have and add, say, 4 to it.
Not in the C++ standard library, but still worth mentioning: boost::format. It will let you write printf-like format strings while still being type-safe.
Example:
#include <boost/format.hpp>
#include <iostream>
#include <string>
struct PersonData
{
std::string name;
int age;
};
PersonData persons[] =
{
{"Gary", 1},
{"Whitney", 12},
{"Josephine ", 101}
};
int main(void)
{
for (auto person : persons)
{
std::cout << boost::format("%-20s %5i") % person.name % person.age << std::endl;
}
return 0;
}
Outputs:
Gary 1
Whitney 12
Josephine 101
struct X
{
const char *s;
int num;
} tab[] = {
{"Gary",1},
{"LongName",23},
{"VeryLongName",456}
};
int main(void)
{
for (int i = 0; i < sizeof(tab) / sizeof(struct X); i++ )
{
// C like - example width 20chars
//printf( "%-20s %5i\n", tab[i].s, tab[i].num );
// C++ like
std::cout << std::setw(20) << std::left << tab[i].s << std::setw(5) << std::right << tab[i].num << std::endl;
}
getchar();
return 0;
}
Related
I wanna compare value that stored in filename[i] and filename[j] and print out the value in filename[i] that do not have the same filename as in filename[j]. I know it is possible to do using set_difference and sort solution but I do not know exactly to write the sort and set_differences code. Here i provide my original code so that u can test it out and more understand what I'm trying to do.
my full code:
#include <string>
#include <iostream>
#include <ctime> //important when to make random filename- srand(time(0))
#include <opencv2\opencv.hpp> //important when using opencv
#include <vector> //when using vector function
using namespace std;
using namespace cv; //important when using opencv
int main(int argc, char* argv[]) {
vector<String> filenames;
int a, i;
srand(time(0)); //seed random filenames - for random filename
// Get all jpg in the folder
cv::glob("C:\\Users\\x\\Documents\\Aggressive\\abc", filenames);
for (size_t i = 0; i < filenames.size(); i++)
{
Mat im = imread(filenames[i]); //read the filename location
std::cout << "\n";
std::size_t found = filenames[i].find_last_of("//\\");
//std:cout << " file: " << filenames[j].substr(found + 1) << '\n'; //display filename and its format (.jpg)
std::string::size_type const p(filenames[i].substr(found + 1).find_last_of('.')); //eg: 2.jpg then it will find the last '.'
std::string file_without_extension = filenames[i].substr(found + 1).substr(0, p); //eg: 2
std::cout << " file : " << filenames[i].substr(found + 1).substr(0, p); //display filename without .jpg
}
cout << "\n";
cout << "There's " << filenames.size() << " files in the current directory.\n" << endl; // total file in the specific directory
cout << "Enter array size: \n";
cin >> a;
for (int j = 0; j < filenames.size(); j++) {
//generate random filename
int index = rand() % filenames.size(); //random based on total of the file in the directory
//cout << filenames[index] << endl; //display the random number but might be redundant
//swap filenames[j] with filenames[index]
string temp = filenames[j];
filenames[j] = filenames[index];
filenames[index] = temp;
}
for (int j = 0; j < a; j++) {
//cout << "Random image selected:" << filenames[j] << endl; //basically to avoid the redundant random filename
Mat im = imread(filenames[j]); //read filename location
std::size_t found = filenames[j].find_last_of("//\\");
//std:cout << " file: " << filenames[j].substr(found + 1) << '\n'; //display filename and its format (.jpg)
std::string::size_type const p(filenames[j].substr(found + 1).find_last_of('.')); //eg: 2.jpg then it will find the last '.'
std::string file_without_extension = filenames[j].substr(found + 1).substr(0, p); //eg: 2
std::cout << " file: " << filenames[j].substr(found + 1).substr(0, p); //display filename without .jpg
string written_directory = "C:/Users/x/Documents/folder/" + filenames[j].substr(found + 1).substr(0, p) + ".jpg"; // write filename based on its original filename.
imwrite(written_directory, im);
}
return 0;
}
In my opinion this is a perfect example of an XY Problem. From you question, from your code and even from the comments, people do not really understand what you want to do. With that I mean, what do you want to achieve?
