On Wed, 19 Sep 2018 10:44:14 +0200, stecdose wrote: > HOW does a 16bit truncate limit to 16384bytes? > > b1 b0 (byte 1, byte 0) > b2 | | (byte 2) > | | | > decimal 81920 = hex 014000 decimal 65536 = hex FFFF size of size_t = 2 > bytes > > 16k is far less than 64k, why does it truncate to 16k? why not 64k if > greater than 64k?
TK Chia tried to allocate 81920 bytes (0x14000), while malloc accepts an unsigned short (16-bits) only. Hence only the lowest 16 bits of 0x14000 were actually fed to malloc(), ie. 0x4000, that is 16K. Mateusz -- FreeDOS is present on the USENET, too! alt.os.free-dos _______________________________________________ Freedos-devel mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/freedos-devel
