ref: 6a9a0cd8f6ee83e6fbd3424c337bacfb8e90502a
dir: /misc.c/
/*
* misc.c: Miscellaneous helpful functions.
*/
#include <assert.h>
#include <ctype.h>
#ifdef NO_TGMATH_H
# include <math.h>
#else
# include <tgmath.h>
#endif
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include "puzzles.h"
char MOVE_UI_UPDATE[] = "";
char MOVE_NO_EFFECT[] = "";
char MOVE_UNUSED[] = "";
void free_cfg(config_item *cfg)
{
config_item *i;
for (i = cfg; i->type != C_END; i++)
if (i->type == C_STRING)
sfree(i->u.string.sval);
sfree(cfg);
}
void free_keys(key_label *keys, int nkeys)
{
int i;
for(i = 0; i < nkeys; i++)
sfree(keys[i].label);
sfree(keys);
}
/*
* The Mines (among others) game descriptions contain the location of every
* mine, and can therefore be used to cheat.
*
* It would be pointless to attempt to _prevent_ this form of
* cheating by encrypting the description, since Mines is
* open-source so anyone can find out the encryption key. However,
* I think it is worth doing a bit of gentle obfuscation to prevent
* _accidental_ spoilers: if you happened to note that the game ID
* starts with an F, for example, you might be unable to put the
* knowledge of those mines out of your mind while playing. So,
* just as discussions of film endings are rot13ed to avoid
* spoiling it for people who don't want to be told, we apply a
* keyless, reversible, but visually completely obfuscatory masking
* function to the mine bitmap.
*/
void obfuscate_bitmap(unsigned char *bmp, int bits, bool decode)
{
int bytes, firsthalf, secondhalf;
struct step {
unsigned char *seedstart;
int seedlen;
unsigned char *targetstart;
int targetlen;
} steps[2];
int i, j;
/*
* My obfuscation algorithm is similar in concept to the OAEP
* encoding used in some forms of RSA. Here's a specification
* of it:
*
* + We have a `masking function' which constructs a stream of
* pseudorandom bytes from a seed of some number of input
* bytes.
*
* + We pad out our input bit stream to a whole number of
* bytes by adding up to 7 zero bits on the end. (In fact
* the bitmap passed as input to this function will already
* have had this done in practice.)
*
* + We divide the _byte_ stream exactly in half, rounding the
* half-way position _down_. So an 81-bit input string, for
* example, rounds up to 88 bits or 11 bytes, and then
* dividing by two gives 5 bytes in the first half and 6 in
* the second half.
*
* + We generate a mask from the second half of the bytes, and
* XOR it over the first half.
*
* + We generate a mask from the (encoded) first half of the
* bytes, and XOR it over the second half. Any null bits at
* the end which were added as padding are cleared back to
* zero even if this operation would have made them nonzero.
*
* To de-obfuscate, the steps are precisely the same except
* that the final two are reversed.
*
* Finally, our masking function. Given an input seed string of
* bytes, the output mask consists of concatenating the SHA-1
* hashes of the seed string and successive decimal integers,
* starting from 0.
*/
bytes = (bits + 7) / 8;
firsthalf = bytes / 2;
secondhalf = bytes - firsthalf;
steps[decode ? 1 : 0].seedstart = bmp + firsthalf;
steps[decode ? 1 : 0].seedlen = secondhalf;
steps[decode ? 1 : 0].targetstart = bmp;
steps[decode ? 1 : 0].targetlen = firsthalf;
steps[decode ? 0 : 1].seedstart = bmp;
steps[decode ? 0 : 1].seedlen = firsthalf;
steps[decode ? 0 : 1].targetstart = bmp + firsthalf;
steps[decode ? 0 : 1].targetlen = secondhalf;
for (i = 0; i < 2; i++) {
SHA_State base, final;
unsigned char digest[20];
char numberbuf[80];
int digestpos = 20, counter = 0;
SHA_Init(&base);
SHA_Bytes(&base, steps[i].seedstart, steps[i].seedlen);
for (j = 0; j < steps[i].targetlen; j++) {
if (digestpos >= 20) {
sprintf(numberbuf, "%d", counter++);
final = base;
SHA_Bytes(&final, numberbuf, strlen(numberbuf));
SHA_Final(&final, digest);
digestpos = 0;
}
steps[i].targetstart[j] ^= digest[digestpos++];
}
/*
* Mask off the pad bits in the final byte after both steps.
