I need to print a filled square in Linux terminal using my C++ program (1cm x 1cm size). I tried to use ASCII 254 (■), but in terminal it print as garbage character. I'm not sure how to print extended ASCII character using c++. Here are two methods I have tried to print extended ASCII. but not succeed.
First method
for(int i=128; i< 255; i++ )
{
std::cout << static_cast<char>(i) << std::endl;
}
Second method
unsigned char temp = 'A'
for(int i=65; i< 255; i++ )
{
std::cout << temp++ << std::endl;
std::wcout << temp << std::endl;
}
Any suggestion or alternative Idea?
Try using the unicode cout << "\u25A0";
http://www.fileformat.info/info/unicode/category/So/list.htm
Or try just:
std::cout << (char)254u;
To get output as you want, try this:
#include<iostream>
#include<windows.h>
using namespace std;
void setconsolecolor(int textColor, int bgColor)
{
SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), (textColor +
(bgColor * 16)));
}
int main()
{
cout<<"The chunk of Blocks in the colors : \n";
cout<<" ";
setconsolecolor(0,9);
cout<<" ";
setconsolecolor(0,4);
cout<<" ";
setconsolecolor(0,8);
cout<<" \n";
setconsolecolor(0,0);
return 0;
}
Try this :
char t = -2;
cout << t;
Like Sebastian Kuczyński has suggested , we could use that to do great stuff like bar graphs , hostogram etc. Its very cool.
Code
printf("\n\nHistogram of Float data\n");
for (i = 1; i <= bins; i++)
{
count = hist[i];
printf("0.%d |", i - 1);
for (j = 0; j < count; j++)
{
printf("%c", (char)254u);
}
printf("\n");
}
Output
Histogram of Float data
0.0 |■■■■■■■■■■■■■■■■■■■■■■
0.1 |■■■■■■■■■■■■■■■■
0.2 |■■■■■
0.3 |■■■■■■■■■■■■■■
0.4 |■■■■■■■■
0.5 |■■■■■■■■■■■■■■■■
0.6 |■■■■■■■■■■
0.7 |■■■■■■■
0.8 |■■■■■■■■■■■■■■■
0.9 |■■■■■■■
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Here is my program, it is using c++ and I done it on MacOS VSCode.
Library:
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <string>
#include <iomanip>
using namespace std;
Function:
float* read_data(int& size){
int a_size = 0;
static float a[10];
float* temp = a;
cout << "Please enter values, and press 'Q' when finished: ";
for (int i = 0; i < size; i++){
cin >> a[i];
if (cin.fail()){
break;
}else{
a_size++;
}
}
cin.clear();
size = a_size;
return temp;
}
Main function:
int main(){
int size1 = 10;
int size2 = 10;
int size3 = 0;
float array[10];
float addarray[10];
float longarray[20];
float* ptr = array;
float* addptr = addarray;
float* longptr = longarray;
cout << fixed << setprecision(2);
ptr = read_data(size1);
cout << setw(15) << "Original array" << setw(5) << "=" << setw(5) << "{ " << *ptr;
for (int i = 0; i < size1; i++){
cout << ", " << *(ptr + i);
}
cout << " }" << endl;
for (int i = 0; i < size1; i++){
*(longptr + size3) = *(ptr + i);
size3++;
}
addptr = read_data(size2);
for (int i = 0; i < size2; i++){
*(longptr + size3) = *(addptr + i);
size3++;
}
cout << setw(15) << "New array" << setw(5) << "=" << setw(5) << "{ " << *longptr;
for (int i = 0; i < size3; i++){
cout << ", " << *(longptr + i);
}
cout << " }" << endl;
return 0;
}
The main objective of the program was to prompt the user for an array, maximum of 10 elements. Prompting the user was done using the float* read_data(int& size) function.
The program would echo or print out the inputted array.
After that, the user was prompt for a second time using the same function to get another array of elements, in this case is a list of float values.
Then, the program would use a dynamic data allocation (DMA) technique to combine the two array into one long array. The new array will be printed out and the program is terminated.
Problem
As I said before, the program are supposed to prompt the user for inputs every time the read_data() were called.
I called the function twice. It did run twice, but failed to prompt the user for input the second time around.
I thought the problem was due to the cin.fail() arguments. That is why I tried to mess with the cin.ignore() and cin.clear() either by removing them or only using one of them.
I am honestly at lost on what is the root of the problem.
