I've been battling with figuring out how to pass a 2D array to a function and I think I've figured it out. My problem now though is for some reason this array (see below) is growing from 25 to 100 and I can't figure out why. I can't pinpoint where it's going haywire.
#include <iostream>
void testFunc(int (&n)[5][5]) {
n[0][0] = 5;
}
int main() {
int arr3[5][5];
// The array is initialized here with all values equaling 8.
for (int i = 0; i < 5; i++) {
for (int j = 0; j < 5; j++) {
arr3[i][j] = 8;
}
}
testFunc(arr3); // Function is called here changing [0][0] to the value 5.
for (int i = 0; i < 5; i++) {
for (int j = 0; j < 5; j++) {
std::cout << arr3[i][j] << ' ';
}
std::cout << '\n' << std::endl;
}
std::cout << sizeof(arr3) << std::endl;
return 0;
}
When I try to write the for-loop with i < sizeof(arr3) I get a size of 100. Not sure why. Where is it getting that value?
Your array size is 5*5 = 25, and int takes 4 bytes. so it becomes 100.
sizeofis not the number of elements. It's the size of the object in chars, that is, in bytes for most modern systems.
Your system has 32-bit (4 char/bytes) ints. Which gives you 5*5*4 = 100.
The size of data is also depend on size of CPU registers. In 32 bit machine the size of int is 4. However in earlier 16 bit machines the size of int is 2.
Related
#include <iostream>
using namespace std;
int* flipArray(int input[], int n)
{
int output[n];
int pos = 0;
for (int i = n-1; i >= 0; i--)
{
output[pos++] = input[i];
}
int* p = output;
for (int k = 0; k < n; k++)
cout << *p-k << endl << endl;
return p;
}
int main()
{
const int SIZE = 5;
int firstArray[SIZE];
for (int n = 0; n < SIZE; n++)
{
firstArray[n] = n+1;
}
int* a;
a = flipArray(firstArray, SIZE);
for (int j = 0; j < SIZE; j++)
cout << *a-j << endl;
cout << endl;
cout << *a << '\t' << *a+1 << '\t' << *a+2;
return 0;
}
I am attempting to flip firstArray using a function that returns a pointer, but I am struggling to understand how accessing an index using a pointer works.
Here is why I am confused:
Within the function flipArray, the following for-loop:
for (int k = 0; k < n; k++)
cout << *p-k << ' ';
prints "5 4 3 2 1" to the console. It was my understanding that I should be accessing an element of a vector with *(p+k), not *(p-k). If I print *(p+k), "5 6 7 8 9" is printed to the console. If I print the array without pointers and using k as the index location, "5 4 3 2 1" is printed to the console.
Within my main function, however, the values of *a which is assigned pointer p from the flipArray function, I do not get the same results:
for (int j = 0; j < SIZE; j++)
cout << *a-j << endl;
prints 5
0
-1
-2
-3 to the console, and
for (int j = 0; j < SIZE; j++)
cout << *a+j << endl;
prints 5
2
3
4
5 to the console.
Further, I thought that the pointer location of *p and the pointer of location of *a should be the same! But when I print the address &p in the function, I get the location of 0x28fde0, and when I print the address of &a in the main, I get the location 0x28fedc. Of course, these were done during the same run.
Could someone tell me where I have gone astray? Thanks!
Thanks to everyone for the informative answers.
I have updated my solution, and it is now returning what I would expect it to. I have a new question about memory leaks and when pointers need to be deleted.
int* flipArray(int input[], int n)
{
int* output = new int[n];
int pos = 0;
for (int i = n-1; i >= 0; i--)
output[pos++] = input[i];
return output;
}
int main()
{
const int SIZE = 5;
int firstArray[SIZE];
for (int n = 0; n < SIZE; n++)
{
firstArray[n] = n+1;
}
int* a;
a = flipArray(firstArray, SIZE);
for (int j = 0; j < SIZE; j++)
cout << a[j] << " "; // can also be written as *(a+j), which is more prone to bugs
delete [] a;
return 0;
}
Will the pointer output be deleted when the function flipArray returns? If not, how should I delete output while also returning it? Is deleting the pointer a in my main function the same thing as deleting output, because they point to the same location?
It has been pointed out that your main problem is coming from the operator precedence. The * operator in *p - k is evaluated before the -. This means that k will be subtracted from the value of the int pointed at by p.
This is a huge pain, which is why the braces pointer[k] are commonly used. There are situations where using pointer arithmetic *(pointer + k) makes more sense, but it can be a source of bugs.
One point to note here: it is always better to use parenthesis even if you are not sure whether or not you need them.
