使用scanf解析字符串时出现奇怪的行为

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.

Answer 1

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.

Answer 2

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).