I've made a simple program that counts matrices, here's the code:
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int result[3] = {0,0,0};
int matrixc[3][6] = {
{0,0,0,0,0,1},
{0,0,0,1,1,1},
{1,0,1,0,0,1}
};
for(int x=0;x <3;x++)
{
for(int y=0;y < 6;y++)
{
result[x] += (matrixc[x][y] * pow(2,6-y));
}
cout << result[x] << endl;
}
}
The output is what I wanted, it is: 2,14,and 82.
But, when I delete the initialization in the integer array of result:
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int result[3]; //deleted initialization
int matrixc[3][6] = {
{0,0,0,0,0,1},
{0,0,0,1,1,1},
{1,0,1,0,0,1}
};
for(int x=0;x <3;x++)
{
for(int y=0;y < 6;y++)
{
result[x] += (matrixc[x][y] * pow(2,6-y));
}
cout << result[x] << endl;
}
}
I got odd outputs: 1335484418,32618, and 65617.
Would you like to explain me why would the output be different between an array with and without an initialization?
Actually, I don't want to initialize all result array, because I have a huge data of matrices.
Is it possible if I use std::vector without initializing all of the result array?
Would you like to explain me why would the output be different between an array with and without an initialization?
Without initialisation, automatic variables aren't initialised. They will have an indeterminate value, depending on what happened to be in the memory they occupy.
Actually, I don't want to initialize all "result" array, because I have a huge data of matrices.
You can zero-initialise the whole array, even if it's huge, like this:
int result[huge] = {};
although, if it is huge, then it shouldn't be an automatic variable. These are typically kept on the stack, which is typically not huge and liable to overflow if you put too much stuff on it.
Is it possible if I use std::vector without initializing all of the "result" array?
Yes, a vector will zero-initialise its elements by default.
Without the initialization, the result array contains undetermined values, i.e. values that could be anything that fits into an int.
As I mentioned in my comment if you do not initialize result it will have undetermined values. Later on your are then adding a value to an unknown value which will still be an unknown value. In this situation you need to initialize your data, can zero initialize like so:
int result[3] = {} ;
Would you like to explain me why would the output be different between an array with and without an initialization?
Seriously? If you don't initialize the array elements they are not initialized!
That means they contains junk data, so when you do result[x] += xxx; you are adding to junk, so you get more junk.
Actually, I don't want to initialize all "result" array, because I have a huge data of matrices.
Then you shouldn't rely on their initial value being zero.
You can do this:
int result[3] = { }; // initialize all elements to zero
Is it possible if I use std::vector without initializing all of the "result" array?
std::vector always initializes its members.
When you delete the initialization of the result array, those locations are initially set to unspecified, arbitrary values. You then add to this arbitrary value within the for loop, and end up those unexpected results. The results may be completely different the next time you run your program.
Also, since result[x] += ... reads from an uninitialized variable, deleting the initialization results in your code having undefined behavior.
If you switch over to using a vector you can zero initialize it as
std::vector<int> result(count); // zero initialized with `count` elements
or
std::vector<int> result; // contains zero elements
result.resize(count); // now it contains `count` elements, all set to 0
Related
I have a double pointer Array of a structure:
typedef struct Position{
int x;
int y;
} Position;
Position** array = (Position**)malloc(sizeof(Position*)*10); //10 elements
array[0] = (Position*)malloc(sizeof(Position*));
array[0]->x = 10;
array[0]->y = 5;
Can I calculate the length of set array and if so, how?
The normal way for arrays does not work :
int length = sizeof(<array>)/sizeof(<array>[0]);
Once you have dynamically allocated an array, there is no way of finding out the number of elements in it.
I once heard of some hacky way to obtain the size of a memory block, (msize) which would allegedly allow you to infer the size of the data within the block, but I would advice against any such weird tricks, because they are not covered by the standard, they represent compiler-vendor-specific extensions.
So, the only way to know the size of your array is to keep the size of the array around. Declare a struct, put the array and its length in the struct, and use that instead of the naked array.
As you marked the question as C++, I would suggest that you use std::vector, then, after you "allocated some memory" (or requested some memory to allocated by std::vector constructor or by using push_back, or resize), you can simply get the size back using by using std::vector::size.
typedef struct Position{
int x;
int y;
} Position;
std::vector<Position> array(10);
array[0].x = 10;
array[0].y = 5;
size_t size = array.size(); // will be 10
Having only a pointer to some memory block, you cannot defer the size of this memory block. So you cannot defer the number of elements in it.
