Enum C++ Get by Index - c++

I was wondering in C++ if I have an enum can I access the value at the second index? For example I have
enum Test{hi, bye};
if I want 'hi', can I do something like Test[0], thanks.

Yes and no. If your Enum does not have explicit values then it is possible. Without an explicit values, enum values are given numeric values 0-N in order of declaration. For example ...
enum Test {
hi, // 0
bye // 1
}
This means that indexes just translates into a literal value.
Test EnumOfIndex(int i) { return static_cast<Test>(i); }
This of course does 0 validation at runtime and as soon as you add an explicit value it will break down. But it will work in the default scenario.

Unless specified otherwise, enums start numbering at 0, and increment by 1 each entry.
enum Test
{
hi, //0
bye, //1
count //2
}
You can cast an int to the type of the enum to get the value you want, such as:
(Test)0;
//or
Test(0);
Which lets you do things like:
for(int i = 0; i < count; i++)
{
DoSomething((Test)i);
}

Enumerations map names to values. In your case, (int)hi would have a value of 0, and (int)bye a value of 1. You can use a cast to get the value of hi:
int myInteger = 0;
Test myValue = (Test)myInteger;
Note, though, that myValue could be an invalid enum value if myInteger is out of range.

No, but you could cast from int
Test test = (Test)0;

Depends what you mean by "I want 'hi'".
If you mean you want the value, then you can get it by casting an int, as others have said.
Casting a numeric literal to enum type is usually pointless - if you know which value you're expecting, you can use the name. That way, the enum can change without breaking your code. I guess it's possible that something really weird is going on, where someone has created an enum, and documented what "3" means but not which enum value it is. But then you'd want to fix the API.
Casting an integer value known at runtime to enum might be helpful if you have serialized data. As long as you know it's in range of the enum, the result is defined.
If you mean you want the string "hi", then you can't have it. Unlike Java, in C++ the names of the values in enumerated types exist only at compile time, not at runtime, and only map in one direction.

Your best option might be something like this:
enum Test{hi = 0, bye};
Then you can simply refer to 'hi' with the number 0, and 'bye' with 1.
Although this really defeats the whole purpose of using an enumeration in the first place.

If you are excepting the value to returned as {Hi or bye} ,then you cannot get the value like that .
i would not suggest this to be done inorder to get the actual value but it can be used as hack
string return_value(int index)
{
string temp = "";
switch (index)
{
case 1: temp = "hi"
break;
case 2: temp = "bye";
break;
defualt :
break;
}
return temp;
}
typecasting to enum would again return the index but you can assign to some other enum variable

#include <iostream>
#define GENERATE_ENUM(ENUM) ENUM,
#define GENERATE_STRING(STRING) #STRING,
#define FOREACH_TEST(ID) ID(hi) ID(bye) ID(good)
enum TEST { FOREACH_TEST(GENERATE_ENUM) };
static const char * Test[] = { FOREACH_TEST(GENERATE_STRING) };
int main() {
printf("%s ",Test[0]);
printf("%s\n",Test[bye]);
for (int i=0; i<2; i++) printf("Test[%d] = %s\n", i, Test[i]); }
compile and run with: g++ enum-test.cpp -o enum-test; ./enum-test
output:
hi bye
Test[0] = hi
Test[1] = bye

