use of atoi() function in c++

Question

when I pass a string variable in the below code, g++ gives an error:

cannot convert 'std::__cxx11::string {aka std::__cxx11::basic_string}' to 'const char*' for argument '1' to 'int atoi(const char*)'

My code is:

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

int main()
{
    string a = "10";
    int b = atoi(a);
    cout<<b<<"\n";
    return 0;
}

But if I change the code to :

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

int main()
{
    char a[3] = "10";
    int b = atoi(a);
    cout<<b<<"\n";
    return 0;
}

It works completely fine.

Please explain why string doesn't work. Is there any difference between string a and char a[] ?

Answer 1

atoi is an older function carried over from C.

C did not have std::string , it relied on null-terminated char arrays instead. std::string has a c_str() method that returns a null-terminated char* pointer to the string data.

int b = atoi(a.c_str());

In C++11, there is an alternativestd::stoi() function that takes a std::string as an argument:

#include <iostream>
#include <string>

int main()
{
    std::string a = "10";
    int b = std::stoi(a);
    std::cout << b << "\n";
    return 0;
}

Answer 2

You need to pass a C style string.

Ie use c_str()

Change

int b = atoi(a);

to

int b = atoi(a.c_str());

PS:

This would be better - get the compiler to work out the length:

char a[] = "10";

Answer 3

atoi() expects a null-terminated char* as input. A string cannot be passed as-is where a char* is expected, thus the compiler error. On the other hand, a char[] can decay into a char* , which is why using a char[] works.

When using a string , call its c_str() method when you need a null-terminated char* pointer to its character data:

int b = atoi(a.c_str());

Answer 4

There is a difference between them. For each one there are different functions:

String

As said before, for string there is the stoi function:

string s("20");
cout << stoi(s) * 2; // output: 40

char*

In the past, atoi used to handle char* conversion.

However, now atoi is replaced by strtol which gets 3 parameters:

1. char* of the characters to parse to long ,
2. char** that returns pointer to after the parsed string,
3. int for the base the number should be parsed from (2, 10, 16, or whatever).

char c[]="20";
char* end;
cout << strtol(c, &end, 16); // output: 32

There are whole bunch of functions like strtol like strtof , strtod , or strtoll which converts to float , double , long , and long long .

The main advantages of those new functions are mostly error-handling, and multi-base support.

The main disadvantage is that there isn't a function that converts to int , only to long (besides other types).

For more details, see https://stackoverflow.com/a/22866001/12893141

Answer 5

according to the documentation of atoi() , the function expects a "pointer to the null-terminated byte string to be interpreted" which basically is a C-style string. std::string is string type in C++ but it have a method c_str() that can return a C-string which you can pass to atoi() .

string a = "10";
int b = atoi(a.c_str());

But if you still want to pass std::string and your compiler supports C++ 11, then you can usestoi()

string a = "10";
int b = stoi(a);