Question !!!! Warning Message!!!!

**her4men** · 02-27-2010

Hi all,

I'm doing a program which is input a text file, and match every punctuation in that text file.

But when i compiled the program, it gave me a warning message which is "passing argument 2 of 'regexec' makes pointer from integer without a cast."

here is my code.

Code:

void process(FILE *fp, int width, int punctuation)
{

    char single_c;

    int rc, i, pos = 0;

    regex_t * myregex = calloc(1, sizeof(regex_t));

    rc = regcomp( myregex, "[^0-9a-zA-Z ]+", REG_EXTENDED | REG_NOSUB ); //regular expression.
                                                                         //case: punctuation

    while ( (single_c = getc(fp) ) != EOF)
    {
        if( 0 == regexec(myregex,(char )single_c, 0 , 0 , 0 ) )
        {
            printf("String %c matches.\n", single_c );
        }
        else
        {
            printf("String %c matches.\n", single_c );
        }
    }
}

Can someone tell me how to fix this problem?

**cas** · 02-27-2010

The second argument to regexec() is a char*, not a char. You must pass a string.

You should also avoid casting. First, casting a char to a char is meaningless. But generally speaking--especially if you're new to C--a cast is the wrong thing to do. They are, of course, useful in some cases, but nowhere near as often as they tend to be used.

**her4men** · 02-28-2010

oh~!! ok!

I got it. Thanks.

By the way, I have found lots example, and most of them use pointer to do converting, for example:

char a;
char *b;

b=&a

I just wondering that is C uses pointer a lot??
Is there other way to convert single char to string?

**laserlight** · 02-28-2010

Originally Posted by her4men

I just wondering that is C uses pointer a lot??

"A lot" is relative, but yes.

Originally Posted by her4men

Is there other way to convert single char to string?

That does not convert a single char to a string. It simply assigns the address of a char to a pointer to char. No strings are involved. If you want to convert a single char to a string, create an array of two chars, and assign the given char to the first element of the array, and a null character to the second element of the array. The array would then contain a string.

**her4men** · 02-28-2010

There's a strange thing appeared in my program

Code:

void process(FILE *fp, int width, int punctuation)
{

    char single_c,
         *temp_char,
         whole_String[width],
         temp_String[width];

    int rc, i, pos = 0;

    regex_t * myregex = calloc( 1, sizeof(regex_t) );

    rc = regcomp( myregex, "[^a-zA-Z0-9 ]+", REG_EXTENDED | REG_NOSUB );
    printf("RC from regcomp() = %d\n", rc); //regular expression.
                                                               //case: punctuation

    while ( ( single_c = getc(fp) ) != EOF && width != pos)
    {
        temp_char = &single_c; //string point to char.
        if( 0 == regexec(myregex, temp_char, 0 , 0 , 0 ) )
        {

            printf("%c\n", single_c );
        }
        else
        {
            printf("abc");
        }

        pos++;
    }

    free(myregex);
}

In the if statement , my program never go into else part.
Suppose when I give ":" to stdin, then this should print out "abc".

Can anyone tell me what my problem is?

**nonpuz** · 02-28-2010

Code:

  temp_char = &single_c; //string point to char.

Assigning a pointer to char (char *) to a char type does not result in a "string" as should be passed to regexec.

But I'm curious is why are you using a regexec against a SINGLE character? Shouldn't you be testing against the entire text of the file, ie a long string (which you would have to allocate enough memory to fit the text in)

**Salem** · 03-01-2010

> temp_char = &single_c; //string point to char.
And where is the \0 to mark the end of the string in this ?

Simply declaring a bunch of variables with the right type isn't enough.

As nonpuz says, testing a single char with a regex is pointless.

It seems to me you need

Code:

char buff[BUFSIZ]
while ( fgets( buff, sizeof buff, fp ) != NULL ) {
  if( 0 == regexec(myregex, buff, 0 , 0 , 0 ) )

}

Oh, and I also find operand reversal to be annoying and counter-productive.
Question 17.4

Your modern compiler will almost certainly warn you (and is certainly capable of warning you) if you write if ( var = 0 )

You would get an error anyway if you did
if ( regexec(myregex, buff, 0 , 0 , 0 ) = 0 )
because the function result isn't an lvalue.

Further, if you find yourself with
if ( var1 = var2 )
then no amount of rearranging the deckchairs will save your program from sinking.

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