I am trying to initialise a 2D Array in C filled with X's except for the index (2,2) which will be filled with the letter 'C'. However, when I run the code below, not only do I get a 'C' at (2,2) but for some reason, I also end up getting a 'C' at the index (1,9) (see output below).
I tried changing the width and height values and realised that it works sometimes. For example, when I make height = 10 and width = 10, I get the correct output with only one 'C' in its proper slot.
I'm quite new to C programming and have no idea why it's producing the correct output sometimes. Any help would be much appreciated!
int width = 10;
int height = 7;
int x = 2;
int y =2;
int limit = 3;
//initialising 2D array
char board[width][height];
for(int i = 0; i < height; i++){//rows
for (int j = 0; j < width; j++){//cols
if(i == y && j == x){
board[y][x] = 'C';
}
else{
board[i][j] = 'X';
}
}
}
//printing 2D array
for(int i = 0; i < height; i++){//rows
for (int j = 0; j < width; j++){//cols
printf("%c ", board[i][j]);
}
printf("\n");
}
You got the array declaration wrong.
Instead of
char board[width][height];
you need
char board[height][width];
/* Rows Cols */
Related
Program is supposed to take every pixel around the pixel and average color values.
It seems to be working fine visually, but obviously it is not as it doesn't pass the test. The only test it passes is 'pixel in the corner'.
I would really appreciate some hints on where the bug might be.
Here is the code:
//iterate through each row and column
for (int i = 0; i < height; i++)
{
for (int j = 0; j < width; j++)
{
/*create variables to define 3x3 box for each pixel
also create variables to store sum of each colour values for all elements of each box*/
int highi = i + 1;
int highj = j + 1;
int counter = 0;
int blue = 0;
int red = 0;
int green = 0;
//iterate through each element of newly created box and add sums of colour values
for(int lowi = i-1 ; lowi <= highi; lowi++)
{
if(lowi < 0)
{
lowi = i;
}
for (int lowj = j-1 ; lowj <= highj; lowj++)
{
if(lowj < 0)
{
lowj = j;
}
blue += image[lowi][lowj].rgbtBlue;
red += image[lowi][lowj].rgbtRed;
green += image[lowi][lowj].rgbtGreen;
counter++;
}
//calculate average of colour values for each pixel
image[i][j].rgbtBlue = blue/counter;
image[i][j].rgbtRed = red/counter;
image[i][j].rgbtGreen = green/counter;
}
The issue was that OP was writing to his image[][] array while reading from it, this causing it to be in error after the first read/write pass.
When creating a 2D array and setting the value of some element, other elements get set too, what could be the reason behind such anomaly?
Following is the code example.
#include <stdio.h>
#define MAX_X 240
#define MAX_Y 2
char grid[MAX_X][MAX_Y];
int main()
{
int i,j,row,col;
col = MAX_X;
row = MAX_Y;
// Init 2D array
for (i = 0; i < row; i++) {
for (j = 0; j < col; j++) {
grid[i][j] = '.';
}
}
grid[0][121] = 'X'; // << [ISSUE HERE] `X` is written into 2 elements instead of 1
// Display 2D array
for (i = 0; i < row; i++) {
for (j = 0; j < col; j++) {
printf("%c", grid[i][j]);
}
printf("\n");
}
}
Note:
Issue showed up on g++, and was reproduced using this snippet on online gdb's compiler
You define grid as grid[240][2] which means 240 rows x 2 columns, and as #wildplasser mention you swap the col and row assignment. It should be:
row = MAX_X;
col = MAX_Y;
I have puzzles that looks like this:
=== ====
=== ===
=== ====
left edge has length from 0 to 10 000 and right also, middle part is from 1 to 10 000.
So the question is if i can build a rectangle? like first puzzle has length of left edge equal to 0 and the last puzzle has right edge of length 0 and in the middle they fit perfectly?
