Function Problems

[awsdert]

Is there any way to make this function not throw an error during compile?

void opt (int number, char *variable) {
	char *name;
	char *type;
	switch (number) {
	case 2: {
		bool *value;
		name = "Option 3";
		value  = false;
		type = "bool";
	}
	break;
	case 1: {
		char *value;
		name = "Option 2";
		value = "Hello";
		type = "string";
	}
	break;
	default: {
		int value;
		name = "Option 1";
		value = 0;
		type = "number";
	}
	break;
	}
	if (variable == "name") return type;
	else if (variable == "type") return type;
	else return value;
}

I only started learning C/C++ recently & I'm doing this just for later reference on multi-type returns.

[guestgulkan]

You can't do this.
If a function has a return type of void, it means that it does not return anything,
so these statement are errors

	if (variable == "name") return type;
	else if (variable == "type") return type;
	else return value;

[Albatross]

*sigh*

Yes, and it's called a template.
http://cplusplus.com/doc/tutorial/templates/

-Albatross

[guestgulkan]

Was that in reply to my answer??

[Albatross]

No, actually... what you said is perfectly valid as far as we both know. Functions declared as void can't return anything in standard c++.

What I think he wanted to know was how to return stuff without being constrained by the type.

-Albatross

[awsdert]

Thanks to both of you, I'll take a look at the link Albotross provided me (and yes your right about what I wanted to return)

[awsdert]

Okay new problem:

template <class option>
option opt (int number, char *variable) {
	char *name;
	char *type;
	switch (number) {
	case 2: {
		bool *value;
		name = "Option 3";
		value = false;
		type = "bool";
	}
	break;
	case 1: {
		char *value;
		name = "Option 2";
		value = "Hello";
		type = "string";
	}
	break;
	default: {
		int value;
		name = "Option 1";
		value = 0;
		type = "number";
	}
	break;
	}
	if (variable == "name") return type;
	else if (variable == "type") return type;
	else return value;
}

Compiler gives this (I omitted the problems from the main function):

Compiling: main.cpp
C:\Users\awsdert\Desktop\SOS\main.cpp: In function `option opt(int, char*)':
C:\Users\awsdert\Desktop\SOS\main.cpp:39: error: `value' was not declared in this scope
Process terminated with status 1 (0 minutes, 0 seconds)
12 errors, 0 warnings

I can't see where I'm going wrong.

[Albatross]

What exactly is this function supposed to do?

-Albatross

[awsdert]

Never mind, I fixed it.

[awsdert]

Astonishing timing - as for what it supposed to do, it's to parse cmdline options into, I just needed to read the data first before implimenting writing to the data.

[Galik]

You can't return different types in the same function even using a template.

C++ is not like weakly typed languages such as Javascript.

Each function can only return one type. In a template function, that one type can vary. But within any particular invocation of the template function, only one type can be returned.

[awsdert]

I'm not after all types at the same time, I'm after a type that's determind by the values I give the function.

[Galik]

Indeed. But that is simply not possible in C++.

[Albatross]

Well... um... void*

-Albatross

[Galik]

You have to remember that when you call your function you need to know the return tybe before you call it:

int var = opt (2, "-size");

// need to declare the type of the variable
// that will accept the return value from the function
// BEFORE calling the function.

[Galik]

void* = evil

[Galik]

typedef void* evil;

[Albatross]

I know, I know, don't get your deques in a knot.

But it would do (theoretically) what he/she wanted after an explicit type cast, no?

-Albatross

[Albatross]

Also, you could declare multiple return types for this function. Like...

*deleted*

Actually, you can't do that without changing something about the variables. I really am short on coffee.

-Albatross

[Galik]

@Albatross

I recommend going to get a large coffee...