This appears to be a false positive. According to:
https://msdn.microsoft.com/en-us/library/9t02bbsx.aspx ...the warning is generated because "If the shift overflowed the 32-bit value, bits are lost." But that's what is actually intended in bitop (it's a 32-bit bit operations library). I believe the error gets flagged by MSVC because of this line in lua_bitop.c, line #96: #define BRET(b) lua_pushnumber(L, (lua_Number)(SBits)(b)); return 1; ...since 'b' resolves to (b << n), and lua_Number is a double, then line #116 ends up being this: static int bit_lshift(lua_State *L) { uint32_t b = barg(L, 1); uint32_t n = barg(L, 2) & 31; lua_pushnumber(L, (double)(int32_t)(b << n)); return 1; } So MSVC doesn't like that lua_pushnumber() line, apparently. But that's the intended behavior, because Lua numbers are doubles. -hadriel On Feb 7, 2015, at 4:54 PM, Bill Meier <wme...@newsguy.com> wrote: > Hadriel: > > MSVC2013 Code Analysis is giving the following warning: > > > ...\ws-git\epan\wslua\lua_bitop.c(116) : warning C6297: Arithmetic overflow: > 32-bit value is shifted, then cast to 64-bit value. Results might not be an > expected value. > > After quick look at the code, my reaction: > Uh Oh... Twisty maze of passages > > So I'm going to punt. > > Can you take a look (or redirect upstream) or whatever ? > > Thanks > > Bill ___________________________________________________________________________ Sent via: Wireshark-dev mailing list <wireshark-dev@wireshark.org> Archives: http://www.wireshark.org/lists/wireshark-dev Unsubscribe: https://wireshark.org/mailman/options/wireshark-dev mailto:wireshark-dev-requ...@wireshark.org?subject=unsubscribe