I am given the number of puzzle i have and their params like this:
6
1 9 2
0 3 1
0 4 1
8 9 0
2 9 0
1 5 0
and result can be any of that:
2
0 3 1
1 5 0
or
3
0 3 1
1 9 2
2 9 0
or
2
0 4 1
1 5 0
But if there is no result i have to printf("no result")
I have to do this in C, I thought about doing some tree and searching it with BFS where vertices would have edge lengths and edge would have middle length and when reached 0 i would go all way up and collect numbers but it's hard to code. So i decided to do recursion but im also stuck:
#include<stdio.h>
int main(){
int a;
scanf("%d", &a);//here i get how many puzzles i have
int tab[a][3];//array for puzzles
int result[][3];//result array
int k = 0;//this will help me track how many puzzles has my result array
for(int i = 0; i < a; i++){//upload puzzles to array
for(int j = 0; j < 3; j++){
scanf("%d", &tab[i][j]);
}
}
findx(0, a, tab, result, k);//start of recursion, because start has to be length 0
}
int findx(int x, int a, int *tab[], int *result[], int k){//i am looking for puzzle with length x on start
for(int i = 0; i < a; i++){
if(tab[a][0] == x){//there i look for puzzles with x length at start
if(tab[a][2] == 0){//if i find such puzzle i check if this is puzzle with edge length zero at the end
for(int m = 0; m < 3; m++){//this for loop add to my result array last puzzle
result[k][m] = tab[a][m];
}
return print_result(result, k);//we will return result go to print_result function
}
else{//if i have puzzle with x length on the left and this is not puzzle which ends rectangle i add this puzzle
//to my result array and again look for puzzle with x equal to end length of puzzle i found there
for(int m = 0; m < 3; m++){
result[k][m] = tab[a][m];
k += 1;
}
findx(tab[a][2], a, tab, result, k);
}
}
}
printf("no result");
}
int print_result(int *result[], int k){
printf("%d", &k);//how many puzzles i have
printf("\n");
for(int i = 0; i < k; i++){//printing puzzles...
for(int j = 0; j < 3; j++){
printf("%d ", &result[i][j]);
}
printf("\n");//...in separate lines
}
}
I have an error that result array can't look like this int result[][3] because of of [] but I don't know how many puzzles I'm gonna use so?... and I have implicit declaration for both of my functions. Guys please help, I dont know much about C and its super hard to solve this problem.
I'm not sure I understand the overall logic of the problem, but you definitely are in need of some variable sized containers for result AND tab. Arrays are fixed size and must be defined at compile time. The following should at least compile without warnings:
#include<stdio.h>
#include<stdlib.h>
void print_result(int (*result)[3], int k){
printf("%d", k);//how many puzzles i have
printf("\n");
for(int i = 0; i <= k; i++){//printing puzzles...
for(int j = 0; j < 3; j++){
printf("%d ", result[i][j]);
}
printf("\n");//...in separate lines
}
}
void findx(int x, int a, int (*tab)[3], int (*result)[3], int k){//i am looking for puzzle with length x on start
for(int i = 0; i < a; i++){
if(tab[i][0] == x){//there i look for puzzles with x length at start
if(tab[i][2] == 0){//if i find such puzzle i check if this is puzzle with edge length zero at the end
for(int m = 0; m < 3; m++){//this for loop add to my result array last puzzle
result[k][m] = tab[i][m];
}
print_result(result, k);//we will return result go to print_result function
return;
}
else{//if i have puzzle with x length on the left and this is not puzzle which ends rectangle i add this puzzle
//to my result array and again look for puzzle with x equal to end length of puzzle i found there
for(int m = 0; m < 3; m++){
result[k][m] = tab[i][m];
k += 1;
///** Increase size of result **/
//int (*newptr)[3] = realloc(result, (k+1) * sizeof(int[3]));
//if (newptr)
// result = newptr;
}
findx(tab[i][2], a, tab, result, k);
}
}
}
printf("no result\n");
}
int main(){
int a;
scanf("%d", &a);//here i get how many puzzles i have
int (*tab)[3] = malloc(a * sizeof(int[3]));//array for puzzles
int (*result)[3] = malloc(a * sizeof(int[3]));//array for puzzles
int k = 0;//this will help me track how many puzzles has my result array
for(int i = 0; i < a; i++){//upload puzzles to array
for(int j = 0; j < 3; j++){
scanf("%d", &tab[i][j]);
}
}
findx(0, a, tab, result, k);//start of recursion, because start has to be length 0
}
Note that I changed the tab and result types to (*int)[3]. Due to order of operations, we need parentheses here. Because they are variable size, they require dynamic memory allocations. In the interest of brevity and readability, I did not check the returned values of malloc or realloc. In practice, you should be checking that the returned pointer is not NULL. Because we are using dynamic memory allocations, we should also use free if you plan on doing anything else with this program. Otherwise, it doesn't really matter because exiting the program will free the resources anyway. You actually don't want to free. because we are passing a pointer by value to findx and the realloc can change the address, it may come back with a different address. Also, take note that I needed to include <stdlib.h> for the dynamic memory allocations.
