Fix undefined behavior caused by left-shifting signed 1LL by 63

According to the C11 standard (ISO/IEC 9899:2011, 6.5.7), left-shifting
a signed value into the sign bit results in undefined behavior if the
result is not representable in the type. Specifically, the expression
1LL << 63 invokes undefined behavior because 1LL is a signed long long,
and 2^63 exceeds its positive range.

Replace 1LL << 63 with 1ULL << 63 to ensure the shift is performed on
an unsigned value, which is well-defined.

Author: visitorckw

app/test64.c

@@ -47,7 +47,7 @@ void task(void)
     }
 
     printf("test signed divide...\n");
-    c = stuff.num | 1LL << 63;
+    c = stuff.num | 1ULL << 63;
     for (i = 0; i < 20; i++) {
         stuff.unum = c / d;
         printf("%x%x / %x%x = %x%x\n", (uint32_t) (c >> 32),
@@ -64,15 +64,15 @@ void task(void)
         stuff.unum <<= 1;
     }
 
-    stuff.unum = 1LL << 63;
+    stuff.unum = 1ULL << 63;
 
     printf("testing right (logical) shifts..\n");
     for (i = 0; i < 64; i++) {
         printf("%x %x\n", stuff.l.high, stuff.l.low);
         stuff.unum >>= 1;
     }
 
-    stuff.unum = 1LL << 63;
+    stuff.unum = 1ULL << 63;
 
     printf("testing right (arithmetic) shifts..\n");
     for (i = 0; i < 64; i++) {