CODEWITHSUNDEEP
c++overloading
Danny
Danny

Reputation: 2673

C++ bool expression as function argument calls wrong overload

In the test program below, calling the checkit method with a bool predicate instead calls the integer overload. Why?

(i == 10) should create a boolean value, which should then trigger the bool overload to be called, no?

Program Output

should be int
  I'm a bool 1
should be bool
  I'm a bool 1
should also be bool
  I'm a int 1

Source

class Overload
{
public:
    void checkIt(bool n) { printf("  I'm a bool %d\n", n);  }
    void checkit(long n) { printf("  I'm a long %ld\n", n); }
    void checkit(int n ) { printf("  I'm a int %d\n", n); }
};

int main(int argc, char *argv[])
{
    Overload obj;

    int i = 10;
    bool b = true;

    printf("should be int\n");
    obj.checkIt(i);

    printf("should be bool\n");
    obj.checkIt(b);

    printf("should also be bool\n");
    obj.checkit( (i == 10) );
}

EDIT

Duh! sorry for the typo.

However, after fixing typo and commenting out the bool method, the int method is called with the bool predicate. Is there a default conversion from bool to int if a more appropriate signature is not found? Strictly speaking, there is no bool signature so I was expecting either a compile or runtime error...

Updated Code

class Overload
{
public:
    //void checkIt(bool n) { printf("  I'm a bool %d\n", n);  }
    void checkIt(long n) { printf("  I'm a long %ld\n", n); }
    void checkIt(int n ) { printf("  I'm a int %d\n", n); }
};

int main(int argc, char *argv[])
{
    Overload obj;

    int i = 10;
    bool b = true;

    printf("should be int\n");
    obj.checkIt(i);

    printf("should be bool\n");
    obj.checkIt(b);

    printf("should also be bool\n");
    obj.checkIt( (i == 10) );
}

Updated Output

should be int
  I'm a int 10
should be bool
  I'm a int 1
should also be bool
  I'm a int 1

Upvotes: 1

Views: 665

Answers (3)

Harindu Dilshan
Harindu Dilshan

Reputation: 174

Yes C++ uses implicit conversion.bool,char is automatically converted into int /short/long or vice versa.

Upvotes: 0

snowynguyen
snowynguyen

Reputation: 318

Remember that function names (and almost everything else) in C++ are CaSe SeNsItIvE.
In your updated code, there are 2 checkIt methods, one for int type and one for long type. You called checkIt 3 times, and the 2 last times you passed to the method a bool value. In C++ bool value automatically casts to int with true == 1 and false == 0 

bool a = 1 // equals to true
bool b = 0 // equals to false

As there are no methods accepting 'bool' type, the value automatically casts to int and calls the int method.

Upvotes: 0

melpomene
melpomene

Reputation: 85757

checkIt isn't overloaded; there's only a bool version.

checkit is overloaded; there's a long and an int version.

Your code calls checkIt twice and checkit once.

Upvotes: 3

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