CODEWITHSUNDEEP
c++arrays
JONNI CAREYN
JONNI CAREYN

Reputation: 97

Array of char arrays C++

I am writing an exchange program and I have a char array that holds another 6 char arrays. The program works fine except for one thing: when I try to print the arrays' values, it prints only the first letter.

Maybe the code isn't right, but I didn't find another way to make this exchange. Actually I think the problem is in pointers that point to that arrays, but again I am not sure.

Here is the code:

#include <iostream>
#include <string>
#include <fstream>
#include <stdlib.h>
using namespace std;

int main(int argc, char* argv[])
{
  cout<<"EXCHANGE POKEMON:"<<endl;
char a[100]="PIKACHU", b[100]="CHARMELEON", c[100]="GEODUDE", d[100]="GYARADOS", e[100]="BUTTERFREE", f[100]="MANKEY", tmp[100];
char t[6] = {*a,*b,*c,*d,*e,*f};

while(1)
{
  for(int i = 0; i < 6; i++)
  {
    cout << i << ") " << t[i] <<endl;
  }

  cout<<endl<<"Choose a pokemon:";

  int x, y;
  cin>>x;

  if(x == -1)
  {
    exit(0);
  }
  cout << "Choose a pokemon to exchange with:";
  cin>>y;

  *tmp = t[x];
  t[x] = t[y];
  t[y] = *tmp;
  }
}

Upvotes: 1

Views: 21065

Answers (2)

rodrigo
rodrigo

Reputation: 98348

With this line:

char t[6];

you are not creating an array of arrays of char. You are creating a simple array of chars. So it is no surprise that every element is a single character.

You probably want:

char *t[6] = {a, b, c, d, e, f};

Note that the *a is actually equivalent to a[0]!

Then the tmp array is used wrong: in C you copy strings with strcpy(), not with the assignment operator. Or alternatively you hold a pointer to the original string.

But you are using C++, so why not just use std::string? Maybe you are just learning arrays and want to do it the hard way?

Upvotes: 5

Heisenbug
Heisenbug

Reputation: 39164

You defined t as an array of char, initialized dereferencing the address of the arrays defined before. That is equivalent to obtaining their first element.

*a is equivalent to a[0]

What you are looking for is an array of char pointer instead:

char *t[6] = {a,b,c,d,e,f};

Upvotes: 2

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