Doubt with operator ==

Question

Doubt with operator ==

Navigation

#1 by (1 votes)
#2 by (0 votes)

1

Having the following operator declared in a string class, would it be correct to use it in a split class in the following way?

Operator:

  bool CCadena::operator==(const CCadena& obj) const {
  return strcmp(m_pText, obj.m_pText) == 0;
}

function:

if(m_nomLocal==nomEquip)
    {
        return true;
    }
    else if(m_nomVisitant==nomEquip)
    {
        return true;
    }
    else
    {
        return false;
    }

I need to compare if the content of the nomlocal and visitor varianles are equal to the parameter nomEquip and return true of being so and false otherwise.

c++

asked by ElPatrón 16.03.2017 в 18:14

source

2 answers

0

It seems to me that m_nomLocal , m_nomVisitant and nomEquip are variables const char * while operator == that defined is used to compare objects CCadena . Your operator == compares members m_pText of CCadena s and therefore does not serve to compare m_local, m_nomVisitant and nomEquip directly.

#include <iostream>
#include <string.h>

class CCadena{
public:
    const char* m_pText;
    CCadena(const char* pText): m_pText(pText){}
    bool operator==(const CCadena& obj) const {
        return strcmp(m_pText, obj.m_pText) == 0;
    }
};

int main(){
    CCadena local("aaaa");
    CCadena equip("bbbb");
    std::cout << (local == equip) << std::endl;

    CCadena visitante("aaaa");
    std::cout << (local == visitante) << std::endl;
}

answered by 16.03.2017 в 19:47

Save file in data / data / package / Configure international domains with WPML

score 1 · Accepted Answer

If you are interested in knowing if nomEquip is equivalent to m_nomLocal or m_nomVisitant and assuming that the three variables are of type CCadena , it would be more natural to leave it like this:

if(m_nomLocal==nomEquip || m_nomVisitant==nomEquip)
{
  return true;
}
else
{
  return false;
}

or like this:

return (m_nomLocal==nomEquip || m_nomVisitant==nomEquip);

The two options are equivalent, of course your method would also work, but it is more repetitive and, therefore, prone to errors.