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I can never understand how to print unsigned long datatype in C.

Suppose unsigned_foo is an unsigned long, then I try:

  • printf("%lu\n", unsigned_foo)
  • printf("%du\n", unsigned_foo)
  • printf("%ud\n", unsigned_foo)
  • printf("%ll\n", unsigned_foo)
  • printf("%ld\n", unsigned_foo)
  • printf("%dl\n", unsigned_foo)

And all of them print some kind of -123123123 number instead of unsigned long that I have.

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7 Answers 7

674

%lu is the correct format for unsigned long. Sounds like there are other issues at play here, such as memory corruption or an uninitialized variable. Perhaps show us a larger picture?

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  • 20
    Oops, %lu worked this time. Thanks. Something else must have happened before and it didn't work.
    – bodacydo
    Commented Jul 9, 2010 at 4:52
  • 1
    @bodacydo: If you've got a bug, it might appear at semi-random... make sure your variable has a valid value before you try printing it.
    – Thanatos
    Commented Jul 9, 2010 at 4:54
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    Even if the variable is uninitialized, there should be no way printf reaches a point where it could print a minus sign when the format specifier was %lu. Technically it's undefined behavior but in reality the variable has an unpredictable value that gets passed to printf which printf then interprets as unsigned. I'm guessing bodacydo's original problem was flow reaching an incorrect printf call instead of the one intended... Commented Jul 9, 2010 at 6:24
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    @Anisha Kaul: %lu is a valid conversion specification, %ul is not. %lu, broken out is: % — starts a "conversion specification"; l — the length modifier, l means "[unsigned] long int"; u — the conversion specifier, u is for an unsigned int to be printed out as decimal. Because we gave the length modifier l, it then accepts an unsigned long int. The letters must be in that order: percent, length, conversion. (There are a few more options, such as width and precision, that you can add. See the man page, as it documents all this precisely!)
    – Thanatos
    Commented Feb 14, 2012 at 7:12
  • 2
    %ul will just print unsigned (with %u), and then the letter "l" verbatim. Just as "%uw" will print unsigned, followed by letter "w". % starts the convspec, u (or some other character, like d, s, c...) ends it.
    – Veky
    Commented Jun 11, 2013 at 3:27
61

For int %d

For long int %ld

For long long int %lld

For unsigned long long int %llu

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  • %lu for unsigned long
  • %llu for unsigned long long
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  • 26
    Yoda convention: "unsigned long" "(l)ong (u)nsigned" is.
    – Reb.Cabin
    Commented Dec 11, 2016 at 22:20
26

Out of all the combinations you tried, %ld and %lu are the only ones which are valid printf format specifiers at all. %lu (long unsigned decimal), %lx or %lX (long hex with lowercase or uppercase letters), and %lo (long octal) are the only valid format specifiers for a variable of type unsigned long (of course you can add field width, precision, etc modifiers between the % and the l).

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    %ld will work fine till the value of std::numeric_limits<unsigned long>::max()/2. Above that %ld will print wrong value(negative value). Commented Aug 22, 2013 at 9:28
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    @KaushikAcharya: No, above that it's UB. And it's arguably even UB before that, since printf is specified to require the exact correct argument types without the allowances that va_arg would have. Commented Dec 2, 2014 at 5:00
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The correct specifier for unsigned long is %lu.

If you are not getting the exact value you are expecting then there may be some problems in your code.

Please copy your code here. Then maybe someone can tell you better what the problem is.

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The format is %lu.

Please check about the various other datatypes and their usage in printf here

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int main()
{
    unsigned long long d;
    scanf("%llu",&d);
    printf("%llu",d);
    getch();
}

This will be helpful . . .

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