Reformatted Tiny 2048.c

gistfile1.txt

M[16],X=16,W,k;

main(){
T(system("stty cbreak"));
puts(W&1?"WIN":"LOSE");
}

K[]={2,3,1};

s(f,d,i,j,l,P){
for(i=4;i--;)
	for(j=k=l=0;k<4;)
		j<4?P=M[w(d,i,j++)],W|=P>>11,l*P&&(f?M[w(d,i,k)]=l<<(l==P):0,k++),l=l?P?l-P?P:0:l:P:(f?M[w(d,i,k)]=l:0,++k,W|=2*!l,l=0);
}

w(d,i,j){
return d?w(d-1,j,3-i):4*i+j;
}

T(i){
for(i=X+rand()%X;M[i%X]*i;i--);
i?M[i%X]=2<<rand()%2:0;
for(W=i=0;i<4;)s(0,i++);
for(i=X,puts("\e[2J\e[H");i--;i%4||puts(""))
printf(M[i]?"%4d|":"    |",M[i]);
W-2||read(0,&k,3)|T(s (1,K[(k>>X)%4]));
}
//[2048] reformated 
Original Version : https://gist.github.com/justecorruptio/9967738
by : Jay Chan

The reason why 3 characters are read is that arrow keys generate 3-byte sequences.

Key	bytes	Hex value
arrow left	0x1b, '[', 'D'	0x1b5b44
arrow right	0x1b, '[', 'C'	0x1b5b43
arrow up	0x1b, '[', 'A'	0x1b5b41
arrow down	0x1b, '[', 'B'	0x1b5b42

When the characters are read, the 3-byte sequence actually is reversed.
So for instance, in the last case ('arrow down'), k is assigned 0x425b1b.

k is then used to decide which of the K-array value to use.
k is right-shifted by 16 bits, reducing it to 0x42, so we have K [ 0x042 % 4 ], which is equivalent to K[2], the 3rd element of K[].

Thus, the statement after the read (0, &k, 3) evaluates to calling s (1, 1).
I wonder why K[] does not have 4 values, one for each arrow key.

What is this mate i dont understand this code can u pls help me? I think w,a,s,d doesnt work for example i put w but numbers doesnt went up why?