The discussion revolves around a programming issue related to sorting names in a vector within a C++ program. Participants explore the behavior of a conditional statement in the context of debugging, focusing on string comparisons and memory addresses.
Participants generally agree on the importance of proper string comparison techniques and the limitations of flowcharts for larger programming tasks. However, there is no consensus on a singular correct method for program design, as opinions vary on the effectiveness of different approaches.
Participants note potential issues with memory management and string handling in C++, highlighting the need for careful consideration of how data is stored and compared.
Programmers and students interested in debugging C++ code, string handling, and effective program design methodologies may find this discussion valuable.
What you're doing is an insertion sort with a binary search instead of iterative comparisons to determine the insertion point ##-## there's a C++ example here: https://www.geeksforgeeks.org/binary-insertion-sort/yungman said:
Actually this is a lot easier than what I am doing. The tricky part is to start the vector with no element. If you look at the subscripts S, M and E in my word file, you'll see they are different when you first push_back into an empty file. As it turn out, I don't even have to create the second page for 6 and 7 elements, you can see the trend in the first page already. That's the reason I separate into 3 different cases, case0, case1 and case2 and have to be treated differently.sysprog said:
That's not usually tricky, the implementation should go something like this:yungman said:
high = array.length - 1. It will be negative if there is no element in the array. It is this kind of small little things that are tricky. You can have the program running correctly until you hit these kind of potholes. It was to the point that I decided to stop and wrote out the chart in the word file. Then it became clear. I literally deleted all the former work and start over new and finished in one day with the chart.pbuk said:
#include <iostream>
#include <cstring>
#include <vector>
#include <iomanip>
using namespace std;
const static int nameLen = 25, phoneLen = 12;
struct Directory { char lastName[nameLen]; char firstName[nameLen]; char phone[phoneLen]; int element; };
std::vector<Directory>DirV;
Directory Dir;
void addName(int&);
void sort_firstName(int&);
void showRange(char[]);
void deleteName(int);
int main()
{
int newName, index = 0, compSize, numDelete;
char more, displayAll, choice;
char selName[25];
count << " Welcome to the directory program. You can add name, contact informations and\n";
count << " it will store in a file. You can display selected names and choose to delete\n";
count << " anyone you want.\n\n";
do {
count << " Please enter 'a' to add names, 's' to dispay complete list. 'r' to show \n";
count << " range of names around a name. 'd' for delete a specific name. 'q' to quit.\n\n";
count << " Please enter what you want to do: "; cin >> choice; count << "\n\n";
switch (choice)
{
case 'a': //Add new names and sort
{ do { count << " Enter last name: "; cin >> Dir.lastName;
//count << " Enter first name: "; cin >> Dir.firstName;
addName(newName);
sort_firstName(newName);
count << " \n\nDo you want to enter another name? "; cin >> more;
}while (tolower(more) == 'y');
break;
};//End case 'a'
case 's': //Display entire list of names and info
{ index = 0;
if(DirV.size() > 0)
{ do { count << " Element #" << DirV[index].element <<
"] is: Last name: " << left << setw(10) << DirV[index].lastName <<
" first name: " << left << setw(10) << DirV[index].firstName << "\n\n";
index++;
} while (index < DirV.size()); break;
} else count << " There is no name in the file.\n\n";
};//End case 's'
case 'r':
{ if (DirV.size() > 0)
{ count << " Enter the first 2 character of the last name to search.";
cin >> selName; count << "\n\n";
showRange(selName);//Display range before and after the name initials.
}else count << " There is no name in the file.\n\n";
break;
};
case 'd'://Delete a name.
{ showRange(selName);
count << " Enter the Element# of the left to be deleted: ";
cin >> numDelete;
deleteName(numDelete);
break;
};//End case 'd'
case 'q':{ count << " Are you sure you want to quit? ";
cin >> choice; count << "\n\n";};
default: count << " Not a valid choice.\n";
}//End switch
} while(choice != tolower('q'));//End choice what to do
return 0;
}//End main()
void addName(int& newName)
{ int size, selCase;
DirV.push_back(Dir);//Push to next available element of DirV.
size = DirV.size();// DirV[midPt].element = midPt; newName = midPt;
int startPt = 0, endPt = size - 2, midPt = (endPt - startPt) / 2;
if((size == 1)||((strcmp(DirV[endPt].lastName, Dir.lastName) <= 0) & (size >= 2)))
selCase = 0;//Don't have to sort or anything, Dir is in last element already
//Case 0 for first entry when size = 1 OR when Dir.lastName is
// >= to last element of DirV(Dir.lastName >= DirV[endPt].lastName.
