How to printf "unsigned long" in C?

Question

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.

Answer 1

First of all, you have to realize that these are just bits stored in the computer, and they are INTERPRETED however you say it should be...

In the below example, I took your example (-123123123), and stored it in an unsigned long long variable foo.

Then, when I tried to print it out, it was interpreted as an unsigned long long (18446744073586428493), which is the exact same bit pattern as that for -123123123.

To see that, I printed it out again using a variety of printf modifiers.

(see the following reference page... http://www.cplusplus.com/reference/cstdio/printf/)

#include <stdio.h>

int main(int argc, const char* argv[]) {

// converting your -123123123 to unsigned long long

unsigned long long foo = (unsigned long long)-123123123;

unsigned long bar = (unsigned long)-123123123; // converting your -123123123 to unsigned long

printf("%llu\n", foo);    // unsigned long long (have to put the ll first, then u for unsigned long long)

  printf("%zu\n", bar);    // unsigned long (have to put z first, then u for unsigned long)

  printf("%ld\n", bar);    // signed long (have to put l first, then d for signed long)

  printf("%d\n", (int)bar);  // signed int (or just say d, for signed int)

 

return 0;

}

//---------------------------------- OUTPUT ------------------------

18446744073586428493

18446744073586428493

-123123123

-123123123

4171844173

Program ended with exit code: 0

Hope that helps.