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something like string to int? didn't solve yet

award982 (129)
Hello!

Im trying to make a program that you tipe in any numbers as a string and the
program sums them, like this:
Enter a number:4865

and then it would display 23, becouse 4+8+6+5=23

I managed to do a 3 number example using getch from conio.h library but its a verry stupid way of doing this, but i just didn't know how to do better.
heres the code:

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#include <cstdlib>
#include <iostream>
#include <conio.h>
#define input getch()
#define space <<" "<<
using namespace std;

int main(int argc, char *argv[])
{
int a,b,c;

cout<<"Input 3 numbers: ";
a = input;
b = input;
c = input;

int d;
int e;
int f;


cout<<endl<<a space b space c<<endl; //test//
//========= a ============//
if(a == 49)
{
     d = 1;
}
if(a == 50)
{
     d = 2;
}
if(a == 51)
{
     d = 3;
}
if(a == 52)
{
     d = 4;
}
if(a == 53)
{
     d = 5;
}
if(a == 54)
{
     d = 6;
}
if(a == 55)
{
     d = 7;
}
if(a == 56)
{
     d = 8;
}
if(a == 57)
{
     d = 9;
}
//=============== b ========================//
if(b == 49)
{
     e = 1;
}
if(b == 50)
{
     e = 2;
}
if(b == 51)
{
     e = 3;
}
if(b == 52)
{
     e = 4;
}
if(b == 53)
{
     e = 5;
}
if(b == 54)
{
     e = 6;
}
if(b == 55)
{
     e = 7;
}
if(b == 56)
{
     e = 8;
}
if(b == 57)
{
     e = 9;
}
//================= c ===========//
if(c == 49)
{
     f = 1;
}
if(c == 50)
{
     f = 2;
}
if(c == 51)
{
     f = 3;
}
if(c == 52)
{
     f = 4;
}
if(c == 53)
{
     f = 5;
}
if(c == 54)
{
     f = 6;
}
if(c == 55)
{
     f = 7;
}
if(c == 56)
{
     f = 8;
}
if(c == 57)
{
     f = 9;
} 
       
int sum = d + e + f;
cout<<"\nSum is: "<<sum<<"!\n";
    
    
    
    
    
    
    system("PAUSE");
    return EXIT_SUCCESS;
}


Please tell me a better way of doing this!
Last edited on
bluezor (298)
You'll have to use a stringstream for that case -
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#include <sstream>

std::string myString = 50235;
int myNumber = 0;

std::stringstream myString_ss (myString);
myString_ss >> myNumber;
//safely puts the number into an int
//it's recommended you check if the string IS an int first..
//or it might cause errors
Last edited on
kbw (9488)
Don't you need to input a string rather than individual numbers?

You can read a line as follows:
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std::string line;
std::getline(std::cin, line);


The you access each letter in the string using:
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for (size_t i = 0; i != line.size(); ++i)
{
    char c = line[i];
}


Then you convert each of those characters to a number and add them up.
Last edited on
award982 (129)
This is not what i had in mind, though i think this is a good way to do it.
How lets say i split an int containing a number 123 into 3 ints, the firts contains the first digit, the second the secont digit etc...
so it would be like this:
int i = 123;
int a,b,c,;

cout<<a (space)<<b(space)<<c(space);

and the program result would be:
1 2 3.

and then i could sum the three ints like this:

int sum = a+b+c;
cout<<sum;

and it would display:
6


EDIT: sorry, i really have to refresh the page before posting
EDIT 2: i dont understand either method and neither works
Last edited on
bluezor (298)
Ahh.. I didn't understand what you were asking.
Well you can make an array the length of the integer, pick off all the individual integers and place in appropriate array slot (hint - use % and /) then another int, sum would be the total of the array.
award982 (129)
And how i should do that?
jRaskell (347)
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string s = "34573456";
int sum = 0;
for(unsigned int x = 0; x < s.size(); x++)
  sum += s[x]-48;

//sum will equal 37


The characters 0-9 have ascii values of 48-57. s[x] is of type char, which automatically converts to type int when used in a context expecting an int. So, subtract 48 from the value, and you get the integer value of that numerical character. A bounding clause could also be used to ignore ascii characters outside of 0-9 ie:

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if(s[x]>=48 && s[x]<=57)
  sum += s[x]-48;

eryar (15)

jRaskell (96) Oct 9, 2009 at 9:29pm 
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 string s = "34573456";
int sum = 0;
for(unsigned int x = 0; x < s.size(); x++)
  sum += s[x]-48;

//sum will equal 37


The characters 0-9 have ascii values of 48-57. s[x] is of type char, which automatically converts to type int when used in a context expecting an int. So, subtract 48 from the value, and you get the integer value of that numerical character. A bounding clause could also be used to ignore ascii characters outside of 0-9 ie:
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2
 
 if(s[x]>=48 && s[x]<=57)
  sum += s[x]-48;


Good idea
gcampton (861)
And how i should do that?


modulus the % symbol is used to get the remainder of integer division, so you can use % and / to break up a number into individual numbers eg.

10 / 10 = 1 this gives you the first number.

like say we have 43 / 10 the answer is 4.
43 % 10, gives you the last number '3'.

EDIT: typo
Last edited on
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