C++ String array sorting

Question

I am having so much trouble trying to figure out the sort function from the C++ library and trying to sort this array of strings from a-z , help please!!

I was told to use this but I cant figure out what I am doing wrong.

// std::sort(stringarray.begin(), stringarray.end());

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main()
{
  int z = 0;
  string name[] = {"john", "bobby", "dear", 
                   "test1", "catherine", "nomi", 
                   "shinta", "martin", "abe", 
                   "may", "zeno", "zack", "angeal", "gabby"};

  sort(name[0],name[z]);

  for(int y = 0; y < z; y++)
  {
    cout << name[z] << endl;
  }
  return 0;
}

@Mattingly: Why would anybody want to use a substantially slower and a lot harder to use approach to sorting? — Dietmar Kühl
– Dietmar Kühl, Commented Aug 17, 2013 at 19:44
@Mattingly: Avoiding the STL is extremely easy. The C++ Standard Library not so much. If that's what you actually meant, why are you avoiding it "like the plague"? That seems really strange. What is your good reason for this? — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Aug 17, 2013 at 22:37

Dietmar Kühl · Accepted Answer · 2013-08-17 19:43:06Z

35

The algorithms use iterator to the beginning and past the end of the sequence. That is, you want to call std::sort() something like this:

std::sort(std::begin(name), std::end(name));

In case you don't use C++11 and you don't have std::begin() and std::end(), they are easy to define yourself (obviously not in namespace std):

template <typename T, std::size_t Size>
T* begin(T (&array)[Size]) {
    return array;
}
template <typename T, std::size_t Size>
T* end(T (&array)[Size]) {
    return array + Size;
}

answered Aug 17, 2013 at 19:43

Dietmar Kühl

155k18 gold badges238 silver badges395 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

P0W · Accepted Answer · 2013-08-18 08:44:52Z

16

int z = sizeof(name)/sizeof(name[0]); //Get the array size

sort(name,name+z); //Use the start and end like this

for(int y = 0; y < z; y++){
    cout << name[y] << endl;
}

Edit :

Considering all "proper" naming conventions (as per comments) :

int N = sizeof(name)/sizeof(name[0]); //Get the array size

sort(name,name+N); //Use the start and end like this

for(int i = 0; i < N; i++){
    cout << name[i] << endl;
}

Note: Dietmar Kühl's answer is best in all respect, std::begin() & std::end() should be used for std::sort like functions with C++11, else they can be defined.

edited Aug 18, 2013 at 8:44

answered Aug 17, 2013 at 19:43

P0W

48.1k9 gold badges78 silver badges124 bronze badges

11 Comments

TemplateRex Over a year ago

for the love of future programmers: at the very least, rename z to N and y to i, so that it at least resembles all other code using index manipulations...

Lightness Races in Orbit Over a year ago

Please don't get array sizes like that.

László Papp Over a year ago

@TemplateRex, for the love of future programmers, at the very least, do not use "N". I renamed it to sname yesterday, and also 'y' to 'i'. Did you like my reply then? :)

László Papp Over a year ago

@Lightness Races in Orbit: pardon ?

TemplateRex Over a year ago

@LaszloPapp I made the comment here because this answer is the accepted one, and promotes an off-beat naming convention that -while not technically wrong, of course- is just making it harder to read

|

olevegard · Accepted Answer · 2013-08-17 21:28:51Z

11

Example using std::vector

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>

int main()
{
    /// Initilaize vector using intitializer list ( requires C++11 )
    std::vector<std::string> names = {"john", "bobby", "dear", "test1", "catherine", "nomi", "shinta", "martin", "abe", "may", "zeno", "zack", "angeal", "gabby"};

    // Sort names using std::sort
    std::sort(names.begin(), names.end() );

    // Print using range-based and const auto& for ( both requires C++11 )
    for(const auto& currentName : names)
    {
        std::cout << currentName << std::endl;
    }

    //... or by using your orignal for loop ( vector support [] the same way as plain arrays )
    for(int y = 0; y < names.size(); y++)
    {
       std:: cout << names[y] << std::endl; // you were outputting name[z], but only increasing y, thereby only outputting element z ( 14 )
    }
    return 0;

}

http://ideone.com/Q9Ew2l

This completely avoids using plain arrays, and lets you use the std::sort function. You might need to update you compiler to use the = {...} You can instead add them by using vector.push_back("name")

edited Aug 17, 2013 at 21:28

answered Aug 17, 2013 at 19:52

olevegard

5,5521 gold badge27 silver badges29 bronze badges

4 Comments

juanchopanza Over a year ago

This is a good answer, but it would be better to use const references in the range based for loop instead of copying the strings.

olevegard Over a year ago

@juanchopanza I agree, I just used auto for simplicity. Also I wasn't entirely sure what it would evaluate to in this case.

juanchopanza Over a year ago

It would evaluate to std::string, so you could use for(const auto& currentName : names) to get const std::string& and avoid the copies.

olevegard Over a year ago

@juanchopanza Okay, thanks. I was thinkging it could evalutate to std::vector<std::string>::iterator guess I'm not that famililar with C++11 yet. Updated my answer to use const &

TemplateRex · Accepted Answer · 2013-08-17 19:55:13Z

6

Your loop does not do anything because your counter z is 0 (and 0 < 0 evaluates to false, so the loop never starts).