For arrays of pointers, however, you could infer the number of elements in it under the following conditions:
make sure that every pointer (except the last one) points to a valid object.
for the last pointer in the array, make sure that it is always NULL.
Then you can derive the length by counting until you reach NULL.
Maybe there are some other similar strategies.
Solely from the pointer itself, however, you cannot derive the number of elements in it.
Old question, but in case someone needs it:
#include <stdio.h>
...
int main()
{
char **double_pointer_char;
...
int length_counter = 0;
while(double_pointer_char[length_counter])
length_counter++;
...
return 0;
}
I am basically looking for some sort of "dynamic" way of passing the size/length of an array to a function.
I have tried:
void printArray(int arrayName[])
{
for(int i = 0 ; i < sizeof(arrayName); ++i)
{
cout << arrayName[i] << ' ';
}
}
But I realized it only considers its bytesize and not how many elements are on the array.
And also:
void printArray(int *arrayName)
{
while (*arrayName)
{
cout << *arrayName << ' ';
*arrayName++;
}
}
This has at least printed me everything but more than what I expected, so it doesn't actually work how I want it to.
I reckon it is because I don't exactly tell it how big I need it to be so it plays it "safe" and throws me some big size and eventually starts printing me very odd integers after my last element in the array.
So I finally got this work around, yet I believe there is something better out there!:
void printArray(int *arrayName)
{
while (*arrayName)
{
if (*arrayName == -858993460)
{
break;
}
cout << *arrayName << ' ';
*arrayName++;
}
cout << '\n';
}
After running the program a few times I realized the value after the last element of the array that I have input is always: -858993460, so I made it break the while loop once this value is encountered.
include <iostream>
include <conio.h>
using namespace std;
// functions prototypes
void printArray (int arrayName[], int lengthArray);
// global variables
//main
int main ()
{
int firstArray[] = {5, 10, 15};
int secondArray[] = {2, 4, 6, 8, 10};
printArray (firstArray,3);
printArray (secondArray,5);
// end of program
_getch();
return 0;
}
// functions definitions
void printArray(int arrayName[], int lengthArray)
{
for (int i=0; i<lengthArray; i++)
{
cout << arrayName[i] << " ";
}
cout << "\n";
}
Thank you very much.
TL;DR answer: use std::vector.
But I realized it [sizeof()] only considers its bytesize and not how many elements are on the array.
That wouldn't be a problem in itself: you could still get the size of the array using sizeof(array) / sizeof(array[0]), but the problem is that when passed to a function, arrays decay into a pointer to their first element, so all you can get is sizeof(T *) (T being the type of an element in the array).
About *arrayName++:
This has at least printed me everything but more than what I expected
I don't even understand what inspired you to calculate the size of the array in this way. All that this code does is incrementing the first object in the array until it's zero.
After running the program a few times I realized the value after the last element of the array that I have input is always: -858993460
That's a terrible assumption and it also relies on undefined behavior. You can't really be sure what's in the memory after the first element of your array, you should not even be accessing it.
Basically, in C++, if you want to know the size of a raw array from within a function, then you have to keep track of it manually (e. g. adding an extra size_t size argument), because of the way arrays are passed to functions (remember, they "decay into" a pointer). If you want something more flexible, consider using std::vector<int> (or whatever type of objects you want to store) from the C++ standard library -- it has a size() method, which does exactly what you want.
1st try
When arrays are passed into functions they decay to pointers. Normally, using sizeof on an array would give you its size in bytes which you could then divide by the size in bytes of each element and get the number of elements. But now, since you have a pointer instead of an array, calling sizeof just gives you the size of the pointer (usually 4 or 8 bytes), not the array itself and that's why this fails.
2nd try
The while loop in this example assumes that your array ends with a zero and that's very bad (unless you really did use a zero as a terminator like null-terminated strings for example do). If your array doesn't end with a zero you might be accessing memory that isn't yours and therefore invoking undefined behavior. Another thing that could happen is that your array has a zero element in the middle which would then only print the first few elements.
3rd try
This special value you found lurking at the end of your array can change any time. This value just happened to be there at this point and it might be different another time so hardcoding it like this is very dangerous because again, you could end up accessing memory that isn't yours.
Your final code
This code is correct and passing the length of the array along with the array itself is something commonly done (especially in APIs written in C). This code shouldn't cause any problems as long as you don't pass a length that's actually bigger than the real length of the array and this can happen sometimes so it is also error prone.
