How to printf “unsigned long” in C?

The Question :

362 people think this question is useful

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.

The Question Comments :

The Answer 1

527 people think this answer is useful

%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?

The Answer 2

33 people think this answer is useful
  • %lu for unsigned long
  • %llu for unsigned long long

The Answer 3

30 people think this answer is useful

For int %d

For long int %ld

For long long int %lld

For unsigned long long int %llu

The Answer 4

23 people think this answer is useful

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).

The Answer 5

9 people think this answer is useful
int main()
    unsigned long long d;

This will be helpful . . .

The Answer 6

9 people think this answer is useful

The format is %lu.

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

The Answer 7

8 people think this answer is useful

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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