Noob, trying to learn C++

[rrowan327]

Hi Folks,
I am hoping for a little bit of direction. The program below is just a test program to create arrays based on input of two numbers. The numbers are only 0 to 9 in each array. They can only be either even or odd number but not mixed. My issue currently is figuring out how to use a struct and a function that allows me to select each array without duplicating the function 5 times (ie: b3,b4,b5,b6,b7).
Any help will be greatly appreciated.
Thanks Rick


Output would be:
b6 = {4,6,8,0}
b7 = {7,9,1,3}

//test program to build array
#include <cstdio>
using namespace std;

struct B {
    int bit[4] = {0};
}b3, b4, b5, b6, b7;

void bLoad6 (int tnum, int bnum){
    for (int x = 0; x<4;x++)
    {
        b6.bit[x] = tnum == bnum ? tnum +=2 : tnum;
        b6.bit[x] = tnum > 9 ? tnum -=10 : tnum;
        b6.bit[x] = tnum == bnum ? tnum +=2 : tnum;
        tnum +=2;
    }
}

void bLoad7 (int tnum, int bnum){
    for (int x = 0; x<4;x++)
    {
        b7.bit[x] = tnum == bnum ? tnum +=2 : tnum;
        b7.bit[x] = tnum > 9 ? tnum -=10 : tnum;
        b7.bit[x] = tnum == bnum ? tnum +=2 : tnum;
        tnum +=2;
    }
}
//Main Program
int main()
{
    //Load array with 4 numbers
    bLoad6(2, 2);
    bLoad7(5, 5); //I don't want to have 5 simple functions. Just one

    // print array b6
    for (int x = 0; x<4;x++)
    {
       printf("bit6[%d] = %d\n", x+1, b6.bit[x]);
    }
    printf("\n\n");
    for (int x = 0; x<4;x++)
    {
       printf("bit7[%d] = %d\n", x+1, b7.bit[x]);
    }
    return 0;
} //End of Main

[lastchance]

@rrowan327,

Could you possibly have another go at explaining what you are trying to achieve.

The following code produces the same output as your program ... but I don't understand what you are asking.

#include <iostream>
using namespace std;

void load( int a[4], int base )
{
   for ( int i = 1; i <= 4; i++ ) a[i-1] = ( base + i * 2 ) % 10;
}


int main()
{
   int odd[4], even[4];
   load( even, 2 );
   load( odd , 5 );

   for ( int i = 0; i < 4; i++ ) cout << "bit6[" << i+1 << "] = " << even[i] << '\n';
   cout << "\n\n";
   for ( int i = 0; i < 4; i++ ) cout << "bit7[" << i+1 << "] = " << odd [i] << '\n';
}

bit6[1] = 4
bit6[2] = 6
bit6[3] = 8
bit6[4] = 0


bit7[1] = 7
bit7[2] = 9
bit7[3] = 1
bit7[4] = 3

[rrowan327]

hi lastchance,

That works great! Much easier than what I was trying to do. I was basically looking for a way for one function to build the 5 different arrays with different sequence of numbers. I never thought of the modulus operator. I have a Looong way to go. LOL

I mod you code slightly to this.

#include <iostream>
using namespace std;

void load( int a[4], int base )
{
   for ( int i = 1; i <= 4; i++ ) a[i-1] = ( base + i * 2 ) % 10;
}


int main()
{
   int b3[4], b4[4], b5[4], b6[4], b7[4];
   load( b3, 1 );
   load( b4, 3 );
   load( b5, 9 );
   load( b6, 2 );
   load( b7, 5 );

   for ( int i = 0; i < 4; i++ ) cout << "bit3[" << i+1 << "] = " << b3[i] << '\n';
   cout << "\n\n";
   for ( int i = 0; i < 4; i++ ) cout << "bit4[" << i+1 << "] = " << b4[i] << '\n';
   cout << "\n\n";
   for ( int i = 0; i < 4; i++ ) cout << "bit5[" << i+1 << "] = " << b5[i] << '\n';
   cout << "\n\n";
   for ( int i = 0; i < 4; i++ ) cout << "bit6[" << i+1 << "] = " << b6[i] << '\n';
   cout << "\n\n";
   for ( int i = 0; i < 4; i++ ) cout << "bit7[" << i+1 << "] = " << b7[i] << '\n';
}

Many Thanks
Rick

[benhart]

High complexity as this might result with a lot of errors among the code and unless you're using some external tools as checkmarx it might going to be a hard job to detect those.
Make sure you try to decrease the amount of complexity in the code.
Good luck.