So I have a function with several parameters that will need to perform several tasks based on these parameters. But when I call this function I may not need it to perform some of the tasks in it.
For example my function has parameters (int x, int bin, int value) but sometimes when I call it I don't want it to evaluate the part of the function using int value. How can I accomplish this? I've heard of using optional arguments which default the argument to 0 if I don't specify it but that is not what I want. What I want is this, if I provide a value for "int value" then I want the part of my function using this value to evaluate, otherwise, ignore it.
I suggest using function overloading:
void foo(int x, int bin) {
//...
}
void foo(int x, int bin, int value) {
foo(x, bin);
// extra stuff using value...
}
But you could also make value a pointer and use nullptr to signify it shouldn't be used.
Here is another option. In this case if the function is called with only two parameters value will be initialized to a sentinel (-1 in this case) and can be checked in the code.
void foo(int x, int bin, int value=-1) {
// x stuff
// bin stuff
if (value != -1) {
// value stuff
}
}
This will only work if there is some invalid value though which could be 0, -1, or the maybe the max value of an int. Function overloading is probably a better option though.
Related
This question already has answers here:
How do I use arrays in C++?
(5 answers)
Closed 6 years ago.
I have looked at all the other posts with a similar topic, and none help, so please don't flag as a duplicate.
I am defining in main() a const int SIZE = 20;. Then, I pass this as an argument to my function, Mode:
int* Mode(int* numbers, int & mode, const int SIZE)
{
int occurences[SIZE];
// Calcualte mode
}
However, I get the error, expression must have a constant value.
My function call (in main) looks like this:
int* occurencesPtr = Mode(numbersPtr, mode, SIZE);
With SIZE being defined at the beginning to the literal 20.
I understand that the error is because the function's version of SIZE only acquires its value when the function is called (?), but I don't know how I could work around this.
I have even tried passing to the function a const int * const SIZEPtr = &SIZE, but that didn't work either. Help?
EDIT: I am not trying to use a variable size!! Notice that I have made SIZE a const everywhere! I just want to use that same SIZE constant to declare my array.
EDIT: Dynamic arrays are not what I need. I just want a normal, named, array, defined with a constant size value passed to the function.
There is a misconception here with what const means, probably because it's a little confusing that this works:
const int SIZE = 20;
int array[SIZE];
but this doesn't:
void foo(const int SIZE) {
int array[SIZE];
// ...
}
const int SIZE = 20;
foo(SIZE);
The issue is that the array size in an array declaration must be a core constant expression. Simplified, that means an expression that's evaluatable at compile time to be a constant. That is true in the first case (you can see that SIZE is the integral constant 20) but that is not true in the second case. There, the SIZE function parameter is just const - in the sense that it is nonmodifiable - and not a core constant expression. You can see the difference in that I can call foo() with something that is clearly unknowable until runtime:
int x;
if (std::cin >> x) {
foo(x);
}
In order to pass an argument into foo, and have that argument be used as an array bound, it is not enough to have it be const - the actual integral value must be encoded into the type (unless you call foo() as constexpr which I'm assuming is not the case here). In which case, you'd have to do something like:
template <int SIZE>
void foo() { ... }
const int SIZE = 20;
foo<SIZE>();
or:
template <int SIZE>
void foo(std::integral_constant<int, SIZE > ) { ... }
const int SIZE = 20;
foo(std::integral_constant<int, SIZE>{} );
or simply have SIZE be a global constant or otherwise accessible to foo() in a way that doesn't have to do with its arguments.
Or, there's always the simple option: use std::vector:
void foo(const int SIZE) {
std::vector<int> v(SIZE);
...
}
I understand that the error is because the function's version of SIZE only acquires its value when the function is called (?), but I don't know how I could work around this.
Option 1
Instead of defining SIZE in main, add a constexpr function. Use the constexpr function instead of passing the size.
constexpr int getSize()
{
return 20;
}
int* Mode(int* numbers, int & mode)
{
int occurences[getSize()];
// ...
}
Option 2
Use std::vector instead of array.
int* Mode(int* numbers, int & mode, int size)
{
std::vector<int> occurences[size];
// ...
}
Option 3
Use a function template.
template <size_t SIZE>
int* Mode(int* numbers, int & mode, int size)
{
int occurences[SIZE];
// ...
}
Option 4
Use a function template and std::array.
template <size_t SIZE>
int* Mode(int* numbers, int & mode, int size)
{
std::array<int, SIZE> occurences;
// ...
