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Compare structure componets in a function

Posted on 1998-05-21
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Last Modified: 2008-03-06
I have a function to compare individual structure componets but it will not compare them. See Function below.
int
local_address(address_t 1_array[], address_t 2_array[], int x)
{
 address_t v;
 int i;
 int status;

 for (i = 0; i < x; ++i)
 if (1_array->address[0] == v.address[i] &&
           1_array->address[1] == v.address[i])
 {
   2_array[i++] = v;
   status = 1;
 }
else
 {
   status = 0;
 }
 return(status);
 
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Question by:moorem
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7 Comments
 
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Expert Comment

by:RONSLOW
ID: 1250891
What is this function trying to do ???
0
 
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Accepted Solution

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RONSLOW earned 100 total points
ID: 1250892

  int local_address(address_t 1_array[], address_t 2_array[], int x) {
    address_t v;

you have non initialized v

    int i;
    int status;

    for (i = 0; i < x; ++i)

a style note .. use {} for multi-line loops even if not strictly necessary
      if (1_array->address[0] == v.address[i] && 
          1_array->address[1] == v.address[i])

you seem VERY confused about arrays and how to dereference them .. we need to talk at length about this !!
      {
        2_array[i++] = v;

this use of i++ here seems very odd

        status = 1;
      }
      else
      {
        status = 0;
      }
    return(status);

status gets set to zero or one throughout the loop .. its value will be the last one set.  This doesn't seem to be what you want to do

  }

You seem to be in a lot of trouble here !!

0
 

Author Comment

by:moorem
ID: 1250893
Hi Ronslow

I have a structure like below.
typedef
{
  int address[4];
  char name[50];
}address_t
The data is read in from a file in the input process and then passed to this function. I need it to compare 1st 2 element
of address & if same store in the other array for ouput.
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LVL 10

Expert Comment

by:RONSLOW
ID: 1250894
What do the int values (address) in the address_t mean?  It doesn't really, matter - just curious.

SO... you want to match the first 2 items of the address[4] within address_t.

I'd imagine this to be done with..

int CompareAddresses (address_t* pAddress1, address_t* pAddress2) {
  int i;
  for (i = 0; i < 2; i++) {
    if (pAddress1->address[i] < pAddress2->address[i]) return -1;
    if (pAddress1->address[i] > pAddress2->address[i]) return +1;
  }
  return 0;
}

This function returns 0 if the first 2 elements of address[] +1 if greater, -1 if less.

Now for your storage .. I assume you are passing in a single new address and want to append it to the end of the array of addresses if it is not already there...

int AddNewAddress (address_t* pAddress, address_t addresses[], int* pCount) {
  int i;
  /* check all the currently stored addresses */
  for (i = 0; i < *pCount; i++) {
    /* if already there then return 0 */
    if (0 == CompareAddresses(pAddress, &addresses[i])) {
      return 0;
    }
  }
  /* wasn't there .. add to the end */
  /* no check on array overflow though */
  addresses[*pCount] = *pAddress;
  /* bump up the count becuase we just add a new address */
  (*pCount)++;
  /* return 1 to indicate we added a new address */
  return 1;
}

Is this what you are trying to do??

0
 

Author Comment

by:moorem
ID: 1250895
Ronslow

What I need to do is compare to of the structures componets againts all data in the array and if they are the same copy that element into an array and pass it back to main for the display function. I can e-mail you a copy of the source code if you will give me a e-mail address. Thanks in advance for your help.
0
 
LVL 10

Expert Comment

by:RONSLOW
ID: 1250896
Thw code I suggested compares a component with all the others in the array and add it if not there (which is what you said you originall said/hinted that you wanted to do).

Now .. you say you want to compare it to each items in the array and a match if any .. this is different

int FindAddress (address_t* pAddress, address_t addresses[], int count) {
  int i;
  /* check all the currently stored addresses */
  for (i = 0; i < count; i++) {
    /* if already there then return 0 */
    if (0 == CompareAddresses(pAddress, &addresses[i])) {
      return i;
    }
  }
  return -1;
}

This function return the array subscript where a match for the given item is found, otherwise it returns -1.  THe calling program can check the return value and take the approrpriate action.

0
 

Author Comment

by:moorem
ID: 1250897
I have tried several of the solutions but it still does not seem to be doing what I think it should do. Maybe it is just me. Thanks for the assistance and now I must move on to the next lab I am a student.
0

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