Instead, if you have access to C++11 (and you really should aim for that!) try to use iterators, e.g. by using the non-member function std::begin() and std::end(), and a range-for loop to display the result:

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main() 
{
    int z = 0;
    string name[] = {"john", "bobby", "dear", "test1", "catherine", "nomi", "shinta", "martin", "abe", "may", "zeno", "zack", "angeal", "gabby"};

    sort(begin(name),end(name));

    for(auto n: name){
         cout << n << endl;
    }
    return 0;    
}

Live example.

edited Aug 17, 2013 at 19:55

answered Aug 17, 2013 at 19:43

TemplateRex

70.9k21 gold badges176 silver badges316 bronze badges

Comments

László Papp · Accepted Answer · 2013-08-17 19:46:06Z

3

This works for me:

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

int main() {
    string name[] = {"john", "bobby", "dear", "test1", "catherine", "nomi", "shinta", "martin", "abe", "may", "zeno", "zack", "angeal", "gabby"};

    int sname = sizeof(name)/sizeof(name[0]);

    sort(name, name + sname);

    for(int i = 0; i < sname; ++i)
        cout << name[i] << endl;

    return 0;
}

answered Aug 17, 2013 at 19:46

László Papp

53.5k40 gold badges118 silver badges142 bronze badges

Comments

nelt22 · Accepted Answer · 2014-05-20 21:23:05Z

2

As many here have stated, you could use std::sort to sort, but what is going to happen when you, for instance, want to sort from z-a? This code may be useful

bool cmp(string a, string b)
{
if(a.compare(b) > 0)
    return true;
else
    return false;
}

int main()
{
string words[] = {"this", "a", "test", "is"};
int length = sizeof(words) / sizeof(string);
sort(words, words + length, cmp);

for(int i = 0; i < length; i++)
    cout << words[i] << " ";
cout << endl;
    // output will be: this test is a 

}

If you want to reverse the order of sorting just modify the sign in the cmp function.

Hope this is helpful :)

Cheers!!!

answered May 20, 2014 at 21:23

nelt22

4101 gold badge5 silver badges14 bronze badges

Comments

zettix · Accepted Answer · 2017-12-27 21:12:35Z

The multiset container uses a red-black tree to keep elements sorted.

// using the multiset container to sort a list of strings.
#include <iostream>
#include <set>
#include <string>
#include <vector>


std::vector<std::string> people = {
  "Joe",
  "Adam",
  "Mark",
  "Jesse",
  "Jess",
  "Fred",
  "Susie",
  "Jill",
  "Fred", // two freds.
  "Adam",
  "Jack",
  "Adam", // three adams.
  "Zeke",
  "Phil"};

int main(int argc, char **argv) {
  std::multiset<std::string> g(people.begin(), people.end()); // """sort"""
  std::vector<std::string> all_sorted (g.begin(), g.end());
  for (int i = 0; i < all_sorted.size(); i++) {
    std::cout << all_sorted[i] << std::endl;
  }
}

Sample Output:

Adam
Adam
Adam
Fred
Fred
Jack
Jess
Jesse
Jill
Joe
Mark
Phil
Susie
Zeke

Note the advantage is that the multiset stays sorted after insertions and deletions, great for displaying say active connections or what not.

rashedcs · Accepted Answer · 2017-05-27 03:49:39Z

0

We can sort() function to sort string array.

Procedure :

At first determine the size string array.
use sort function . sort(array_name, array_name+size)
Iterate through string array/

Code Snippet

#include<bits/stdc++.h>
using namespace std;

int main()
{
    ios::sync_with_stdio(false);

    string name[] = {"john", "bobby", "dear", "test1", "catherine", "nomi", "shinta", "martin", "abe", "may", "zeno", "zack", "angeal", "gabby"};

    int len = sizeof(name)/sizeof(name[0]);

    sort(name, name+len);

    for(string n: name)
    {
         cout<<n<<" ";
    }
    cout<<endl;

    return 0;
}

answered May 27, 2017 at 3:49

rashedcs

3,7632 gold badges43 silver badges44 bronze badges

Comments

David Elliman · Accepted Answer · 2013-08-17 19:53:25Z

-2

My solution is slightly different to any of those above and works as I just ran it.So for interest:

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>

using namespace std;

int main() {
  char *name[] = {"john", "bobby", "dear", "test1", "catherine", "nomi", "shinta", "martin", "abe", "may", "zeno", "zack", "angeal", "gabby"};
  vector<string> v(name, name + 14);

  sort(v.begin(),v.end());
  for(vector<string>::const_iterator i = v.begin(); i != v.end(); ++i) cout << *i << ' ';
  return 0;
}

answered Aug 17, 2013 at 19:53

David Elliman

1,3998 silver badges15 bronze badges

2 Comments

Lightness Races in Orbit Over a year ago

You don't get warnings about the deprecated conversions between char const* and char*? String literals are immutable. I also wouldn't accept that magic number in a code review.

Lightness Races in Orbit Over a year ago

(In fact, the deprecated conversion was removed in C++11, making this program ill-formed!)

Collectives™ on Stack Overflow

C++ String array sorting

9 Answers 9

Comments

11 Comments

4 Comments

Comments

Comments

Comments

Comments

Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

9 Answers 9

Comments

11 Comments

4 Comments

Comments

Comments

Comments

Comments

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related