Search String table

Ingo Foerster
Ingo Foerster used Ask the Experts™
on
Hello,

i have a list of thousand of different IDs. A sample:
static struct
{
  TCHAR org_openbddb[41];
  TCHAR cross_openbddb[41];
} openbddb_matrix[] = {
/* The ISO 639 language codes.
 */
	{_T("ec62b87bfd5585ac8be5c7fb0cbba1ce6a29c0bd"), _T("ec62b87bfd5585ac8be5c7fb0cbba1ce6a29c0bd")}, 


};

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I need to search the value "org_openbddb" in the list and return the "cross_openbddb" value if found. If nothing found, return "". Any ide how to do this?
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ozo
Most Valuable Expert 2014
Top Expert 2015

Commented:
in c++, you can use std::map
jkr
Top Expert 2012

Commented:
If you need a quick lookup that you have to perform often, I'd suggest storing these structs in a 'std::map' to look them up.

#include <map>
using namespace std;

static struct
{
  TCHAR org_openbddb[41];
  TCHAR cross_openbddb[41];
} openbddb_matrix[] = {
/* The ISO 639 language codes.
 */
	{_T("ec62b87bfd5585ac8be5c7fb0cbba1ce6a29c0bd"), _T("ec62b87bfd5585ac8be5c7fb0cbba1ce6a29c0bd")}, 

{NULL, NULL} // I assume this is at the end
};
 
// ...
typedef std::string<TCHAR> tstring;

map<tstring,tstring> map_lang_code;

// store all in map
int n = 0;
while(openbddb_matrix[n].org_openbddb) {

  map_lang.code.insert(map<tstring,tstring>::value_type(openbddb_matrix[i].org_openbddb,openbddb_matrix[i].cross_openbddb));
  ++n;
}

TCHAR* lookup_org_value(const TCHAR* val) {

  // lookup
  map<tstring,tstring>::iterator i;

// e.g. like: i = map_lang_code.find(_T("ec62b87bfd5585ac8be5c7fb0cbba1ce6a29c0bd"));

  i = map_lang_code.find(val);

  if (i != map_lang_code.end()) return i->second; // found it

  return NULL;  // nothing found
}

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Top Expert 2012
Commented:
On the other hand, if you only need to do that once or twice, a simple loop will do:

TCHAR* lookup_org_value(const TCHAR* val) {

  int n = 0;
  while(openbddb_matrix[n].org_openbddb) {

     if (!_tcscmp(openbddb_matrix[n].org_openbddb,val)) return openbddb_matrix[n].cross_openbddb; // found it
     ++n;
  }

  return NULL; // nothing found
}

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If you use C++11 or if you can use Boost, the unordered_map can be faster (better in O() terms). However, it probably should be measured if it matters. (As you probably use Visual Studio, the older versions named the unordered_map as hash_map.)
Top Expert 2016

Commented:
ec62b87bfd5585ac8be5c7fb0cbba1ce6a29c0bd

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the codes seem to be 20 unsigned 8-bit integers (bytes) in hex notation. if so, you could spare more than half the size if you convert them to binary integers. if your TCHAR is a wide character the saving would be more than 3/4 of the size needed for strings.

to use a std::map with the 20 bytes you probably would define a little class like that:

class LangCode
{
     unsigned char codes[20];
public:
     LangCode() { memset(codes, 0, sizeof(codes)); }
     LangCode(TCHAR scod[]) 
     { 
        assert(_tcslen(scod) == 40); 
        char szt[3] = { '\0' };
        for (int i = 0, j = 0; i < 40; i += 2, j++)
        {
             szt[0] = (char)scod[i];
             szt[1] = (char)scod[i+1];
             std::istringstream iss(szt);
             iss >> std::hex >> codes[j];
             assert(iss);
        }
     }
 
     bool operator<(const LangCode& lc) const
     { return memcmp(codes, lc.codes, sizeof(codes)) < 0; } 
};

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with that you could add and find members like

    TCHAR * psz1 = L"ec62b87bfd5585ac8be5c7fb0cbba1ce6a29c0bd";
    TCHAR * psz2 = L"23b87bfd5585ac8fffc7fb0cba1ba1ce6a200000";
    std::map<LangCode, LangCode> maplc;

    maplc[psz1] = psz2;
    ...
    std::map<LangCode, LangCode>::iterator f;
    if ((f = maplc.find(psz1)) != maplc.end())
    {
        std::cout << "found" << std::endl;
    }
    else
    {
        std::cout << "not found" << std::endl;
    }

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you see the map would make an implicit conversion when using TCHAR strings.

if you read the codes from a database you would fill the map when loading the data.

if you have all the mapping code pairs in a 'matrix' like you posted you could do

struct LCPair
{
    LangCode lc1;
    LangCode lc2;
};

LCPair matrix[] = 
{ 
{_T("ec62b87bfd5585ac8be5c7fb0cbba1ce6a29c0bd"),
    _T("23b87bfd5585ac8fffc7fb0cba1ba1ce6a200000") }, 
{_T("ff62b87bfd5325ac8be5c7fb0cbba1ce6a29c0bd"),
    _T("11111afd5585ac8fffc7fb0cba1ba1ce6a200000")}, 
....
};

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after that you could fill the map by code like

for (size_t n = 0; n < sizeof(matrix)/sizeof(matrix[0]); ++n)
     maplc[matrix[n].lc1] = matrix[n].lc2;

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alternatively, you could sort the matrix by the first member and then make a binary search on the sorted matrix instead of using a map.

Sara
Ingo FoersterOwner, CTO

Author

Commented:
Great. Thank you.

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