printf the ascii char % sometimes gives c and not %

Question

I am a little bit confused about how printf treats ascii characters.

When I print the character % like this, I get a correct response like "ascii %", and this is fine.

printf("ascii %%   \n");
printf("ascii \x25 \n");
printf("ascii %c   \n", 0x25);

ascii % 

And I can put them on the same line like this, and I get "ascii % %", and that is also fine.

printf("ascii %c \x25 \n", 0x25);

ascii % %

But I can't do that in the other order since then I get c and not %, like this "ascii %c"

printf("ascii \x25 %c \n", 0x25);

ascii %c

What is happening?

However I noticed that it seems like printf treats \x25 like the normal % sign, since if I add another % directly after the output (\x25%) it becomes what I expect.

printf("ascii \x25% %c \n", 0x25);

ascii % %

But then I also noticed that printing a single % also seems to work, but I did not expect it to.

printf("ascii %  \n");

ascii %

Why did this work, I thought that a single % was not a valid input into printf... Could someone clarify how printf is supposed to work?

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Note: I am using the default gcc on Ubuntu 12.04 (gcc version 4.6.3).

Answer 1

In C strings syntax, the escape character \ starts an escape sequence that the compiler uses to generate the actual string what will be included into the final binary. Hence within C source code "%" and "\x25" will generate exactly the same character strings in the final binary.

Regarding the single %, this is an malformed format string. It results in undefined behavior. Therefore

printf("ascii \x25 %c \n", 0x25);

is actualy exactly the same as

printf("ascii % %c \n", 0x25);

It results in undefined behavior. There's nothing more to do, trying to understand what's happening with single percent characters is a waste of time.