string to double

[beacarpa]

I would like to convert the string without losing the decimal part and obtainig the same accuracy...

this is my code:

#include "stdafx.h"

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>

using namespace std;

int _tmain(int argc, char* argv[]){

string param = "1.65"; 
double L3;

L3=atof(param.c_str());
 
}

If I use atoi L3 = 1.00000; and if I use atof L3 = 1.6499999999; Any idea??

Thank you!!

[kevinkjt2000]

You need to understand how decimals on a computer are stored to answer this question.
http://stackoverflow.com/questions/14972107/how-does-a-computer-store-decimal-point-values
Specifically maybe read up on the IEEE 754-1985 and the standard that superseded it IEEE 754-2008. Those documents describe how most modern computers deal with floating point numbers.

In short, computers can not represent every value exactly.

[Smac89]

Might just be your compiler, but this displays correctly on ideone:
http://ideone.com/WP8Hyx

Alternatively, you can use iomanip to specify the precision:

#include <iostream>
#include <cstdlib>
#include <iomanip>
using namespace std;

int main() {
	char decimal[] = {"1.65"};
	double L = atof(decimal);
	cout << setprecision(3) << atof(decimal) << " " << L << endl;
	return 0;
}

[Chervil]

The problem isn't with the conversion from string to double. If you put
double L3 = 1.65; the result is the same.

1 2 3 4 5 6 7 8 9 10 11 12 13

string param = "1.65"; double L3; L3 = atof(param.c_str()); double a = 1.65; cout.precision(16); cout << fixed; cout << "L3 = " << L3 << endl; cout << "a = " << a << endl; cout << "L3 - a = " << L3 - a << endl;

L3 = 1.6499999999999999 a = 1.6499999999999999 L3 - a = 0.0000000000000000