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function that uses size of array??

P: n/a
I have a function that takes two pointers to an array. The first point
to the first element while the other points to the last element.
int nums[] = { 1, 2, 3, 4, 5, 7, 8, 9};
int* result;
int end = (sizeof(nums)/4)-1;
int s = 7;

result = ::myfind(nums, nums + end, s);

the implementation goes like:
int* myfind(int* arr_start, int* arr_end, int& s) {
int arr_size = (arr_end - arr_start)+1;
int* result = arr_end;

for (int i = 0; i < arr_size; i++) {
if (*(arr_start+i) == s)
{
result = arr_start+i;
}
}

return result;

}

I have been told that elements in an array does not necessary lie after
each other in memory. So if I wanted a char array or string array
instead I guess the above function would not work?

In that case how do I make sure that I access the right elements in the
right order in an array and that I can calculate its size with:

int arr_size = (arr_end - arr_start)+1;
Apr 26 '07 #1
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5 Replies


P: n/a
I have been told that elements in an array does not necessary lie
after each other in memory. So if I wanted a char array or string
array instead I guess the above function would not work?
For an array, they do. For a std::list, they don't.
The function would work for a std::vector.
In that case how do I make sure that I access the right elements in
the right order in an array and that I can calculate its size with:
If elements do not come in line, how do you know that *(++pFirst) will
point to the second element?

It's prefectly OK to use this for arrays. However, since we are 2007
and this is C++, please _do_ use std::vector instead.
Apr 26 '07 #2

P: n/a
Gernot Frisch wrote:
>I have been told that elements in an array does not necessary lie
after each other in memory. So if I wanted a char array or string
array instead I guess the above function would not work?

For an array, they do. For a std::list, they don't.
The function would work for a std::vector.
>In that case how do I make sure that I access the right elements in
the right order in an array and that I can calculate its size with:

If elements do not come in line, how do you know that *(++pFirst) will
point to the second element?

It's prefectly OK to use this for arrays. However, since we are 2007
and this is C++, please _do_ use std::vector instead.

ok but will I get the correct size of a std::list with integers if I use:
int* myfind(int* arr_start, int* arr_end, int& s) {
int arr_size = (arr_end - arr_start)+1;

does arr_size not only give the right result if elements are placed
after each other in memory?
Apr 26 '07 #3

P: n/a
desktop wrote:
:: Gernot Frisch wrote:
:::: I have been told that elements in an array does not necessary lie
:::: after each other in memory. So if I wanted a char array or string
:::: array instead I guess the above function would not work?
:::
::: For an array, they do. For a std::list, they don't.
::: The function would work for a std::vector.
:::
:::: In that case how do I make sure that I access the right elements in
:::: the right order in an array and that I can calculate its size with:
:::
::: If elements do not come in line, how do you know that *(++pFirst)
::: will point to the second element?
:::
::: It's prefectly OK to use this for arrays. However, since we are 2007
::: and this is C++, please _do_ use std::vector instead.
:::
:::
::
:: ok but will I get the correct size of a std::list with integers if I
:: use:
::
::
:: int* myfind(int* arr_start, int* arr_end, int& s) {
:: int arr_size = (arr_end - arr_start)+1;
::
::
::
:: does arr_size not only give the right result if elements are placed
:: after each other in memory?

Yes, but with std::list and std::vector you don't have to compute the size
yourself, just call the container's size() member function.

Also, you might not need the size at all, as the containers also provide
begin() and end() functions!
Bo Persson


Apr 26 '07 #4

P: n/a
On Apr 26, 10:52 am, desktop <f...@sss.comwrote:
I have a function that takes two pointers to an array. The first point
to the first element while the other points to the last element.

int nums[] = { 1, 2, 3, 4, 5, 7, 8, 9};
int* result;
int end = (sizeof(nums)/4)-1;
An integer on my system is not 4 bytes.
int s = 7;

result = ::myfind(nums, nums + end, s);

the implementation goes like:

int* myfind(int* arr_start, int* arr_end, int& s) {
int arr_size = (arr_end - arr_start)+1;
int* result = arr_end;

for (int i = 0; i < arr_size; i++) {
if (*(arr_start+i) == s)
{
result = arr_start+i;
}
}

return result;

}

I have been told that elements in an array does not necessary lie after
each other in memory. So if I wanted a char array or string array
instead I guess the above function would not work?

