C++ multiple returns from single function - c++

Question

still pretty new to C++.
Had to write a function in class to count the number of each digit in a multi-dimensional array.
Now I didn't think you could return multiple int values from a single called function and then use all of these returns in a text based answer, so I attempted to return a different value depending on the value requested as parameter for each digit.
The code below is unnecessarily long and I'm still receiving the following errors.
main-1-3.cpp: In function 'int main()':
main-1-3.cpp:12:21: error: too few arguments to function 'int
count_numbers(int(*)[4], int)'
count_numbers(array);
^
main-1-3.cpp:7:12: note: declared here
extern int count_numbers(int array[4][4], int);
With a bit of debugging I could likely fix up these compile errors, but I feel like my method is extremely convoluted and was wondering if it were possible to call the function as:
count_number(array[4][4])
(Removing the need for the Q parameter) and then have count_numbersreturn all ten of the number values and the just output them as text like << ",3:" << three << instead of << ",8:" << count_numbers(array, 8) <<
Anyway, the function.cpp and main.cpp are below.
If anyone could point me in the right direction it would greatly appreciated. Just need to know the correct method so I can condense this code.
function.cpp
#include <iomanip>
#include <locale>
#include <sstream>
#include <string>
#include <iostream>
int count_numbers(int array[4][4], int Q)
{
int X=0;
int Y=0;
int zero=0;
int one=0;
int two=0;
int three=0;
int four=0;
int five=0;
int six=0;
int seven=0;
int eight=0;
int nine=0;
while(X<4)
{
if(array[X][Y]==0)
{
zero=zero+1;
}
if(array[X][Y]==1)
{
one=one+1;
}
if(array[X][Y]==2)
{
two=two+1;
}
if(array[X][Y]==3)
{
three=three+1;
}
if(array[X][Y]==4)
{
four=four+1;
}
if(array[X][Y]==5)
{
five=five+1;
}
if(array[X][Y]==6)
{
six=six+1;
}
if(array[X][Y]==7)
{
seven=seven+1;
}
if(array[X][Y]==8)
{
eight=eight+1;
}
if(array[X][Y]==9)
{
nine=nine+1;
}
Y++;
if(Y==4)
{
Y=0;
X++;
}
}
if(Q==0)
{
return zero;
}
if(Q==1)
{
return one;
}
if(Q==2)
{
return two;
}
if(Q==3)
{
return three;
}
if(Q==4)
{
return four;
}
if(Q==5)
{
return five;
}
if(Q==6)
{
return six;
}
if(Q==7)
{
return seven;
}
if(Q==8)
{
return eight;
}
if(Q==9)
{
return nine;
}
}
main.cpp
#include <iomanip>
#include <locale>
#include <sstream>
#include <string>
#include <iostream>
extern int count_numbers(int array[4][4], int);
int array[4][4] = { {1,2,3,4} , {1,2,3,4} , {1,2,3,4} , {1,2,3,4}};
int main()
{
count_numbers(array);
std::cout << ",0:" << count_numbers(array, 0) << ",1:" << count_numbers(array, 1) << ",2:" << count_numbers(array, 2) << ",3:" << count_numbers(array, 3) << ",4:" << count_numbers(array, 4) << ",5:" << count_numbers(array, 5) << ",6:" << count_numbers(array, 6) <<",7:" << count_numbers(array, 7) << ",8:" << count_numbers(array, 8) << ",9:" << count_numbers(array, 9) << std::endl;
}
PS. Ignore incorrect indentation its just from pasting to this site
PPS. Thanks for any assistance.
EDIT
Thank you "Vlad From Moscow" for the assistance.
My initial (terrible) code would have worked if I'd simply removed the (unintentional) count_number(array); call from main.cpp
However the for loop system proposed by Vlad allowed me to shrink the code by 80%. It now looks like this:
int count_numbers(int array[4][4], int Q)
{
int ans=0;
for(int i=0; i<4; i++)
{
for(int j=0; j<4; j++)
{
ans += array[i][j] ==Q;
}
}
return ans;
}
Thank you for the assistance which was great, I began coding on Python so it was my lack of understanding of loops in C++ which was the real problem here.
Anyway, problem solved..

Answer 1

This call
count_numbers(array);
does not make sense and moreover is invalid because the function requires two arguments instead of one.
The function itself can be written the following way
const size_t N = 4;
size_t count_number( const int ( &a )[N][N], int value )
{
size_t n = 0;
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < N; j++ )
{
n += a[i][j] == value;
}
}
return n;
}
and called like
std::cout << "0: " << count_number( array, 0 )
<< ", 1: " << count_number( array, 1 )
<< ", 2: " << count_number( array, 2 )
<< ", 3: " << count_number( array, 3 )
<< ", 4: " << count_number( array, 4 )
<< ", 5: " << count_number( array, 5 )
<< ", 6: " << count_number( array, 6 )
<< ", 7: " << count_number( array, 7 )
<< ", 8: " << count_number( array, 8 )
<< ", 9: " << count_number( array, 9 )
<< std::endl;
Here is a demonstrative program
#include <iostream>
const size_t N = 4;
size_t count_number( const int ( &a )[N][N], int value )
{
size_t n = 0;
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < N; j++ )
{
n += a[i][j] == value;
}
}
return n;
}
int main()
{
int array[N][N] =
{
{ 1, 2, 3, 4 } ,
{ 1, 2, 3, 4 } ,
{ 1, 2, 3, 4 } ,
{ 1, 2, 3, 4 }
};
bool first = true;
for ( int value = 0; value < 10; ++value )
{
size_t n = count_number( array, value );
if ( n )
{
std::cout << ( first ? first = false, "" : ", " )
<< value << ": " << n;
}
}
std::cout << std::endl;
}
Its output is
1: 4, 2: 4, 3: 4, 4: 4
A more general approach can look the following way
#include <iostream>
#include <iterator>
#include <algorithm>
template <typename InputIterator, typename T>
size_t count_number( InputIterator first,
InputIterator last,
const T &value )
{
size_t n = 0;
for ( ; first != last; ++first )
{
n += std::count( std::begin( *first ), std::end( *first ), value );
}
return n;
}
int main()
{
const size_t N = 4;
int array[N][N] =
{
{ 1, 2, 3, 4 } ,
{ 1, 2, 3, 4 } ,
{ 1, 2, 3, 4 } ,
{ 1, 2, 3, 4 }
};
bool first = true;
for ( int value = 0; value < 10; ++value )
{
size_t n = count_number( std::begin( array ), std::end( array ), value );
if ( n )
{
std::cout << ( first ? first = false, "" : ", " )
<< value << ": " << n;
}
}
std::cout << std::endl;
}
The program output will be the same as it is shown above.
If the array contains only digits that you need to count then the function can look like
const size_t N = 4;
void count_digits( const int ( &a )[N][N], size_t ( &digits)[10] )
{
for ( size_t &x : digits ) x = 0;
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < N; j++ )
{
++digits[a[i][j]];
}
}
}
To call the function you need to declare in main an array like for example
size_t digits[10];

Answer 2

you can return int[] or event better since you are using c++ return vector<int>.
In your function you can replace one, two, .. with something like
vector<int> result(10);
This will create a vector with 10 entries, all of them 0.
Then replace thing like three = three + 1 with result[3] = result[3] + 1 or even more condensed result[3]++
Then at the end you can write ... << ",0:" << result[0] << ... or even do a for loop like:
for (int i = 0; i < 10; ++i) count << "," << i << ":" << result[i];
You can do all this with int[] as well if that's required but you need to take care of allocating and deallocating the memory. vector will simplify your code a lot.