Related

Correct way to convert an enum to value

I work on a project that has a lot of enums defined like this:
enum BundleSize
{
BUNDLE_SIZE_5 = 5,
BUNDLE_SIZE_10 = 10,
BUNDLE_SIZE_107 = 107,
};
I wrote a function to get the value from it like:
int convertBundleSizeEnumToVal(BundleSize b)
{
switch(b)
{
case BUNDLE_SIZE_5: return 5; break;
case BUNDLE_SIZE_10: return 10; break;
case BUNDLE_SIZE_107: return 107; break;
default: // handle appropriately and error out
}
}
I realize I don't really need a converter (method 1 and 2 both work below):
BundleSize b = getRandomBundleSize();
printf("The size is %d\n", convertBundleSizeEnumToVal(b)); // method 1
printf("The size is %d\n", b); // method 2
But I feel that method 1 is still the "correct" way to do this, because in the future something like this might come along:
enum BundleSize
{
BUNDLE_SIZE_5 = 5,
BUNDLE_SIZE_5_POINT_5 = 6,
BUNDLE_SIZE_10 = 10,
BUNDLE_SIZE_107 = 107,
};
Which now effectively breaks method 2 without an obvious runtime error, whereas method 1 will atleast catch the problem in the default switch case.
I'm curious about what people think is the right way to handle these conversions.
You can use the static_cast operator:
static_cast<int>(b)
IMO, this is "correct". You shouldn't be defining functions to convert every possible enum variant to an integer, especially with large enums.
Actually, you are using simple enum, so you can simply use your enum variable as a value.
For enum class (strict version of enum) you should use static_cast to convert your enum to value.
You can simplify your function to:
int convertBundleSizeEnumToVal(BundleSize b)
{
switch(b)
{
case BUNDLE_SIZE_5:
case BUNDLE_SIZE_10:
case BUNDLE_SIZE_107:
return b;
default: // handle appropriately and error out
}
}
Method 2, is broken because the underlying type of enum is not necessarily int, and so the format specifier %d may be wrong and behaviour of the program may be undefined.
You can cast the enum value to int first, and it will be correct as long as you don't use higher values than can be represented by int. You can use either static_cast or explicit cast. Or you can use an implicit conversion:
int value = getRandomBundleSize();
printf("The size is %d\n", b);
Regarding which you should use, well it depends on what you need. If you think that enum values added later should not be converted by convertBundleSizeEnumToVal but should result in an error, then method 1 is indeed superior.
If you wish all enum values to be converted, then the function is error prone, as you would have to remember to add a case each time a new value is added. In this case method 2 is superior.

How to display the enum value for something within a class?

I am currently building a poor version of the game "Battleship" and have to use an array of Enums to display the board. For my header I have created:
enum class PlayerPiece {
AIRCRAFT,
BATTLESHIP,
CRUISER,
SUBMARINE,
PATROL,
EMPTY,
};
class Board {
public:
PlayerPiece playerBoard[100];
PlayerPiece enemyBoard[100];
void reset();
void display() const;
};
When I get to my source code, I try displaying the board as numbers. As of right now the board is EMPTY after I run my reset command. But after I want to display the array, I get an error saying "no operator << matches these operands ....", I understand that means I need to overload the << command to display properly, but why doesn't it just display the '5' that was assigned? Isn't that the point of Enums? What I have so far is:
void Board::reset(){
for (int i = 0; i < 100; ++i){
playerBoard[i] = PlayerPiece::EMPTY;
enemyBoard[i] = PlayerPiece::EMPTY;
};
}
void Board::display() const{
for (int i = 0; i < 100; ++i){
cout << playerBoard[i] << endl; //
};
}
I have made other codes where I don't have to overload the << operator to display the number attached with ENUM. Am I missing something? Any help would be appreciated.
If you want to see the number associated with the scoped enum type, use a static_cast like this:
cout << static_cast<int>(playerBoard[i]) << endl;
Normal (unscoped) enums don't need this cast, as their types implicitly cast to int, or whatever underlying type you specified. That's probably why this hasn't happened to you before.
If you remove the class and write enum with Unscoped enumeration
enum PlayerPiece {
AIRCRAFT,
BATTLESHIP,
CRUISER,
SUBMARINE,
PATROL,
EMPTY,
};
You can print the number you wanted.
The difference between scoped and unscoped(from cplusplus.com):
Before C++11, all enums were just, basically, integers. And you could use them like that. It made it too easy to give bad values to functions expecting a restricted set of values. For example:
1
2
3
4
5
6
7
8
9
10
enum round_mode { round_half_up, round_half_down, round_bankers };
double round( double x, round_mode = round_half_up )
{
...
};
int main()
{
double x = round( 2.5, 42 );
}
Edit & Run
It compiles, but it isn't pretty.
With C++11, the compiler now knows all kinds of things about your enums, and doesn't let you blithely mix them with incorrect values.
Essentially, it promotes an enum to a first-class object -- it isn't just an integer.
The other issue is that the name for each enumeration bleeds into the containing scope. So the following would be a name conflict:
1
2
enum color_masks { red = 0xFF0000, green = 0x00FF00, blue = 0x0000FF };
int red = 0xFF0000;
You can't have both the identifier 'red' as an enum value and as an integer variable name in the same scope.
While the example here is contrived, it isn't too far off from things that happen all the time -- and that programmers have to take pains to avoid.
(Some identifier names are common. For example, 'max'. If you #include , there's a 'max' macro in there, which plays havoc if you also #include and try to use the 'max' function, or #include and try to find the numeric_limit ::max(). I know that's a macro problem, but it's the first name conflict I could come up with...)