}
You're confusing things. A constant expression has nothing to do with const (at least not that much) ;).
let's think we are the compiler and face this function:
void foo(const int SIZE) { }
The constmerely says "we are not able to change the function-local variable SIZE inside the function body.
We need to compile it without assuming that SIZE is compile time constant. Why?
Because there is noone stoping us from doing something like:
int i{};
std::cin >> i;
foo(i);
You can pass any (matching/convertible) value to a by value const function argument.
What should happen when the compiler assumed the value passed to foo was a compile time constant expression?
If you want to pass compile time constants, use templates and while you're at it use std::array instead of T[N]:
template<std::size_t N>
void foo()
{
std::array<int, N> occurences;
}
const isn't doing what you think it's doing in your Mode function.
When const is used in function definition, const is simply telling the compiler that the function will not change the argument declared const inside of the scope of it's function. But that does not make the argument a constant, it is actually called a constant expression. Some compilers enforce this, others do not, and so will allow you to change const expressions (arguments passed with const keyword).
In order to use a globally accessible constant value which you can use, like SIZE, you'll need to declare a global constant before the function is called; which could be declared outside of main(), or at least outside the scope of all other functions but main(), if you must declare all inside main. Pass the global constant to the Mode function just as you would any other variable.
Oh, and, main() needs a return type.
I've edited the code to meet your specific constraints.
Here is a variation on your original code:
int main(){
//Declare constants first.
const int SIZE = 20; /*Could declare here instead.*/
//Declare variables next.
int *intPtr = 0; // to hold the pointer passed from Mode.
int *numbersPointer = 0;
int mode = 0;
//Define Mode (using OP's code.)
int* Mode(int* numbers, int & mode, const int size){
int occurences[size];
// Calculate mode
}
/*Now use constants, variables, and functions.*/
intPtr = Mode(numbersPointer, mode, SIZE); //Call mode.
return 0;
}
Lets say I have a function
void doStuff(vector<int> &a, int b, vector<int> &c> {
c = vector<int>(a.size());
for (int i = 0; i < a.size(); i++) {
c[i] = a[i] + b;
}
}
obviously, upon seeing the function, we know that "c" is the output.
For anybody who hasn't seen the function definition though, it remains a mystery unless i name c something like "output_c". Maybe I'm just being vein but I don't like naming things "ouput_xxx", is there any syntax candy for letting the user of the function know that its supposed to be the output?
Syntax, by itself, can be a guide to indicate which one is an input argument and which one is an output argument. However, an output argument can also serve as an input argument too. You cannot tell that by just looking at the signature.
Examples:
int foo(int arg); // The argument is copy by value. It can only be an input argument.
void foo(std::vector<int> const& arg); // The argument is by const&.
// It can only be an input argument.
void foo(std::vector<int>& arg); // The argument is by &. It can be:
// 1) an output argument.
// 2) an input and output argument.
// 3) an input argument (bad practice)
You could add a preprocessor directive:
#define OUT
and put it in the parameter list like so:
void doStuff(vector<int> &a, int b, OUT vector<int> &c) ...
I think I've seen some APIs do something like this. That way it is explicitly stated in the function signature but you don't have to modify the variable names. The code is also unchanged at compile time since OUT is not defined to be anything, it is just a defined symbol.
I think, though, I would rely on your own documentation when writing the function and/or return-by-value instead of doing something like this. You could also make use of the const keyword to flag a parameter that is guaranteed not to change - that's what the syntax is designed for.
So I just had a thought, is it possible to return a parameter sent when a function is called. And if it is, is this considered fine or is it bad style?
Example:
int main()
{
...
int value = 1;
value = Foo(value);
...
}
int Foo(int i)
{
i = i * 2;
return (i);
}
As the parameter is being passed in and returned by value, this is fine - there is an implicit copy occurring when you call the function and when it returns.
For example
int value=1,other=0;
other=Foo(value);
other is now 2, value will still be 1
If you were passing in a reference or pointer then you would potentially run risks.
e.g. if the signature of Foo was
int Foo( int &i )
Then after the code chunk I used above, both other and value would be 2
There's no problem with "returning a parameter" in your example. You are not really "returning a parameter" at all. You are simply using the parameter in the argument expression of return. It is the result of that expression (the value of i) that gets returned, not the parameter itself.
