%p formats to 8 hex digits for a 64-bit binary

Question

I have a 64-bit binary (modfied version of sqlite, but this shouldn't mattrr):

> file /home/aromanov/IdeaProjects/sqlite/sqlite3
/home/aromanov/IdeaProjects/sqlite/sqlite3: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.32, BuildID[sha1]=0x39a2db352d3bc451ed621ad0588eec3008df034b, not stripped

produced with GCC 4.7.4. However, debugging code in it outputs just 8 hex digits (32 bits) for %p in format string, where I expected 16. Is this normal?

There is no output and no code producint any output in your question. Without both it's hard to tell what's wrong. — Gerhardh, Jan 19 '17 at 13:36
Please provide a reference to the C standard where a specific formatting is required. Hint: the documentations of `printf` will already answer your question. Please do some research before asking. — too honest for this site, Jan 19 '17 at 13:57

score 1 · Accepted Answer · answered Jan 19 '17 at 13:40

1

At least on linux/glibc, the %pformat specifier does not do left padding with zeroes. Thus a pointer to e.g. address 0x000000000000ffff is formatted simply as "0xffff". You can use %.16p if you want 16 digits in the output.

answered Jan 19 '17 at 13:40

nos

223,662
58
417
506

2

AFAIK `"%.16p"` is gnu specific. – Jabberwocky Jan 19 '17 at 13:43
1

Looking at the POSIX specification of [`printf()`](http://pubs.opengroup.org/onlinepubs/9699919799/functions/printf.html), there is nothing in the wording about the `p` conversion specification to indicate that the flags, width and precision are not allowed, though since the format it implementation-defined, the effect of those modifiers is likewise unclear. – Jonathan Leffler Jan 19 '17 at 15:26

%p formats to 8 hex digits for a 64-bit binary

1 Answers1