You do have a second problem:
Here you are declaring output on the stack as a local variable, then you are returning output. When you return back to the previous stack frame, this pointer will be pointing to a decallocated buffer:
int* flipArray(int input[], int n)
{
int output[n]; // allocated on the stack
int pos = 0;
for (int i = n-1; i >= 0; i--)
{
output[pos++] = input[i];
}
int* p = output;
for (int k = 0; k < n; k++)
cout << *p-k << endl << endl;
return p; // this stack frame ends.
}
This means the contents of the buffer can be overwritten if the space the buffer is using is reallocated. Use new to allocate on the heap:
int* output = new int[n];
make sure to call delete on the pointer when you are done using it.
This bug can even present security vulnerabilities in your applications, so make sure you know when to allocate on the heap in C++.
Update:
Question: When this function returns, the array still exists in memory, and it's location is stored in the pointer a. Does returning the value output delete it? If not, will deleting the pointer a when I am done with it in the main function serve the same purpose?
When you delete the pointer, the memory pointed to that pointer is deallocated and the pointer is left dangling. A reference to a deleted pointer is pointing at memory that is technically free, which is bad. If the allocator library decides that it wants to reuse that space, your buffer, which is now in free space, will be reallocated. This means your buffer will lose all data integrity and the data inside of it cannot be trusted.
A common practice is to assign pointers to NULL when you are done using them. This way your program will crash and you will know where your bug is:
int* p = new int[10];
...
delete p;
p = NULL;
...
p[0] = 0; // this will now crash because you are accessing NULL.
for (int k = 0; k < n; k++)
cout << *p-k ;
It was my understanding that I should be accessing an element of a vector with *(p+k), not *(p-k).
Your understanding is right.You are not accessing the array here.p points to the first element 5 and every time k is substracted from it, which gives you 5-0 5-1 5-2 and so on which is equivalent to the filpped array.So if you want to access the array using the pointer
for (int k = 0; k < n; k++)
cout << *(p+k) ;// Note the paranthesis
for (int k = 0; k < n; k++)
cout << *p-k << endl << endl;
What this code is doing is completely different from what you think it does.
*p - k will be processed like
*p = 5 - 0 = 5
*p = 5 - 1 = 4
and so on not *(p+k) or *(p-k)
For your understanding :
int a[5] = { 1,2,6,4,5};
In order to access 3rd element in the array you do
a[2] = *(a+2)
and not
*a + 2
*(a + 2) = 6 and *a + 2 = 1 + 2 = 3
Take care of not returning the pointer to the local variable which will lead to undefined behavior
I'm trying to create a magic square that will print four different grid sizes (5x5, 7x7, 9x9, 15x15). The error I'm having is the array magsquare within the function tells me it needs a constant integer. (I can't use pointers) This is a class assignment.
#include <iostream>
#include <iomanip>
using namespace std;
void magicSquare(int n){
int magsquare[n][n] = { 0 }; /*THIS is the error with [n][n]*/
int gridsize = n * n;
int row = 0;
int col = n / 2;
for (int i = 1; i <= gridsize; ++i)
{
magsquare[row][col] = i;
row--;
col++;
if (i%n == 0)
{
row += 2;
--col;
}
else
{
if (col == n)
col -= n;
else if (row < 0)
row += n;
}
}
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
cout << setw(3) << right << magsquare[i][j];
}
cout << endl;
}
}
int main(){
int n = 5;
magicSquare(n);
return 0;
}
Indentation may look incorrect, but it's right. Sorry.
The failure is because standard C++ cannot allocate dynamically sized array on the stack, as you are trying to do.
int magsquare[n][n];
As far as magicSquare is concerned n is only known at runtime and for an array to be allocated on the stack it's size must be known at compile time.
Use a 15 x 15 array.
int magsquare[15][15];
As long as you know this is the largest you'll ever need, you should be ok.
Alternatives (which you've already said you can't use)
Use new to declare a 2d array of the required dimensions. (Remember to delete[] it though)
Use std::vector
It may also be a good idea to add a check that n values over 15 or under 1 are rejected, otherwise you'll face undefined behaviour if any values outside of 1-15 are passed into the function.
