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what auto concealed

technologist (607)
What type does auto stand for here in lines 24 and 40? How does it appear in the code if it were written without auto?

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#include<iostream>

int add(int x, int y)
{
	return x + y;
}

int subtract(int x, int y)
{
	return x - y;

}

int multiply(int x, int y)
{
	return x * y;
}

int divide (int x, int y)
{
	return x/y;
}

auto getArithmeticFunction(char op)
{
	switch(op)
	{
	case '+' : return &add;
	case '-' : return &subtract;
	case '/' : return &divide;
	case '*' : return &multiply;
	}
}

int main()
{
	char op;
	std::cout<<"input operator :";
	std::cin>>op;
	auto func = getArithmeticFunction(op);
	std::cout << func(10,5) << std::endl;


	return 0;
}


Thx
Last edited on
Duthomhas (13129)
The type of the functions it returns is:

    int (*)(int, int)
JLBorges (13770)
> How does it appear in the code if it were written without auto?

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using function_type = int(int,int) ; // function_type is the type of the function

function_type* getArithmeticFunction(char op) // return pointer to function of type int(int,int)
{
	switch(op)
	{
	case '+' : return &add;
	case '-' : return &subtract;
	case '/' : return &divide;
	case '*' : return &multiply;
        default: return nullptr ; // *** repeat: avoid undefined behaviour when 'op' is not one of the above
                                  // http://www.cplusplus.com/forum/general/190962/#msg922593
	}
}

int main()
{
	char op;
	std::cout<<"input operator :";
	std::cin>>op;

	// auto func = getArithmeticFunction(op);
	// std::cout << func(10,5) << std::endl;
        
        function_type* func = getArithmeticFunction(op);
        if( func != nullptr ) std::cout << func(10,5) << '\n';
}
technologist (607)
Hi,

Yours is a more straightforward way of dealing with auto in this case. I also found this to be useful, a typedef: which is similar to your tweak as well?

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#include<iostream>

typedef int (*arithmeticFcn)(int, int);


int add(int x, int y)
{
	return x + y;
}

int subtract(int x, int y)
{
	return x - y;

}

int multiply(int x, int y)
{
	return x * y;
}

int divide (int x, int y)
{
	return x/y;
}

arithmeticFcn getArithmeticFunction(char op)
{
	switch(op)
	{
	case '+' : return &add;
	case '-' : return &subtract;
	case '/' : return &divide;
	case '*' : return &multiply;
        default: return nullptr ;
	}
}

int main()
{
	char op;
	std::cout<<"input operator :";
	std::cin>>op;
	 arithmeticFcn func = getArithmeticFunction(op);
	std::cout << func(10,5) << std::endl;


	return 0;
}
Last edited on
JLBorges (13770)
There is no difference between the type alias declaration using arithmeticFcn = int(*)(int,int);
and the typedef declaration typedef int (*arithmeticFcn)(int, int);
technologist (607)
Awesome, thx.
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