I have a struct:
struct KeyPair
{
int nNum;
string str;
};
Let's say I initialize my struct:
KeyPair keys[] = {{0, "tester"},
{2, "yadah"},
{0, "tester"}
};
I would be creating several instantiations of the struct with different sizes. So for me to be able to use it in a loop and read it's contents, I have to get the number of elements in a struct. How do I get the number of elements in the struct? In this example I should be getting 3 since I initialized 3 pairs.
If you're trying to calculate the number of elements of the keys array you can simply do sizeof(keys)/sizeof(keys[0]).
The point is that the result of sizeof(keys) is the size in bytes of the keys array in memory. This is not the same as the number of elements of the array, unless the elements are 1 byte long. To get the number of elements you need to divide the number of bytes by the size of the element type which is sizeof(keys[0]), which will return the size of the datatype of key[0].
The important difference here is to understand that sizeof() behaves differently with arrays and datatypes. You can combine the both to achieve what you need.
http://en.wikipedia.org/wiki/Sizeof#Using_sizeof_with_arrays
sizeof(keys)/sizeof(*keys);
If you're trying to count the elements of the array, you can make a macro
#define NUM_OF(x) (sizeof(x)/sizeof(x[0]))
You mean the count of elements in keys ? In such a case you can use int n = sizeof(keys)/sizeof(keys[0]);
In C++ it is generally not possible to do this. I suggest using std::vector.
The other solutions work in your specific case, but must be done at compile time. Arrays you new or malloc will not be able to use those tricks.
If you're trying to calculate the number of elements of the keys array
you can simply do sizeof(keys)/sizeof(keys[0]).
This can not be a general good solution, due to structure padding.
Related
I create an array of size int arr[50]; but I will insert value in it during compile time , like my solution will insert 10 values in it after performing some function (different amount of values can come) , Now in second part of my program I have to loop through the array like it should iterate <= total values of array like in int arr[50] my program save 10 values , it should iterate to it only 10 times but how I can get that there is only 10 values in that array.
arr[50]=sum;
for (int ut=0; ut<=arr[100].length();ut++)
Though i know ut<=arr[100].length() is wrong , but its just assumption , that function will work if I solve condition in this way.
Edit:
I know we can use vector , but I am just looking that type of thing using array.
Thanks for response
First of all, the array you show is not a "Dynamic Array". It's created on the stack; it's an automatic variable.
For your particular example, you could do something like this:
int arr[50];
// ... some code
int elem_count = sizeof(arr) / sizeof(arr[0]);
In that case, the sizeof(arr) part will return the total size of the array in bytes, and sizeof(arr[0]) would return the size of a single element in bytes.
However, C-style arrays come with their share of problems. I'm not saying never use them, but keep in mind that, for example, they adjust to pointers when passed as function arguments, and the sizeof solution above will give you an answer other than the one you are looking for, because it would return sizeof(int*).
As for actual dynamically allocated arrays (where all what you have is the pointer to that array), declared as follows:
int *arr = new int[50];
// ... do some stuff
delete [] arr;
then sizeof(arr) will also give you the size of an int* in bytes, which is not the size you are looking for.
So, as the comments suggested, if you are looking for a convenient random access container where you want to conveniently and cheaply keep track of the size, use a std::vector, or even a std::array.
UPDATE
To use a std::array to produce equivalent code to that in your question:
std::array<int, 50> arr;
and then use it like a normal array. Keep in mind that doing something like arr[100] will not do any bounds checking, but at least you can obtain the array's size with arr.size().
like array.length in java is there any built in method in c++ to findout size of an array?
I know about length(). but it only works for strings only ...
And i tried this ...
int a[10];
a[0]=1;
a[1]=2;
print(sizeof(a)/size(a[0]))
but it gives output as 10 but is there a way getting only 2 as output
If you're using C++, don't use arrays, use std::vector instead (especially if you need the count of currently held items, not the container's capacity). Then you can write:
std::vector<int> vec;
vec.push_back(1);
vec.push_back(2);
printf("%d\n", vec.size());
int a[10];
declares an array of 10 ints; sure, you're only initialising the first two, but the other 8 are still there, they're just (probably) filled with junk at the moment.
To do what you want, you should use a std::vector instead. You can then do this:
std::vector<int> a;
a.push_back(1);
a.push_back(2);
std::cout << a.size() << std::endl; // prints 2
Arrays in C/C++ do not store their lengths in memory, so it is impossible to find their size purely given a pointer to an array. Any code using arrays in those languages relies on a constant known size, or a separate variable being passed around that specifies their size.
