Optimize hot loops with direct access, LUTs, and closed-form formulas

1 files changed, 24 insertions, 9 deletions

diff --git a/src/osc/clipboard.c b/src/osc/clipboard.c
index a5ac59d..4f550ba 100644
--- a/src/osc/clipboard.c
+++ b/src/osc/clipboard.c
@@ -3,21 +3,36 @@
 #include <stdlib.h>
 #include <string.h>
 
-static int b64_val(char c)
+static const int8_t b64_lut[256] = {
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 62, -1, -1, -1, 63,
+    52, 53, 54, 55, 56, 57, 58, 59, 60, 61, -1, -1, -1, -1, -1, -1,
+    -1,  0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14,
+    15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, -1, -1, -1, -1, -1,
+    -1, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40,
+    41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, -1, -1, -1, -1, -1,
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
+    -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1,
+};
+
+static inline int b64_val(char c)
 {
-    if (c >= 'A' && c <= 'Z') return c - 'A';
-    if (c >= 'a' && c <= 'z') return c - 'a' + 26;
-    if (c >= '0' && c <= '9') return c - '0' + 52;
-    if (c == '+') return 62;
-    if (c == '/') return 63;
-    return -1;
+    return b64_lut[(unsigned char)c];
 }
 
 static bool b64_decode(const char *s, size_t n, char *out, size_t *out_len)
 {
     size_t pad = 0;
-    while (pad < n && s[n - pad - 1] == '=')
-        pad++;
+    if (n > 0 && s[n - 1] == '=') {
+        pad = (n > 1 && s[n - 2] == '=') ? ((n > 2 && s[n - 3] == '=') ? 3 : 2) : 1;
+    }
     n -= pad;
     if (pad > 2 || (pad && n % 4 + pad != 4))
         return false;