int array..

[Rooting]

hi friends..

i have an array same this

int array[] = {225, 335};
int codeguru[2];

i would to insert every bite of array[0]
in codeguru array...

if i do
strcpy , show me an error becouse the imput is
a char* not a int..

how i can??..

output of variables..
codeguru[0] = 2
codeguru[1] = 2
codeguru[2] = 5


thanks!!

[exterminator]

See memcpy

EDIT : sorry, you need to convert the input int array element into string (see FAQs for ways to do that) and take each character and store into the second int array. Another way could be to take out each digits (units/tens/hundreds etc places) out of the int directly and assign as needed into the int array.

[Rooting]

ohh thanks a lot..
whith your help i progress to define the array in string..
but i don't know how now i convert the string in an int array..

#include <string>
#include <sstream>
#include <iostream>

using namespace std;

template <class T>
std::string to_string(T t, std::ios_base & (*f)(std::ios_base&))
{
  std::ostringstream oss;
  oss << f << t;
  return oss.str();
}

int main(int argc, char *argv[])
{ 
    

   int array[] = {112, 113};     
        string s1;
    int numeri;
    
   s1 = to_string<int>(array[0], dec);
   
   cout << s1; 
   return 0;

}

thanks!

[cvogt61457]

Here's a cheap, bad, hack

memcpy((char*)array, (char*)codeguru, 2*sizeof(int))

Why do you want to refer to the data in bytes? Both are ints already.
No need to change to a different data type.

for( int i=0; i<2; i++)
    codeguru[i] = array[i]

[Rooting]

Why do you want to refer to the data in bytes? Both are ints already.
No need to change to a different data type.

for( int i=0; i<2; i++)
    codeguru[i] = array[i]

in this mode i assign array sae of codeguru..

int his array..
int array[] = {112, 113};

i would that codeguru[] contain the single number of array[0]

so the output of variables..

codeguru[0] //contain 1
codeguru[1] //contain 1
codeguru[2] // contain 2

how can i do?

[cvogt61457]

You seem to mixing your types.
Is this a string or an int?

112 decimal = 0x70
This takes only 1 byte of the 4 bytes for an int

array[0] = 0x70
array[1] = 0
array[2] = 0
array[3] = 0

if this is a string, then "1" is actually 0x31 - not 0x01

[cvogt61457]

If your talking about a string reprsentation of an integer value, then you
would have this for a "112"

char myString[size];
myString[0] = 0x31  // "1"
myString[1] = 0x31  // "1"
myString[2] = 0x32  // "2"
myString[3] = 0x00  // null terminator

String data is actually stored with the least significate byte in the lowest
byte in storage. The start of the string is on the left side.

However, since you are working in C++, I would use std:string. It makes
working with strings very easy. It does all the string managment.

[Philip Nicoletti]

Is this what you are looking for ?

#include <string>
#include <sstream>
#include <iostream>

using namespace std;

template <typename T1, typename T2>
T2 Convert(const T1 & t1)
{
  std::stringstream ss;
  ss << t1;
  T2 t2 = T2();
  ss >> t2;
  return t2;
}

int main(int argc, char *argv[])
{ 
    

   int array[] = {112, 113};    
   
   string s = Convert<int,string>(array[0]);

   int * arr2 = new int[s.size()];

   for (int i=0; i<s.size(); ++i)
   {
       arr2[i] = Convert<char,int>(s[i]);

       cout << arr2[i] << "\n";
   }

   delete [] arr2;
   
   return 0;

}

[Rooting]

ohh yeahhh!!


thanks a lot dear Philip Nicoletti...
i study it to understand the structure...

thanks a lot!!..