CODEWITHSUNDEEP
c++functionarguments
CIMPLE
CIMPLE

Reputation: 13

Can I use non-default datatype as argument for functions in 1 line?

I think the title of question is little bit confusing, so I'll explain that.

Let's see this very simple code.

#include <iostream>

class C {
public:
    int val;

    C(){
        val = 2;
    }

    void changeVal(int i){
        val = i;
    }

};

void printout(int val){
    std::cout << "int val : " << val << std::endl;
}

void printout(C c){
    std::cout << "class val : " << c.val << std::endl;
}

int main()
{
    C c;
    printout(1);
    printout(c);    

    //printout(C());    // ok, I can understand it.
    //printout(C().changeVal(0)); /// ?????
    return 0;
}

As you can see, function 'printout' is for printing out the input argument. My question is, when I use the int value(default datatype), then I just put in the function real number '1', however, when I use my class instance('class C'), then I have to declare my class instance before the function.

So, is there any way to make this kind of non-default datatype for function argument in 1-line?

The actual situation is, I have to use the 4x4 matrix as argument for some function. For this reason, I have to make some matrix, and initialize(make zero) that, and use it. But if I can do this same job with just 1 line, my source code will be more clear than now.

That's my question. I hope you could understand my question. Thank you.

Upvotes: 0

Views: 94

Answers (1)

juanchopanza
juanchopanza

Reputation: 227418

You can pass a temporary:

printout(C());

Note that, since you don't need a copy of the C parameter, it would make more sense to pass it by const reference:

void printout(const C& c) { ... }

Upvotes: 2

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