Is there any way to convert a long into a string without losing the precision of the long integer? When I try to concatenate a long integer with a string it loses the precision.
Have you tried to use the formatfloat() instruction?
Long to String Conversion_Campbellsci论坛问答
If I try to use that it will convert the long into a float and it will lose precision.
Just prior to the release of the last OS the instruction FormatLong was added. It is not yet documented in CRBasic help file and does not show up in the instruction list, but it is in the OS and the CRBasic Precompiler (which means if you type it in correctly, it should compile if you have the most recent OS & precompiler). I have limited information on it and have done NO testing, so I am not fully certain how it works. Here is the info I do have :)
FormatLong uses the specifiers d or i in order to convert the long argument into a string decimal value,or x or X in order to convert the long argument into a string hexadecimal value, or o in order to convert the argument into a string base 8 (octal).
This site is a good reference for both the long and float conversions:
http://www.cplusplus.com/reference/clibrary/cstdio/fprintf/
(Note, referring to the above reference, we support the float specifiers with formatFloat (G,g,e,E,f) and the integer specifiers (i,d,o,x,X) with formatLong. We currently do not support char (c) or string (s) specifiers. We also support only one specifier in the format string.)
Note that for both FormatFloat and FormatLong, the
%[flags][width][.precision][length]specifier
can occur in the middle of the format string, as in "I am %d years old"
Here are some examples:
str(1) = formatLong(255,"%d")
str(2) = formatLong(255,"%X")
str(3) = formatLong(255,"%x")
str(4) = formatLong(255,"%#X")
str(5) = formatLong(255,"%#x")
str(6) = formatLong(255,"%5d")
str(7) = formatLong(255,"%05d")
str(8) = formatLong(255,"%05x")
str(9) = formatLong(255,"%05X")
str(10) = formatLong(255,"%o")
str(11) = formatLong(65,"I am %d years old")
str(12) = formatLong(255,"&H%x") '&HFF (255)
Public number as Long
number = str(12) '"&Hff" is 255
I hope this helps. Regards,
Dana W.
Thanks for the information Dana.
I have an unrelated question to this topic. Is there a way to calculate the checksum of a numeric array using the checksum function or any other function? It seems it only lets me calculate the checksum on a string.