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Whats the difference between %ul and %lu C format specifiers? %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
c++ - printf and %llu vs %lu on OS X - Stack Overflow I can try to use macros to try to choose the correct string (#define LU either %llu or %lu, and in the process uglifying the printf strings a bit) but on the Mac I've got a 64-bit word size (so _LP64 would be defined and UINTPTR_MAX != 0xffffffff) and yet it still uses long long for the 64 bit int types
c# - What does this regexp mean - \p {Lu}? - Stack Overflow Therefore, \p{Lu} will match an uppercase letter that has a lowercase variant And, the opposite \p{Ll} will match a lowercase letter that has an uppercase variant Concisely, this would match any lowercase uppercase that has a variant from any language:
LU decomposition error in statsmodels ARIMA model I know there is a very similar question and answer on stackoverflow (here), but this seems to be distinctly different I am using statsmodels v 0 13 2, and I am using an ARIMA model as opposed to a
How to printf unsigned long in C? - Stack Overflow @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
printf - Difference between %zu and %lu in C - Stack Overflow 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
printf format specifiers for uint32_t and size_t - Stack Overflow 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 (long is the biggest type in C89, size_t is very unlikely to be larger)
c - Why do I get %lu when I try to print a u64 variable with %llu . . . When I print the number using the format specifier "%llu", what is printed is "%lu" I also compare the value I get from atoll or strtoll with the expected value and it is smaller, which I guess shows that an overflow has occurred Why does an overflow occur if the number fits in a u64 variable? The number for example is 946688831000