How to search an array problem

jenmaz · #1 15-Nov-2004, 20:13

Hi there,
I am working on some problems and I am learning to search arrays.

Here is the scenerio or problem.

Given*:
an int*variable* k ,
an int*array* currentMembers that has been declared* and initialized*,
an int*variable* nMembers that contains the number of elements* in the array*,
an int*variable* memberID that has been initialized*, and
an int*variable* isAMember , write code that assigns* 1 to isAMember if the value* of memberID can be found in currentMembers , and that assigns* 0 to isAMember otherwise. Use only k , currentMembers , nMembers , and isAMember .

here is my answer...Can anyone tell me what I am doing wrong?

CPP / C++ / C Code:

for (k = 0; k < nMembers; k++)
   {
      if (currentMembers[k]=currentMembers[memberID])
         {
         currentMembers[isAMember] = 1;
         }
         else 
         {
         currentMembers[isAMember] = 0;
         }
  }

nkhambal · #2 15-Nov-2004, 21:20

One obvious mistake is in if condition. For comparison you need to use "==" (is equal to) instead of "=" which means assignment.

Also, I don't think you have correct logic there in the loop. You need to compare "memberID" with current value of currentMembers[k]. The correct program would be.

CPP / C++ / C Code:

for (k = 0; k < nMembers; k++)
   {
      if (currentMembers[k]==memberID)
         {
         isAMember = 1;
         }
         else 
         {
         isAMember = 0;
         }
  }

jenmaz · #3 16-Nov-2004, 12:46

Thanks,

This makes sense but how do I make it check both ends of the array? Right now it only checks one end?

Quote:

Originally Posted by nkhambal

One obvious mistake is in if condition. For comparison you need to use "==" (is equal to) instead of "=" which means assignment.

Also, I don't think you have correct logic there in the loop. You need to compare "memberID" with current value of currentMembers[k]. The correct program would be.

CPP / C++ / C Code:

for (k = 0; k < nMembers; k++)
   {
      if (currentMembers[k]==memberID)
         {
         isAMember = 1;
         }
         else 
         {
         isAMember = 0;
         }
  }

nkhambal · #4 16-Nov-2004, 14:26

I didn't get it. Could you pls elaborate a bit

. If possible with some example.

Thanks,

WaltP · #5 17-Nov-2004, 01:58

Quote:

Originally Posted by jenmaz

Thanks,

This makes sense but how do I make it check both ends of the array? Right now it only checks one end?

Actually, it checks every character. It's just that the if statement resets isAMember each time thru the loop so at the end of the loop isAMember is a value based only on currentMembers[nMembers-1]

CPP / C++ / C Code:

for (k = 0; k < nMembers; k++)
{
    if (currentMembers[k]==memberID)
    {
        isAMember = 1;
    }
    else 
    {
        isAMember = 0;
    }
}

I assume you want isAMember to be 1 if memberID is in the array somewhere. If so, try this:

CPP / C++ / C Code:

isAMember = 0;    // initialize to member not found
for (k = 0; k < nMembers; k++)
{
    if (currentMembers[k] == memberID)
    {
        isAMember = 1;    // member is found
    }
}

Now isAMember will be zero if memberID is not in the array, or 1 if it is.

Version 2:

CPP / C++ / C Code:

isAMember = -1;    // initialize to member not found
for (k = 0; k < nMembers; k++)
{
    if (currentMembers[k] == memberID)
    {
        isAMember = k;    // member is found
    }
}

Now isAMember will be -1 if memberID is not in the array, or > than -1 if the member is in the array. In fact, isAMember will be the index of the member in currentMembers so it will point to the actual member.

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