The heart of the issue is that you want users to enter 'Q', a char, into your float variable.
Here's a small example:
#include <iostream>
int main() {
int a;
std::cin >> a;
if (std::cin.fail()) {
std::cout << "Fail.\n";
} else {
std::cout << "All clear.\n";
}
std::cin >> a;
std::cout << (2 * a) << '\n';
}
Output:
~/tmp
❯ ./a.out
Q
Fail.
0
~/tmp
❯ ./a.out
3
All clear.
3
6
So, you can see that you were on the right track. The issue is that if std::cin did fail, you are now responsible to clean up your mess.
std::cin.clear() is a good start. It resets the fail flag bits. But the stream is still in a bad state. What you haven't done is clean it up completely.
You're calling std::cin.ignore(), but leaving the parameter list empty.
Something like this is best practice:
#include <iostream>
#include <limits>
int main() {
int a;
std::cin >> a;
if (std::cin.fail()) {
std::cout << "Fail.\n";
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
std::cin.clear();
} else {
std::cout << "All clear.\n";
}
std::cin >> a;
std::cout << (2 * a) << '\n';
}
Output:
~/tmp
❯ ./a.out
Q
Fail.
3
6
While that should fix address the question, your code is still fundamentally broken.
EDIT
Here's a mini-code review:
#include <iostream>
#include <algorithm> // These includes
#include <cmath> // are
#include <cstdlib> // not
#include <string> // used
#include <iomanip>
using namespace std; // Bad practice
// You should not attempt to return a C-style array
float* read_data(int& size){ // Why is the size being passed by reference?
int a_size = 0;
static float a[10]; // Not doing what you think
float* temp = a;
cout << "Please enter values, and press 'Q' when finished: ";
for (int i = 0; i < size; i++){ // Formatting; should be ") {"
cin >> a[i];
if (cin.fail()){ // Addressed above; root cause of your question
break;
}else{
a_size++;
}
}
cin.clear();
size = a_size;
return temp;
}
// While I assume the goal of the assignment is to get familiar with pointers,
// This assignment is trivial with vectors.
int main(){
int size1 = 10; // If declared const/constexpr, you'd only need one **
int size2 = 10; // These are bad names
int size3 = 0; // Front loading declarations is bad practice.
float array[10]; // ** And could use that variable here to avoid the magic number
float addarray[10];
float longarray[20]; // Not dynamic per your requirements.
float* ptr = array; // Still not dynamic, also unnecessary given the prior declarations
float* addptr = addarray;
float* longptr = longarray;
cout << fixed << setprecision(2);
ptr = read_data(size1);
// The first cout is not good, what do you think *ptr will print?
cout << setw(15) << "Original array" << setw(5) << "=" << setw(5) << "{ " << *ptr;
for (int i = 0; i < size1; i++){
cout << ", " << *(ptr + i); // The pointers can still use array syntax
}
cout << " }" << endl;
for (int i = 0; i < size1; i++){
*(longptr + size3) = *(ptr + i);
size3++;
}
addptr = read_data(size2);
for (int i = 0; i < size2; i++){
*(longptr + size3) = *(addptr + i);
size3++;
}
// As soon as you start repeating yourself, consider writing a function.
// The logic will only live in one place, and only need to be changed in
// one place.
cout << setw(15) << "New array" << setw(5) << "=" << setw(5) << "{ " << *longptr;
for (int i = 0; i < size3; i++){
cout << ", " << *(longptr + i);
}
cout << " }" << endl;
return 0;
}
A lot going on that is less than great. The biggest flub appears to be the static array. static in this scenario extends the lifetime of the variable until the end of the program. So, when you call this function the second time, you overwrite the first array with the second because there's only ever one array for every call of that function. In the future, I would refrain from grabbing random code online and just dumping it in your program unless you know what it does.
Now, let's move on to a working solution.
The first thing I'm going to do is state the requirements as I understood them.
Have the user enter data to fill two arrays of floats.
Each array should hold a maximum of ten elements.
The user enters "Q" to indicate that they are finished entering data.
Print the first array.
Tack the second array on to the end of the first, in a dynamic fashion.
Print the combined array.
Anytime you get an assignment, your first task should be to restate the problem in your own words. This will clarify requirements and you will demonstrate to yourself that you understand the problem to be solved.
