?How to pass row/col of 2D array into function

#include <stdio.h>
#define ROWS 5
#define COLS 10
void pop_arr(int r,int c,int a[r][c]);
void pr_arr(int r,int c,int a[r][c]);
void manipulate_arr(int r,int c,int a[r][c]);
int is_it_neven(int *val);
//https://otvet.mail.ru/question/211934162
//more logical, maybe, just to compare two neighboring rows by noneven elements
//counter, and switch them, if order is wrong
//i don't know syntaxys of passing row(out of 2D array)into function
//if i to calculate neven elements in all rows instead, once i switch two
//rows, that calulation stops making sense, as order will be different
//otherwise i need different containers to keep in every row of 2D array
//additional element, which contains counter of non even elements in row
//THAT zadacha is quite misleading, it forces me to think in wrong direction first!
//SOMETIME, LIKE 10 days ago i had the same dilemma, just left it as it was
//and totally forgot about it, can;t even think about what was that zadacha i got
//stuck with

//poka proga beret po 1 elementu stroki, proverjaet, esli on nechetnyj - summiruet!
//xochu, htob brala vsju stroku, i odnim maxom proverjala!

int main()
{
    int i,j,k,l, arr[ROWS][COLS],counter[ROWS];
    pop_arr(ROWS,COLS,arr);
    pr_arr(ROWS,COLS,arr);

    if(is_it_neven(&arr[0][3]))
        printf("it's even number!");

    //calculates how many non even numbers contains each row
    for(i=0;i<ROWS;i++)
    {
        counter[i]=0;
        for(j=0;j<COLS;j++)
        {
            if(is_it_neven(&arr[i][j]))//esli nechetnoe?
               (counter[i])++;
        }
    }
    //print how many non even numbers contains each row
    for(i=0;i<ROWS;i++)
    {
        printf("%d st row contains %d non even numbers\n",i,counter[i]);
    }

    return 0;
}
void pop_arr(int r,int c,int a[r][c])
{
    int i,j;
    for(i=0;i<r;i++)
    {
        for(j=0;j<c;j++)
        {
            a[i][j]= rand()%20;
        }
    }
}
void pr_arr(int r,int c,int a[r][c])
{
    int i,j;
    for(i=0;i<r;i++)
    {
        for(j=0;j<c;j++)
        {
            printf("%d\t",a[i][j]);
        }
        printf("\n");
    }
}
int is_it_neven(int *val)
{
    return (*val%2)? 1:0;
}