Additional Issues
Your functions print_results and findx are not declared when you call them in main. Your function either need to be above main or have "function prototypes" above main.
In the printfs you do not need the &. You do not want to send the address of the variable to printf. You want to send what will actually be printed.
Now what?
The program still does not provide you with the correct results. It simply outputs 0 as the result every time. This should at least give you a starting point. By changing this line in print_results:
for(int i = 0; i < k; i++){//printing puzzles...
to
for(int i = 0; i <= k; i++){//printing puzzles...
I was at least able to output 0 0 0. This seems more correct because if k is 0, we don't loop at all.
#include<stdio.h>
void findx(int x, int a, int tab[a][3], int result[200000][3], int puzzlesinresult) { //i am looking for puzzle with length x on start
for (int i = 0; i < a; i++) {
if (tab[i][0] == x) { //there i look for puzzles with x length at start
if (tab[i][2] == 0) { //if i find such puzzle i check if this is puzzle with edge length zero at the end
for (int m = 0; m < 3; m++) { //this for loop add to my result array last puzzle
result[puzzlesinresult][m] = tab[i][m];
}
return print_result(result, puzzlesinresult); //we will return result go to print_result function
} else { //if i have puzzle with x length on the left and this is not puzzle which ends rectangle i add this puzzle
//to my result array and again look for puzzle with x equal to end length of puzzle i found there
while (result[puzzlesinresult - 1][2] != tab[i][0] && puzzlesinresult > 0) {
puzzlesinresult -= 1;
}
int isusedpuzzle = 0;
for (int j = 0; j < puzzlesinresult; j++) {
if (result[j][0] == tab[i][0] && result[j][1] == tab[i][1] && result[j][2] == tab[i][2]) {
isusedpuzzle = 1;
} else {
//pass
}
}
if (isusedpuzzle == 0) {
for (int m = 0; m < 3; m++) {
result[puzzlesinresult][m] = tab[i][m];
}
puzzlesinresult += 1;
findx(tab[i][2], a, tab, result, puzzlesinresult);
}
}
}
}
}
void print_result(int result[200000][3], int puzzlesinresult) {
printf("%d\n", puzzlesinresult + 1); //how many puzzles i have
for (int i = 0; i < puzzlesinresult + 1; i++) { //printing puzzles...
for (int j = 0; j < 3; j++) {
printf("%d ", result[i][j]);
}
printf("\n"); //...in separate lines
}
exit(0);
}
int main() {
int a;
scanf("%d", & a); //here i get how many puzzles i have
int tab[a][3]; //array for puzzles
int result[100][3]; //result array
int puzzlesinresult = 0; //this will help me track how many puzzles has my result array
for (int i = 0; i < a; i++) { //upload puzzles to array
for (int j = 0; j < 3; j++) {
scanf("%d", & tab[i][j]);
}
}
for (int i = 0; i < a; i++) { //here i delete puzzles that doesnt contribute anything like 1 x 1,2 x 2,..
if (tab[i][0] == tab[i][2] && tab[i][0] != 0) {
for (int p = i; p < a; p++) {
for (int j = 0; j < 3; j++) {
tab[p][j] = tab[p + 1][j];
}
}
}
}
findx(0, a, tab, result, puzzlesinresult); //start of recursion, because start has to be length 0
printf("NONE");
}
This returns sometimes correct result. If you can find when this program fails I would rly appreciate sharing those cases with me :)
I'm trying to search a 3x3 2d array diagonally, like this:
I want to check if all boxes in the diagonal have the same value. Here is how I try to do it:
thisOne = board[0][2]; //set to 'X'
for(i = 0; i<3; i++) {
for(j = 3; j>0; j--){
if(board[i][j-1] != thisOne) {
thisOne= '\0';
}
}
}
//since all boxes were 'X', thisOne is still set to 'X'
if(thisOne != '\0') {
winner = thisOne;
printf("vinnare på nördöst\n");
}
So after running this code, winner should be 'X', if all boxes are X's. But the code does not do that, why is that?
