Drawing a bar graph in C - arrays

The user should input the elements of an array(maximum 20 of them). The task is to draw a bar graph using |, *, and - by using these elements. A line of minuses should be drawn instead of zero, and * instead of last member of an array. If the number is 3 for example, two | and one * above them should be drawn. The problem that I have with my code is that it sometimes takes too long to execute, draws one instead of two minuses between the array elements, and sometimes doesn't draw the line with minuses.
#include <stdio.h>
int main() {
int n[20];
int i,j,counter=0,input=0,min,max;
printf("Elements of the array: \n");
while(input!=1000){
scanf("%d", &input);
n[counter]=input;
counter++;
}
counter--;
min=n[0];
max=n[0];
for(i=1;i<counter;i++){
if(n[i]>max) max=n[i];
if(n[i]<min) min=n[i];
}
for(i=0;i<max-min+1;i++){
for(j=0;j<2*counter;j++){
if(j%2==0 && n[j/2]==max-i) printf("*");
else if(i==max) printf("-");
else if(j%2==0 && n[j/2]>0 && i<max && n[j/2]>max-i) printf("|");
else if(j%2==0 && n[j/2]<0 && i>max && n[j/2]<max-i) printf("|");
else printf(" ");
}
printf("\n");
}
}// Input:4 -3 7 0 -1 1000 Expected output:
*
|
|
* |
| |
| |
| |
---------*-----
| *
|
*
My output:
*
|
|
* |
| |
| |
| |
------*---
| *
|
*
Input: 5 4 3 2 1 2 3 4 5 Expected output:
* *
| * * |
| | * * | |
| | | * * | | |
| | | | * | | | |
---------------------------
My output:
* *
| * * |
| | * * | |
| | | * * | | |
| | | | * | | | | //