What makes this interesting is the "Q" to quit. Note that I'm using double quotes. We will read all of our data as strings, and we have to convert to float as needed.
std::stof() exists, but it's not as simple as just calling it. That function can throw exceptions, and we want to ensure that only actual float values get converted.
So we'll wrap that call in a function of our own. Here's an example:
#include <exception>
#include <iostream>
#include <string>
class bad_user_input : public std::exception {
const char* what() const noexcept override {
return "User input was not valid.";
}
};
float convert_string_to_float(const std::string& val) {
std::size_t marker = 0;
float f = 0.0f;
try {
f = std::stof(val, &marker);
} catch(...) {
throw bad_user_input();
}
// Was the entire value entered a float?
if (marker != val.length()) {
throw bad_user_input();
}
// Getting here means a valid float was entered.
return f;
}
int main() {
float f = 0.0f;
// Test the function
try {
f = convert_string_to_float("5.6");
} catch(...) {
std::cerr << "Issue.\n";
}
std::cout << f << "\n\n";
f = 0.0f;
try {
f = convert_string_to_float("5.6cat");
} catch(...) {
std::cerr << "Issue.\n";
}
std::cout << f << "\n\n";
f = 0.0f;
try {
f = convert_string_to_float("Q");
} catch(...) {
std::cerr << "Issue.\n";
}
std::cout << f << '\n';
}
Output:
❯ ./a.out
5.6
Issue.
0
In convert_string_to_float
stof: no conversion
Issue.
0
Now that we can convert strings to floats, with guarantees, we are ready to read values. To test that we're reading correctly, we also want to be able to print our arrays:
#include <exception>
#include <iostream>
#include <string>
class bad_user_input : public std::exception {
const char* what() const noexcept override {
return "User input was not valid.";
}
};
float convert_string_to_float(const std::string& val) {
std::size_t marker = 0;
float f = 0.0f;
try {
f = std::stof(val, &marker);
} catch(...) {
throw bad_user_input();
}
// Was the entire value entered a float?
if (marker != val.length()) {
throw bad_user_input();
}
// Getting here means a valid float was entered.
return f;
}
void read_array_from_user(float* arr, int& size, const int maxCapacity, std::istream& sin = std::cin) {
size = 0;
std::string input;
while (size < maxCapacity && std::getline(sin, input)) {
if (input == "Q" || input == "q" || input.length() == 0) {
return;
}
try {
arr[size] = convert_string_to_float(input);
} catch(const std::exception& e) {
std::cerr << "Bad input. Reason: " << e.what() << "\nTry again.\n";
continue;
}
++size;
}
}
void print_array(float* arr, int size, std::ostream& sout = std::cout) {
sout << "{ ";
for (int i = 0; i < size; ++i) {
sout << arr[i] << (i == size - 1 ? " " : ", ");
}
sout << "}\n";
}
int main() {
constexpr int maxArrayCapacity = 10;
float arrayOne[maxArrayCapacity];
int arrayOneSize = 0;
read_array_from_user(arrayOne, arrayOneSize, maxArrayCapacity);
print_array(arrayOne, arrayOneSize);
}
Output:
~/tmp
❯ ./a.out
1
2
3
4
q
{ 1, 2, 3, 4 }
~/tmp took 3s
❯ ./a.out
1
2
Q
{ 1, 2 }
~/tmp took 2s
❯ ./a.out
1
2
3
4
5
6
{ 1, 2, 3, 4, 5, 6 }
~/tmp took 4s
❯ ./a.out
1
2
3
4
5
6
7
8
9
0
{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 }
~/tmp took 5s
❯ compilecpp tmp_repair.cpp
~/tmp
❯ ./a.out
1
2
3
4
5
6
7
8
9
0
{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 }
~/tmp took 4s
❯ ./a.out
1
2
3
q
{ 1, 2, 3 }
~/tmp took 2s
❯ ./a.out
1
2b
Bad input. Reason: User input was not valid.
Try again.
2
3
Q
{ 1, 2, 3 }
~/tmp took 3s
❯ ./a.out
1
2
3
Q
{ 1, 2, 3 }
~/tmp took 2s
❯ ./a.out
1
2
3
{ 1, 2, 3 }
So, we can read an array with a maximum of 10 values. The reading is robust enough to handle typos or obviously bad input. But we don't have to read 10 values, it's the maximum. We keep track of the actual size ourselves.
Take note on how simple the main() function is because we put our sub-tasks into their own functions.
