a functions that returns an array

[codepantry]

Environemtnt: Visual C++ on Windows Xp

I am new somewhat new to C++ and would like to some advise and
correction to my code below. I am simply trying to write function that
generates an array and returns it to the caller for output and further
use. I have read bits sugessting that thios can be done using pointers,
structs and or vectors but no workin solution yet. Thanks in
advance...

int generateInteger s(int searchList[])
/*This function generates a thousand integers held in an array*/
{
//searchList[1000];
for(int i=0; i<1000;++i)
{
searchList[i]=(i+1)*3;
}
return searchList;
}
int main()
{
int searchList[1000];
generateInteger s(searchList);

for(int i=0; i<1000;++i) {

cout << searchList << endl;
}

}

[Gianni Mariani]

co********@gmai l.com wrote:

Environemtnt: Visual C++ on Windows Xp

I am new somewhat new to C++ and would like to some advise and
correction to my code below. I am simply trying to write function that
generates an array and returns it to the caller for output and further
use. I have read bits sugessting that thios can be done using pointers,
structs and or vectors but no workin solution yet. Thanks in
advance...

int generateInteger s(int searchList[])
This defines a function that returns a SINGLE "int" by value. You're
passing in a pointer to an array of int (no size given).

Why are you returning an int ? The code below does initialize the array
except you're trying to return a pointer where you otherwise declared to
return an int by value.
/*This function generates a thousand integers held in an array*/
{
//searchList[1000];
for(int i=0; i<1000;++i)
{
searchList[i]=(i+1)*3;
}
return searchList;
}

This code would do what you're looking for.

void generateInteger s(int * searchList, unsigned length)
{
for(int i=0; i<length;++i)
{
searchList[i]=(i+1)*3;
}
}

The function above is more appropriate - it simply sets the values into
the array passed in.

// simply for determining the number of elements in an array....
template <typename w_Type, int w_N>
inline int CountElements( w_Type (&i_array)[ w_N ] )
{
return w_N;

}

int main()
{
int searchList[1000];
generateInteger s(searchList, CountElements( searchList ) );

for(int i=0; i<CountElements ( searchList );++i) {

cout << searchList[i] << endl;
}

}

However, if you REALLY want to return a value (lots of memory copies
that degrade performance and not a good thing to do), you can do this:

#include <vector>
std::vector<int > generateInteger s(int count)
{
std::vector<int > searchList(coun t);

for(int i=0; i<count;++i)
{
searchList[i]=(i+1)*3;
}
return searchList;
}
int main()
{
int count = 1000;
std::vector<int > searchList = generateInteger s(count);

for(int i=0; i<searchList.si ze(); ++i)
{
std::cout << searchList[i] << std::endl;
}
}

The example above does more like what you described. It returns a vector
and probably makes 2 copies before you're done with the process. You
really want to avoid that if you can.

Finally, you can use a vector passed in by reference to eliminate the
copies (similar to what you did in your original example).

#include <vector>
void generateInteger s(std::vector<i nt> & searchList)
{
for(int i=0; i<searchList.si ze();++i)
{
searchList[i]=(i+1)*3;
}
}
int main()
{
std::vector<int > searchList(1000 );
generateInteger s(searchList);

for(int i=0; i<searchList.si ze(); ++i)
{
std::cout << searchList[i] << std::endl;
}
}