You need to check only diagonal cells instead of checking all the cells.
You are not breaking/exiting the check loop when the first not matching char is retrieved.
Moreover your nested for does not what you guess: inner one loops into all columns of each row, but you want to che only the diagonal values...
You can easily a simple while
int i=0;
int j=2;
while ((i<3) && (j>=0) && (board[i][j] == thisOne))
{
i++;
j--;
}
// if i<3 the diagonal is not full of thisOne char
if ( i < 3)
{
}
As said by #BLUEPIXY, the problem is that the j loop is nested inside the i loop. So for every iteration in the i loop, the j loop runs 3 times on every column, instead of just working on the minor diagonal. There are several ways to fix this, although the most optimal way would be to use only one single loop and only one variable i.
for(i=0;i<3;i++) {
if(board[i][2-i]!=thisOne) {
thisOne='\0'
break;
}
}
To achieve your goal you'd simply need to decrement X iterator & Y iterator when going through your array.
Here is a simple example:
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int arr[3][3];
int it_y;
int it_x;
it_y = 0;
it_x = 2;
arr[0][0] = 0;
arr[0][1] = 1;
arr[0][2] = 2;
arr[1][0] = 3;
arr[1][1] = 4;
arr[1][2] = 5;
arr[2][0] = 6;
arr[2][1] = 7;
arr[2][2] = 8;
while (it_x < 3 && it_x >= 0)
{
printf("[%d][%d]: '%d'\n", it_y, it_x, arr[it_y][it_x]);
--it_x;
++it_y;
}
return EXIT_SUCCESS;
}
You can do like
for(int row=0,col=2; row<3; row++,col--)
{
if(board[row][col] != thisOne)
{
thisOne= '\0';
}
}
You can only check diagonal elements like this
for(i = 0, j = 3-1; i < 3; i++, j--) {
if(board[i][j] != thisOne) {
thisOne = '\0';
}
}
I need to find the duplicate elements in a two dimensional array.
route_ptr->route[0][1] = 24;
route_ptr->route[0][2] = 18;
route_ptr->route[1][1] = 25;
route_ptr->route[2][1] = 18;
route_ptr->route[3][1] = 26;
route_ptr->route[3][2] = 19;
route_ptr->route[4][1] = 25;
route_ptr->route[4][2] = 84;
Those are my data; the duplicate entries of route[2][1] (duplicate of route[0][2]) and route[4][1] (duplicate of route[1][1]) has to be found.
The solution is the duplicate 'i' value of route[i][j] which is 2 & 4 from this example.
please guide me.
#include <stdio.h>
struct route{
int route[6][6];
int no_routes_found;
int count_each_route[6];
};
int main() {
struct route *route_ptr, route_store;
route_ptr=&route_store;
int i,j,k;
// the data
route_ptr->route[0][1] = 24;
route_ptr->route[0][2] = 18;
route_ptr->route[1][1] = 25;
route_ptr->route[2][1] = 18;
route_ptr->route[3][1] = 26;
route_ptr->route[3][2] = 19;
route_ptr->route[4][1] = 25;
route_ptr->route[4][2] = 84;
route_ptr->count_each_route[0]=3;
route_ptr->count_each_route[1]=2;
route_ptr->count_each_route[2]=2;
route_ptr->count_each_route[3]=3;
route_ptr->count_each_route[4]=3;
route_ptr->no_routes_found=5;
//// process
for (i = 0; i <(route_ptr->no_routes_found) ; i++)
{
for (j = 1; j < route_ptr->count_each_route[i]; j++)
{
printf("\nroute[%d][%d] = ", i, j);
printf("%d",route_ptr->route[i][j]);
}
}
}
The solution expected is:
route[0][1] is compared by route [0][2] i.e [24 !=18]
route[0][1] and route [0][2] is compared by route[1][1] i.e [24 && 18 !=25]
route[0][1] and route[0][2] and route[1][1] is compared by route[2][1] i.e [ 24&&18&&25 is compared by 18, there is a matching element,
save the newcomer 'i' value which matches to the existence and drop it for next checking]
break the 'i' loop
route[0][1], route[0][2], route[1][1] is now compared route[3][1]
route[0][1], route[0][2], route[1][1] ,[3][1] is now compared route[3][2]
route[0][1], route[0][2], route[1][1] ,[3][1] ,[3][2] is now compared to route [4][1] i.e [ now there is a match to route[1][1], so save the newcomer 'i' value and break the 'i' loop
So i values [2 and 4] are duplicate, and that is my expected result of my code.
