convert string to int array

[kz07]

Is there any way to convert String s = "1234" to int s[4] = {1,2,3,4}?
and how can i check if it does not contain any non-numeric characters?

thanks=)

[MrSpigot]

To test the string for non-numerics - loop through the string and call standard library function isdigit() on each character.

To convert to the array - loop through the (numeric) string, for each character subtract '0' and put the result in the corresponding element of the array.

[hmortensen]

Hi kz07,

MrSpigot explained it very well, but I thought I would add some code, as I would do it..

#include "stdafx.h"
#include <iostream>
#include <string>

using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{

	string s = "0123456789017893469178234691";
	int* numbers = new int[s.length()];
	char currentChar;

Now we have the variables we need.
First we chesk if the string holds anything other than numbers, if not then add the currentChar to the numberlist.

for(int i = 0;i<s.length();i++)
	{
		currentChar = s.at(i);

		if(currentChar<(char)'0' || currentChar>(char)'9')
		{	cout << "error in input string" << endl;
			return -1;
		}
		else
		{
			numbers[i]= (int)currentChar-'0';
		}

	}

Now you should have a int array containing all the numbers.

for(int i = 0;i<s.length();i++)
	{
		cout << "index " << i << " = " << numbers[i] << endl;
	}

	delete numbers;
	return 0;
}

Hope this added to the explanation by MrSpigot.

Have FUN++,

[kz07]

thanks!! i get it now.. =)

[kz07]

wait..
this is my code:

#include <iostream>
#include <string>
#include <fstream>
#include <istream>


using namespace std;

void listTitles();
void displayItem(int []);
void addNewItem(int []);
void save(int [], string, string);

int main(){
    
    
    string name = "NAME ";
    char ans = ' ';    
    int num;
    int pin[4] = {1,2,3,4} ;
    char currentChar;
    int i;
    
    
    cout << "Engineering Journal - " << name << endl << endl;
    
    for(i = 0;i < num.length; i++){
          cout << "Enter your four digit pin: ";
          cin >> num;
          currentChar = num.at(i);

            if(currentChar<(char)'0' || currentChar>(char)'9'){
                     cout << "Invalid Pin" << endl;
             }
    }
//....

im still lost.. =(

[mrnutty]

" num.at(i) "

num is an integer and not an array. This does not work.
I think what you are trying to do is something like this :

int digits[4] = {0};
int num;
cin >> num;
int i  = 0;
while(num)
{
     digits[i] = num%10; //get the last digits
     num /=10; //remove the last digit
}

Although not tested, I think that will extract the digits from num into
an int array. Then you can convert it into char if you want.

[kz07]

ok.. ill try if it works.
coz, what im trying to do, is to store the pin(which is 1234) in a string first.. then get the user input and check if it doesnt contain any non numeric characters.. after that, ill store that string into an integer array..

[tux4life]

I guess you mean all characters other than 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 when you say: any non-numeric character.

Why not just get your number like this:

int num;
cout << "Enter a number: ";

if( !(cin >> num) )
{
    // Reading a number failed, most likely due the fact the user
    // didn't enter a valid number.
    
    /*
         Put your code to handle such a situation here
    */
}

or use a stringstream:

stringstream ss;
char cnum[] = "1234";
int inum

ss << cnum;

if( !(ss >> inum) )
{
    // Again: extracting the number failed
    // Most likely due the fact that you're trying to read
    // something else than a number.
}

You need to include the sstream header in order to make the above snippet compile, you can do this by adding the following include directive to your program: #include <sstream> .

[tux4life]

if(currentChar<(char)'0' || currentChar>(char)'9'){
   cout << "Invalid Pin" << endl;
}

Explicitly casting to a char isn't even needed here, the following code is equivalent:

if(currentChar<'0' || currentChar>'9'){
   cout << "Invalid Pin" << endl;
}

[kz07]

thanks again! =)