Inverted arguments in scanf()

Question

I was (quickly) writing some code and accidently inverted the arguments in scanf():

char i[] = \"ABC1\\t\";
scanf(i, \"%s\");

Comp

Answer 1

I suppose it shouldn't.

int scanf ( const char * format, ... );

i was normally converted to a const char*, all the rest parameters are just "ellipsis" and cannot be checked at compile time.

Answer 2

Ha! I found it. Hitting gcc with the -Wformat=2 flag caught it.

Posting the info for reference of others:

Here's the list of flags I found

-Wformat Check calls to printf and scanf, etc., to make sure that the arguments supplied have types appropriate to the format string specified...

I had assumed -Wall had -Wformat in it, which it does, but the really important part about what I just found:

-Wformat is included in -Wall. For more control over some aspects of format checking, the options -Wformat-y2k, -Wno-format-extra-args, -Wno-format-zero-length, -Wformat-nonliteral, -Wformat-security, and -Wformat=2 are available, but are not included in -Wall.

Answer 3

The manual entry for scanf (man scanf) gives the prototype:

int scanf(const char *format, ...);

A char[] is just a special type of char *, so the first argument is satisfied. Secondary arguments are evaluated at runtime (if I recall), so they aren't even considered by the compiler here. From the compiler's prospective, this is a fine call to the function given its prototype.

Also, the compiler never checks whether you are trying to write to invalid locations. The great (or terrible) thing about C is that it will let you do more or less what you want, even if what you want is a bad idea.