Another solution
Another solution would be to use std::vector, a container which along with keeping track of its size, also allows you to add as many elements as you want, i.e. the size doesn't need to be known at runtime. So you could do something like this:
#include <iostream>
#include <vector>
#include <cstddef>
void print_vec(const std::vector<int>& v)
{
std::size_t len = v.size();
for (std::size_t i = 0; i < len; ++i)
{
std::cout << v[i] << std::endl;
}
}
int main()
{
std::vector<int> elements;
elements.push_back(5);
elements.push_back(4);
elements.push_back(3);
elements.push_back(2);
elements.push_back(1);
print_vec(elements);
return 0;
}
Useful links worth checking out
Undefined behavior: Undefined, unspecified and implementation-defined behavior
Array decay: What is array decaying?
std::vector: http://en.cppreference.com/w/cpp/container/vector
As all the other answers say, you should use std::vector or, as you already did, pass the number of elements of the array to the printing function.
Another way to do is is by putting a sentinel element (a value you are sure it won't be inside the array) at the end of the array. In the printing function you then cycle through the elements and when you find the sentinel you stop.
A possible solution: you can use a template to deduce the array length:
template <typename T, int N>
int array_length(T (&array)[N]) {
return N;
}
Note that you have to do this before the array decays to a pointer, but you can use the technique directly or in a wrapper.
For example, if you don't mind rolling your own array wrapper:
template <typename T>
struct array {
T *a_;
int n_;
template <int N> array(T (&a)[N]) : a_(a), n_(N) {}
};
You can do this:
void printArray(array<int> a)
{
for (int i = 0 ; i < a.n_; ++i)
cout << a.a_[i] << ' ';
}
and call it like
int firstArray[] = {5, 10, 15};
int secondArray[] = {2, 4, 6, 8, 10};
printArray (firstArray);
printArray (secondArray);
The key is that the templated constructor isn't explicit so your array can be converted to an instance, capturing the size, before decaying to a pointer.
NB. The wrapper shown isn't suitable for owning dynamically-sized arrays, only for handling statically-sized arrays conveniently. It's also missing various operators and a default constructor, for brevity. In general, prefer std::vector or std::array instead for general use.
... OP's own attempts are completely addressed elsewhere ...
Using the -858993460 value is highly unreliable and, in fact, incorrect.
You can pass a length of array in two ways: pass an additional parameter (say size_t length) to your function, or put a special value to the end of array. The first way is preferred, but the second is used, for example, for passing strings by char*.
In C/C++ it's not possible to know the size of an array at runtime. You might consider using an std::vector class if you need that, and it has other advantages as well.
When you pass the length of the array to printArray, you can use sizeof(array) / sizeof(array[0]), which is to say the size in bytes of the whole array divided by the size in bytes of a single element gives you the size in elements of the array itself.
More to the point, in C++ you may find it to your advantage to learn about std::vector and std::array and prefer these over raw arrays—unless of course you’re doing a homework assignment that requires you to learn about raw arrays. The size() member function will give you the number of elements in a vector.
In C/C++, native arrays degrade to pointers as soon as they are passed to functions. As such, the "length" parameter has to be passed as a parameter for the function.
C++ offers the std::vector collection class. Make sure when you pass it to a function, you pass it by reference or by pointer (to avoid making a copy of the array as it's passed).
#include <vector>
#include <string>
void printArray(std::vector<std::string> &arrayName)
{
size_t length = arrayName.size();
for(size_t i = 0 ; i < length; ++i)
{
cout << arrayName[i] << ' ';
}
}
int main()
{
std::vector<std::string> arrayOfNames;
arrayOfNames.push_back(std::string("Stack"));
arrayOfNames.push_back(std::string("Overflow"));
printArray(arrayOfNames);
...
}
Consider this piece of code.
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector <int *> test;
vector <int *> v;
int *a = new int;
int *b = new int;
*a = 1;
*b = 2;
v.push_back (a);
v.push_back (b);
for (int i = 0; i < 2; ++i)
{
int n = *v[i];
test.push_back (&n);
}
cout << *test[0] << " " << *test[1] << endl;
delete a;
delete b;
return 0;
}
The problem's statement is:
"Given this code, answer the following questions:
Why does "test" vector contain only 2's?
How can we change for loop to copy properly (only code inside for loop)?"
I couldn't answer any of these questions, so a little bit of help will be appreciated.
Thanks in advance.
That code introduces dangling pointers. The body of the loop looks like this:
{
int n = *v[i];
test.push_back (&n);
}
The local variable n loses scope as soon as the loop body ends, so the pointer &n is now a dangling pointer. If it happens that test contains only 2's, that's just what randomly came out of what is undefined behavior.
