Hello,
14.7 is of type 'double'.
If you want to use it as float, then simply write 14.7f.
HCS08 might matter, as it offers an option to treat double as float.
Assuming you are using float with 32bits and double with 64bits, then your example will give following code.
if you do
foo==14.7
then left side is float, right side is double. According to ANSI the comparison is done with
(double)foo==14.7
(64bit comparison
while the one with the cast is performed on float.
You can easily see this from the assembly code:
6: int bar1(void) { return foo==14.7;}bar1:00000000 450000 LDHX #foo00000003 CD0000 JSR DLONG00000006 95 TSX 00000007 CD0000 JSR DCMP_RC0000000A 40 NEGA 0000000B 2D66 BMS *+104 ;abs = 0x00730000000D 6666 ROR 102,X0000000F 6666 ROR 102,X00000011 66A7 ROR 167,X00000013 082703 BRSET 4,0x27,*+6 ;abs = 0x001900000016 8C CLRH 00000017 5F CLRX 00000018 81 RTS 00000019 AE01 LDX #0x010000001B 8C CLRH 0000001C 81 RTS 7: int bar2(void) { return foo==(float)14.7;}bar2:0000001D 320002 LDHX foo00000020 653333 CPHX #0x333300000023 2608 BNE *+10 ;abs = 0x002D00000025 320000 LDHX foo00000028 65416B CPHX #0x416B0000002B 2703 BEQ *+5 ;abs = 0x00300000002D 8C CLRH 0000002E 5F CLRX 0000002F 81 RTS 00000030 AE01 LDX #0x0100000032 8C CLRH 00000033 81 RTS In the first case the float variable is converted to a double, then compared.
In the second case the compiler simply can do a binary compare.
As a side comment: testing on equality for float/double might something to do only carefully.
So the question is not if the the float representation of 14.7 is the same, but if 14.7f is represented the same way as 14.7 (double).
Hope this helps,
BK