what is the value of an array element if I don't assign it myself

what would be the result if I wrote this
int array1[2];
cout << array1[0] ;
and how can I do this pseudocode :
if array1[0] doesn't have a value then assign its value to 1
I'm using C++ on DevCPP
The elements of array are uninitialized, and it is undefined behaviour to read them before writing to them. Your program is ill-formed. There is no way to "check" for this; it is your responsibility to write a correct program.
The initial value of unassigned array values is undefined (unless the array element type is a class/struct, in which case the default constructor will be called for each array element). In your first example, the behavior is undefined since you have not initialized the array element before using it.
If you want to retain an "unassigned" status then you need to use a class that encapsulates this, for example using the nullable pattern for value types.
Consider using Boost.Optional: you'd declare the array as boost::optional<int> array1[2]; and then you can test if (array1[0]) to see if that particular element has a value.
There is one point that the answers I'm seeing thus far seem to have missed. It depends on where your array is defined.
If the array is local to a function, like:
int f() {
int array1[2];
cout << array1[0] ;
}
...then the other answers are correct: the content of array1 contains unspecified values, and your attempt to read the value and send it to cout gives undefined behavior.
On the other hand, you may have defined array1 as a global variable:
int array1[2];
int f() {
cout << array1[0];
}
In this case, the content of array1 is required to be initialized to 0 for any arithmetic type (or NULL for an array of pointers). In a case like this, writing out the value in array1[0] is perfectly fine and gives defined results -- it must be 0. In this case, there is no any way to tell whether an element of an array containing the value 0 has that value because it was automatically initialized to 0, or was assigned that value later.
If you really need to know whether a value has been written to a variable, it's possible to write a class that will do that:
template <class T>
class value {
T val;
bool assigned;
public:
value(T const init=T()) : assigned(false), val(init) {}
value &operator=(T const &t) {
assigned = true;
val = t;
}
operator T() { return val; }
bool was_assigned() { return assigned; }
};
// ...
value<int> array2[2];
if (!array2[0].was_assigned())
array2[0] = 1;
It's usually easier and more efficient to just define the type to always start out initialized to a known value, so you never really care about whether it's been assigned to or not though. In short, although this supports what you've asked for, my immediate reaction is that there's probably a better/cleaner way to accomplish your ultimate goal. Before you even consider using something like this, I'd strongly recommend stepping back from what you're doing, and trying to figure out if there's a better way to do it. My guess is that there is/will be (and if you can't find it, you might want to ask another question, telling us about why you're trying to do this, to see if somebody can see a more direct way to accomplish your goal).
As far I remember that depend on the OS
As other users said, you need to initialize a then use a for loop to test each value one by one and modify them, if they fulfill a condition, I leave you a C snippet:
/* Variable declaration and initialization to 0s (You can use another value as default )*/
int a[ 5 ] = { 0 };
/* If the array[ i ] has 0 as value */
for( i = 0; i < 5; i++){
if ( a[ i ] == 0 ){
a[ i ] = 1;
}
}
If you don't initialise the element yourself, the element will obtain the value from the memory location it is stored on now (and will most probably convert it to its data type). Consider this program :
#include <iostream>
using namespace std;
int foo(int A[])
{
cout << A[0] << A[1];
}
int main()
{
int array[2];
foo(array);
}
This will give the output 00.
But now consider this code :
int main()
{
int array[2];
cout << array[0] << array[1];
}
It will give some random integer output. This is so because the uninitialised array picks up the value stored on the memory location it now occupies. You can check its memory adress by &array[0] and print it in different data types for some thought provoking questions.
eg: cout << &array[0] << char(array[0]) << bool(array[0]) etc.