Related

Segmentation Core dumped

| |0|1|2|3|4|5|6|7|8|9|
|0| | | | | | | | | | |
|1| | | | | | | | | | |
|2| | | | | | | | | | |
|3| | | | | | | | | | |
|4| | | | | | | | | | |
|5| | | | | | | | | | |
|6| | | | | | | | | | |
|7| | | | | | | | | | |
|8| | | | | | | | | | |
|9| | | | | | | | | | |
I am trying to make a grid that looks like this. I currently keep getting Segmentation Fault Core dumped whenever I try to run the executable. The compiler is also not showing any errors. I am not sure how to print the numbers inside the grid either. Below is the code I currently have (just a part of the whole assignment). Any help is greatly appreciated.
void displayBoard(Cell board[BOARD_HEIGHT][BOARD_WIDTH], Player * player)
{
/* TODO */
int i,j;
char grid[BOARD_HEIGHT + 2][BOARD_WIDTH + 2];
for(i = 0; i < BOARD_HEIGHT; i++)
{
for(j = 0; j < BOARD_WIDTH; j++)
{
grid[BOARD_HEIGHT + 2][BOARD_WIDTH + 2] = '|';
printf("%c", grid[BOARD_HEIGHT + 2][BOARD_WIDTH + 2]);
printf("%s", EMPTY_OUTPUT);
}
printf("\n");
}
}
The BOARD_HEIGHT and BOARD_WIDTH are variables defined in a header file.
#ifndef BOARD_H
#define BOARD_H
#include "helpers.h"
#include "player.h"
#define BOARD_WIDTH 10
#define BOARD_HEIGHT 10
typedef enum cell
{
EMPTY,
BLOCKED,
PLAYER
} Cell;
#define EMPTY_OUTPUT " "
#define BLOCKED_OUTPUT "*"
Cell BOARD_1[BOARD_HEIGHT][BOARD_WIDTH];
Cell BOARD_2[BOARD_HEIGHT][BOARD_WIDTH];
typedef enum playerMove
{
PLAYER_MOVED,
CELL_BLOCKED,
OUTSIDE_BOUNDS
} PlayerMove;
void initialiseBoard(Cell board[BOARD_HEIGHT][BOARD_WIDTH]);
void loadBoard(Cell board[BOARD_HEIGHT][BOARD_WIDTH],
Cell boardToLoad[BOARD_HEIGHT][BOARD_WIDTH]);
Boolean placePlayer(Cell board[BOARD_HEIGHT][BOARD_WIDTH], Position position);
PlayerMove movePlayerForward(Cell board[BOARD_HEIGHT][BOARD_WIDTH],
Player * player);
void displayBoard(Cell board[BOARD_HEIGHT][BOARD_WIDTH], Player * player);
#endif
this second part is in the header file.
The full code can be found here a1
Your bug is on the line where you index grid beyond the end of the array:
grid[BOARD_HEIGHT + 2][BOARD_WIDTH + 2] = '|'; // Bug
This line writes beyond the end of the grid array defined locally as:
char grid[BOARD_HEIGHT + 2][BOARD_WIDTH + 2];
Note that the valid indexes into grid[][] go from 0 to BOARD_HEIGHT+1, and 0 to BOARD_WIDTH+1. Your code is accessible one past this in both dimensions.
This results in corrupting the stack, so when the displayBoard function returns, it has corrupted the stack and you get a segmentation fault. Removing this line eliminates the seg fault, but the code still needs work for function correctly.
You should reference the board array, and use indexes i and j, not hardcoded constants.
Here is a working version where I index into the board array, store the current cell value in currentCell, then display it based on a switch statement. This produces the following output given an empty board filled with zeroes:
scott> gcc -g -ogeraldTest -O0 board.c
scott> geraldTest
| |0|1|2|3|4|5|6|7|8|9|
|0| | | | | | | | | | |
|1| | | | | | | | | | |
|2| | | | | | | | | | |
|3| | | | | | | | | | |
|4| | | | | | | | | | |
|5| | | | | | | | | | |
|6| | | | | | | | | | |
|7| | | | | | | | | | |
|8| | | | | | | | | | |
|9| | | | | | | | | | |
Here is the code for main() and printBoard, both in board.c. Note that printBoard is passed the board by value, but it would be better to pass it by reference.
#include <stdio.h>
#include "board.h"
void displayBoard(Cell board[BOARD_HEIGHT][BOARD_WIDTH], Player * player)
{
int i,j;
// First display the header at the top.
printf("| |");
for(j=0; j<BOARD_WIDTH;j++)
printf("%d|",j);
printf("\n");
for(i = 0; i < BOARD_HEIGHT; i++)
{
// Display each row number
printf("|%d|",i);
for(j = 0; j < BOARD_WIDTH; j++)
{
// Bug: grid[BOARD_HEIGHT + 2][BOARD_WIDTH + 2] = '|';
Cell currentCell = board[i][j];
switch(currentCell) {
case BLOCKED:
printf("%s",BLOCKED_OUTPUT);
break;
case PLAYER:
printf("P");
break;
default:
printf("%s",EMPTY_OUTPUT);
break;
}
printf("|");
// This code is wrong
// printf("%c|", grid[BOARD_HEIGHT + 2][BOARD_WIDTH + 2]);
// printf("%s", EMPTY_OUTPUT);
}
printf("\n");
}
}
int main(int argc, char ** argv)
{
Cell myBoard[BOARD_HEIGHT][BOARD_WIDTH] = {0};
Player bob = 0;
displayBoard(myBoard, &bob);
}
I took your board.h and defined the missing types in order to get it to compile. You can likely just use your own headers as is.
#ifndef BOARD_H
#define BOARD_H
// #include "helpers.h"
// #include "player.h"
#define BOARD_WIDTH 10
#define BOARD_HEIGHT 10
typedef int Boolean; // Added
typedef int Player; // Added
typedef struct { // Added
int row;
int col;
} Position;
typedef enum cell
{
EMPTY,
BLOCKED,
PLAYER
} Cell;
#define EMPTY_OUTPUT " "
#define BLOCKED_OUTPUT "*"
Cell BOARD_1[BOARD_HEIGHT][BOARD_WIDTH];
Cell BOARD_2[BOARD_HEIGHT][BOARD_WIDTH];
typedef enum playerMove
{
PLAYER_MOVED,
CELL_BLOCKED,
OUTSIDE_BOUNDS
} PlayerMove;
void initialiseBoard(Cell board[BOARD_HEIGHT][BOARD_WIDTH]);
void loadBoard(Cell board[BOARD_HEIGHT][BOARD_WIDTH],
Cell boardToLoad[BOARD_HEIGHT][BOARD_WIDTH]);
Boolean placePlayer(Cell board[BOARD_HEIGHT][BOARD_WIDTH], Position position);
PlayerMove movePlayerForward(Cell board[BOARD_HEIGHT][BOARD_WIDTH],
Player * player);
void displayBoard(Cell board[BOARD_HEIGHT][BOARD_WIDTH], Player * player);
#endif

C programming : how to create a game board with tokens in each cell?

