I am working on writing a simple linear line calculator. For example, a user can enter two equations (strings) such as y=5x+3 and y=-3x+6. The most basic feature of this calculator is that it will return the intersection point of these two lines.
The obstacle I can't seem to figure out is how to parse the string into two pieces of data: the slope, and the y-intercept. This is a simple calculator, so the format of both lines will be y=mx+b, however, both the slope and/or y-intercept may be non-integer numbers (i.e. floats).
I came across a function in the string library called stod, which converts a number in a string to a numerical value (am I understanding this correctly?).
My question is, will this function do the job? If so, how exactly do I use the "idx" parameter? I don't quite understand it.
If this isn't going to work, how can I parse this user-entered data?
- both equations are strings (y=mx+b)
- m and b have private variables dedicated in storing the decimal value (i.e. double m_ and double b_ are private member variables)
2 Answers
This is how the idx parameter works:
#include <string> #include <iostream> int main(void) { std::string data = "y=5.9568x+3.14"; //say you have a string like this.. double y, x, m, b; y = 0; x = 0; std::size_t offset = 0; //offset will be set to the length of characters of the "value" - 1. m = std::stod(&data[2], &offset); //So we want to get the value "5.9568 b = std::stod(&data[offset + 3]); //When we reach this line, offset has a value of 6 std::cout<<b; return 0; } So now you're asking why does it have a value of 6? Well because:
5.9568 is exactly: 6 characters in length. Thus on the next line when we do
b = std::stod(&data[offset + 3]);
we are actually feeding it a pointer to address of x + 3.. and that turns out to be right at the beginning of the 3.14.
In other words it's equivalent to:
std::stod(&data[9]);
So that idx parameter is actually the index/length of the double in characters within the string. If the string is:
str = "3.14159"
Then std::stod(str, &idx) will make idx equal to: 6.
if the string is:
str = "y = 1024.789" then std::stod(&str[4], &idx) will make idx equal to: 8 STARTING FROM &str[4]..
Here's something simple with no error checking to get you started:
Assuming your input string is always exactly of the form y=mx+b and you wish to parse it to obtain the numerical values of m and b you can first tokenize the string with y, =, x, and as delimiters.
An example of a tokenizing function can be found here. Here it is reproduced:
void tokenize(const std::string &str, std::vector<std::string> &tokens, const std::string &delimiters) { // Skip delimiters at beginning. std::string::size_type lastPos = str.find_first_not_of(delimiters, 0); // Find first "non-delimiter". std::string::size_type pos = str.find_first_of(delimiters, lastPos); while (std::string::npos != pos || std::string::npos != lastPos) { // Found a token, add it to the vector. tokens.push_back(str.substr(lastPos, pos - lastPos)); // Skip delimiters. Note the "not_of" lastPos = str.find_first_not_of(delimiters, pos); // Find next "non-delimiter" pos = str.find_first_of(delimiters, lastPos); } } The first argument is the string to tokenize, the second is a reference to a vector<string> which the function will put the tokens in, and the third argument is a string containing all the delimiter characters. You can use it with the delimiters mentioned above like this:
string s = "y=-3x + 10"; vector<string> tokens; tokenize(s, tokens, "y=x "); For the example string above tokens will contain the following strings: -3, +, and 10.
Now you can iterate over tokens and call stod() on each token. You can put the results of stod() in a vector<double>:
vector<double> doubles; for (vector<string>::iterator iter = tokens.begin(); iter != tokens.end(); ++iter) { try { doubles.push_back(stod(*iter)); // size_t* idx is an optional argument } catch (...) { // handle exceptions here. stod() will throw an exception // on the "+" token but you can throw it away } } Now doubles should have exactly 2 elements -- one for the slope and another for the intercept. Assuming the slope came first (the string was of the form y=mx+b instead of y=b+mx) then you can extract them from doubles:
double m = doubles[0]; double b = doubles[1]; Parsing the initial string is more complicated if the user is allowed different forms like y=b+mx (in that case the intercept came first), and much more complicated if the user can enter even stranger (but valid) forms like x*m+b=y (now you can't just assume that the number before the x character is the slope). It's not clear from your question exactly what alternate forms are considered valid, but nonetheless this should get you started.
Finally, as to your question about *idx, stod() puts into it the position of the first character after the number it parsed. This allows you to easily parse multiple numbers in a single string by skipping the number that was just parsed. Using the example at your reference link with some added comments:
std::string orbits ("365.24 29.53"); std::string::size_type sz; // alias of size_t double earth = std::stod (orbits,&sz); // sz now holds the position of the first character after 365.24, which is whitespace // the next call to stod() will start from the sz position double moon = std::stod (orbits.substr(sz)); 
