Drawing a chess board with SDL2's SDL_RenderDrawRect - c++

Question

I am fairly new to SDL2 and as my first project I wanted to just create a chess board. This has proven to be harder than I thought.
I have tried lots of different ways to draw the fields of the chess board with SDL_RenderDrawRect, this is the current state:
#include <SDL2/SDL.h>
int main()
{
bool quit = false;
SDL_Event event;
SDL_Init(SDL_INIT_EVERYTHING);
SDL_Window *window = SDL_CreateWindow("Chess", SDL_WINDOWPOS_UNDEFINED,
SDL_WINDOWPOS_UNDEFINED, 680, 680, 0);
SDL_Renderer *render = SDL_CreateRenderer(window, -1,
SDL_RENDERER_ACCELERATED | SDL_RENDERER_PRESENTVSYNC);
for (int x = 0; x > 3; x++) {
for (int y = 0; x > 8; x++) {
SDL_Rect rect;
rect.x = x*10;
rect.y = y*10;
rect.w = 128;
rect.h = 128;
SDL_SetRenderDrawColor(render, 159, 84, 8, 255);
SDL_RenderFillRect(render, &rect);
}
}
SDL_RenderPresent(render);
while (!quit) {
SDL_WaitEvent(&event);
switch (event.type) {
case SDL_QUIT:
quit = true;
break;
}
}
SDL_Quit();
return 0;
}
No rectangles want to show up when I use SDL_RenderDrawRect in a loop. Any ideas why that is?
Cheers!

Answer 1

There are multiple errors in your for loops.
You have to change your conditions in your for loops. Both iteration variables(x and y) start with 0, but your condition to run the foor loop is that they are greater than 3 or 8, so they wont get executed. Change it to less than the value(< instead of >)
You have to change the iteration variable of your second for loop. You create y but the condition and the value change is for x
Your Rect is 128x128 big, but you multiply your x and y from your loops by 10, that means your Rectangles will overlap. You have to multiply it by at least 128.
If you look at a Chess Board, beginning at the top left and going from left to right, every second field is colored. To implement that you have to start in the top left corner, and iterate over every cell, and then jump to the next row. Every second field, you have to draw a Rect.
This is a slightly edited version of your algorithm:
int startPos = 0;
for (int y = 0; y < 8; y++) {
for (int x = startPos; x < 3; x+=2) {
SDL_Rect rect;
rect.x = x * 129;
rect.y = y * 129;
rect.w = 128;
rect.h = 128;
SDL_SetRenderDrawColor(render, 159, 84, 8, 255);
SDL_RenderFillRect(render, &rect);
}
startPos = 1 - startPos;
}
Explanation:
In the first row, the first field will be colored, because startPos is 0.
Since every row has the opposite start of its predecessor, we have to change startPos as soon as the last cell of a row is drawn.
Since we only need to draw every second cell, the x has to be increased by two every iteration
Hope this helped you a little bit

Answer 2

Although ur problem is solved, leaving this for someone who face such issue
(anyhow its quite easy to make the board, I wrote this code)
this will create a standard 8x8 chess board
void Create_Board() {
SDL_Rect Box = { 0,0,80,80}; COORD position = { 0 };
bool toggle_color = 0;
for (int i = 0; i < 8; i++)
{
for (int j = 0; j < 8; j++)
{
if (toggle_color)
SDL_SetRenderDrawColor(renderer, 0, 0, 0, 255);
else
SDL_SetRenderDrawColor(renderer, 255, 255, 255, 255);
SDL_RenderFillRect(renderer, &Box);
Box.x += 80;
if(j != 7) //needed for chess color pattern
toggle_color = !toggle_color;
}
Box.x = 0;
Box.y += 80;
}
}