Explore the classic joy of Tetris with my C-coded implementation!
Discover the magic behind the falling blocks, intricate code, and timeless gaming experience.
Let's bring the nostalgia to life! 🚀
First, let's imagine the gameboard and its tetrimnoes as matrices:
After that, we can easily imagine how the code of the tetrominoes would be:
/*This Struct is used to save information about the tetromino*/
typedef struct {
char **boyut;
int en;
int boy;
int x;
int y;
int tmp_en;
} Parca;
/*get a tetromino*/
Parca eldeParca;
/*define matrices for each tetromino*/
const Parca parcalar[PARCALAR] = {
{(char *[]){(char[]){1,1,0},(char[]){1,1,0}}, 2, 2},
{(char *[]){(char[]){1,0,0}}, 1, 1},
{(char *[]){(char[]){1,1,0},(char[]){0,1,1},(char[]){0,0,0}}, 3, 2},
{(char *[]){(char[]){1,1,1},(char[]){0,1,0},(char[]){0,0,0}}, 3, 2},
{(char *[]){(char[]){1,1,1},(char[]){0,0,0},(char[]){0,0,0}}, 3, 1},
{(char *[]){(char[]){1,1,1},(char[]){1,0,0},(char[]){0,0,0}}, 3, 2},
};So, how much easy it is to check the full rows and overwrite the above ones?
void skorlama( int r, int c, char oyun[r][c])
{
int i, j, sum, count=0;
for(i=0 ; i<r ; i++){
sum = 0;
for(j=0 ; j<c ; j++) {
sum+=oyun[i][j];
}
if(sum == c ){
count++;
int l, k;
for(k = i;k >=1;k--)
for(l=0 ; l<c ; l++)
oyun[k][l]=oyun[k-1][l];
for(l=0 ; l<c ; l++)
oyun[k][l]=0;
}
}
skor += 100*count*i;
}Last bu not least, we should print or render the board and the tetromino.
As always, while 0s indicates empty areas 1s will indicate full ones.
However, The tetromino that the player contols should be displayed in a different way so we print it as a buffer.
oid printleme( int r, int c, char oyun[r][c])
{
char Buffer[r][c];
int i, j;
for ( i = 0; i < r; i++)
{
for ( j = 0; j < c; j++)
{
Buffer[i][j] = 0;
}
}
for(i = 0; i < eldeParca.en ;i++){
for(j = 0; j < eldeParca.en ; j++){
if(eldeParca.boyut[i][j])
Buffer[eldeParca.x+i][eldeParca.y+j] = eldeParca.boyut[i][j];
}
}
.
.
.
/*check main.c for the full code.*/
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