atoi() only giving 0 result - c++

Question

#include <iostream>
#include <math.h>
#include <iomanip>
#include <sstream>
#include <stdio.h>
#include <string>
#include <stdlib.h>
using namespace std;
int main()
{
ostringstream str;
double num = pow(2,1000);
int sum = 0;
str << setprecision(1000) << num;
string here = str.str();
cout << here << "\n\n";
/*for(int i = 0; i < here.length(); i++)
{
sum += atoi(&here[i]);
}*/
cout << atoi(&here[0]);
cout << atoi(&here[1]);
cout << atoi(&here[2]);
}
output:
10715086071862673209484250490600018105614048117055336074437503883703510511249361
22493198378815695858127594672917553146825187145285692314043598457757469857480393
45677748242309854210746050623711418779541821530464749835819412673987675591655439
46077062914571196477686542167660429831652624386837205668069376
000
Why all 0s?

Answer 1

Going out on a limb here and assuming you don't actually want to use std::atoi. If you want to sum every digit in a string, you want to convert the digit character into its digit value. The quickest way to do this is to subtract the character constant '0'. In your loop, simply use:
for(int i = 0; i < here.length(); i++)
{
sum += here[i] - '0';
}
This is possible because subtracting the '0' from the various characters in the string results in the numeric value that the character represents.
'0' - '0' == 0
'1' - '0' == 1
'2' - '0' == 2
//etc
'9' - '0' == 9
As far as I can remember, the C++ standard does not force any particular encoding, but it does specify that the digit characters must be contiguous so while the above is safe when the string contains only digits, the subtraction on other characters that may appear in the string will throw off your result:
'E' - '0' == ???
'.' - '0' == ???
'+' - '0' == ???

Answer 2

That's how std::atoi indicates an error. In this case, the error is that the numeric value in the array is larger than the largest possible integer (which is technically undefined behavior with atoi, but your implementation apparently treats it as any other error)

Answer 3

atoi converts a string to an integer (probably 32-bits or 64-bits on your platform).
The number you have stored in here is larger than INT_MAX, so atoi returns zero:
On success, the function returns the converted integral number as an int value. If no valid conversion could be performed, a zero value is returned.
EDIT: actually, didn't even read my own link carefully enough, apparently it's undefined behavior in this case
There is no standard specification on what happens when the converted value would be out of the range of representable values by an int.
from www.cplusplus.com

Answer 4

'here[0]' returns the first character of 'here' as a char.
'&here[0]' returns the address of 'here[0]'. You're not wanting the address. '&' is for getting the address of a variable.
std::atoi(here[0]) returns the first character of here as a char, and converts that char to an int... or would, if 'atoi' handled chars. It doesn't - it handles arrays of chars. Giving it one char probably wouldn't compile.
std::atoi(&here[0]) compiles, but isn't what you want. atoi will keep reading chars until it reaches a null character.
It means that given the string "567321":
std::atoi(&here[0]) would return "987654321"
std::atoi(&here1) would return "87654321"
std::atoi(&here2) would return "7654321"
std::atoi(&here[3]) would return "654321"
... and so on.
If you are really wanting to sum all the numbers, and are required to use std::atoi(), then you can do it using std::string::substr():
for(int i = 0; i < here.length(); i++)
{
std::string subString = here.substr(i,1); //Returns a substring starting at 'i', and including 1 character
sum += atoi(subString.c_str());
}
A better way is to use the method #dreamlax posted... but if you are learning about strings and std::atoi, learning about std::string::substr() is important to know.
If you are using C++11, you'd rewrite it using std::stoi:
for(int i = 0; i < here.length(); i++)
{
std::string subString = here.substr(i,1); //Returns a substring starting at 'i', and including 1 character
sum += std::stoi(subString);
}