It is a vague guess that you want to copy a specified number of random selected JPEG files from one directory to the other. And that you want to show the filenames of the files that will not be copied.
Let me give you some examples, what is the reason for all this confusion.
First and most important, you do not show the full code. Definitions and variable types and functions are misssing an. This is also not a Minimum, Reproducable Example. And the description in your question is hard to understand.
I have two set of array
You have "two set array"? Do you mean, you have 2 [std::set][3] of [std::array][3]. Or maybe you have simply 2 [std::vector][3] of std::string. From what we can see in the code, we could assume a std::vector<std::string>>, but we do not know, because you did not show the feinition of "filenames".
Then, you are talking about "2" something. But we do see only one "filenames". So, 2 or 1?
in a comment you are writing
in the array 2 i had a random filename based on the size of array that an user entered
My guess is that you do not want to have a random filename, but you want to select filenames with a random index from the first vector and put it into a 2nd vector? But we can see only 1 vector "filenames" where you do some random swapping activity.
Then you have written
imread is actually to read the whole file in the folder of directory
This function is very important, what does it do? And what do you mean by "read the file"? Do you mean "filename", so the name of the file? Or the contents of the file? And what is the meaning of "folder of directory"? All filenames in one folder? Or subfolder of a directory entry?
So now my objective is to print out all the file that do not have same filename in the array 2
Again, do we really have 2 arrays(vector)? are they different?
And then, where do you copy the files?
So, you see, it is very hard to understand. Even, if people would like to help you, they cannot, because they do not understand you. Better to show a link to your original home work. Then people can help you. Members here on Stack Overflow want to help. But please allow them to do so.
Here I give you an abstract example for the random selection problem and set_difference problem:
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <random>
int main() {
// Define 2 Vectors for filenames
// This vector is an example for files that could be in a specified directory
std::vector<std::string> fileNamesInDirectory{"8.jpg","5.jpg", "6.jpg", "9.jpg", "1.jpg", "4.jpg", "2.jpg", "3.jpg", };
// Print the filenames as information for the user
for (size_t i = 0U; i < fileNamesInDirectory.size(); ++i) {
std::cout << fileNamesInDirectory[i] << "\n";
}
// Next: Select randomly a given number of filenames from the above vector
// So, first get the number of selections. Inform the user
std::cout << "\nEnter a number of filenames that should be copied randomly. Range: 1-"<< fileNamesInDirectory.size()-1 << "\t";
size_t numberOfSelectedFileNames{};
std::cin >> numberOfSelectedFileNames;
// Check for valid range
if (numberOfSelectedFileNames == 0 || numberOfSelectedFileNames >= fileNamesInDirectory.size()) {
std::cerr << "\n*** Error. Wrong input '" << numberOfSelectedFileNames << "'\n";
}
else {
// Copy all data from fileNamesInDirectory
std::vector<std::string> selection{ fileNamesInDirectory };
// Shuffle the data randomly: Please see here: https://en.cppreference.com/w/cpp/algorithm/random_shuffle
std::random_device rd;
std::mt19937 g(rd());
std::shuffle(selection.begin(), selection.end(), g);
// Resize to the number, given by the user
selection.resize(numberOfSelectedFileNames);
// Now we have a random list of filenames
// Show, what we have so far. Now, because we are learning, we will use the range based for
std::cout << "\n\nOriginal file names:\n";
for (const std::string& s : fileNamesInDirectory) std::cout << s << "\n";
std::cout << "\n\nRandomly selected file names:\n";
for (const std::string& s : selection) std::cout << s << "\n";
// Sort both vectors
std::sort(fileNamesInDirectory.begin(), fileNamesInDirectory.end());
std::sort(selection.begin(), selection.end());
// Show again to the user:3
std::cout << "\n\nOriginal file names sorted:\n";
for (const std::string& s : fileNamesInDirectory) std::cout << s << "\n";
std::cout << "\n\nRandomly selected file names sorted:\n";
for (const std::string& s : selection) std::cout << s << "\n";
// Now, find out the difference of both vectors, meaning, what will not be selected and later copied
std::vector<std::string> difference{};
// Calculate the difference with a std::algorithm: https://en.cppreference.com/w/cpp/algorithm/set_difference
std::set_difference(fileNamesInDirectory.begin(), fileNamesInDirectory.end(), selection.begin(), selection.end(), std::back_inserter(difference));
std::cout << "\n\nThe following file names have not been selected:\n";
for (const std::string& s : difference) std::cout << s << "\n";
}
return 0;
}
If you are more advanced then you can and will use functions from the C++ filesystem library. That will make life easier . . .
