Search results
May 25, 2014 · 1 Answer. Sorted by: 59. %lu is correct, while %ul is incorrect. A printf format specifier follows the form %[flags][width][.precision][length]specifier. u is a specifier meaning "unsigned decimal integer". l is a length modifier meaning "long". The length modifier should go before the conversion specifier, which means %lu is correct.
Jul 9, 2010 · 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 ...
Jul 29, 2022 · What is the difference between %zu and %lu in string formatting in C? %lu is used for unsigned long values and %zu is used for size_t values, but in practice, size_t is just an unsigned long. CppCheck complains about it, but both work for both types in my experience. Is %zu just a standardized way of formatting size_t because size_t is commonly ...
Aug 25, 2012 · One, explicit cast your variable: printf("%lu", (unsigned long)i); Two, use the correct format specifier: printf("%d", i); Also, there are problems with assigning an unsigned long to an int - if the long contains a too big number, then it won't fit into the int and get truncated. answered Aug 25, 2012 at 7:32.
size_t existed at least since C89 but the respective format specifier %zu (specifically the length modifier z) was added to the standard only since C99. So, if you can't use C99 (or C11) and had to print size_t in C89, you just have to fallback to other existing types, such as: printf("%lu\n", (unsigned long)n); answered Dec 21, 2016 at 13:54.
Dec 31, 2012 · Obviously in C++ the preferred solution is to use iostreams instead of printf as then the problem disappears. But you can always just cast the value passed to printf to make the type always correct: printf("%llu", static_cast<unsigned long long>(value)); answered Dec 28, 2012 at 18:40. Mark B. 96.1k 10 110 196.
The Win2k/Win9x issue is likely due to unsigned long long datatype being relatively new (at the time, with the C99 standard) but C compilers (including MinGW/GCC) utilizing the old Microsoft C runtime which only supported the C89 spec. I only have access to really old and reasonably recent Windows API docs.
Dec 17, 2013 · what does lu mean in this context: I have found nothing about it. Thank you! It's an unsigned long literal. It means that the literal is forced to be of type unsigned long. Read about integer literals, especially look at "The type of the literal" table on that page. I didn't know you can't switch the letters.
Oct 19, 2016 · All that's needed is that the format specifiers and the types agree, and you can always cast to make that true. long is at least 32 bits, so %lu together with (unsigned long)k is always correct: uint32_t k; printf("%lu\n", (unsigned long)k); size_t is trickier, which is why %zu was added in C99. If you can't use that, then treat it just like k ...
Jan 10, 2021 · I understand that %zd is the suggested way to format the sizeof result. Not quite. Use %zu for size_t (or %zX, %zx or %zo) Use %zd for ssize_t [1] Since sizeof returns a size_t, you want %zu. %zd is a de facto standard, not a de jure standard. There is none of the latter for printing ssize_t, which is a type defined by POSIX (not C).
