I ran the following code segment in C:
printf("%%%\n");
I got the output "%" (without quotes). Can anyone explain what exactly happened? Why we got only one % sign in result?
Asked
Active
Viewed 72 times
-2
-
Related: http://stackoverflow.com/questions/1860159/how-to-escape-the-sign-in-cs-printf – dragosht Sep 25 '14 at 09:18
-
Read [printf(3)](http://man7.org/linux/man-pages/man3/printf.3.html) & wikipage on [undefined behavior](http://en.wikipedia.org/wiki/Undefined_behavior) – Basile Starynkevitch Sep 25 '14 at 09:18
-
actually, I got two `%%` sign – rakib_ Sep 25 '14 at 09:18
-
@rakib: I ran the code on Codeblocks13.12 using GNU GCC Compiler, & It gave only one % sign. Which compiler are you using? – Prakhar Awasthi Sep 25 '14 at 09:27
-
My gcc version is gcc (GCC) 4.8.3 20140624 (Red Hat 4.8.3-1). – rakib_ Sep 25 '14 at 09:28
1 Answers
4
%% will print %. %\n is not a valid conversion specifier.
You should always enable warnings. See the following:
warning: unknown conversion type character 0xa in format [-Wformat=]
printf("%\n");
^
As noted in the comments, this is undefined behavior because according to the C11 standard, it is undefined behavior if:
— An invalid conversion specification is found in the format for one of the formatted input/ouptut functions [...]
-
-
@rakib Because %\n is not a valid format specifier. Because You should always enable warnings. – Pascal Cuoq Sep 25 '14 at 09:27
-
I ran the code on Codeblocks13.12 using GNU GCC Compiler, & It gave only one % sign with no warnings.. I even ran the printf("%\n"); but it also did not give any warnings.. I think warnings are enabled in my computer (coz I got some warnings in different questions).. Can you explain why is this happening? – Prakhar Awasthi Sep 25 '14 at 09:28
-
-