使用atoi将字符数组转换为int可以得到不同的值

Question

Trying to convert a character array to int data type in C++11. Why is the int value i different than the character array value ? Is this because 51,093,843,802 is greater than 2,147,483,647 ?

// Example program
#include <iostream>
#include <string>

int main()
{
    char value[] = "51093843802";
    printf("%s\n", value);
    int i = std::atoi(value);
    printf("%d\n", i);
}

Output:

51093843802
-445763750

Answer 1

As you said, the value 51093843802 can't (usually) fit in int , and overflowing a signed integer is undefined behavior (and thus a huge no no) in C++.

You need to use long long int (or just long int will do if you're on a system where that's 64 bits ):

#include <iostream>
#include <string>

int main()
{
    char value[] = "51093843802";
    std::cout << value << '\n';
    long long i = std::stoll(value);
    std::cout << i << '\n';
}

Output:

 51093843802 51093843802 

Notice that I changed i 's datatype to be long long , std::atoi to std::stoll and printf to std::cout , to be more in line with C++ best practises.

Answer 2

int is too small to store 51093843802 ( range of values ). Use atoll (to long long) and replace i's type to int64_t :

char value[] = "51093843802";
printf("%s\n", value);
int64_t i = std::atoll(value);
printf("%ld\n", i);

Using long with atol is not safe because on Windows long size is 4 bytes even on 64bits system.