Reading the second array only requires a few lines in our main() now.
int main() {
constexpr int maxArrayCapacity = 10;
float arrayOne[maxArrayCapacity];
int arrayOneSize = 0;
read_array_from_user(arrayOne, arrayOneSize, maxArrayCapacity);
print_array(arrayOne, arrayOneSize);
float arrayTwo[maxArrayCapacity];
int arrayTwoSize = 0;
read_array_from_user(arrayTwo, arrayTwoSize, maxArrayCapacity);
print_array(arrayTwo, arrayOneSize);
}
Easy-peasy there. Now we need to combine the arrays "dynamically."
We know the sizes of our two arrays, so we know the size of the final array. Now we just need to allocate the memory on the heap. I'm not going to use new, which is what I assume you're supposed to do. My reasoning is that C++ has had better methods for managing dynamic allocations since 2011.
#include <memory>
// ...
int main() {
constexpr int maxArrayCapacity = 10;
float arrayOne[maxArrayCapacity];
int arrayOneSize = 0;
read_array_from_user(arrayOne, arrayOneSize, maxArrayCapacity);
print_array(arrayOne, arrayOneSize);
float arrayTwo[maxArrayCapacity];
int arrayTwoSize = 0;
read_array_from_user(arrayTwo, arrayTwoSize, maxArrayCapacity);
print_array(arrayTwo, arrayTwoSize);
int combinedArraySize = arrayOneSize + arrayTwoSize;
// This is considered dynamic; it's allocated on the heap
auto combinedArray = std::make_unique<float[]>(combinedArraySize);
int idx = 0;
for (int i = 0; i < arrayOneSize; ++i) {
combinedArray[idx] = arrayOne[i];
++idx;
}
for (int i = 0; i < arrayTwoSize; ++i) {
combinedArray[idx] = arrayTwo[i];
++idx;
}
print_array(combinedArray.get(), combinedArraySize);
}
Output:
~/tmp
❯ ./a.out
1
2
q
{ 1, 2 }
3
4
5
q
{ 3, 4, 5 }
{ 1, 2, 3, 4, 5 }
Hopefully, the most gnarly stuff is over where you get the user input, mostly due to the requirement that users can enter a letter to signal they're done. And that makes sense since now you have to do type conversions. Other languages like python make it possible with a lot less LOC (Lines Of Code) on your part. The task itself is not difficult, but ensuring users behave is a whole other beast. If you're allowed to assume well-behaved input will always be provided, you can simplify the code quite a bit.
I've written this code that writes coordinates to a text file. I then use excel to plot these coordinates and it creates a rectangle with circles in it.
It works completing fine, but I want to optimize it. I thought I could optimize it by creating another function that has the syntax for the coordinates so I can call the function rather than writing it out every time and repeating the code. However, I am not sure how to do this, I have tried but its not working so some help would be really appreciated.
#include <iostream>
#include <fstream>
#include <math.h>
using namespace std;
#define PI 3.14159265
//this is the function i am trying to create
double coordinate_syntax(double x_coordinate, double y_coordinate){
return map << x_coordinate << ' ' << y_coordinate << endl;
}
int build_environment_map(){
double x0[5]={0.8, 0.8, 1.7, 2.2, 2.25};
double y0[5]={1.5, 2.25, 1.0, 1.3, 0.25};
double r=0.125, wallLength=2.5;
double RD=PI /180;
ofstream map;
map.open("EnvironmentMap.txt");
// The following code is to create a charger based on its centre.
for (int i=0; i<360; i=i+20) {
map << x0[4] + r*cos(i*RD) << ' ' << y0[4] + r*sin(i*RD) << endl;
//coordinate_syntax( x0[4] + r*cos(i*RD), y0[4] + r*sin(i*RD) );
}
for (int i=0; i<360; i=i+10) {
map << x0[0] + r*cos(i*RD) << ' ' << y0[0] + r*sin(i*RD) << endl;
//coordinate_syntax(x0[0]+ r*cos(i*RD), y0[0]+ r*sin(i*RD));
}
for (double j=0 ; j<wallLength; j=j+0.01){
map << j << ' ' << 0 << endl;
//coordinate_syntax(j,0);
}
for (double x=0.3; x<0.55; x=x+0.01){
map << x << ' ' << 0.3 << endl;
//coordinate_syntax(x, 0.3);
}
map.close();
return 0;
}
int main(int argc, char **argv){
int i;
//calling the function
i = build_environment_map();
return 0;
}
there is code.