If you want to "properly" copy the data over to test, you can change the for loop body to this:
{
int* n = new int;
*n = *v[i];
test.push_back (n);
}
Please take the "properly" with a grain of salt...
You push two the same pointers to n into test array. n equals the last element of your first array. Note that after control flow exited the loop, all pointers to n become invalid. So, in fact your test array contains invalid pointers, not pointers to 2s.
You should create a copy of each integer:
int* n = new int(*v[i]);
test.push_back (n);
Note also that you have memory leak here. Each int created using new should be later destroyed using delete.
The first question is a trick question: The vector contains pointers to a variable that no longer exists, and dereferencing that could cause pretty much any output. I imagine on some machines and compilers it prints all 2s however.
I can't understand what the exercise is trying to do (why does it use vectors of pointers for example) so I can't really help with how to solve the problem.
One way you could do it is by making test store by value:
First change the test vector to vector <int> test;
Then change the push_back to something like test.push_back (n); and finally the print statements to remove the now-unneeded * operators.
EDIT for comment:
First, I'm suspect of this book: It shouldn't be demonstrating undefined behavior or raw pointers to single builtin types. But you can change your loop body if you want:
for (int i = 0; i < 2; ++i)
{
int* n = new int;
*n = *v[i];
test.push_back (&n);
}
Note that both this will cause a memory leak unless you later delete those pointers, a problem that storing by value eliminates.
1) I think that the premise of the question is faulty. The loop adds two elements to test, each contains the address of the automatic variable n, the scope of which is limited to the body of the loop. It's not guaranteed that n will be allocated the same memory location in both passes through the loop, but I suppose that it's likely that most compilers will reuse the same location in both passes.
Moreover, n is out of scope at the output statement. So referencing the pointers in test to those memory locations is undefined. Again, there's a good chance that they will still contain the values assigned in the loop.
So, only if the same location gets reused for n in the second pass of the loop and that location has not been overwritten at the time the output statement is executed, will the output be "2 2". There is no guarantee of either of these premises.
2) To get the output "1 2" without changing anything outside the loop, one could change the definition of n to int& n = *v[i], which would be a single character change from the given code, though the end result is rather strange.
A simpler solution would be to eliminate the temporary n and simply test.push_back(v[i]).
I am trying to create an array that generates random values, then assign a pointer to that array in order to use it in other functions.
Question 1: Is this the right approach?
Question 2: When I run the code below, my pointer function generates values inconsistent with what the actual array's value is. What am I doing wrong?
int size = 100;
int theray[size];
for(int i=0; i<size; i++)
{
theray[i] = (rand()%100);
}
//Output array
cout<<"The array: ";
for(int j=0; j<size; j++)
{
cout<<theray[j]<<" ";
}
cout<<endl;
int (*parray)[100] = &theray;
cout<<"The array pointer: ";
for(int k=0; k<size; k++)
{
cout<<*parray[k]<<" ";
}
Question 1: is this the right approach?
No. The right approach is to use std::vector<int> if size is not known at compile time1, and std::array<int, size> if it is2. There is no need for pointers here.
void foo(const std::vector<int>& v)
{
// do stuff with v
}
...
std::vector<int> v(size); // vector with size elements
// do something with v
// pass v to a function
foo(v);
Question 2: when I run the code below, my pointer function generates values inconsistent with what the actual array's value is. What am I doing wrong?
If you use C++ idioms you won't even encounter this problem, so I consider the question moot. However, in your case you have a problem of operator precedence: be explicit about applying de-reference * before access []:
cout<< (*parray)[k] << " ";
1 As shown in the example, you can use an std::vector as a fixed size array, where the size need not be known at runtime. Just bear in mind that it is possible to change it's size after construction.
2In your example, size is not a compile time constant so you cannot use std::array. However, if you had declared it as const int size = 100; then it would be considered a compile time constant.
Your code is a bit off in three ways. First, there is no need to use &theray. Array names already reference a memory address. You can simply assign the pointer to theray. Second, you're declaring an array of 100 pointers. Based on your description, it sounds like you just want one pointer that points to the array. Your declaration should just be int *parray instead of int *parray [100]. Finally, once you have a pointer to the array, you can access elements of the array the same way you would with the original array, only with the name of the pointer, instead of the name of the array. Try changing your last block of code (starting with the pointer declaration to this:
int *parray;
parray = theray;
cout<<"The array pointer: ";
for(int k=0; k<size; k++)
{
cout<<parray[k]<<" ";
}
Question 1
Is this the right approach?
Usually not. It depends on what you are trying to achieve.
