Nested loops creating pattern (not diamond) in C - c

Question

So for my assignment I have to take the inputs of length and width and print out patterns of "*" based on the inputs. The minimum height is 7 and only goes up by odd integers and width is any multiple of 6.
The basic format of the output using a height of 7 and width of 12:
************
************
*** ***
*** ***
*** ***
************
************
So basically the first and last 2 lines are straight through the entire width, with the odd numbered rows containing 3 asterisks followed by 3 spaces, until it reaches the end of the width. The even numbered rows start off with 3 spaces.
I've figured out how to print the first two lines using the following code:
do
{
printf("*");
++i;
}while(i<width);
printf("\n");
do
{
printf("*");
++j;
}while(j<=width);
printf("\n");
But for the life of me, I cannot come up with the correct way to use basic nested loops to print out the inside pattern. I asked a programmer friend who is unfamiliar with C but wrote up a basic program in Java. I don't know Java and have tried to translate it but notice some big discrepancies in the logic between the two languages that is causing me headaches. Here is his code:
// LOGGING
var consoleLine = "<p class=\"console-line\"></p>";
console = {
log: function (text) {
$("#console-log").append($(consoleLine).html(text));
}
};
// PATTERN PARAMETERS
var rows = 6;
var cols = 7;
// hard code a space so html respects it
var space = " "
console.log("cols: " + cols + " rows: " + rows);
for (y = 0; y < rows; ++y) {
var line = "";
for (x = 0; x < cols; ++x) {
// First two and last two rows do not have patterns and just print filled
if (y == 0 || y == 1 || y == rows - 1 || y == rows - 2) {
line += "*";
} else {
if (y % 2 == 0) {
// Even row
line += x % 6 < 3 ? "*" : space;
} else {
// Odd row
line += x % 6 >= 3 ? "*" : space;
}
}
}
console.log(line);
}
Please help me or point me in the right direction!! I've searched online but can't seem to find a solution that's worked yet!
Edit- forgot to mention that all "printf" uses can only print one character at a time... Such as a single *
Edit edit- I GOT IT WORKING!!!! Thank you all so, so much for your input and guidance! Here's what I have that is working perfectly:
for (y = 0; y < height; ++y)
{
printf("\n");
for (x = 0; x < width; ++x)
{
// First two and last two rows do not have patterns and just print filled lines
if (y == 0 || y == 1 || y == height - 1 || y == height - 2)
{
printf("*");
}
else
{
if (y % 2 == 0)
{
if(x%6<3)
{
printf("*");
}
else
{
printf(" ");
}
} else {
// Odd row
if(x%6>=3)
{
printf("*");
}
else
{
printf(" ");
}
}
}
}
printf("\n");

Answer 1

Write a function with 3 arguments n,a,b that prints n groups of 3 of each argument a and b alternately. You can call this function to print the 4 different kinds of lines. You can make a loop to print the middle section repeatedly. Have fun!

Answer 2

A simpler alternative:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv) {
int row, col, height = atoi(argv[1]), cols = atoi(argv[2]);
for (row = 0; row < height; row++) {
for (col = 0; col < cols; col++) {
putchar(row < 2 || row >= height - 2 ||
col % 6 / 3 == row % 2 ? '*' : ' ');
}
putchar('\n');
}
}