#include "pch.h"
#include <algorithm>
#include <iostream>
#include <vector>
#include <stdlib.h>
using namespace std;
vector<int> SearchInt(vector<int> vec, int num) {
vector<int> temp(2);
sort(begin(vec), end(vec));
int j = 0;
for (int i : vec) {
if (i > num) {
temp[0] = i;
temp[1] = j;
return { temp };
}
//cout << i << " !>= " << num << endl ;
j++;
}
cout << "NO";
exit(0);
}
int main()
{
int n;
cin >> n;
vector<int> nums(n, 0);
vector<int> NewNums(n, 0);
for (int i = 0; i < n; i++) {
cin >> nums[i];
}
if (n != nums.size()) {
cout << "://";
return 0;
}
sort(begin(nums), end(nums));
NewNums[1] = nums[nums.size() - 1];
nums.erase(nums.begin() + nums.size() - 1);
NewNums[0] = nums[nums.size() - 1];
nums.erase(nums.begin() + nums.size() - 1);
for (int j = 2; j <= NewNums.size() - 1; j++) {
NewNums[j] = SearchInt(nums, NewNums[j-1]- NewNums[j-2])[0];
nums.erase(nums.begin() + SearchInt(nums, NewNums[j] - NewNums[j - 1])[1]);
}
if (NewNums[NewNums.size()-1] < NewNums[NewNums.size() - 2] + NewNums[0]) {
cout << "YES" << endl;
for (int i : NewNums) {
cout << i << " ";
}
return 0;
}
else {
cout << "NO";
return 0;
}
}
His task is to check whether it is possible from the given Each number is less than the sum of the two adjacent ones.
(each number is less than both of two adjacent ones)
But there is a problem - with a large number of numbers, the code takes too long. Please help me to optimize it, or just give some advice.
numbers cаn not be null.
time limit: 3.0 s
n <= 500000
You are given n numbers a1, a2,…, an. Is it possible to arrange them in a circle so that each number is strictly less than the sum of its neighbors?
For example, for the array [1,4,5,6,7,8], the left array satisfies the condition, while the right array does not, since 5≥4 + 1 and 8> 1 + 6.
Input data
The first line contains one integer n (3≤n≤105) - the number of numbers.
The second line contains n integers a1, a2,…, an (1≤ai≤109) - the numbers themselves. The given numbers are not necessarily different.
Output
If there is no solution, print "NO" on the first line.
If it exists, print "YES" on the first line. After that, on the second line print n numbers - the elements of the array in the order in which they will stand on the circle. The first and last elements you print are considered neighbors on the circle. If there are multiple solutions, output any of them. You can print a circle starting with any of the numbers.
First I'll only briefly analyze technical shortcomings of your code - without analyzing its meaning. After that I'll write my solution of the problem you defined.
Performance problems of your code are due to some strange decisions:
(1) passing std::vector<int> by value and not by reference to SearchInt function - this implies allocating and copying of the whole array on each function invocation,
(2) call SearchInt two times per loop iteration in function main instead of only one,
(3) sort array within each invocation of SearchInt - it is already sorted before the loop.
To be honest your code feels ridiculously time-consuming. I'm only wondering if that was your intention to make it as slow as you possibly can...
I will not analyze correctness of your code according to problem description. To be honest even after fixing technical shortcomings your code seems to me utterly sub-optimal and quite incomprehensible - so it is just easier to solve the problem from scratch to me.
The answer to the problem as defined is YES if the biggest number is smaller than the sum of the second big and the third big and NO otherwise - this follows from the fact that all numbers are positive (in range 1 - 109 according to newly found problem description). If the answer is YES then to make a circle that satisfies the problem description you just need in a sorted sequence of input numbers switch places of the biggest number and the next big one - that's all.
