Calling a char equivalent function with integer in c++ using Function Overloading

Question

I am overloading a function it just prints the value which it received as an argument . Here is the code

#include<iostream>
using namespace std;
void show(int c)
{
    cout<<"Int C : "<<c<<endl;
}
void show(char c)
{
    cout<<"char C : "<<c<<endl;
}
void show(float c)
{
    cout<<"Float C : "<<c<<endl;
}
void show(string c)
{
    cout<<"String C : "<<c<<endl;
}
void show(const char* c)
{
    cout<<"char *C : "<<c<<endl;
}
main()
{
    string s("Vector");
    show(25);
    show('Z');
    show("C++");
    show("Hello");
    show(4.9f);
    show(s);
}

Here show(65) calls the integer function.

Is there any possibility that I can simply call show(65) to print the ASCII equivalent value 'A' i.e calling show(char) without calling show(int)

WhiZTiM · Answer 1 · 2017-01-02T14:05:23.140

4

Is there any possibility that I can simply call show(65) to print the ASCII equivalent value 'A' i.e calling show(char) without calling show(int)

Yes. You convert it to a char

show(static_cast<char>(65));

or convert it using a "function-style cast" fashion.

show(char(65));

Note: This will work only if your environment uses ASCII, because C++ does not mandate ASCII encoding for characters. Though systems using a different encoding are rare.

edited Jan 02 '17 at 14:05

answered Jan 02 '17 at 13:36

WhiZTiM

21,207
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Would *"function-style cast"* be the phrase you're after ? – Quentin Jan 02 '17 at 13:40
1

@Quentin, Ahh yes. Thanks! fixed. – WhiZTiM Jan 02 '17 at 13:42
`show(static_cast(65));` what does this static cast mean and what is a function style cast ? – Anjaneyulu Jan 02 '17 at 13:44
2

@Anjaneyulu - Comments aren't the place for an introduction. [Pick a good book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) and go from there. – StoryTeller - Unslander Monica Jan 02 '17 at 13:50

ban · Answer 2 · 2017-01-02T14:30:20.510

1

Or you can try by creating variable with char type and u can pass it to Show function.

Ex :

char test = 'Z';

Show(test);

It will works!

edited Jan 02 '17 at 14:30

answered Jan 02 '17 at 13:47

ban

71
7

the ascii encoding of `'Z'` is not `65`. – StoryTeller - Unslander Monica Jan 02 '17 at 13:52
I dont think he was receiving 65 output for 'Z'. He mentioned there at the end of the post like " show(65) to print the ASCII equivalent value 'A' " – ban Jan 02 '17 at 14:26

score 0 · Answer 3 · answered Jan 02 '17 at 15:20

No, it won't work because:

You already have a function which takes a number as an argument.
You have to manually convert it to char:show(char(65)).
Even if you convert it to char, it would only work if you're in an environment which uses ASCII (it's not the only encoding system in C++).

If you already know what char you want you can simply do this:

show('A');

Calling a char equivalent function with integer in c++ using Function Overloading

3 Answers3