How To Check How Much Characters Are Between Certain Characters

[Sunny101]

I have a test soon and want to make sure I'm doing these right.

If we have something like this:

code a[10]="la#sdas$lf"


How would you make sure there are atleast two characters between # and $.

[Sunny101]

bump?

[TheIdeasMan]

If you want to use char arrays, you can write your own find function which loops through the array & returns the position of the found char. Do this for both search chars then subtract.

If you use a C++ string, you can use the find algorithm to do the same thing.

http://www.cplusplus.com/reference/algorithm/find/

Even though the example is for vectors, you can do the same thing for strings.

Hope all goes well.

[Sunny101]

That's what i've been trying to do, however, the math isn't correct.
Basically I do something like this:

char code [10] ;
cout << "Enter a word with @ first some words then # in it:";
cin.getline(code,10);

/*kind@oftwo#la*/

//Make sure there are atleast two characters between @ and #

//Find position of @
for (int i = 0; i < 10; i ++)
{
    string[i] = @;
}
//Find position of #
for (int j = 0; j < 10; j ++)
{
    string[j] = #;
}

int place;
place = string[j] - string[i]; // to find how much characters

This doesn't work though..hmmm 

[TheIdeasMan]

for (int i = 0; i < 10; i ++)
{
    string[i] = @;  //assignment
}

This doesn't work because the assignment has to be a small int (0 to 255) or a char with single quotes.

What you need is:

unsigned short Position@ = 0;

for (int i = 0; i < 10; i ++)
{
    if (string[i] == '@')  //comparison, note the single quote for a char 
          Position@ = i;
}

Hope this helps

[vlad from moscow]

It would be much more useful if you would show a code snip that can be compiled. Where is the magic name string defined in your code?!!!

If you are using a charater array it is natural to use standard C functions for character arrays

char *p, *q;
int n = 0;

if ( ( p = strchr( code, '@' ) ) && ( q = strchr( code, '#' ) ) ) n = abs( p - q ) - 1;

[vlad from moscow]

It would be more correctly to write the following way

char *p, *q;
int n = 0;

if ( ( p = strchr( code, '@' ) ) && ( q = strchr( code, '#' ) ) ) 
{
   if ( q < p ) std::swap( p, q );
   while ( *p == *( p + 1 ) ) ++p;
   n = abs( p - q ) - 1;
}

[vlad from moscow]

And one more solution.:)

char *p, *q;
int n = 0;

char s[] = "##A#B@@";
if ( ( p = strchr( s, '@' ) ) && ( q = strchr( s, '#' ) ) ) 
{
	if ( q < p ) std::swap( p, q );
	p = std::find( std::reverse_iterator<char *>( q ), std::reverse_iterator<char *>( p ), *p ).base();
	n = abs( p - q );
}