Quick question. I'm doing my portfolio C++ question for a uni assignment. It is standard deviation. My question is to do with reading multiple strings from a file, see text here;
Design and write a c++ program that reads a set of scores from the file scored.dat, and outputs their mean and standard deviation to cout.
I'm not going to bother with the actual equation, I have that part nearly sorted. My query is based directly on outputting the text read from the file into strings. For example, if the document had these three scores:
10
15
11
Instead of outputting the text as it is, it would put them into three strings;
Score_One (Which would be 10)
Score_Two (Which would be 15)
Score_Three (Which would be 11)
I hope I am making sense here guys. Thanks.
you need to do something like this:
int raw_score_one = 11; //you have already set this but for the sake of clarity
std::stringstream output;
output << raw_score_one;
std::string Score_One = output.str();
for each score...
Here is a solution which was just fun to write and which hardly a solution I would expect to show up. It may have some entertainment and/or educational value as well. The basic idea is that values a written as
std::cout << 10 << 15 << 11 << reset << '\n';
std::cout << 1 << 2 << 3 << reset << '\n';
To achieve this, a bit of machinery is needed but it isn't that bad, really. The code is blow:
#include <locale>
#include <iostream>
#include <algorithm>
static int index(std::ios_base::xalloc());
static std::string const names[] = { "One", "Two", "Three", "Four", "Five" };
static std::string const score("Score_");
static std::string const other(" (Which would be ");
std::ostream& reset(std::ostream& out)
{
out.iword(index) = 0;
return out;
}
struct num_put
: std::num_put<char>
{
iter_type do_put(iter_type to, std::ios_base& fmt, char_type fill,
long v) const {
to = std::copy(score.begin(), score.end(), to);
if (fmt.iword(index) < 5) {
to = std::copy(names[fmt.iword(index)].begin(),
names[fmt.iword(index)].end(), to);
++fmt.iword(index);
}
else {
throw std::runtime_error("index out of range!");
}
to = std::copy(other.begin(), other.end(), to);
to = this->std::num_put<char>::do_put(to, fmt, fill, v);
*to++ = ')';
*to++ = ' ';
return to;
}
};
int main()
{
std::cout.imbue(std::locale(std::locale(), new num_put));
std::cout << 10 << 15 << 11 << reset << '\n';
std::cout << 1 << 2 << 3 << reset << '\n';
}
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.