Advantage of array of enum over constant and macro? [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Should I use #define, enum or const?
Advantage and disadvantages of #defines vs. constants?
How enum will be more useful than #define and const.
memory and code visibilty point of view and readability point of view.
can I convert (type cast) enum to array of int, If I have taken all value within integer.
Example:
class MapPlt_clContainer {
public:
enum tagVectorType {
enVTAll = 0,
enVTPolygon,
enVTLines
}
};
tVoid MapPlt_clRender::vDrawLineClass( MapPlt_clContainer::tagVectorType* )
While calling function enum pass
vDrawLineClass( ClassArray_Group ); //Working
While calling array base address pass
int test[3] =
{
5,
6,
7,
};
vDrawLineClass( test); //Not Working
Error!!
Should it type cast it automatically? or it is compiler dependent. In my case it is giving error.
enum is a separate type unlike #define and the language (with the help of compiler) will help you ensure you are not mixing values of different types (even if they are of the same numerical value).
Additionally the value of an enum is available to the debugger whereas the original meaning of the #define is lost during the pre-processing time (before the code generation has even begun).
Type-casting an enum to an int is an automatic built-in process while the opposite conversion is trickier as not all the int values could be valid for your particular enum.
Modern compilers will also warn you if you have used all the possible enum's values in a switch statement that has no default clause, something that cannot be checked for #defines
If you are using an enum as an integer in C++ you have a smell. An enum defines a type, and only the values of that type should be used. (I realize this isn't enforced and the enum can be interpreted as an int, but with C++ it generally shouldn't).
Also, a big pet peeve of mine: Don't put "Type" in the name for an enum in C++. The values of an enum are not "types" (in the C++ sense of the word). As soon as you start doing template code, you will HATE all the enums with the word Type in their type name.
Also, any time you are trying to typecast in your design, you are doing it wrong. That is an awful smell in C++. You shouldn't have to do it, and you certainly shouldn't design it into your code (i.e. use it as a "feature").
Finally, this part:
int test[3] =
{
5,
6,
7,
};
vDrawLineClass( test); //Not Working
This is a straight up ugly hack. Do what you say, say what you do:
MapPlt_clContainer::tagVectorType test[3] =
{
MapPlt_clContainer::enVTAll,
MapPlt_clContainer::enVTPolygon,
MapPlt_clContainer::enVTLines
};
vDrawLineClass( test);
In addition to the points made in other answers, I would like to add the following:
If you have multiple types and you need to iterate on them, you will have to use an array of constants, which will be something like this:
const int states[] = {STATE_1,STATE_2, STATE_3, STATE_4 };
int numStates = sizeof(states)/sizeof(state[0]);
for (int i = 0; i < numStates; i++) {
// Do something with states[i]..
}
With enumerations, this can be simplified as
enum states{cState_1 = 0, cState_2, cState_3, cState_4, cNumStates};
for (int i = 0; i < numStates; i++) {
// do something with i
}

Why can't I increment a variable of an enumerated type?