*/
if (bits % 8)
bmp[bits / 8] &= 0xFF & (0xFF00 >> (bits % 8));
}
}
/* err, yeah, these two pretty much rely on unsigned char being 8 bits.
* Platforms where this is not the case probably have bigger problems
* than just making these two work, though... */
char *bin2hex(const unsigned char *in, int inlen)
{
char *ret = snewn(inlen*2 + 1, char), *p = ret;
int i;
for (i = 0; i < inlen*2; i++) {
int v = in[i/2];
if (i % 2 == 0) v >>= 4;
*p++ = "0123456789abcdef"[v & 0xF];
}
*p = '\0';
return ret;
}
unsigned char *hex2bin(const char *in, int outlen)
{
unsigned char *ret = snewn(outlen, unsigned char);
int i;
memset(ret, 0, outlen*sizeof(unsigned char));
for (i = 0; i < outlen*2; i++) {
int c = in[i];
int v;
assert(c != 0);
if (c >= '0' && c <= '9')
v = c - '0';
else if (c >= 'a' && c <= 'f')
v = c - 'a' + 10;
else if (c >= 'A' && c <= 'F')
v = c - 'A' + 10;
else
v = 0;
ret[i / 2] |= v << (4 * (1 - (i % 2)));
}
return ret;
}
char *fgetline(FILE *fp)
{
char *ret = snewn(512, char);
int size = 512, len = 0;
while (fgets(ret + len, size - len, fp)) {
len += strlen(ret + len);
if (ret[len-1] == '\n')
break; /* got a newline, we're done */
size = len + 512;
ret = sresize(ret, size, char);
}
if (len == 0) { /* first fgets returned NULL */
sfree(ret);
return NULL;
}
ret[len] = '\0';
return ret;
}
int getenv_bool(const char *name, int dflt)
{
char *env = getenv(name);
if (env == NULL) return dflt;
if (strchr("yYtT", env[0])) return true;
return false;
}
/* Utility functions for colour manipulation. */
static float colour_distance(const float a[3], const float b[3])
{
return (float)sqrt((a[0]-b[0]) * (a[0]-b[0]) +
(a[1]-b[1]) * (a[1]-b[1]) +
(a[2]-b[2]) * (a[2]-b[2]));
}
void colour_mix(const float src1[3], const float src2[3], float p, float dst[3])
{
int i;
for (i = 0; i < 3; i++)
dst[i] = src1[i] * (1.0F - p) + src2[i] * p;
}
void game_mkhighlight_specific(frontend *fe, float *ret,
int background, int highlight, int lowlight)
{
static const float black[3] = { 0.0F, 0.0F, 0.0F };
static const float white[3] = { 1.0F, 1.0F, 1.0F };
float db, dw;
int i;
/*
* New geometric highlight-generation algorithm: Draw a line from
* the base colour to white. The point K distance along this line
* from the base colour is the highlight colour. Similarly, draw
* a line from the base colour to black. The point on this line
* at a distance K from the base colour is the shadow. If either
* of these colours is imaginary (for reasonable K at most one
* will be), _extrapolate_ the base colour along the same line
* until it's a distance K from white (or black) and start again
* with that as the base colour.
*
* This preserves the hue of the base colour, ensures that of the
* three the base colour is the most saturated, and only ever
* flattens the highlight and shadow to pure white or pure black.
*
* K must be at most sqrt(3)/2, or mid grey would be too close to
* both white and black. Here K is set to sqrt(3)/6 so that this
* code produces the same results as the former code in the common
* case where the background is grey and the highlight saturates
* to white.