I'm trying to fill an array with numbers 1111 to 8888, with each integer in the number being between 1 and 8 in c++. However, when I run it, it's only outputting large negative numbers indicating an error. I honestly have clue what the error is so it would be appreciated if you could help me out. Thanks!
int fillArray()
{
int arrayPosition;
int guesses[4096];
arrayPosition = 0;
for (int i = 1; i <= 8; i++)
for (int j = 1; j <= 8; j++)
for (int k = 1; k <= 8; k++)
for (int m = 1; m <= 8; m++)
{
guesses[arrayPosition] = ((i * 1000) + (j * 100) + (k *10) + m);
cout << guesses[arrayPosition];
arrayPosition++;
}
return guesses[4096];
}
Your return type is wrong. int fillArray(), but you're trying to return an int[4096] that was declared on the stack... What you're actually doing with return guesses[4096]; is returning the first memory location after your array in memory, which is probably just garbage, hence your issue with large negative numbers.
You can fix it by allocating your array in the heap, and returning a pointer to the start of that array:
int * fillArray()
{
int arrayPosition;
int * guesses = new int[4096];
// other stuff stays the same...
return guesses;
}
However, since your function is called fillArray, it would make more sense to pass in an array and fill it rather than creating the array in the function. (If you wanted to do that, might call it something like make_1_to_8_array instead, to make it more clear that you're constructing something that will need to be deleted later.) Giving an int* as the first argument would allow you to pass in the base address of your array that you want filled:
void fillArray(int * guesses)
{
int arrayPosition;
// other stuff stays the same...
}
Or, if you want to verify that the you're using an array of the exact size:
void fillArray(int (&guesses)[4096])
{
int arrayPosition;
// other stuff stays the same...
}
Note that the function now returns void since you just update the array that was passed in, and you don't need to return anything new.
Your for-loops look correct, but your array handling is off, as is highlighted by other answers.
It is more usual in C++ to use std::vector and to pass this in by reference as an argument. This saves you having to handle memory allocations and deallocations. Here's an example, including the output in the for-loops:
#include <iostream>
#include <vector>
int fillArray(std::vector<int>& guesses)
{
for (int i = 1; i <= 8; i++)
for (int j = 1; j <= 8; j++)
for (int k = 1; k <= 8; k++)
for (int m = 1; m <= 8; m++)
{
guesses.push_back((i * 1000) + (j * 100) + (k * 10) + m);
std::cout << guesses.back() << std::endl;
}
return guesses.back();
}
int main()
{
std::vector<int> guesses;
std::cout << fillArray(guesses) << std::endl;
}
You are creating your array locally then attempting to return it. If you try printing (to debug) out the result of your array prior to returning, you will see it is ok. However, once you return, the array is no linger valid. Try passing in an array into your function instead.
I'm having trouble understanding what the difference between these two code snippets is:
// out is of type char* of size N*D
// N, D are of type int
for (int i=0; i!=N; i++){
if (i % 1000 == 0){
std::cout << "i=" << i << std::endl;
}
for (int j=0; j!=D; j++) {
out[i*D + j] = 5;
}
}
This code runs fine, even for very big data sets (N=100000, D=30000). From what I understand about pointer arithmetic, this should give the same result:
for (int i=0; i!=N; i++){
if (i % 1000 == 0){
std::cout << "i=" << i << std::endl;
}
char* out2 = &out[i*D];
for (int j=0; j!=D; j++) {
out2[j] = 5;
}
}
However, the latter does not work (it freezes at index 143886 - I think it segfaults, but I'm not 100% sure as I'm not used to developing on windows) for a very big data set and I'm afraid I'm missing something obvious about how pointer arithmetic works. Could it be related to advancing char*?
EDIT: We have now established that the problem was an overflow of the index (i.e. (i*D + j) >= 2^32), so using uint64_t instead of int32_t fixed the problem. What's still unclear to me is why the first above case would run through, while the other one segfaults.
N * D is 3e9; that doesn't fit in a 32 bit int.
When using N as size of array, why use int?
does a negative value of an array has any logical meaning?
what do you mean "doesn't work"?
just think of pointers as addresses in memory and not as 'objects'.
char*
void*
int*
are all pointers to memory addresses, and so are exactly the same, when are defined or passes into a function.
char * a;
int* b = (char*)a;
void* c = (void*)b;
a == b == c;
The difference is that when accessing a, a[i], the value that is retrieved is the next sizeof(*a) bytes from the address a.
And when using ++ to advance a pointer the address that the pointer is set to is advanced by
sizeof(pointer_type) bytes.
Example:
char* a = 1;
a++;
a is now 2.
((int*)a)++;
a is now 6.
Another thing:
char* a = 10;
char* b = a + 10;
&(a[10]) == b
because in the end
a[10] == *((char*)(a + 10))
so there should not be a problem with array sizes in your example, because the two examples are the same.
EDIT
Now note that there is not a negative memory address so accessing an array with a signed negative value will convert the value to positive.
int a = -5;
char* data;
data[a] == data[MAX_INT - 5]
For that reason it might be that (when using sign values as array sizes!) your two examples will actually not get the same result.