I have a enumerated type StackID, and I am using the enumeration to refer to an index of a particular vector and it makes my code easier to read.
However, I now have the need to create a variable called nextAvail of type StackID. (it actually refers to a particular stackID ). I tried to increment it but in C++, the following is illegal:
nextAvail++;
Which sort of makes sense to me ... because there's no bounds checking.
I'm probably overlooking something obvious, but what's a good substitute?
I also want to link to this question.
I'm probably overlooking something obvious, but what's a good substitute?
Overloading operator++:
// Beware, brain-compiled code ahead!
StackID& operator++(StackID& stackID)
{
#if MY_ENUMS_ARE_CONTIGUOUS && I_DO_NOT_WORRY_ABOUT_OVERFLOW
return stackID = static_cast<StackID>( ++static_cast<int>(stackID) );
#else
switch(stackID) {
case value1 : return stackID = value2;
case value2 : return stackID = value3;
...
case valueN : return stackID = value1;
}
assert(false);
return stackID; // some compilers might warn otherwise
#endif
}
StackID operator++(StackID& stackID, int)
{
StackID tmp(stackID);
++stackID;
return tmp;
}
Because enumerations do not have to be contiguous. E.g. take this example:
enum Colors {
cRed, // = 0
cBlue, // = 1
cGreen = 3
}
What should happen in this scenario?
Colors color = cBlue;
Colors other = color++;
Should other be cGreen or should it be 2. In that case it's not a valid enumeration member anymore. What about this?
Colors color = cGreen;
Colors other = color++;
Should other be cRed (wrap around) or 4?
As you can see, being able to increment enumeration values introduces a whole lot of questions and complicates the simple mechanism that they intend to be.
If all you care about is the integer value being incremented, then simply cast to int and increment that.
Casting back and forth to/from int is of course the obvious solution, then you make clear that you understand that the addition is happening "outside" the enum:
nextAvail = static_cast<StackID>(static_cast<int>(nextAvail) + 1);
Why not store nextAvail as an int instead if you're going to do arithmetic operations on it?
Another option would be to wrap the enum in your own type and overload operator ++ for it (which also could wrap around or something for instance).
An enumeration is semantically supposed to represent a set of distinct related, values.
So you could have
enum Colour {RED, GREEN, BLUE};
But that should be equivalent to:
enum Colour {GREEN, BLUE, RED};
The problem is that if you increment an enum then those representations are not the same. GREEN++ in the first case is not the same as GREEN++ in the second.
Making your program dependent on the declaration of the enum is a recipe for disaster - maintainers may assume that the order of the enum doesnt matter, introducing many silent bugs.
Very Simple:
nextAvail = (StackID)(nextAvail + 1);
Enums are going to be type int, so you can cast them. Is this what you're trying to do?
int ndx = (int) StackID.SomeValue;
...
++ndx;
This is going to make someone very confused down the line, of course.
It occurs to me that you're using an enum where you should be using const, or even #define. enum is most appropriate when you have arbitrary values (where the exact value is not meaningful).
I've overloaded the ++/-- operator in this way:
enum STATE {STATE_1, STATE_2, STATE_3, STATE_4, STATE_5, STATE_6};
// Overload the STATE++ operator
inline STATE& operator++(STATE& state, int) {
const int i = static_cast<int>(state)+1;
state = static_cast<STATE>((i) % 6);
return state;
}
// Overload the STATE-- operator
inline STATE& operator--(STATE& type, int) {
const int i = static_cast<int>(type)-1;
if (i < 0) {
type = static_cast<STATE>(6);
} else {
type = static_cast<STATE>((i) % 6);
}
return type;
}
With respect to oprator++, $5.2.6/1 states- "The type of the operand shall be an arithmetic type or a pointer to a complete object type."
StackID does not fit the bill here. It is of enumeration type.
One option is like this
$5.7/1 - "For addition, either both operands shall have arithmetic or enumeration type, or one operand shall be a pointer to a completely defined object type and the other shall have integral or enumeration type."
enum Possibility {Yes, No, Maybe};
Possibility operator++(Possibility const& r){
return Possibility(r + 1); // convert r to integer, add 1, convert back to Enum
}
int main(){
Possibility p = Yes;
Possibility p1 = ++p;
}
I'm quite happy with this C plus C++ solution for a for loop incrementing an enum.
for (Dwg_Version_Type v = R_INVALID; v <= R_AFTER; v++)
=>
int vi;
for (Dwg_Version_Type v = R_INVALID;
v <= R_AFTER;
vi = (int)v, vi++, v = (Dwg_Version_Type)vi)
The other solutions here are not C backcompat, and quite large.