I am creating a game board with tokens on it for an exercise.
Its a 9x9 board which I have already created but I am having problems with adding the tokens into each cell. Whenever I add in my code for the tokens it screws up the format of my game board.
I am fairly new to C, so sorry if this might be a stupid question, but I have tried searching and fixing it myself but had no luck so far. Any help would be appreciated.
Here's my code for the gameboard:
int x,y;
for (x=0; x<9; x++)
{
printf (" +");
for (y=0; y<9; y++)
printf("---+");
printf ("\n%d", x+1);
printf (" |");
for (y=0; y<9; y++)
printf(" |");
printf ("\n");
}
printf (" +");
for (y=0; y<9; y++)
for (y=0; y<9; y++)
printf("---+");
printf ("\n");
This is my code to display the tokens:
for(x=0; x<9; x++)
{
printf(" ");
switch(board[y][x])
{
case PEG:
printf("o");
break;
case HOLE:
printf(".");
break;
case INVALID:
printf(" ");
}
}
I need "o" to display in all cells, and "." display only in the middle cell.
This is the board that I have created:
+---+---+---+---+---+---+---+
| | | | | | | |
+---+---+---+---+---+---+---+
| | | | | | | |
+---+---+---+---+---+---+---+
| | | | | | | |
+---+---+---+---+---+---+---+
| | | | | | | |
+---+---+---+---+---+---+---+
| | | | | | | |
+---+---+---+---+---+---+---+
| | | | | | | |
+---+---+---+---+---+---+---+
You can't first print the board outline/borders, then print the contents. You need to do them "interleaved", so that you don't mess up the current cursor position.
One way of doing this is to iterate for more than the board's dimensions, i.e. instead od 9x9, iterate over an 11x11 area, and use if to check if the current location is in the board or on the border.
Check the code below:
#include <stdio.h>
int main(void) {
int a[9][9];
int i,j,k;
for(i=0;i<9;i++)
printf("+---");
printf("+\n");
memset(a,0,sizeof(a));
for(i=0;i<9;i++)
{
for(j=0;j<9;j++)
{
if( i == 4 && i==j )
printf("| . ");
else
printf("| %d ",a[i][j]);
}
printf("|\n");
for(k=0;k<9;k++)
printf("+---");
printf("+\n");
}
return 0;
}
Output:
+---+---+---+---+---+---+---+---+---+
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+---+---+---+---+---+---+---+---+---+
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+---+---+---+---+---+---+---+---+---+
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+---+---+---+---+---+---+---+---+---+
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+---+---+---+---+---+---+---+---+---+
| 0 | 0 | 0 | 0 | . | 0 | 0 | 0 | 0 |
+---+---+---+---+---+---+---+---+---+
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+---+---+---+---+---+---+---+---+---+
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+---+---+---+---+---+---+---+---+---+
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+---+---+---+---+---+---+---+---+---+
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
+---+---+---+---+---+---+---+---+---+