One can argue that the "undesirable" property of your code sample is the fact that you are modifying the parameter inside the function, i.e. you are using the parameter as an ordinary local variable. There's nothing formally wrong with it, but sometimes people prefer to preserve the original parameter values throughout the function body. I.e. from that point of view your function would look better as
int Foo(int i)
{
return i * 2;
}
or as
int Foo(int i)
{
int i2 = i * 2;
return i2;
}
but, again, it is not really about "not returning a parameter", but rather about leaving the original value of i untouched inside the function.
There's no problem with doing that and it makes it very clear what's going on.
That's one valid approach to do this, but you might also like the idea of passing by reference:
int main()
{
...
int value = 1;
Foo(value);
...
}
void Foo(int &i)
{
i = i * 2;
}
The drawback to this approach is that you have to pass what's called an lvalue into the function-- basically, something that can be on the left side of an assignment statement, which here means a variable. A call with a literal or temporary, such as Foo(2), will fail to compile. The way you had written it originally will instead do an implicit copy by value into the local scope of the Foo function. Note that the return value is now also void.
Technically, there is no problem, but semantically, it is not advisable: in most cases the input of the function and the return value of the function are not the same, so you are reusing the variable to mean something different. It is clearer in next example
int main()
{
double i = 5;
i = getSquareSurface(i); // i was a length and is now a surface
}
This should be:
int main()
{
double length = 5;
double surface = getSquareSurface(length);
}
Of course, there are cases like the addOne() or in this case the Foo() function where the meaning doesn't change.
EDIT: thanks for all the speedy responses, I have a much better understanding of this concept now. Also, I'll try to make my error messages more clear next time.
EDIT: updated with my newest code. the error happens on line 18. Also, I'm beginning to wonder if my latest issue has to do with the original class itself?
I'm trying to teach myself classes and objects in C++. I did it once by just declaring a void function, outputting something on the screen, calling the object in main and everything worked fine.
Now, I wanted to expand upon this and make a simple addition thing. However, I get a couple errors on Code Blocks:
error: invalid use of non-static member function 'int Addition::add(int, int)'
error: no matching function for call to 'Addition::add()'
Here's my code:
#include <iostream>
using namespace std;
class Addition {
public:
int add (int x, int y) {
int sum;
sum=x+y;
return sum;
}
};
int main()
{
int num1;
int num2;
int ans=addobj.add(num1,num2);
Addition addobj;
addobj.add(num1,num2);
cout<<"Enter the first number you want to add"<<endl;
cin>>num1;
cout<<"Enter the second number you want to add"<<endl;
cin>>num2;
cout<<"The sum is "<<ans<<endl;
}
One of the most important things, a developer should learn to do is to read compiler's messages. It's clear enough:
error: no matching function for call to 'Addition::add()'
Your function in your class is
int add (int x, int y)
it takes 2 arguments and you pass none:
addobj.add();
You have 2 options:
create and initialize x and y inside your main and pass them as arguments
make add without parameters, create x and y inside add's body, as their values are taken from user input.
In this case, as the function's name is add, I'd chose the first option:
declare int x, y; inside your main
read the user input inside the main (the part, where you use cin and cout)
pass the x and y as arguments to add like this: addobj.add( x, y );
store the result (if needed), like this: int result = addobj.add( x, y );
You declared a method add(int, int) that takes two integers as arguments; you have to supply those arguments when you call it. It would be nice to print the returned value, as well:
Addition addobj;
std::cout << addobj.add(1, 2) << std::endl;
Your add function takes two arguments, yet you call it with none, so no matching function could be found. You must call the function as it was declared, i.e.,
addobj.add(1, 2);
Your function takes two arguments and yet you call it without providing them. You need to provide the two integer arguments that your function requires. To be useful you should store the result too. Something like this
int a = 1;
int b = 2;
int result = addjobs.add(a,b);
int add (int x, int y=1)
int main ()
{
int result1 = add(5);
int result2 = add(5, 3);
result 0;
}
VS
int add (int x, int y)
int main ()
{
int result1 = add(5, 1);
int result2 = add(5, 3);
result 0;
}
What is the advantage of using the default function parameter, in term of execution speed, memory usage and etc? For beginner like me, I sometimes got confused before I realized this usage of default function parameter; isn't it coding without default function parameter made the codes easier to read?
Your add function is not a good example of how to use defaulted parameters, and you are correct that with one it is harder to read.