else
{if((strcmp(Dir.lastName, DirV[startPt].lastName) <= 0) && (size > 1))
selCase = 1;//If Dir smaller than DirV[0] AND (size >1)
else
{ startPt = 0, endPt = size - 2; midPt = (endPt - startPt) / 2;
do { if (strcmp(Dir.lastName, DirV[midPt].lastName) <= 0)
{ endPt = midPt; midPt = (startPt + endPt) / 2; }
else {startPt = midPt; midPt = (startPt + endPt) / 2; }
}while (startPt != midPt);
selCase = 2;
}//else selCase = 2
}//End if((strcmp(Dir.lastName, DirV[startPt].lastName) <= 0) && (size > 1))
switch(selCase)
//(size = 1) AND Dir.lastName => DirV[endPt].lastName
{case 0: { DirV[size - 1].element = size - 1; newName = size - 1; break;}
case 1://Dir.lastName <= DirV[0].lastName
{ int i;
for(i = 1; i <= (size - 1); i++)
{ DirV[size - i] = DirV[size - i - 1];
DirV[size - i].element = size - i;
DirV[size - i - 1].element = size - i - 1;
}//End for
DirV[size - i] = Dir; newName = size - i;
DirV[size - i].element = size - i;
break;
}//End case 1
case 2:
{ int j;
for(j = 1; j <= (size - endPt - 1); j++)
{ DirV[size - j] = DirV[size - j - 1];
DirV[size - j].element = size - j;
DirV[size - j - 1].element = size - j - 1;
}
DirV[size - j] = Dir; newName = size - j;
DirV[size - j].element = size - j;
break;
}//end case 2
}//End switch
}
void sort_firstName(int& newName)//DirV[newName] is the newly added name.
{ int beginLname = newName, endLname = newName;//start out all equal
int bubbleDown, bubbleUp;
Directory Temp;//Temporary structure variable to store structure Directory
Temp = DirV[newName];
bool doneUp = false, doneDown = false, moveUp = false;
bubbleUp = newName; bubbleDown = newName;
while ((bubbleUp > 0) & !doneUp)//Prevent bubbleUp going negative
{if((strcmp(DirV[newName].lastName, DirV[newName - 1].lastName) == 0)
& (strcmp(DirV[newName].firstName, DirV[newName - 1].firstName) < 0))
{DirV[newName] = DirV[newName - 1]; DirV[newName].element = newName;
DirV[newName - 1] = Temp; DirV[newName - 1].element = newName - 1;
newName--;bubbleUp--; moveUp = true;}
else (doneUp = true);
}//END while((bubbleUp > 0)& !doneUp)
if (moveUp == false)
{while ((bubbleDown < (DirV.size()-1)) & !doneDown)
{ if((strcmp(DirV[newName].lastName, DirV[newName + 1].lastName) == 0)
&(strcmp(DirV[newName].firstName, DirV[newName + 1].firstName) > 0))
{ DirV[newName] = DirV[newName + 1]; DirV[newName].element = newName;
DirV[newName + 1] = Temp; DirV[newName + 1].element = newName + 1;
newName++; bubbleDown++; }
else doneDown = true;
}//END while (decrement < bubbleUp)
}
}
void showRange(char selName[25])
{
int size = DirV.size(), stpt = 0, edpt = size - 1;
int mdpt = (stpt + edpt) / 2, index, comp;
do //search to within range using 2 characters in selName
{ comp = strncmp(selName, DirV[mdpt].lastName, 2);
if (comp != 0) {if(comp > 0){stpt = mdpt; mdpt = (stpt + edpt)/2;}
else {edpt = mdpt; mdpt = (stpt + edpt) / 2;}}
} while ((comp != 0) && (stpt + 2 <= edpt));//Matching DirV[mdpt]
for(index = -2; index <=3; index++)
{if(((mdpt + index) >= 0) && ((mdpt + index) <= (size - 1)))
{ count << " Element #" << DirV[mdpt + index].element <<
"] is: Last name: " << left << setw(10) << DirV[mdpt + index].lastName <<
" first name: " << left << setw(10) << DirV[mdpt + index].firstName << "\n\n";}
}//End for loop to display range of names if it is valid.
}//End showRange
void deleteName(int numDelete)
{ char sure;
int edpt = DirV.size() - 1;
count << " Is this what you chose?\n";
count << " Element #" << DirV[numDelete].element <<
"] is: Last name: " << left << setw(10) << DirV[numDelete].lastName <<
" first name: " << left << setw(10) << DirV[numDelete].firstName << "\n\n";
count << " Are you sure you want to delete this:\n\n"; cin >> sure;
if (tolower(sure) == 'y')
{
for(int i = 0; i < (edpt - numDelete); i++)
{ DirV[numDelete + i] = DirV[numDelete + i + 1];
DirV[numDelete + i].element = numDelete + i;}
DirV.pop_back();
}
}Definitely agree with jedishrfu - But I Would suggest you to read these articles ;jedishrfu said: