C++ Error: Line 1440 Expression: string subscript out of range - c++

Question

The program builds and runs, however after entering the first integer and pressing enter then the error pop up box appears, then after pressing ignore and entering the second integer and pressing enter the pop up box appears and after pressing ignore it returns the correct answer. I am at my wits end with this can somebody help me fix the pop up box thing.
#include "stdafx.h"
#include <iostream>
#include <cctype>
#include <string>
using namespace std;
#define numbers 100
class largeintegers {
public:
largeintegers();
void
Input();
void
Output();
largeintegers
operator+(largeintegers);
largeintegers
operator-(largeintegers);
largeintegers
operator*(largeintegers);
int
operator==(largeintegers);
private:
int integer[numbers];
int len;
};
void largeintegers::Output() {
int i;
for (i = len - 1; i >= 0; i--)
cout << integer[i];
}
void largeintegers::Input() {
string in;
int i, j, k;
cout << "Enter any number:";
cin >> in;
for (i = 0; in[i] != '\0'; i++)
;
len = i;
k = 0;
for (j = i - 1; j >= 0; j--)
integer[j] = in[k++] - 48;
}
largeintegers::largeintegers() {
for (int i = 0; i < numbers; i++)
integer[i] = 0;
len = numbers - 1;
}
int largeintegers::operator==(largeintegers op2) {
int i;
if (len < op2.len) return -1;
if (op2.len < len) return 1;
for (i = len - 1; i >= 0; i--)
if (integer[i] < op2.integer[i])
return -1;
else if (op2.integer[i] < integer[i]) return 1;
return 0;
}
largeintegers largeintegers::operator+(largeintegers op2) {
largeintegers temp;
int carry = 0;
int c, i;
if (len > op2.len)
c = len;
else
c = op2.len;
for (i = 0; i < c; i++) {
temp.integer[i] = integer[i] + op2.integer[i] + carry;
if (temp.integer[i] > 9) {
temp.integer[i] %= 10;
carry = 1;
} else
carry = 0;
}
if (carry == 1) {
temp.len = c + 1;
if (temp.len >= numbers)
cout << "***OVERFLOW*****\n";
else
temp.integer[i] = carry;
} else
temp.len = c;
return temp;
}
largeintegers largeintegers::operator-(largeintegers op2) {
largeintegers temp;
int c;
if (len > op2.len)
c = len;
else
c = op2.len;
int borrow = 0;
for (int i = c; i >= 0; i--)
if (borrow == 0) {
if (integer[i] >= op2.integer[i])
temp.integer[i] = integer[i] - op2.integer[i];
else {
borrow = 1;
temp.integer[i] = integer[i] + 10 - op2.integer[i];
}
} else {
borrow = 0;
if (integer[i] - 1 >= op2.integer[i])
temp.integer[i] = integer[i] - 1 - op2.integer[i];
else {
borrow = 1;
temp.integer[i] = integer[i] - 1 + 10 - op2.integer[i];
}
}
temp.len = c;
return temp;
}
largeintegers largeintegers::operator*(largeintegers op2) {
largeintegers temp;
int i, j, k, tmp, m = 0;
for (i = 0; i < op2.len; i++) {
k = i;
for (j = 0; j < len; j++) {
tmp = integer[j] * op2.integer[i];
temp.integer[k] = temp.integer[k] + tmp;
temp.integer[k + 1] = temp.integer[k + 1] + temp.integer[k] / 10;
temp.integer[k] %= 10;
k++;
if (k > m) m = k;
}
}
temp.len = m;
if (temp.len > numbers) cout << "***OVERFLOW*****\n";
return temp;
}
using namespace std;
int main() {
int c;
largeintegers num1, num2, result;
num1.Input();
num2.Input();
num1.Output();
cout << " + ";
num2.Output();
result = num1 + num2;
cout << " = ";
result.Output();
cout << "\n\n";
num1.Output();
cout << " - ";
num2.Output();
result = num1 - num2;
cout << " = ";
result.Output();
cout << "\n\n";
num1.Output();
cout << " * ";
num2.Output();
result = num1 * num2;
cout << " = ";
result.Output();
cout << "\n\n";
c = num1 == num2;
num1.Output();
switch (c) {
case -1:
cout << " is less than ";
break;
case 0:
cout << " is equal to ";
break;
case 1:
cout << " is greater than ";
break;
}
num2.Output();
cout << "\n\n";
system("pause");
}

Answer 1

It seems you are falling victim to the difference between C-style strings and C++ strings. C-style strings are a series of chars followed by a zero (or null) byte. C++ strings are objects that contain a series of characters (usually char, but eventually this will be an assumption you should break) and that know their own length. C++ strings can contain null bytes in the middle of themselves without problem.
To loop through all of the characters of a C++-style string, you can do one of a number of things:
You can use the .size() or .length() members of a string variable to find the number of characters in it, as in for (int i=0; i<str.size(); i++) { char c = str[i];
You can use .begin() and .end() to get iterators to the beginning and end of the string, respectively. A for loop in the form for (std::string::iterator it=str.begin(); it!=str.end(); ++it) will loop you through the members of the string by accessing *it.
If you're using C++11, you can use the for loop construct as follows: for (auto c: str) where c will be of the type of a character of the string str.
In the future, to solve problems like these, you can try using the debugger to see what happens when your program crashes or hits an exception. You likely would find that inside of largeintegers::Input() you running into either a memory access violation or some other problem.
Finally, as a future-looking criticism, you should not use C-style arrays (where you say int integer[ numbers ];) in favor of using C++-style containers, such as vector. A vector is a series of objects (such as ints) that can expand as needed.