String problems c++

[mpg]

Really I dont' understand excuse me.As I say I live in Italy and I don't understand well what you write becuase i didn't study at school some operators that you write or I dont understand for my english not perfect.
So can you write the complete code ?
The result mus t be : abcd e efgh aebfcgdh

#include <cstdlib>
#include <ctime>
#include <cmath>
#include <iostream>
using namespace std;
int ConcAltStringhe(string a,string b)
// es. abcd e efgh aebfcgdh
{

    int dim=a.length();
    int dim1=b.length();
   // int dim2=c.length();
    int i=0;
    string result;

    if(dim==dim1)
    {
        for(int i=0;i<dim;i++)
        {
          ??????????????????
        }
    }
    return result;

    }
int main()
{
  srand(time(NULL));


  string a;
  cin>>a;
  string b;
  cin>>b;
  cout<<ConcAltStringhe(a,b);



    return 0;
}

[salem c]

Seriously.

result += a[i];
result += b[i];

[mpg]

Doesn't go.
There is a mistake on "return result"....

#include <iostream>
using namespace std;
int ConcAltStringhe(string a,string b)
// es. abcd e efgh aebfcgdh
{

    int dim=a.length();
    int dim1=b.length();
   // int dim2=c.length();
    int i=0;
    string result;

    if(dim==dim1)
    {
        for(int i=0;i<dim;i++)
        {
         result += a[i];
         result += b[i];
        }
    }
    return result;

    }
int main()
{
  srand(time(NULL));


  string a;
  cin>>a;
  string b;
  cin>>b;
  cout<<ConcAltStringhe(a,b);



    return 0;
}

[dhayden]

You've declared ConcAltStringhe as returning an int:
int ConcAltStringhe(string a,string b)
but at line 21, it returns a string:
return result;
So you need to change the declaration to indicate that it returns a string.

can you write the complete code ?

You'll find that people here don't like to do that. Programming is like football: the only way to get good is to practice.

It's also like math: every letter and every symbol mean something specific You really have to understand the meaning. You can't write a+b instead of a-b in math and expect to solve an equation. In the same way, you can't declare a function as returning an int and then try to return a string. You have read the error message and try to understand what it says.

[mpg]

Here is but now the mistake is in "string ConcAltStringhe( string a, string b)"

#include <cstdlib>
#include <ctime>
#include <cmath>
#include <iostream>

string ConcAltStringhe( string a, string b)
// es. abcd e efgh aebfcgdh
{

    int dim=a.length();
    int dim1=b.length();
    //int dim2=c.length();
    int i=0;
    string result;

    if(dim==dim1)
    {
        for(int i=0;i<dim;i++)
        {
           result += 'a[i]';
           result += 'b[i]';
          // b[i]= result
        }
    }
    return result;

    }
int main()
{
  srand(time(NULL));


  string a;
  cin>>a;
  string b;
  cin>>b;
  cout<<ConcAltStringhe(a,b);



    return 0;
}

[dhayden]

           result += 'a[i]';
           result += 'b[i]';

should be:

           result += a[i];
           result += b[i];

Single quotes are used to express a character literal. So 'a' is the ASCII letter a and '3' is the ASCII character 3 (NOT the number 3). So the compiler doesn't understand 'a[i]' because the quote makes it think you're entering a character literal but a[i] isn't a character a character literal.

Without the single quotes, a[i] is an expression that evaluates to the character at position i within the string a. That's what you want.

Pay attention to the text of the error messages. On my compiler (g++) the error message is:

foo.cxx:20:22: warning: multi-character character constant [-Wmultichar]
            result += 'a[i]';
                      ^~~~~~

So it's confused that you have mutiple characters to express a character constant.

You also need to add:
#include <string>
near the top to get the definition of the string class.

Finally, since string, cin and cout are all defined within a "namespace", you should add
using namespace std;
after your last #include line. Note that for complex programs, it's frowned upon to use the entire namespace, but for a beginner program I think it's okay.

When I make these changes to your code, it works.

[mpg]

Excuse me for the part "if the strings have different lengths, the program returns the empty string" how must I do in the code?

#include <cstdlib>
#include <ctime>
#include <cmath>
#include <iostream>

string ConcAltStringhe( string a, string b)
// es. abcd e efgh aebfcgdh
{

    int dim=a.length();
    int dim1=b.length();
    string result;

    if(dim==dim1)
    {
        for(int i=0;i<dim;i++)
        {
           result += 'a[i]';
           result += 'b[i]';
        
        }
    }
    return result;

    }
int main()
{



  string a;
  cin>>a;
  string b;
  cin>>b;
  cout<<ConcAltStringhe(a,b);



    return 0;
}

[mpg]

Dhayden can you explain me :
"
result += a[i];
result += b[i];
return result

and how result give me the solution aebfcgdh ?
I wish really understand th mechanism, I undestrand the the solution but I will never be arrived at this solution , so I wish the undestrand better than I can.
Many thanks!!!

[jonnin]

+ for a string is concatenate. It is set up to accept both strings and single letters.

so empty string + 'a' = "a".
"abc"+"def" = "abcdef".

so he is taking 1 letter from each input and putting it on the end of the string each loop iteration.


what do you get if you have abc and def? iteration 1 takes a and d, so its ad .. iteration two takes be, adbe … etc.

[mpg]

I understand thanks, I wish to konw the part of the question of how "if the strings have different lengths, the program returns the empty string"
So I write "abcd " and "edfghi" that have different length I must have string empy, how I can I do this in the code?

[jonnin]

he did that. he gets both lengths, and if they are not equal, it just returns result which was empty (it skips the loop).

int dim=a.length();
    int dim1=b.length();
    string result;

    if(dim==dim1)
    {
        for(int i=0;i<dim;i++)
        {
           result += 'a[i]';
           result += 'b[i]';
        
        }
    }
    return result;

[salem c]

> result += 'a[i]';
> result += 'b[i]';
WTF.
Now you're just being a troll if this is your regressive step from previously working code.

[dhayden]

I don't think I can help you any more. You should find someone who can help you face-to-face. There are just too many things that you're having trouble understanding. See if you can talk to the teacher, a TA or a fellow student.

Sorry,
Dave