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sprintf with long?

Holger Warzelhan

Hello,

I am trying to create a string with a variable of type long, using sprintf.

char command [20];
long ofs = 363080;

sprintf (command, "VAL=%d", ofs);
The output string I get is this:
VAL=-30136
which means that the long variable was converted into a short.
Does sprintf not support long types?

Thank you for your help!
Holger

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  • 0

    In C51, it is very important that you match the type of the actual parameters to the type of the format specifier; eg,
    %d is for a signed int;
    %u is for an unsigned int;
    %Ld is for a signed long;
    %Lu is for an unsigned long.

    I guess C166 is similar and you need to add the 'L' (not case-sensitive)

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  • 0

    In C51, it is very important that you match the type of the actual parameters to the type of the format specifier; eg,
    %d is for a signed int;
    %u is for an unsigned int;
    %Ld is for a signed long;
    %Lu is for an unsigned long.

    I guess C166 is similar and you need to add the 'L' (not case-sensitive)

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  • 0 in reply to Andrew Neil

    Thank you for your help!
    I only looked in a standard C-manual, where there's no such thing as an L.
    Of course I didn't think of checking the Keil manual first... sorry for that :)
    Holger

  • 0 in reply to Holger Warzelhan

    I only looked in a standard C-manual, where there's no such thing as an L

    Hmm... I think you need to throw that manual away and get a proper one!

    My ancient K&R (2nd Edition, 1988) says,
    "Between the % and the conversion character there may be ... an h if the integer is to be printed as a short, or l if as a long"

    It doesn't mention if the h and l are case-sensitive, but in C51 they certainly aren't (I prefer to use the uppercase L, as lowercase l looks too much like the digit 1 in many fonts)

    Note that K&R also says that printf, "will get confused, and you will get wrong answers, if there are not enough arguments or if they are the wrong type" (my emphasis)