*/
const float k = sqrt(3)/6.0F;
if (lowlight >= 0) {
db = colour_distance(&ret[background*3], black);
if (db < k) {
for (i = 0; i < 3; i++) ret[lowlight*3+i] = black[i];
if (db == 0.0F)
colour_mix(black, white, k/sqrt(3), &ret[background*3]);
else
colour_mix(black, &ret[background*3], k/db, &ret[background*3]);
} else {
colour_mix(&ret[background*3], black, k/db, &ret[lowlight*3]);
}
}
if (highlight >= 0) {
dw = colour_distance(&ret[background*3], white);
if (dw < k) {
for (i = 0; i < 3; i++) ret[highlight*3+i] = white[i];
if (dw == 0.0F)
colour_mix(white, black, k/sqrt(3), &ret[background*3]);
else
colour_mix(white, &ret[background*3], k/dw, &ret[background*3]);
/* Background has changed; recalculate lowlight. */
if (lowlight >= 0)
colour_mix(&ret[background*3], black, k/db, &ret[lowlight*3]);
} else {
colour_mix(&ret[background*3], white, k/dw, &ret[highlight*3]);
}
}
}
void game_mkhighlight(frontend *fe, float *ret,
int background, int highlight, int lowlight)
{
frontend_default_colour(fe, &ret[background * 3]);
game_mkhighlight_specific(fe, ret, background, highlight, lowlight);
}
static void memswap(void *av, void *bv, int size)
{
char tmpbuf[512];
char *a = av, *b = bv;
while (size > 0) {
int thislen = min(size, sizeof(tmpbuf));
memcpy(tmpbuf, a, thislen);
memcpy(a, b, thislen);
memcpy(b, tmpbuf, thislen);
a += thislen;
b += thislen;
size -= thislen;
}
}
void shuffle(void *array, int nelts, int eltsize, random_state *rs)
{
char *carray = (char *)array;
int i;
for (i = nelts; i-- > 1 ;) {
int j = random_upto(rs, i+1);
if (j != i)
memswap(carray + eltsize * i, carray + eltsize * j, eltsize);
}
}
void draw_rect_outline(drawing *dr, int x, int y, int w, int h, int colour)
{
int x0 = x, x1 = x+w-1, y0 = y, y1 = y+h-1;
int coords[8];
coords[0] = x0;
coords[1] = y0;
coords[2] = x0;
coords[3] = y1;
coords[4] = x1;
coords[5] = y1;
coords[6] = x1;
coords[7] = y0;
draw_polygon(dr, coords, 4, -1, colour);
}
void draw_rect_corners(drawing *dr, int cx, int cy, int r, int col)
{
draw_line(dr, cx - r, cy - r, cx - r, cy - r/2, col);
draw_line(dr, cx - r, cy - r, cx - r/2, cy - r, col);
draw_line(dr, cx - r, cy + r, cx - r, cy + r/2, col);
draw_line(dr, cx - r, cy + r, cx - r/2, cy + r, col);
draw_line(dr, cx + r, cy - r, cx + r, cy - r/2, col);
draw_line(dr, cx + r, cy - r, cx + r/2, cy - r, col);
draw_line(dr, cx + r, cy + r, cx + r, cy + r/2, col);
draw_line(dr, cx + r, cy + r, cx + r/2, cy + r, col);
}
char *move_cursor(int button, int *x, int *y, int maxw, int maxh, bool wrap,
bool *visible)
{
int dx = 0, dy = 0, ox = *x, oy = *y;
switch (button) {
case CURSOR_UP: dy = -1; break;
case CURSOR_DOWN: dy = 1; break;
case CURSOR_RIGHT: dx = 1; break;
case CURSOR_LEFT: dx = -1; break;
default: return MOVE_UNUSED;
}
if (wrap) {
*x = (*x + dx + maxw) % maxw;
*y = (*y + dy + maxh) % maxh;
} else {
*x = min(max(*x+dx, 0), maxw - 1);
*y = min(max(*y+dy, 0), maxh - 1);
}
if (visible != NULL && !*visible) {
*visible = true;
return MOVE_UI_UPDATE;
}
if (*x != ox || *y != oy)
return MOVE_UI_UPDATE;
return MOVE_NO_EFFECT;
}
/* Used in netslide.c and sixteen.c for cursor movement around edge. */
int c2pos(int w, int h, int cx, int cy)
{
if (cy == -1)
return cx; /* top row, 0 .. w-1 (->) */
else if (cx == w)
return w + cy; /* R col, w .. w+h -1 (v) */
else if (cy == h)
return w + h + (w-cx-1); /* bottom row, w+h .. w+h+w-1 (<-) */
else if (cx == -1)
return w + h + w + (h-cy-1); /* L col, w+h+w .. w+h+w+h-1 (^) */
assert(!"invalid cursor pos!");
return -1; /* not reached */
}
