使用循环填充的数组* AND交换

Question

I did this program in class and I'm trying to recreate it for an exam coming up. The program is supposed to be an array[2][10] and is supposed to output numbers in this order:

        1 3 5 7 9 11 13 15 17 19
        0 2 4 6 8 10 12 14 16 18

I'm really lost on this and I could really use any help.

#include<iostream>
#include <string>
#include <cstring>
using namespace std;


void out(int n[2][10]);
void fillit(int n[2][10]);
int main(){
int nums[2][10];
fillit(nums);
out(nums);
}
void fillit(int n[2][10]){
n[0][0] = 1;
n[1][0] = 0;

for (int i = 1; i < 10; i++){
    n[0][i] = n[0][i] + 2;
    n[1][i] = n[0][i] + 2;
}
}
void out(int n[2][10]){
for (int r = 0; r <= 1; r++){
    for (int c = 0; c <= 9; c++){
        cout << n[r][c];
    }
    cout << endl;
}
}

Update I have the program successfully filling the array but now I need the program to swap row 1 with row 2 and then output the new array. Ie 0 2 4 6 8 10 12 14 16 18 1 3 5 7 9 11 13 15 17 19

Answer 1

void fillit(int n[2][10]){

 for (int i = 0; i < 10; i++){
    n[0][i] = (i * 2 ) + 1;
    n[1][i] = i * 2;
 }
}

Answer 2

#include<iostream>
#include <string>
#include <cstring>
using namespace std;


void out(int n[2][10]);
void fillit(int n[2][10]);
int main(){
int nums[2][10];
fillit(nums);
out(nums);
}
void fillit(int n[2][10]){
n[0][0] = 1;
n[1][0] = 0;

for (int i = 1; i < 10; i++){
    n[0][i] = n[0][i-1] + 2;
    n[1][i] = n[0][i-1] + 2;
}
}
void out(int n[2][10]){
for (int r = 0; r <= 1; r++){
    for (int c = 0; c <= 9; c++){
        cout << n[r][c];
    }
    cout << endl;
}
}

Answer 3

How about this, its shorter?

int nums[2][10];
for (int i = 0; i < 20; i++)
{
    nums[(i%2)^1][i/2] = i;
}

Answer 4

I noticed that second array elements are even numbers, and the first array corresponding elements are one bigger (and thus odd) ... so this answer accomplishes using ONLY addition. Might be easier to understand.

void fillit(int n[2][10])
{
    int even = 0; // start value

    for (size_t i = 0; i < 10; i++)
    {
       int odd  = even + 1;
       n[0][i] = odd;
       n[1][i] = even;  
       even += 2;
    }
}

---

I noticed that you tagged this problem with C++. Perhaps you should try vectors.

For small vectors, you simply declare with initial values in curly-braces, making it easy to define the matrix limits [2] and [10]. In this example, I initialized using easy to recognize values, so you can tell the 2x10 matrix has not yet been filled. Without this init, the values contained will be random noise.

std::vector<std::vector<int>> n {
   {  11,  12,  13,  14,  15,  16,  17,  18,  19, 20 }, // row 1
   {  21,  22,  23,  24,  25,  26,  27,  28,  29, 30 }  // row 2
   //  1    2    3    4    5    6    7    8    9  10   <-- col
}; 

Yes, you could write in the target value of your output, but then fillit() would not be needed.

For passing the 2x10 vector to fillIt(), remember that you need to mark the matrix as a reference. The following emphasizes that n[0] contains odd numbers, and n[1] contains 10 even numbers.

//                vector x vector        v--reference
void fillIt(std::vector<std::vector<int>>& n)
{
   int even = 0;
   for (size_t c = 0; c < 10; ++c) // row major {
      int odd = even + 1;
      n[0][c] = odd;
      n[1][c] = even;
      even += 2;
   }
}

For test, I recommend just passing the 2x10 vector to a new function "showIt()".

// do not modify--vvvvv                 do not copy--v
void vec2x10Show (const std::vector<std::vector<int>>& n)
{
   // header
   std::cout << "\nCOL->";
   for (int i=0; i<10; ++i) std::cout <<  std::setw(3) << i+1 << " ";

   std::cout << "\nr1:  "; //                              r  c
   for (int c=0; c<10; ++c) std::cout << std::setw(3) << n[0][c] << " ";

   std::cout << "\nr2:  "; //                              r  c
   for (int c=0; c<10; ++c) std::cout << std::setw(3) << n[1][c] << " ";

   std::cout << std::endl;
}

Note the hard coded '10' in each loop (a magic number). Using vectors, this magic number is not necessary because the vector includes this information (your next assignment).