Here is my code for that (for slightly relaxed input format - I'm not checking if number of items is on a separate line and that all items are on the same line - but all correct inputs will be parsed just fine):
#include <set>
#include <iostream>
int main()
{
std::multiset<unsigned> input_set;
unsigned n;
if( !( std::cin >> n ) )
{
std::cerr << "Input error - failed to read number of items." << std::endl;
return 2;
}
if( n - 3U > 105U - 3U )
{
std::cerr << "Wrong number of items value - " << n << " (must be 3 to 105)" << std::endl;
return 2;
}
for( unsigned j = 0; j < n; ++j )
{
unsigned x;
if( !( std::cin >> x ) )
{
std::cerr << "Input error - failed to read item #" << j << std::endl;
return 2;
}
if( x - 1U > 109U - 1U )
{
std::cerr << "Wrong item #" << j << " value - " << x << " (must be 1 to 109)" << std::endl;
return 2;
}
input_set.insert(x);
}
std::multiset<unsigned>::const_reverse_iterator it = input_set.rbegin();
std::multiset<unsigned>::const_reverse_iterator it0 = it;
std::multiset<unsigned>::const_reverse_iterator it1 = ++it;
if( *it0 >= *it1 + *++it )
{
std::cout << "NO (the biggest number is bigger than the sum of the second big and the third big numbers)" << std::endl;
return 1;
}
std::cout << "YES" << std::endl;
std::cout << "Circle: " << *it1 << ' ' << *it0;
do
{
std::cout << ' ' << *it;
}
while( ++it != input_set.rend() );
std::cout << std::endl;
return 0;
}
The basic problem is that we (in pairs) have been tasked with creating a program to - amongst other things - read in a txt file and display it as a sudoku board as part of our introduction to coding at Uni.
We have completed most of the code and have got it working sufficiently well. However, we are struggling with what I would consider to be a basic aspect of this. The code reads in the file and saves it to the classes private char 9*9 matrix. We have run a debugging print-out to check that the values are being saved to the matrix and this is indeed the case. The issue is with the getMatrixVal(int x, int y) function which we think is returning a pointer rather than the character. We have tried returning a pointer from the function and have tried saving to a pointer and then to a char variable. Unfortunately, we am not excellent on pointer's.
With the recent suspensions to University timetables, we are temporarily unable to get the advice and support we require and, hence, we are applying here for advice.
Also, we are using the qt programming software on a raspberry pi - could this have anything to do with our problems?
#include "grid.h"
#include <QDebug>
#include <QChar>
#include <QFile>
grid::grid(char matrix_value, const int x, const int y) {
matrix[x][y] = matrix_value;
//char c = matrix[x][y];
//qDebug() << QChar(c);
}
void grid::load_file() {
QFile file("/home/pi/Documents/ELEC1204/P6/SUDOKU.txt");
char character, c;
int i, j;
if(!file.open(QFile::ReadOnly)) {
qDebug() << "Error in opening file";
return;
} //Print error if message if unable to open file
for(i = 0; i < 9; i++) {
for(j = 0; j < 9; j++) {
SKIP: //Related to 'goto'
file.read(&character, sizeof(char)); //Read ini individual characters
if(character == '\xd' || character == '\xa')
goto SKIP;
//Skip character related to new lines
//Goes to 'SKIP' point in program
else if(character == 'X')
grid('_', i, j);
//Save 'X's as '_'
else
grid(character, i, j);
//Save numbers as they are
c = getMatrixVal(i, j);
qDebug() << "i = " << i << "| j = " << j << "| Character is: " << QChar(character) << "| Matrix character is: " << QChar(c);
//Print-out for debugging
}
}
file.close();
//Close file
}
// Possibly useful source:
// https://doc.qt.io/qt-5/qfile.html
// https://forum.qt.io/topic/60240/reading-file-byte-by-byte/5
char grid::getMatrixVal(const int x, const int y) {
char c = matrix[x][y];
//qDebug() << "Matrix[" << x << "][" << y << "] is: " << QChar(c);
return c;
}
void grid::printMatrix() {
for(int i = 0; i < 9; i++) {
for(int j = 0; j < 9; j++) {
qDebug() << matrix[i][j];
}
}
}
//Prints out matrix
//Debugging function
There is nothing wrong with getMatrixVal: it returns a character from the right place in matrix.
Your call grid(character, i, j) constructs a new grid instance that is unrelated to the current one.