int c2diff(int w, int h, int cx, int cy, int button)
{
int diff = 0;
assert(IS_CURSOR_MOVE(button));
/* Obvious moves around edge. */
if (cy == -1)
diff = (button == CURSOR_RIGHT) ? +1 : (button == CURSOR_LEFT) ? -1 : diff;
if (cy == h)
diff = (button == CURSOR_RIGHT) ? -1 : (button == CURSOR_LEFT) ? +1 : diff;
if (cx == -1)
diff = (button == CURSOR_UP) ? +1 : (button == CURSOR_DOWN) ? -1 : diff;
if (cx == w)
diff = (button == CURSOR_UP) ? -1 : (button == CURSOR_DOWN) ? +1 : diff;
if (button == CURSOR_LEFT && cx == w && (cy == 0 || cy == h-1))
diff = (cy == 0) ? -1 : +1;
if (button == CURSOR_RIGHT && cx == -1 && (cy == 0 || cy == h-1))
diff = (cy == 0) ? +1 : -1;
if (button == CURSOR_DOWN && cy == -1 && (cx == 0 || cx == w-1))
diff = (cx == 0) ? -1 : +1;
if (button == CURSOR_UP && cy == h && (cx == 0 || cx == w-1))
diff = (cx == 0) ? +1 : -1;
debug(("cx,cy = %d,%d; w%d h%d, diff = %d", cx, cy, w, h, diff));
return diff;
}
void pos2c(int w, int h, int pos, int *cx, int *cy)
{
int max = w+h+w+h;
pos = (pos + max) % max;
if (pos < w) {
*cx = pos; *cy = -1; return;
}
pos -= w;
if (pos < h) {
*cx = w; *cy = pos; return;
}
pos -= h;
if (pos < w) {
*cx = w-pos-1; *cy = h; return;
}
pos -= w;
if (pos < h) {
*cx = -1; *cy = h-pos-1; return;
}
assert(!"invalid pos, huh?"); /* limited by % above! */
}
void draw_text_outline(drawing *dr, int x, int y, int fonttype,
int fontsize, int align,
int text_colour, int outline_colour, const char *text)
{
if (outline_colour > -1) {
draw_text(dr, x-1, y, fonttype, fontsize, align, outline_colour, text);
draw_text(dr, x+1, y, fonttype, fontsize, align, outline_colour, text);
draw_text(dr, x, y-1, fonttype, fontsize, align, outline_colour, text);
draw_text(dr, x, y+1, fonttype, fontsize, align, outline_colour, text);
}
draw_text(dr, x, y, fonttype, fontsize, align, text_colour, text);
}
/* kludge for sprintf() in Rockbox not supporting "%-8.8s" */
void copy_left_justified(char *buf, size_t sz, const char *str)
{
size_t len = strlen(str);
assert(sz > 0);
memset(buf, ' ', sz - 1);
assert(len <= sz - 1);
memcpy(buf, str, len);
buf[sz - 1] = 0;
}
/* Returns a dynamically allocated label for a generic button.
* Game-specific buttons should go into the `label' field of key_label
* instead. */
char *button2label(int button)
{
/* check if it's a keyboard button */
if(('A' <= button && button <= 'Z') ||
('a' <= button && button <= 'z') ||
('0' <= button && button <= '9') )
{
char str[2];
str[0] = button;
str[1] = '\0';
return dupstr(str);
}
switch(button)
{
case CURSOR_UP:
return dupstr("Up");
case CURSOR_DOWN:
return dupstr("Down");
case CURSOR_LEFT:
return dupstr("Left");
case CURSOR_RIGHT:
return dupstr("Right");
case CURSOR_SELECT:
return dupstr("Select");
case '\b':
return dupstr("Clear");
default:
fatal("unknown generic key");
}
/* should never get here */
return NULL;
}
char *make_prefs_path(const char *dir, const char *sep,
const game *game, const char *suffix)
{
size_t dirlen = strlen(dir);
size_t seplen = strlen(sep);
size_t gamelen = strlen(game->name);
size_t suffixlen = strlen(suffix);
char *path, *p;
const char *q;
if (!dir || !sep || !game || !suffix)
return NULL;
path = snewn(dirlen + seplen + gamelen + suffixlen + 1, char);
p = path;
memcpy(p, dir, dirlen);
p += dirlen;
memcpy(p, sep, seplen);
p += seplen;
for (q = game->name; *q; q++)
if (*q != ' ')
*p++ = tolower((unsigned char)*q);
memcpy(p, suffix, suffixlen);
p += suffixlen;
*p = '\0';
return path;
}
/*
* Calculate the nearest integer to n*sqrt(k), via a bitwise algorithm
* that avoids floating point.
