printf() precision

Question

char buf[10];
int counter, x = 0;
snprintf (buf, sizeof buf , "%.100d%n", x, &counter); 
printf("Counter: %d\n", counter)

I am learning about precision with printf. With %.100d%n, the precision here gives 100 digits for rendering x.

What I don't understand is why would the counter be incremented to 100, although only 10 characters are actually written to the buffer?

The 100 you're getting is the size that is actually needed to printf the whole thing without truncation. Anyway I'd not use %n alltogether (it's not implemented on all platforms), but rather use the return value of snprintf. — Jabberwocky
– Jabberwocky, Commented Mar 5, 2019 at 14:53
@Jabberwocky %n isn't implemented on windows unless you use gcc -D__USE_MINGW_ANSI_STDIO=1 — Jean-François Fabre
– Jean-François Fabre ♦, Commented Mar 5, 2019 at 14:54
it's the design of %n specifier, which can also be used as a security hole — Jean-François Fabre
– Jean-François Fabre ♦, Commented Mar 5, 2019 at 14:54
@Jean-FrançoisFabre %n being a security hole isn't really a strong argument. See also stackoverflow.com/questions/54957216/… It seems more like just another Microsoftian non-portable incompatibility. — Andrew Henle
– Andrew Henle, Commented Mar 5, 2019 at 15:05

pmg · Accepted Answer · 2019-03-05 14:58:52Z

1

The ten bytes written to buf are 9 spaces and 1 '\0' (zero terminator), counter is assigned the value 100 for 99 spaces and 1 '0' (digit zero).

   buf <== "         "
%.100d <== "          .....   0"

Note that buf is incomplete: it does not have a '0'.

edited Mar 5, 2019 at 14:58

answered Mar 5, 2019 at 14:54

pmg

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1 Comment

Jean-François Fabre Over a year ago

I think OP wants to know why the counter gets the required value when the buffer is too short.

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