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What is the argument for printf that formats a long?

The printf function takes an argument type, such as %d or %i for a signed int. However, I don't see anything for a long value.


r
rogerdpack

Put an l (lowercased letter L) directly before the specifier.

unsigned long n;
long m;

printf("%lu %ld", n, m);

printf("%ld", ULONG_MAX) outputs the value as -1. Should be printf("%lu", ULONG_MAX) for unsigned long as described by @Blorgbeard below.
Actually, you should change it to be %ld, to be more harmonic with OP question.
Yep Dr Beco; further, just %l triggers warning: unknown conversion type character 0x20 in format [-Wformat]
k
kiritsuku

I think you mean:

unsigned long n;
printf("%lu", n);   // unsigned long

or

long n;
printf("%ld", n);   // signed long

D
Dave Dopson

On most platforms, long and int are the same size (32 bits). Still, it does have its own format specifier:

long n;
unsigned long un;
printf("%ld", n); // signed
printf("%lu", un); // unsigned

For 64 bits, you'd want a long long:

long long n;
unsigned long long un;
printf("%lld", n); // signed
printf("%llu", un); // unsigned

Oh, and of course, it's different in Windows:

printf("%l64d", n); // signed
printf("%l64u", un); // unsigned

Frequently, when I'm printing 64-bit values, I find it helpful to print them in hex (usually with numbers that big, they are pointers or bit fields).

unsigned long long n;
printf("0x%016llX", n); // "0x" followed by "0-padded", "16 char wide", "long long", "HEX with 0-9A-F"

will print:

0x00000000DEADBEEF

Btw, "long" doesn't mean that much anymore (on mainstream x64). "int" is the platform default int size, typically 32 bits. "long" is usually the same size. However, they have different portability semantics on older platforms (and modern embedded platforms!). "long long" is a 64-bit number and usually what people meant to use unless they really really knew what they were doing editing a piece of x-platform portable code. Even then, they probably would have used a macro instead to capture the semantic meaning of the type (eg uint64_t).

char c;       // 8 bits
short s;      // 16 bits
int i;        // 32 bits (on modern platforms)
long l;       // 32 bits
long long ll; // 64 bits 

Back in the day, "int" was 16 bits. You'd think it would now be 64 bits, but no, that would have caused insane portability issues. Of course, even this is a simplification of the arcane and history-rich truth. See wiki:Integer


Just to clarify: there are more architectures than x86 and x64, and on those architectures, char, short, int, long and long long have different meanings. For example, a 32 bit mcu with 16 bit memory alignment could use: char=short=int=16 bit; long=32 bits; long long = 64 bits
@AndresR - absolutely, and it matters very deeply to people who do embedded programming (I built a kernel module for a smoke alarm once). Pity those poor souls when they try to use code that wasn't written for portability.
I'm not so sure that "most platforms have a long that's of size 32" is still true. E.g., I'm on Oracle Linux x86_64/amd64, and with nvcc a long is 8 bytes.
"Oh, and of course, it's different in Windows". Is there a cross-platform solution?
@interestedparty333 Yes, I'm >95% sure that in Linux long has the same size as the word/pointer size (so 32-bits on 32-bit Linux and 64-bits on 64-bit Linux). If you look inside the Linux kernel's code or inside Linux drivers, they usually store pointers in long or unsigned long variables. Actually, a few weeks ago a colleague of mine who ported some code from Windows to Linux had to go and change all our longs to uint32_t because we would constantly misuse them when developing under Windows.
k
krato

It depends, if you are referring to unsigned long the formatting character is "%lu". If you're referring to signed long the formatting character is "%ld".


k
D
Dolan Antenucci

In case you're looking to print unsigned long long as I was, use:

unsigned long long n;
printf("%llu", n);

For all other combinations, I believe you use the table from the printf manual, taking the row, then column label for whatever type you're trying to print (as I do with printf("%llu", n) above).


A
Andrew O'Reilly

I think to answer this question definitively would require knowing the compiler name and version that you are using and the platform (CPU type, OS etc.) that it is compiling for.


Your answer should be a comment and not an answer by itself.