Printing two 2D arrays side by side

I've just started doing a unit of programming in C and my first assignment is to make a simple console version of the classic Battleship game.
A part of this - and the part I'm having trouble with at the moment - is displaying information to the player at each turn.
This is what we've been asked to have our display look like:
And, with a minimum of fuss, I've got mine looking like this:
(My Player grid is empty at the moment because its 2D array is empty, but if I use an array with something in it, it's fine.)
My header file:
/* Header files. */
#include <stdio.h>
#include <stdlib.h>
#define UNKNOWN ' '
#define SIZE 10
/* Function prototypes. */
void init(char playerHidden[SIZE][SIZE], char playerReveal[SIZE][SIZE],
char computHidden[SIZE][SIZE], char computReveal[SIZE][SIZE]);
void displayKnownInfo(char playerReveal[SIZE][SIZE],
char playerHidden[SIZE][SIZE],
char computReveal[SIZE][SIZE]);
My C source file:
#include "bship.h"
int main(void)
{
/* Stores player ship position information secret to opponent. */
char playerHidden[SIZE][SIZE];
/* Stores player ship position information known to opponent. */
char playerReveal[SIZE][SIZE];
/* Stores computer ship position information secret to opponent. */
char computHidden[SIZE][SIZE];
/* Stores computer ship position information known to opponent. */
char computReveal[SIZE][SIZE];
init(playerHidden, playerReveal,
computHidden, computReveal);
displayKnownInfo(playerReveal, playerHidden, computReveal);
return EXIT_SUCCESS;
}
/****************************************************************************
* Function init() initialises every cell in the four grids to a safe default
* value. The UNKNOWN constant is used for the initialisation value.
****************************************************************************/
void init(char playerHidden[SIZE][SIZE], char playerReveal[SIZE][SIZE],
char computHidden[SIZE][SIZE], char computReveal[SIZE][SIZE])
{
/*Variables i and j for each dimension of the Arrays*/
int x,y;
/*For each increment BETWEEN 0 and 'SIZE', firstly for i;*/
for(y=0; y<SIZE; y++)
{
/*And then for j;*/
for(x=0; x<SIZE; x++)
{
/*Populate that cell with the constant UNKNOWN*/
playerHidden[x][y]=UNKNOWN;
playerReveal[x][y]=UNKNOWN;
computHidden[x][y]=UNKNOWN;
computReveal[x][y]=UNKNOWN;
}
}
}
/****************************************************************************
* Function displayKnownInfo() presents revealed information about the game in
* the format below. In this example, both contestants have made five
* guesses.
* As you can see, the computer player got lucky with all five guesses and has
* sunk the human players' aircraft carrier. The identity of the ship was
* revealed when the aircraft carrier was HIT the fifth time.
* The human player has been less lucky. The first four guesses were a MISS.
* However, the fifth guess was a HIT on the computer players' submarine. The
* human player does not yet know the identity of this ship yet as it is still
* afloat.
* All other squares are still UNKNOWN.
*
* Player | Computer
* 1 2 3 4 5 6 7 8 9 0 | 1 2 3 4 5 6 7 8 9 0
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* a |A| | | | | | | | | | | a | | | | | | | | | | |
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* b |A| | | | | | | | | | | b | | | | | | | | | | |
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* c |A| | | | | | | | | | | c | | | | | | |=| | | |
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* d |A| | | | | | | | | | | d | | |x| | | | | | | |
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* e |A| | | | | | | | | | | e | | | | | | | | | | |
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* f | | | | | | | | | | | | f | | | | |=| | | | | |
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* g | | | | | | | | | | | | g | | | | | | | | | | |
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* h | | | | | | | | | | | | h | |=| | | | |=| | | |
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* i | | | | | | | | | | | | i | | | | | | | | | | |
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* j | | | | | | | | | | | | j | | | | | | | | | | |
* +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+
* Aircraft Carrier (5/5) | 0/5 ships sunk.
* Battleship (0/4) | 1 hits.
* Destroyer (0/3) | 4 misses.
* Frigate (0/3) |
* Submarine (0/2) |
****************************************************************************/
void displayKnownInfo(char playerReveal[SIZE][SIZE],
char playerHidden[SIZE][SIZE],
char computReveal[SIZE][SIZE])
{
/*Ints for stepping through the arrays*/
int i,j;
/*First row identifier*/
char row='a';
/*Printing first few lines.*/
printf(" Player | Computer\n");
printf(" 1 2 3 4 5 6 7 8 9 0 | 1 2 3 4 5 6 7 8 9 0\n");
printf(" +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+\n");
printf(" %c |", row);
/*Loop through the arrays*/
for(i=0; i<SIZE; i++)
{
for(j=0; j<SIZE; j++)
{
/*Print the char at [i][j] with a '|' afterwards.*/
printf("%c|",playerReveal[i][j]);
}
/*After reaching column '0' on the display, increment the row identifier*/
row++;
/*And print the 'spacer' row.*/
printf("\n +-+-+-+-+-+-+-+-+-+-+ | +-+-+-+-+-+-+-+-+-+-+");
/*If the current row identifier is less than 'k', we want to print it*/
if(row<'k')
{
printf("\n %c |",row);
}
else
{
printf("\n");
}
/*And continue until the array has been printed in full.*/
}
}
}
Eventually, the PlayerHidden array will be overlaid on the PlayerReveal array when a player's ship has been sunk and its identity 'revealed', but for now I'm just wanting to get the computer half of the display working.
Just add this to your code after the second loop where you prints stuff:
printf(" %c |", row);
for(j=0; j<SIZE; j++)
{
/*Print the char at [i][j] with a '|' afterwards.*/
printf("%c|",computReveal[i][j]);
}
So your code looks like:
//...
/*Loop through the arrays*/
for(i=0; i<SIZE; i++)
{
for(j=0; j<SIZE; j++)
{
/*Print the char at [i][j] with a '|' afterwards.*/
printf("%c|",playerReveal[i][j]);
}
printf(" %c |", row);
for(j=0; j<SIZE; j++)
{
/*Print the char at [i][j] with a '|' afterwards.*/
printf("%c|",computReveal[i][j]);
}
//...
Your problem is that you're only printing out the player portion. Take a look at the code:
for(i=0; i<SIZE; i++)
{
for(j=0; j<SIZE; j++)
{
/*Print the char at [i][j] with a '|' afterwards.*/
printf("%c|",playerReveal[i][j]);
}
// ...
}
You need to have another loop where you print out the computerReveal array
for(i=0; i<SIZE; i++)
{
for(j=0; j<SIZE; j++)
{
/*Print the char at [i][j] with a '|' afterwards.*/
printf("%c|",playerReveal[i][j]);
}
for(j=0; j<SIZE; j++)
{
/*Print the char at [i][j] with a '|' afterwards.*/
printf("%c|",computReveal[i][j]);
}
// ...
}
That should do it. You're printing out the full width row separators, but not the portion of the array that creates the cell separators

