Can someone tell me how this is compiling? (sqrt is retuning a string instead)

Question

Here's my code. This is boggling my mind.

#include <iostream>
#include <sstream>
#include <set>
#include <cmath>
#include <cstdlib>
#include "list.h"
#include "stack.h"
#include <limits>
#define PI 3.1415926535897932384626433832795

class RPN : public Stack<float> {
      public:
             std::string sqrt(float n); 
};

std::string RPN::sqrt(float n){
    std::string x;
    x = sqrt(3.0);
    std::ostringstream ss;
    ss << n;
    return (ss.str());
}

Yes that is compiling. sqrt is returning a string. trying to use a double or a float throws a strange error. Can anyone tell me what is going on? I've never had this before. The funny thing, is I'm actually converting to a string later on, but I doubt this will compile any where else...

postfix.cpp: In member function ‘std::string RPN::sqrt(float)’:
postfix.cpp:161:13: error: cannot convert ‘std::string {aka std::basic_string<char>}’ to ‘float’ in assignment

edit: posted wrong compiler error at first.

edit2: line 161 is n=sqrt(n); I have even tried double x = sqrt(n) and many other methods; Oh and when I print out the retrned string in my method posted above, I get a seg fault(obv..)

std::string RPN::sqrt(float n) {
    n = sqrt(n);
    std::ostringstream ss;
    ss << n;
    return (ss.str());
}

Answer 1

The line x = sqrt(3.0); is calling your RDN::sqrt() method which returns a string. I think you're trying to call the sqrt() function in cmath. I would suggest renaming your method to something else. Alternatively, you might be able to call std::sqrt(3.0)

Answer 2

Let's look at the code more closely

std::string RPN::sqrt(float n){
    std::string x; // temporary string variable

    // calling sqrt with 3.0? What?
    // This call actually would make this function be recursive 
    // (it would hide ::sqrt), making the assignment possible 
    // to compile (as sqrt returns a string)
    // This also means that the function will 
    // eventually cause a stack overflow as there is no break case.
    x = sqrt(3.0); 

    std::ostringstream ss; // temporary string stream
    ss << n; // putting x in the string stream

    // returning the string value of the string stream
    // (i.e. converting x to a string)
    return (ss.str()); 
}

In other words, there is no compile errors, but if you ran that code, you will get a run-time error.

EDIT:

Try n = ::sqrt(n) (or n = std::sqrt(n) if you #include <cmath>) instead of n = sqrt(n), as that will just call your own function that you are defining, since your function will mask global scope.

n = sqrt(n) makes your function recursive, and not compile.