*
* (It would probably be OK in practice to use floating point, but I
* felt like overengineering it for fun. With FP, there's at least a
* theoretical risk of rounding the wrong way, due to the three
* successive roundings involved - rounding sqrt(k), rounding its
* product with n, and then rounding to the nearest integer. This
* approach avoids that: it's exact.)
*/
int n_times_root_k(int n_signed, int k)
{
unsigned x, r, m;
int sign = n_signed < 0 ? -1 : +1;
unsigned n = n_signed * sign;
unsigned bitpos;
/*
* Method:
*
* We transform m gradually from zero into n, by multiplying it by
* 2 in each step and optionally adding 1, so that it's always
* floor(n/2^something).
*
* At the start of each step, x is the largest integer less than
* or equal to m*sqrt(k). We transform m to 2m+bit, and therefore
* we must transform x to 2x+something to match. The 'something'
* we add to 2x is at most floor(sqrt(k))+2. (Worst case is if m
* sqrt(k) was equal to x + 1-eps for some tiny eps, and then the
* incoming bit of m is 1, so that (2m+1)sqrt(k) =
* 2x+2+sqrt(k)-2eps.)
*
* To compute this, we also track the residual value r such that
* x^2+r = km^2.
*
* The algorithm below is very similar to the usual approach for
* taking the square root of an integer in binary. The wrinkle is
* that we have an integer multiplier, i.e. we're computing
* n*sqrt(k) rather than just sqrt(k). Of course in principle we
* could just take sqrt(n^2k), but we'd need an integer twice the
* width to hold n^2. Pulling out n and treating it specially
* makes overflow less likely.
*/
x = r = m = 0;
for (bitpos = UINT_MAX & ~(UINT_MAX >> 1); bitpos; bitpos >>= 1) {
unsigned a, b = (n & bitpos) ? 1 : 0;
/*
* Check invariants. We expect that x^2 + r = km^2 (i.e. our
* residual term is correct), and also that r < 2x+1 (because
* if not, then we could replace x with x+1 and still get a
* value that made r non-negative, i.e. x would not be the
* _largest_ integer less than m sqrt(k)).
*/
assert(x*x + r == k*m*m);
assert(r < 2*x+1);
/*
* We're going to replace m with 2m+b, and x with 2x+a for
* some a we haven't decided on yet.
*
* The new value of the residual will therefore be
*
* k (2m+b)^2 - (2x+a)^2
* = (4km^2 + 4kmb + kb^2) - (4x^2 + 4xa + a^2)
* = 4 (km^2 - x^2) + 4kmb + kb^2 - 4xa - a^2
* = 4r + 4kmb + kb^2 - 4xa - a^2 (because r = km^2 - x^2)
* = 4r + (4m + 1)kb - 4xa - a^2 (b is 0 or 1, so b = b^2)
*/
for (a = 0;; a++) {
/* If we made this routine handle square roots of numbers
* significantly bigger than 3 or 5 then it would be
* sensible to make this a binary search. Here, it hardly
* seems important. */
unsigned pos = 4*r + k*b*(4*m + 1);
unsigned neg = 4*a*x + a*a;
if (pos < neg)
break; /* this value of a is too big */
}
/* The above loop will have terminated with a one too big. So
* now decrementing a will give us the right value to add. */
a--;
r = 4*r + b*k*(4*m + 1) - (4*a*x + a*a);
m = 2*m+b;
x = 2*x+a;
}
/*
* Finally, round to the nearest integer. At present, x is the
* largest integer that is _at most_ m sqrt(k). But we want the
* _nearest_ integer, whether that's rounded up or down. So check
* whether (x + 1/2) is still less than m sqrt(k), i.e. whether
* (x + 1/2)^2 < km^2; if it is, then we increment x.
*
* We have km^2 - (x + 1/2)^2 = km^2 - x^2 - x - 1/4
* = r - x - 1/4
*
* and since r and x are integers, this is greater than 0 if and
* only if r > x.
*
* (There's no need to worry about tie-breaking exact halfway
* rounding cases. sqrt(k) is irrational, so none such exist.)
*/
if (r > x)
x++;
/*
* Put the sign back on, and convert back from unsigned to int.
*/
if (sign == +1) {
return x;
} else {
/* Be a little careful to avoid compilers deciding I've just
* perpetrated signed-integer overflow. This should optimise
* down to no actual code. */
return INT_MIN + (int)(-x - (unsigned)INT_MIN);
}
}
/* vim: set shiftwidth=4 tabstop=8: */