In that case how do I make sure that I access the right elements in the
right order in an array and that I can calculate its size with:

int arr_size = (arr_end - arr_start)+1;
What padding a particular compiler might give to some type is not your
concern.
sizeof() will return the size including the padding, if any.
Both arrays and std::vector hold elements in contiguous memory space.
The problem is you can't assume that an integer has 4 bytes, let the
compiler/platform decide that.
On mine, its 8 bytes (but i don't care).

The term 'end' does not depict the last element, it points to one past
the last element.
So once you've reached the end, you are no longer in the container.
back() usually returns the last element.

You are aware that your array has 8 elements, right? Its got no 6.
Replace all T's with int if templates disturb you.

#include <iostream>

template< typename T >
T* find_ptr(T* const pbegin, T* const pend, const T& t)
{
size_t size = pend - pbegin;
std::cout << "size = " << size;
std::cout << std::endl;
for(T* p = pbegin; p != pend; ++p)
{
if(t == *p)
return p;
}
return pend;
}

int main()
{
int nums[] = { 1, 2, 3, 4, 5, 7, 8, 9 };
// p_end is one past the end of container
int* const p_end = &nums[ sizeof(nums)/sizeof(int) ];

int* p_result = find_ptr(&nums[0], p_end, 7);
// if p_end wasn't returned, something was found
if( p_end != p_result )
{
std::cout << "found at: " << p_result;
std::cout << "\tvalue: " << *p_result;
std::cout << std::endl;
} else {
std::cout << "p_end reached, element not found.\n";
}
}

/*
size = 8
found at: 0x7fff7ef0e554 value: 7
*/

/* another way... pass the whole array by reference
template< typename T, const size_t Size >
T* find_ref(T (& array)[ Size ], const T& t)
{
size_t size = sizeof(array)/sizeof(T);
std::cout << "size = " << size;
std::cout << std::endl;
for(size_t i = 0; i < size; ++i)
{
if(t == array[i])
return &array[i];
}
// one past the last element
return &array[Size];
}
*/
Apr 26 '07 #5

P: n/a
On Thu, 26 Apr 2007 16:52:03 +0200, desktop <ff*@sss.comwrote in
comp.lang.c++:
I have a function that takes two pointers to an array. The first point
to the first element while the other points to the last element.
int nums[] = { 1, 2, 3, 4, 5, 7, 8, 9};
int* result;
int end = (sizeof(nums)/4)-1;
This is a very, very dumb thing to do. What if you compile this on a
platform where sizeof(int) is not 4? There are platforms in use today
where it is 1, 2, or 8, and quite possible others.

Much, much smarter to do:

int end = (sizeof nums / sizeof *nums);

Then if you decide to change nums from an array of int to an array of
short or long or double, you won't have to touch this line.
int s = 7;

result = ::myfind(nums, nums + end, s);

the implementation goes like:
int* myfind(int* arr_start, int* arr_end, int& s) {
int arr_size = (arr_end - arr_start)+1;
int* result = arr_end;

for (int i = 0; i < arr_size; i++) {
if (*(arr_start+i) == s)
{
result = arr_start+i;
}
}

return result;

}

I have been told that elements in an array does not necessary lie after
each other in memory. So if I wanted a char array or string array
instead I guess the above function would not work?
Who told you this? What are the qualifications of this person that
you should believe him/her? If the person actually said this, and you
are not just misinterpreting what he/she said, than the person is very
unqualified to be considered an authority on C++.

--
Jack Klein
Home: http://JK-Technology.Com
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Apr 27 '07 #6

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