However, this not true for all functions. Consider std::vector::resize, which looks something like:
template<class T>
struct vector_imitation {
void resize(int new_size, T new_values=T());
};
Here, resizing without providing a value uses T(). This is a very common case, and I believe almost everyone finds the one-parameter call of resize easy enough to understand:
vector_imitation<int> v; // [] (v is empty)
v.resize(3); // [0, 0, 0] (since int() == 0)
v.resize(5, 42); // [0, 0, 0, 42, 42]
The new_value parameter is constructed even if it is never needed: when resizing to a smaller size. Thus for some functions, overloads are better than defaulted parameters. (I would include vector::resize in this category.) For example, std::getline works this way, though it has no other choice as the "default" value for the third parameter is computed from the first parameter. Something like:
template<class Stream, class String, class Delim>
Stream& getline_imitation(Stream &in, String &out, Delim delim);
template<class Stream, class String>
Stream& getline_imitation(Stream &in, String &out) {
return getline_imitation(in, out, in.widen('\n'));
}
Defaulted parameters would be more useful if you could supply named parameters to functions, but C++ doesn't make this easy. If you have encountered defaulted parameters in other languages, you'll need to keep this C++ limitation in mind. For example, imagine a function:
void f(int a=1, int b=2);
You can only use the given default value for a parameter if you also use given defaults for all later parameters, instead of being able to call, for example:
f(b=42) // hypothetical equivalent to f(a=1, b=42), but not valid C++
If there is a default value that will provide correct behavior a large amount of the time then it saves you writing code that constantly passes in the same value. It just makes things more simple than writing foo(SOME_DEFAULT) all over the place.
It has a wide variety of uses. I usually use them in class constructors:
class Container
{
// ...
public:
Container(const unsigned int InitialSize = 0)
{
// ...
}
};
This lets the user of the class do both this:
Container MyContainer; // For clarity.
And this:
Container MyContainer(10); // For functionality.
Like everything else it depends.
You can use it to make the code clearer.
void doSomething(int timeout=10)
{
// do some task with a timeout, if not specified use a reasonable default
}
Is better than having lots of magic values doSomething(10) throughout your code
But be careful using it where you should really do function overloading.
int add(int a)
{
return a+1;
}
int add(int a,int b)
{
return a+b;
}
As Ed Swangren mentioned, some functions have such parameters that tend to have the same value in most calls. In these cases this value can be specified as default value. It also helps you see the "suggested" value for this parameter.
Other case when it's useful is refractoring, when you add some functionality and a parameter for it to a function, and don't want to break the old code. For example, strlen(const char* s) computes the distance to the first \0 character in a string. You could need to look for another characted, so that you'll write a more generic version: strlen(const char* s, char c='\0'). This will reuse the code of your old strlen without breaking compatibility with old code.
The main problem of default values is that when you review or use code written by others, you may not notice this hidden parameter, so you won't know that the function is more powerful than you can see from the code.
Also, google's coding style suggests avoiding them.
A default parameter is a function parameter that has a default value provided to it. If the user does not supply a value for this parameter, the default value will be
used. If the user does supply a value for the default parameter, the user-supplied value is used.
In computer programming, a default argument is an argument to a function that a programmer is not required to specify. In most programming languages, functions may take one or more arguments. Usually, each argument must be specified in full (this is the case in the C programming language)
Advantages of using default parameter, as others have pointed out, is indeed the "clarity" it brings in the code with respect to say function overloading.
But, it is important to keep in mind the major disadvantage of using this compile-time feature of the language: the binary compatibility and default function parameter does not go hand in hand.
For this reason, it is always good to avoid using default params in your API/interfaces classes. Because, each time you change the default param to something else, your clients will need to be recompiled as well as relinked.
Symbian has some very good C++ design patterns to avoid such BC.
Default parameters are better to be avoided.
let's consider the below example
int DoThis(int a, int b = 5, int c = 6) {}
Now lets say you are using this in multiple places
Place 1: DoThis(1);
Place 2: DoThis(1,2);
Place 3: DoThis(1,2,3);
Now you wanted to add 1 more parameter to the function and it is a mandatory field (extended feature for that function).
int DoThis(int a, int x, int b =5, int c=6)
Your compiler throws error for only "Place 1". You fix that. What about other others?
Imagine what happens in a large project? It would become a nightmare to identify it's usages and updating it rightly.
Always overload:
int DoThis(int a) {}
int DoThis(int a, int b {}
int DoThis(int a, int b, int c) {}
int DoThis(int a, int b, int c, int x) {}