In an array of 10 ints, when it is declared, memory is allocated for 10 int values. even if you initialize just two, the rest of it contains some junk values and the memory remains allocated.
If you want the used size, your best bet is to use std::vector.
if you want to know the number of elements in an array you can do this
int array[3] = {0, 1, 2};
int arraylength = sizeof(array)/ sizeof(*array);
Sure. It's name is vector::size. It doesn't apply to C-style arrays, only to std::vector. Note that Java's Array class is also not a C-style array.
I want to implement a function that gets as a parameter a dimension "n" of an array of integers. This function also gets values "k_1, k_2, ..., k_n" defining the size of the array. Then this function will fill this n-dimensional array.
How do I implement this efficiently with C++?
For example for n = 3 I would use
vector < vector < vector < int > > > array;
But I don't know the dimension at compile time.
Use a one-dimensional array, and fake the other dimensions using multiplication of offsets for indexing, and you can pass the dimension sizes in by vector, i.e.
std::vector<int> create_md_array(const std::vector<int> & dimensions)
{
int size = std::accumulate(dimensions.begin(), dimensions.end(), 1, std::multiplies<int>());
return std::vector<int>(size);
}
You have a couple of choices. You can implement it yourself, basically multiplying the coordinates by the sizes to linearize the multi-dimensional address, and just have a simple std::vector<whatever> to hold the data.
Alternatively, you could use std::valarray and friends to accomplish the same. It has a set of classes that are specifically intended for the kind of situation you describe -- but they're used so rarely that almost nobody understands them. Writing the code yourself stands a good chance of being easier for most people to read and understand.
For a hw assignment, we are to code a reduce routine that looks like:
int reduce(long array[], int size)
//Where array is the array to reduce, and size is the size of the array.
Using STL. My initial thoughts were to create a set, put all items in the set with a comparison, but then I realized that the set I would create would never be usable since the function returns the size of the new set, but not the set itself to be used. So I'm not sure how I'd go about using the STL to reduce an array. Any thoughts? Thanks.
Edited:
Sorry, reduce is just to reduce the array into a sorted array without duplicates.
E.g. {4, 4, 2, 1} -> {1, 2, 4}
Sort the array using std::sort, then apply std::unique on it to remove duplicates. std::unique works only on sorted arrays. Just to simplify matters here is how you get begin and end of a native array:
long* begin = array;
long* end = array + size;
Once you have these two things, you can apply standard algorithms easily. Also, if you need to return the new size(used elements in the array):
long* end_after_unique = unique(...);
return end_after_unique - array;
std::map only allow a single entry and will sort the items for you automatically. The "second" value in your case is a don't care.
std::map<INT32,INT32> mymap;
mymap[i] = array[i];//inserts if not already present
im storing some settings for objects in an array. the id's of objects are used as the key. the id's start from 100000 and go up. if i was to input data for an object with id 100 000, would cpp automatical create 99999 blank key entries starting from 0?
Array size is determined when you create an array.
To access object at index 100 000 you need to have array of at least that size, which answers your question.
If the array is smaller you will access memory at
array begin address + (index*object
size)
which is not a good thing. E.g. the following will print some data but it is a data that are stored at that point in memory and it's outside of your array (not a good thing):
string arr[3];
cout << arr[5] << endl;
Assuming you are talking about standard array like:
string arr[10];
Array's size is specified when you compile it, for example you can't do:
string arr[]; // this will fail to compile, no size specified
But you do:
string arr[] = {"1","2","3"}; // array size is 3
string arr1[3]; // array size is 3
string arr2[3] = {"1"}; // array size is 3
If you want to map extra parameters for object you are better off using std::map like:
class person {};
std::map<person*,int> PersonScore;
This assumes that the additional parameters are not logically part of the object otherwise you would just add them to the object.
Maybe you want somthing along the lines of:
class ArrayPlus100k {
Item underlyingArray[NUM_ELEMENTS];
public:
Item& operator [] (int i) { return underlyingArray[i-100000]; }
// etc.
}
If you truely mean an array, and by key you mean index, then subtracting 100,000 from your index will provide you with a zero based array index. There will be no unused entries.
There may be a better container than a flat array. Choosing the right data structure depends on what you are trying to do. If you are storing objects using a key, you might want to use a std::map<key, value>.
What happens depends entirely on the data structure you choose to use. If you use a map, only the items you insert will take up space in memory. If you use new to allocate an actual array, then you will want to allocate only enough space for for the items you want to store. In that case, adjust your index by subtracting 100,000.
No, it will not create 0-99999, but rather start from 100000 to your array size.
For example, if you declare the following:
int arr[5];
Starting from arr[2], you can store up to arr[7].
I hope you understand...