Converting string to long

Question

I'm trying to convert a string to long. It sounds easy, but I still get the same error. I tried:

include <iostream>
include <string>    

using namespace std;

int main()
{
  string myString = "";
  cin >> myString;
  long myLong = atol(myString);
}

But always the error:

.../main.cpp:12: error: cannot convert 'std::string {aka std::basic_string<char>}' to 'const char*' for argument '1' to 'long int atol(const char*)'

occured. The reference says following:

long int atol ( const char * str );

Any help?

Answer 1

Try

long myLong = std::stol( myString );

The function has three parameters

long stol(const string& str, size_t *idx = 0, int base = 10);

You can use the second parameter that to determine the position in the string where parsing of the number was stoped. For example

std::string s( "123a" );

size_t n;

std::stol( s, &n );

std::cout << n << std::endl;

The output is

3

The function can throw exceptions.

Answer 2

只需写长myLong = atol(myString.c_str());

Answer 3

atol() requires a const char* ; there's no implicit conversion from std::string to const char* , so if you really want to use atol() , you must call the std::string::c_str() method to get the raw C-like string pointer to be passed to atol() :

// myString is a std::string
long myLong = atol(myString.c_str());

A better C++ approach would be using stol() (available since C++11), without relying on a C function like atol() :

long myLong = std::stol(myString);

Answer 4

atol gets as parameter a const char* (C-style string), but you are passing as parameter a std::string . The compiler is not able to find any viable conversion between const char* and std::string , so it gives you the error. You can use the string member function std::string::c_str() , which returns a c-style string, equivalent to the contents of you std::string . Usage:

string str = "314159265";
cout << "C-ctyle string: " << str.c_str() << endl;
cout << "Converted to long: " << atol(str.c_str()) << endl;