Version 1
for (int i=0; i!=N; i++) // i starts at 0 and increments until N. Note: If you ever skip N, it will loop forever. You should do < N or <= N instead
{
if (i % 1000 == 0) // if i is a multiple of 1000
{
std::cout << "i=" << i << std::endl; // print i
}
for (int j=0; j!=D; j++) // same as with i, only j is going to D (same problem, should be < or <=)
{
out[i*D + j] = 5; // this is a way of faking a 2D array by making a large 1D array and doing the math yourself to offset the placement
}
}
Version 2
for (int i=0; i!=N; i++) // same as before
{
if (i % 1000 == 0) // same as before
{
std::cout << "i=" << i << std::endl; // same as before
}
char* out2 = &out[i*D]; // store the location of out[i*D]
for (int j=0; j!=D; j++)
{
out2[j] = 5; // set out[i*D+j] = 5;
}
}
They are doing the same thing, but if out is not large enough, they will both behave in an undefined manner (and likely crash).
I've just started learning C++ so I'm fairly sure the answer may be a simple one. As a test I'm just setting up an array and then wanting to print out the array by looping through it.
My code is below. It prints out my array as expected but then prints out a load of other numbers below it. What are these numbers and where are they coming from? I suspect that 'sizeof' isn't the best to use. All of the examples i've found are alot more complicated than I need. In any case I am interested to understand the extra numbers. Any insight available?
int age[4];
age[0]=23;
age[1]=34;
age[2]=65;
age[3]=74;
for (int i = 0; i <= sizeof(age); i++)
cout << age[i] << endl;
return 0;
...output:
23
34
65
74
4
2147307520
0
2293608
4198582
1
3084992
3085608
-1
2293592
1980179637
-725187705
-2
sizeof gives the size of an object in bytes. If the array elements are larger than one byte (as int usually is), the number will be larger than the array size.
One way to get the number of elements in an array is to divide by the size of an element:
for (size_t i = 0; i < sizeof(age)/sizeof(age[0]); i++)
std::cout << age[i] << '\n';
(note that you also need < rather than <=, or you'll still step off the end).
Another way is to pass a reference to the array to a function template, instantiated for the array size:
template <typename T, size_t size>
void print(T (&array)[size])
{
for (size_t i = 0; i < size; ++i)
std::cout << array[i] << '\n';
}
print(age);
Yet another way is to use a std::vector or std::array instead of a plain array:
std::array<int, 4> age;
age[0]=23;
age[1]=34;
age[2]=65;
age[3]=74;
for (size_t i = 0; i < age.size(); ++i)
std::cout << age[i] << '\n';
sizeof(age) == sizeof(int) * number_of_elements ==>
number_of_elements = sizeof(age) / sizeof(int)
Then your code becomes:
for (int i = 0; i < sizeof(age)/sizeof(age[0]); ++i)
cout << age[i] << endl;
In C++ you may write a function to calculate the size for you(doesn't work with dynamic arrays):
template <class T, std::size_t size>
std::size_t array_size( T(&arr)[size] )
{
return size;
}
If you are up to C++11, you could go with for-each loop:
for(int element : age){
....
}
Also, free-function form of std::begin and std::end can do the job:
for(auto b = std::begin(age); b != std::end(age); ++b){
....
}
Sould be i < 4 because sizeof(age) is 16 on a 32 bits machine.
sizeof(age) is the number of bytes of age, not the number of elements.
Divide it by the size of an element of age to get that:
for (int i = 0; i < sizeof(age) / sizeof(*age); i++)
cout << age[i] << endl;
Note: for dynamic arrays, you have to store the size of the array separately:
std::size_t size = 4; // size_t corresponds to maximum size an array can hold
int* age = new int[size];
for (int i = 0; i < size; i++)
cout << age[i] << endl;
The other numbers are garbage past the end of the array.
age[10] is undefined behavior, which is essentially garbage numbers.
Sizeof is age array is 16 bytes I.e. Sizeof(int) * 4. You need array length.
Since sizeof(age) returns 16, you have your 4 values plus 12 ones whose value comes from the memory that is right after your array. Values in those memory segment is random, depending on what has been stored there right before you launched your program. If you used a memory check tool, you would have had an error since this memory is probably not allocated for your program.
As the other ones said, you should probably giving the exact number of element in your array as an additional variable.
const int COUNT = 4;
int age[COUNT];
age[0]=23; age[1]=34; age[2]=65; age[3]=74;
for (int i = 0; i < COUNT; ++i)
{
cout << age[i] << endl;
}