Strange behavior when parsing a string using scanf

Question

I'm encountering rather strange behavior when preforming a sscanf. Currently working on a windows 7 machine in c.

I have the following:

if( sscanf( str, "%1[a-zA-Z]%31[a-zA-Z+.-]%n", &scheme[ 0 ], &scheme[ 1 ], &num_chars ) >= 1 )
  {
  return( num_chars );
  }

The str variable is a large input string with potentially larger then 32 characters. The scheme variable is declared as an argument to the wrapping function call, it's a 32 character array.

I can easily do this with a couple of scanfs or two separate variables. I was just curious as to why this doesn't work as is.

Edit:
At the time I executed this and the error occurred str contained "tel-net" (was testing the '-') and it resulted in the scheme string having basically no usable characters.

Solution:
I figured out what the problem was, it was actually not a scanf issue at all.

This is how i declared the scheme variable:

IOP_uri_scheme_type   * scheme_str;

IOP_uri_scheme_type was declared as follows:

typedef char    IOP_uri_scheme_type[ IOP_URI_MAX_SCHEME_SZ ];  // Size = 32

The problem was the indexing, scheme[ 1 ] was actually jumping the entire block (all 32 bytes) rather then a character like i was expecting. So technically the scanf was written correctly to begin with (minus the %n thing).

One possible way i can solve this is by casting scheme as a (char *) first or directly manipulating the pointer value, de-referencing it, or just not using a pointer which i don't need anyways.

Thanks for everyone's help.

When posting a question , instead of just saying "getting strange behaviour" - it's best to describe exactly what behaviour you are getting, and what you were expecting (preferably with an exact example for `str` that demonstrates the unexpected output). — M.M, May 14 '14 at 23:11

score 3 · Answer 1 · answered May 14 '14 at 22:48

3

It appears that you are trying to use regular expressions inside sscanf. As far as I know, sscanf does not have any support for regular expressions.

answered May 14 '14 at 22:48

merlin2011

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You can do things with scanf that are similiar to regular expressions... In fact this was based off of one. I have similar expressions scattered across my code (parsing a uri) however this is the only instance where I used the same output variable twice, I believe that is where the problem lies. This compiles and runs, just doesnt populate the output string properly – Justin Youngheim May 14 '14 at 22:55
@user3638657, Take a look at [this answer](http://stackoverflow.com/a/15664770/391161). You might be able to do some tricks that look like it, but I'm not sure if the expression in your question falls under those tricks. – merlin2011 May 14 '14 at 22:58

score 0 · Answer 2 · answered May 14 '14 at 23:37

Here is a test suite I made for this case (with size reduced for readability):

#include <stdio.h>

int main()
{
    char str[] = "tel-net";
    char scheme[13] = { 0 };
    int num_chars;
    int result = sscanf( str, "%1[a-zA-Z]%11[a-zA-Z+.-]%n",
                            &scheme[ 0 ], &scheme[ 1 ], &num_chars );

    printf("result = %d\n", result);
    printf("scheme = '%s'\n", scheme);

    printf("scheme = ");
    for (int ii = 0; ii < sizeof scheme; ++ii)
        printf("%02x ", (unsigned char)scheme[ii]);
    printf("\n");

    if ( result == 2 )
        printf("num_chars = %d\n", num_chars);

    return 0;
}

where the output is:

result = 2
scheme = 'tel-net'
scheme = 74 65 6c 2d 6e 65 74 00 00 00 00 00 00
num_chars = 7

Can you post your output?

Note that your program has a bug, since the %n will not be processed if the second [ fails. You can only return num_chars if the return value is exactly 2.

Regarding the "regular expressions": according to the C standard it is implementation-defined what happens when you use a hyphen inside the [ ] specifier like this. Your compiler (plus C library etc.) may or may not support the usage you are trying. Check your compiler's documentation of scanf to see what it says about this case.

NB. I originally posted an answer saying it was undefined to read into overlapping objects - however I think that is actually false, and it is fine because the arguments are processed in order (and the standard does not say that it it is undefined).

scheme contains the letters 't' '\0' and then garbage after i execute a run through. — Justin Youngheim, May 15 '14 at 13:09
I'll also mention i can use two separate variables as output, so: sscanf( str, "%1[a-zA-Z]%31[a-zA-Z+.-]", &scheme1[ 0 ], &scheme2[ 1 ] ); That works as expected, with the 't' in scheme1 and the 'garbage' + "el-net" in scheme2. — Justin Youngheim, May 15 '14 at 13:19
OK, looks like you solved your problem (`scheme` was actually not a 32-character array, it was a pointer to one). — M.M, May 15 '14 at 21:06

Strange behavior when parsing a string using scanf

2 Answers2