Let me examplify my problem , I've a function like:
void printer(int width, int hight){
for(int i=0;i<width;++i) std::cout<<" & ";
for(int i=0;i<hight;++i) std::cout<<" ^ ";
std::cout<<std::endl;
}
my problem is function printer should always output of both for loop in same width
e.g:
output could look (width 5):
&^
&&&^
or there is anyway that i print any of (from above code) for loop's output in constant width independent of no of times for loop executes
Question is unclear. Are you looking for something like the following ?
void printer(int amps, int carets, int overallWidth){
for (int i = amps + carets; i < overallWidth; i++) std::cout<<" "; // leading padding
for (int i=0;i<amps;++i) std::cout<<"&";
for (int i=0;i<carets;++i) std::cout<<"^";
std::cout<<std::endl;
}
The change was just to add a loop for outputting the padding. (also changed the parameters name for clarity)
printer(1,1,5);
printer(3,1,5);
would then produce the output shown in example
&^
&&&^
and of course, rather than being passed as a parameter, the overallWidth variable could be hardcoded; an implicit constant of the printer() function.
Edit:
The snippet above stayed very close to that of the question. There are however more idiomatic approaches, for example the following "one liner", which uses one of the string constructor overloads to produce the strings of repeated characters, and iomanip's setw() to produce the padding.
void printer(int amps, int carets, int overallWidth){
std::cout << setiosflags(ios::right) << setw(overalWidth)
<< string(amps, '&') + string (carets, '^')
<< std::endl;
}
Look into <iomanip>. Using cout you can specify a width.
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
cout << setiosflags(ios::left) << setw(10) << "Hello"
<< setiosflags(ios::right) << setw(20) << "World!";
return 0;
}
This question already has answers here:
How to concatenate a std::string and an int
(25 answers)
Closed 6 years ago.
int i = 4;
string text = "Player ";
cout << (text + i);
I'd like it to print Player 4.
The above is obviously wrong but it shows what I'm trying to do here. Is there an easy way to do this or do I have to start adding new includes?
With C++11, you can write:
#include <string> // to use std::string, std::to_string() and "+" operator acting on strings
int i = 4;
std::string text = "Player ";
text += std::to_string(i);
Well, if you use cout you can just write the integer directly to it, as in
std::cout << text << i;
The C++ way of converting all kinds of objects to strings is through string streams. If you don't have one handy, just create one.
#include <sstream>
std::ostringstream oss;
oss << text << i;
std::cout << oss.str();
Alternatively, you can just convert the integer and append it to the string.
oss << i;
text += oss.str();
Finally, the Boost libraries provide boost::lexical_cast, which wraps around the stringstream conversion with a syntax like the built-in type casts.
#include <boost/lexical_cast.hpp>
text += boost::lexical_cast<std::string>(i);
This also works the other way around, i.e. to parse strings.
printf("Player %d", i);
(Downvote my answer all you like; I still hate the C++ I/O operators.)
:-P
These work for general strings (in case you do not want to output to file/console, but store for later use or something).
boost.lexical_cast
MyStr += boost::lexical_cast<std::string>(MyInt);
String streams
//sstream.h
std::stringstream Stream;
Stream.str(MyStr);
Stream << MyInt;
MyStr = Stream.str();
// If you're using a stream (for example, cout), rather than std::string
someStream << MyInt;
For the record, you can also use a std::stringstream if you want to create the string before it's actually output.
cout << text << " " << i << endl;
Your example seems to indicate that you would like to display the a string followed by an integer, in which case:
string text = "Player: ";
int i = 4;
cout << text << i << endl;
would work fine.
But, if you're going to be storing the string places or passing it around, and doing this frequently, you may benefit from overloading the addition operator. I demonstrate this below:
#include <sstream>
#include <iostream>
using namespace std;
std::string operator+(std::string const &a, int b) {
std::ostringstream oss;
oss << a << b;
return oss.str();
}
int main() {
int i = 4;
string text = "Player: ";
cout << (text + i) << endl;
}
In fact, you can use templates to make this approach more powerful:
template <class T>
std::string operator+(std::string const &a, const T &b){
std::ostringstream oss;
oss << a << b;
return oss.str();
}
Now, as long as object b has a defined stream output, you can append it to your string (or, at least, a copy thereof).
Another possibility is Boost.Format:
#include <boost/format.hpp>
#include <iostream>
#include <string>
int main() {
int i = 4;
std::string text = "Player";
std::cout << boost::format("%1% %2%\n") % text % i;
}
Here a small working conversion/appending example, with some code I needed before.
#include <string>
#include <sstream>
#include <iostream>
using namespace std;
int main(){
string str;
int i = 321;
std::stringstream ss;
ss << 123;
str = "/dev/video";
cout << str << endl;
cout << str << 456 << endl;
cout << str << i << endl;
str += ss.str();
cout << str << endl;
}
the output will be:
/dev/video
/dev/video456
/dev/video321
/dev/video123
Note that in the last two lines you save the modified string before it's actually printed out, and you could use it later if needed.
For the record, you could also use Qt's QString class:
#include <QtCore/QString>
int i = 4;
QString qs = QString("Player %1").arg(i);
std::cout << qs.toLocal8bit().constData(); // prints "Player 4"
cout << text << i;
One method here is directly printing the output if its required in your problem.
cout << text << i;
Else, one of the safest method is to use
sprintf(count, "%d", i);
And then copy it to your "text" string .
for(k = 0; *(count + k); k++)
{
text += count[k];
}
Thus, you have your required output string
For more info on sprintf, follow:
http://www.cplusplus.com/reference/cstdio/sprintf
cout << text << i;
The << operator for ostream returns a reference to the ostream, so you can just keep chaining the << operations. That is, the above is basically the same as:
cout << text;
cout << i;
cout << "Player" << i ;
cout << text << " " << i << endl;
The easiest way I could figure this out is the following..
It will work as a single string and string array.
I am considering a string array, as it is complicated (little bit same will be followed with string).
I create a array of names and append some integer and char with it to show how easy it is to append some int and chars to string, hope it helps.
length is just to measure the size of array. If you are familiar with programming then size_t is a unsigned int
#include<iostream>
#include<string>
using namespace std;
int main() {
string names[] = { "amz","Waq","Mon","Sam","Has","Shak","GBy" }; //simple array
int length = sizeof(names) / sizeof(names[0]); //give you size of array
int id;
string append[7]; //as length is 7 just for sake of storing and printing output
for (size_t i = 0; i < length; i++) {
id = rand() % 20000 + 2;
append[i] = names[i] + to_string(id);
}
for (size_t i = 0; i < length; i++) {
cout << append[i] << endl;
}
}
There are a few options, and which one you want depends on the context.
The simplest way is
std::cout << text << i;
or if you want this on a single line
std::cout << text << i << endl;
If you are writing a single threaded program and if you aren't calling this code a lot (where "a lot" is thousands of times per second) then you are done.
If you are writing a multi threaded program and more than one thread is writing to cout, then this simple code can get you into trouble. Let's assume that the library that came with your compiler made cout thread safe enough than any single call to it won't be interrupted. Now let's say that one thread is using this code to write "Player 1" and another is writing "Player 2". If you are lucky you will get the following:
Player 1
Player 2
If you are unlucky you might get something like the following
Player Player 2
1
The problem is that std::cout << text << i << endl; turns into 3 function calls. The code is equivalent to the following:
std::cout << text;
std::cout << i;
std::cout << endl;
If instead you used the C-style printf, and again your compiler provided a runtime library with reasonable thread safety (each function call is atomic) then the following code would work better:
printf("Player %d\n", i);
Being able to do something in a single function call lets the io library provide synchronization under the covers, and now your whole line of text will be atomically written.
For simple programs, std::cout is great. Throw in multithreading or other complications and the less stylish printf starts to look more attractive.
You also try concatenate player's number with std::string::push_back :
Example with your code:
int i = 4;
string text = "Player ";
text.push_back(i + '0');
cout << text;
You will see in console:
Player 4
You can use the following
int i = 4;
string text = "Player ";
text+=(i+'0');
cout << (text);
If using Windows/MFC, and need the string for more than immediate output try:
int i = 4;
CString strOutput;
strOutput.Format("Player %d", i);