Got something against index zero, zero?
I also don't see the point of the pointer shenanigans.
It's a general safety thing to initialize all your data. You know, to zero or something.
The algorithm you suggest in your solution is rather hard to be faithful to, but this will find your duplicates. You have to walk through the entire array, in both dimensions, twice.
This will also match all the zeroes in your data, so you could add an exception to ignore routes values of zero.
//Cycling through the array the first time.
for (i = 0; i < 6 ; i++)
{
for (j = 0; j < 6; j++)
{
//Cycling through the array the second time
for (x = 0; x < 6 ; x++)
{
for (y = 0; y < 6; y++)
{
if(i==x && j==y)
continue;
if(routestore.route[i][j] == routestore.route[x][y])
printf("You have a match [%d][%d] = [%d][%d]", i, j, x,y);
}
}
}
}
Ok, so if you only want to see matches once, ie [0][2] == [2][1] but not [2][1] == [0][2], then you can do something like what I have below. This one made me scratch my head. Usually, when it's a simple list of items, you initialize the inner loop to the value of the outer loop, plus one. But you can't quite do that when it's a 2D array. So I gave up and made a super-lame hack-job. I'm a big fan of brute forcing things when possible. I'd normally tell you not to use pointers like this.
Now... this will still have multiple hits if you have three similar values. If that irks you then you need to start building a list and comparing hits against that as you walk through the data.
#include <stdio.h>
#include <string.h>
struct route{
int route[6][6];
int no_routes_found;
int count_each_route[6];
};
int lameAddOneAlternative(int *i, int *j)
{
if((*j)<6)
{
(*j)++;
return 1;
}
else if (*i<6)
{
(*i)++;
(*j) = 0;
return 1;
}
return 0;
}
int main(int argc, char **argv)
{
struct route routeStore;
int i,j,x,y;
memset(routeStore.route,0,sizeof(int)*36);
// the data
routeStore.route[0][1] = 24;
routeStore.route[0][2] = 18;
routeStore.route[1][1] = 25;
routeStore.route[2][1] = 18;
routeStore.route[3][1] = 26;
routeStore.route[3][2] = 19;
routeStore.route[4][1] = 25;
routeStore.route[4][2] = 84;
//Cycling through the array the first time.
for (i = 0; i < 6 ; i++)
{
for (j = 0; j < 6; j++)
{
x=i;
y=j;
//Cycling through the array the second time
while(lameAddOneAlternative(&x,&y))
{
if(routeStore.route[i][j] == 0 )
continue;
if(routeStore.route[i][j] == routeStore.route[x][y])
printf("You have a match [%d][%d], [%d][%d] == %d\n", i, j, x,y, routeStore.route[i][j] );
}
}
}
}
for (i = 0; i <(route_ptr->no_routes_found) ; i++)
{
for (j = 1; j < route_ptr-> count_each_route[i]; j++)
{
for (x = 0; x < (route_ptr->no_routes_found) ; x++)
{
for (y = 0; y < route_ptr-> count_each_route[x]; y++)
{
if(i==x && j==y)
continue;
if(route_ptr->route[i][j] == route_ptr->route[x][y])
printf("You have a match [%d][%d] = [%d][%d]\n", i, j, x,y);
}
}
}