CODEWITHSUNDEEP
c++
Lithium
Lithium

Reputation: 223

Output of large numbers

I need to use a high Number in my program. Here is an minimum example that shows the problem. It just reads in a number and gives out the number again.

using namespace std;
#include <iostream>
#include <stdlib.h>

int main(int argc, char* argv[])
{
    double 
    N = atoi(argv[1]);
    cout << N << endl;
    return 0;
}

Everything works fine till 10^9 If I take 10^10 as input the result is 1.41007e+09. I just don't know why this is the case. I already tried the following:

Upvotes: 0

Views: 1112

Answers (3)

pragnesh
pragnesh

Reputation: 1250

The input 10^10 goes beyond the maximum limit of an integer that is 2147483648 and 10^9 is well within the limit.

Following code can be useful.

using namespace std;
#include <iostream>
#include <stdlib.h>

int main(int argc, char* argv[])
{
    long long N = atoll (argv[1]);
    cout << N << endl;
    return 0;
}

Upvotes: -1

Arpit
Arpit

Reputation: 775

Using atoi on the number greater than INT_MAX will cause the resultant value to start from the negative value INT_MIN.

Try using atoll function. It is in . But this is defined in C++11. So, if your compiler is not up-to-date than it will throw an error.

If the value is again quite large (greater than 8 bytes), then I would suggest you to use external libraries.

Upvotes: 0

M.M
M.M

Reputation: 141586

Using atoi when the resulting value would be outside the bounds of int (i.e. greater than INT_MAX or less than INT_MIN) causes undefined behaviour.

This is a good reason to not use atoi. The alternatives strtol and strtoul from the C library have well-defined behaviour for all inputs.

The C++11 function std::stoi also has well-defined behaviour (it throws an exception if the value is out of range). istream::operator>>(int&) is also well-defined but it has some finicky details relating to the fact that streams can't "look ahead".

Upvotes: 2

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