For high level semantics you'd in most cases use std::vector<int> or, if the size is fixed and you are using C++11, std::array<int, size>. If you actually have to go down to the pointer level, you'd usually write it like this:
int *parray = theray;
cout<<"The array pointer: ";
for(int k=0; k<size; k++)
{
cout<<parray[k]<<" ";
}
This works because arrays will degrade to pointers, and the […] subscripts work on these pointers just like they work on the original arrays.
Question 2
When I run the code below, my pointer function generates values inconsistent with what the actual array's value is, what am I doing wrong?
*parray[k] gets interpreted as *(parray[k]) while you intend to use it as (*parray)[k].
Question 1: is this the right approach?
No. Use std::vector<> for arrays whose size can change dynamically (at run-time). Prefer avoiding pointers and manual memory management.
Question 2: when I run the code below, my pointer function generates values inconsistent with what the actual array's value is. What am I doing wrong?
First of all, the fact of creating pointers so you can pass the array to a function. This is not necessary. Here is how I would use classes from the C++ Standard Library to write that program (in C++11):
#include <vector>
#include <algorithm>
#include <iostream>
#include <iterator>
// Sample function that prints the vectors's content
void foo(std::vector<int> const& v)
{
copy(begin(v), end(v), std::ostream_iterator<int>(std::cout, " "));
}
int main()
{
// Populate the vector...
size_t sz = 10;
std::vector<int> v(sz);
generate(begin(v), end(v), [] () { return rand() % 100; });
// Pass it to a function...
foo(v);
}
I am attempting to write a template/class that has a few functions, but I'm running into what seems like a rather newbie problem. I have a simple insert function and a display values function, however whenever I attempt to display the value, I always receive what looks like a memory address(but I have no idea), but I would like to receive the value stored (in this particular example, the int 2). I'm not sure how to dereference that to a value, or if I'm just completely messing up. I know that vectors are a better alternative, however I need to use an array in this implementation - and honestly I would like to gain a more thorough understanding of the code and what's going on. Any help as to how to accomplish this task would be greatly appreciated.
Example Output (running the program in the same way every time):
003358C0
001A58C0
007158C0
Code:
#include <iostream>
using namespace std;
template <typename Comparable>
class Collection
{
public: Collection() {
currentSize = 0;
count = 0;
}
Comparable * values;
int currentSize; // internal counter for the number of elements stored
void insert(Comparable value) {
currentSize++;
// temparray below is used as a way to increase the size of the
// values array each time the insert function is called
Comparable * temparray = new Comparable[currentSize];
memcpy(temparray,values,sizeof values);
// Not sure if the commented section below is necessary,
// but either way it doesn't run the way I intended
temparray[currentSize/* * (sizeof Comparable) */] = value;
values = temparray;
}
void displayValues() {
for (int i = 0; i < currentSize; i++) {
cout << values[i] << endl;
}
}
};
int main()
{
Collection<int> test;
int inserter = 2;
test.insert(inserter);
test.displayValues();
cin.get();
return 0;
}
Well, if you insist, you can write and debug your own limited version of std::vector.
First, don't memcpy from an uninitialized pointer. Set values to new Comparable[0] in the constructor.
Second, memcpy the right number of bytes: (currentSize-1)*sizeof(Comparable).
Third, don't memcpy at all. That assumes that Comparable types can all be copied byte-by-byte, which is a severe limitation in C++. Instead:
EDIT: changed uninitialized_copy to copy:
std::copy(values, values + currentSize - 1, temparray);
Fourth, delete the old array when it's no longer in use:
delete [] values;
Fifth, unless the code is going to make very few insertions, expand the array by more than one. std::vector typically increases its size by a factor of 1.5.
Sixth, don't increment currentSize until the size changes. That will change all those currentSize-1s into currentSize, which is much less annoying. <g>
Seventh, an array of size N has indices from 0 to N-1, so the top element of the new array is at currentSize - 1, not currentSize.
Eighth, did I mention, you really should use std::vector.
This line is wrong:
memcpy(temparray,values,sizeof values);
The first time this line is run, the values pointer is uninitialized, so it will cause undefined behavior. Additionally, using sizeof values is wrong since that will always give the size of a pointer.
Another issue:
temparray[currentSize] = value;
This will also cause undefined bahavior because you have only allocated currentSize items in temparray, so you can only access indices 0 to currentSize-1.
There is also an error in your array access.
temparray[currentSize/* * (sizeof Comparable) */] = value;
Remember that arrays start at index zero. So for an array of length 1, you would set temparray[0] = value. Since you increment currentSize at the top of the insert function, you will need to do this instead:
temparray[currentSize-1] = value;