Instead, just assign directly to matrix: (or use a setter function)
matrix[i][j] = character;
I am working on an assignment for school using hash tables in a structure program. Part of the assignment is writing a hash table composed of 20 primary buckets and 10 overflow buckets, each with 3 slots composed of a key and data field to disk and then restoring from it. Here is what I have so far:
#include <iostream>
#include <fstream>
#include <iomanip>
#include <stdio.h>
#include <string.h> // for strcpy()
using namespace std;
typedef char STR10[10+1];
typedef char STR20[20+1];
struct SLOT
{
STR10 key;
STR20 data;
};
struct BUCKET
{
SLOT entry[3];
int count;
BUCKET* overflow;
};
struct HASHTABLE
{
BUCKET pBkt[20];
BUCKET oBkt[10];
};
void WriteHTtoDisk (HASHTABLE ht, char *HashDisk);
void ReportHT (HASHTABLE ht, char * when);
int main()
{
int maxP = 20;
int maxO = 10;
int maxS = 3;
HASHTABLE ht;
STR10 mKey;
STR20 mData;
FILE * inFile;
inFile = fopen("DATAIN.dat","rb");
if (inFile == NULL)
{
cout << " DATAIN file access error ... \n";
cout << " Terminating application ... \n ";
cout << " Press any key ... \n ";
return -100;
}
char crLF;
while (!feof(inFile))
{
fscanf(inFile,"%10c%20c\n",mKey,mData);
mKey[10] = mData[20] = 0; // add string terminators
printf(" MyKey: %10s\n MyData: %20s\n",mKey,mData);
cin.ignore(80,'\n'), cin.get();
//InsertIntoHT (ht, mKey, mData);
}
fclose(inFile);
WriteHTtoDisk(ht, "hashTable.dat");
ReportHT (ht,"BEFORE");
return 0;
}
void WriteHTtoDisk (HASHTABLE ht, char *HashDisk)
{
FILE * HASHDISK = fopen(HashDisk, "rb");
int maxBkt = 30;
int maxSlot = 3;
for (int i = 0; i < maxBkt; i++)
{
for (int j = 0; j < maxSlot; j++)
{
fwrite(ht.pBkt[i].entry[j].key,11,sizeof(maxSlot),HASHDISK);
fwrite(ht.pBkt[i].entry[j].data,21,sizeof(maxSlot),HASHDISK);
}
}
}
void ReportHT (HASHTABLE ht, char * when)
{
int maxB = 30;
int maxS = 3;
cout << "Hash Table \n" << "Verification Report \n" << when << " Restoration" << endl;
for (int b = 0; b < maxB; b++)
{
cout << "Bucket " << (b+1) << endl;
if (b < 20)
{
for (int i = 0; i < maxS; i++)
{
cout << setw(3) << "Slot " << (i+1) << ": " << ht.pBkt[b].entry[i].key << setw(3) << ht.pBkt[b].entry[i].data << endl;
}
}
else
{
for (int i = 0; i < maxS; i++)
{
cout << setw(3) << "Slot " << (i+1) << ": " << ht.oBkt[b].entry[i].key << setw(3) << ht.oBkt[b].entry[i].data << endl;
}
}
}
}
The code compiles with no problems, but when I inspect the file, I find that it is all just gibberish and weird symbols. The data I am using was previously extracted from another file and I want to save it in the format in which it was inserted. I am sure the issue is with the lines with fwrite (I am not that experienced with C syntax as I am with C++).
The data was in the DATAIN.dat file like this:
TATUNG CO.EL PR. LONG BEACH CA
KAMERMAN LCIRRUS BEAVERTON, OR
QUADRAM COLOACH AV NORCROSS GE
AST RESEARALTON AV IRVINE CA
I am expecting the new file to look like this:
TATUNG CO.
EL PR. LONG BEACH CA
KAMERMAN L
CIRRUS BEAVERTON, OR
QUADRAM CO
LOACH AV NORCROSS GE
AST RESEAR
ALTON AV IRVINE CA
Any help would be greatly appreciated. Thank you.
It looks like your code doesn't initialize or even use the member count. When a hash bucket is empty, the count should indicate it. In C++ it's easy to implement: just add = 0 to its definition:
struct BUCKET
{
SLOT entry[3];
int count = 0;
BUCKET* overflow;
};
Also, when writing the bucket's data to a file, use the count and don't assume that all the entries in the bucket are filled.
for (int j = 0; j < ht.pBkt[i].count; j++)
...
Also, write only the required number of bytes. fwrite accepts two parameters: the size of the data elements to write and their number. Here, the size is 11 or 21, and the number is 1, because each fwrite call can only write one string to your file.
fwrite(ht.pBkt[i].entry[j].key,11,1,HASHDISK);
fwrite(ht.pBkt[i].entry[j].data,21,1,HASHDISK);
By the way, since you have a STR10 type, you can avoid magic numbers and write sizeof(STR10) instead of 11. This way, when you change the length of your string, your code will still work.