C++ function to print a set - c++

Question

Does anyone have an idea as to why
#include <set>
#include <string>
#include <iostream>
template <typename T>
std::string set2Str(const std::set<T> & S)
{
std::string retstr = "{";
typename std::set<T>::const_iterator it(S.begin()), offend(S.end());
if (it != offend)
retstr.push_back(*it++);
while (it != offend)
{
retstr.push_back(',');
retstr.push_back(*it++);
}
retstr.push_back('}');
return retstr;
}
int main()
{
std::set<int> mySet = {1, 5, 9, 69};
std::cout << set2Str(mySet);
}
is outputting
{,, ,E}
????
Also, is there a more elegant way of writing the function set2Str? The fencepost problem with the commas makes my procedure ugly.

Answer 1

When your algorithm does this:
retstr.push_back(*it++);
The values being punched into your target string are treated as (and converted if possible to) char. But {1, 5, 9, 69} doesn't contain char; it contains int. The result is treating them as ASCII code points, See this table, and pay particular attention to the dec value of each character therein. Note the value for E, for example.
This is one of the many purposes std::ostringstream was made for, and has the benefit of allowing anything that can be written to a character stream be representable, including utilizing custom insertion operators.
#include <iostream>
#include <sstream>
#include <string>
#include <set>
#include <tuple>
template<typename T, typename... Args>
std::string set2str(const std::set<T,Args...>& obj)
{
std::ostringstream oss;
oss << '{';
auto it = obj.cbegin();
if (it != obj.cend())
{
oss << *it++;
while (it != obj.cend())
oss << ',' << *it++;
}
oss << '}';
return oss.str();
}
// custom class to demonstrate custom insertion support
class Point
{
friend std::ostream& operator <<(std::ostream& os, const Point& pt)
{
return os << '(' << pt.x << ',' << pt.y << ')';
}
private:
double x,y;
public:
Point(double x, double y) : x(x), y(y) {}
// used by std::less<> for set ordering
bool operator <(const Point& pt) const
{
return std::tie(x,y) < std::tie(pt.x,pt.y);
}
};
int main()
{
std::set<int> si = { 1,2,3,4,5 };
std::set<double> sd = { 1.1, 2.2, 3.3, 4.4, 5.5 };
std::set<char> sc = { 'a', 'b', 'c', 'd', 'e' };
std::set<unsigned> empty;
std::cout << set2str(si) << '\n';
std::cout << set2str(sd) << '\n';
std::cout << set2str(sc) << '\n';
std::cout << set2str(empty) << '\n';
// using custom class with implemented ostream inseter
std::set<Point> pts { {2.2, 3.3}, {1.1, 2.2}, {5.5, 4.4}, {1.1, 3.3} };
std::cout << set2str(pts) << '\n';
return EXIT_SUCCESS;
}
Output
{1,2,3,4,5}
{1.1,2.2,3.3,4.4,5.5}
{a,b,c,d,e}
{}
{(1.1,2.2),(1.1,3.3),(2.2,3.3),(5.5,4.4)}

Answer 2

You will need to convert *it++ to a string - for example using std::to_string. Otherwise, it might be a char and gets printed as such.
For example, in your case, you just add 69, which is the Ascii Code for E. More likely, you want "69", which you would get with std::to_string(69).

Answer 3

You declared the type to be in an int. The numbers are being read as ASCII. They aren't being converted to a string type.
retstr.push_back(*it++);
this line sees the ANSI/ASCII capital E not the characters '6' and ' 5'

Answer 4

Using sprintf which you might not like but a way.
push_back of an int to a string is your problem.
Actually this won't work if your type is not an int so you would need to detect the type used. Or better would be to use stringstream and convert that way.
#include <set>
#include <string>
#include <iostream>
template <typename T>
std::string set2Str(const std::set<T> & S)
{
std::string retstr = "{";
char buffer[20] = {0};
typename std::set<T>::const_iterator it(S.begin()), offend(S.end());
if (it != offend) {
sprintf(buffer, "%u", *it++);
retstr.append(buffer);
}
while (it != offend)
{
retstr.push_back(',');
sprintf(buffer, "%u", *it++);
retstr.append(buffer);
}
retstr.push_back('}');
return retstr;
}
int main()
{
std::set<int> mySet; // // = {1, 5, 9, 69}; - didn't work with my compiler
mySet.insert(1);
mySet.insert(5);
mySet.insert(9);
mySet.insert(69);
std::cout << set2Str(mySet);
}