*s=f[random()%2](*s++);*s=f[random()%2](*s++), s has its value modified and its prior value read to determine the new value:
*s=f[random()%2](*s++)
^^^*s=f[random()%2](*s++)
^if(i++) would be invalid because i is modified and it's prior value is read for a purpose other than determining the value to be stored. this is obviously not what the authors of the standard intended.
random() anyway so the point is int i=0;
i++ && i++ || i++ && i++i after the above fragment executes?
i++ I would say that the value of i is read once and used to determine both the new value of i and the value of the expression itself. This is OK with my interpretation.
int i = 0;
i++ && i++ || i++ && i++
^ ^ ^1 ^
0 Don't eval 2
i = 0
...
i is 1, incremented
i++ && i++ || i++ && i++
^ ^ ^ ^
0, ,/ i is 2, incremented
next skipped
i incremented
[code]
should be 3
New one:
[code]
i = 1
sizeof (i++) && i++ || i++ int count = 0;
int m = 0; /* Start with the first offset modifier. */
for (m = 0; count<4 && m < 8; m++)
{
if (!m%2) count = 1; /* The count is for one of the four directions. */
int xo = 0,yo = 0,i = 1; /* Offsets and incrementer. */
for (;(x+xo>=0) && (x+xo<WIDTH) /* Make sure we've not */
&& (y+yo>=0) && (y+yo<HEIGHT) /* gone off the grid. */
&& grid[x+xo][y+yo]==pl; /* Every single one needs to be ==pl. */
i++, xo=i*mod[m].x, yo=i*mod[m].y, /* Update offsets. */
count++);/* Increase count. */
}
&& grid[x+xo][y+yo]==pl; /* Every single one needs to be ==pl. */