How to design bar chart in c

I have created the program and generate code below
#include <stdio.h>
int main(){
int a,i;
scanf("%d",&a);
while(a!=-1){
if(a>=0 && a<=80){
for(i=a;i<=a;i++)
printf("|");
printf("%d\n");
scanf("%d",&a);
}
}
with input 1 3 4 4 5 5 -1, it should be display a bar chart like
|
| | |
| | | |
| | | |
| | | | |
| | | | |
but in that codes ,display
|
|
|
|
|
|
can anybody explain why its not working?
for(i=a;i<=a;i++)
^ maybe you want 0 here?
printf("%d\n"); also looks wrong, %d is a formate string uses to print value of int. To print simply newline char it should be just printf("\n");

Double array Strings interconnect in output - C Programming

I have this code:
#include <stdio.h>
int main(void) {
char p[5][5];
int i,j;
for(i=1;i<=1;i++){
for(j=1;j<=2;j++) {
printf("\nInput Product code %d of day %d: ", j,i);
scanf("%s", &p[i][j]);
}
}
for(i=1;i<=1;i++){
for(j=1;j<=2;j++) {
printf("\n\tProduct code %s day %d", &p[i][j],i);
}
}
}
This code outputs:
Input Product code 1 of day 1: hello
Input Product code 2 of day 1: hi
Product code hhi day 1
Product code hi day 1
Does anyone know why hello and hi are interconnecting instead of printing hello in the first one?
I have researched a lot on strings but none seem to work better than this one.
p[][] is not a double array of strings but a double array of chars. It has 5 rows and 5 columns like this:
p[row][column] =
Row| Column -> |
v | 0 | 1 | 2 | 3 | 4 |
| 0 | | | | | |
| 1 | | | | | |
| 2 | | | | | |
| 3 | | | | | |
| 4 | | | | | |
The first scanf() is called when i = 1, j = 1 and copies the string "hello" leaving p[][] like this ('\0' is the NUL character that terminates strings):
| 0 | 1 | 2 | 3 | 4 |
| 0 | | | | | |
| 1 | | H | e | l | l |
| 2 | o |\0 | | | |
| 3 | | | | | |
| 4 | | | | | |
The next time it is called, i = 1, j = 2:
| 0 | 1 | 2 | 3 | 4 |
| 0 | | | | | |
| 1 | | H | h | i |\0 |
| 2 | o |\0 | | | |
| 3 | | | | | |
| 4 | | | | | |
One way you could write this:
#include <stdio.h>
#define MAX_CODE_LEN 80 /* or something */
int main(void)
{
char *code[5][5] = {0},
currentCode[MAX_CODE_LEN] = {0};
int day = 0,
codeNum = 0;
/* start loops at 0 */
for ( day = 0; day < 5; day++ ) {
for ( codeNum = 0; codeNum < 5; codeNum++ ) {
int len = 0;
printf("\nInput Product code %d of day %d: ", codeNum, day);
scanf("%s", currentCode);
/* len doesn't include NUL but malloc needs it */
len = strlen(currentCode);
/* yoda style so compiler catches assignment instead of equality */
if ( 0 == len ) {
/* if the user didn't enter a code move onto the next day */
code[day][codeNum] = NULL;
break;
}
len = len >= MAX_CODE_LEN? MAX_CODE_LEN - 1: len;
code[day][codeNum] = malloc(len * sizeof(char) + 1);
strcpy(code[day][codeNum], currentCode);
}
}
for ( day = 0; day < 5; day++ ) {
for ( codeNum = 0; codeNum < 5; codeNum++ ) {
if ( NULL == code[day][codeNum] ) {
/* no more codes for today */
break;
}
printf("\n\tProduct code %s day %d", code[day][codeNum], day);
}
}
return 0;
}
char p[5][5];
this is a matrix of charachters and not matrix of strings
and with your scanf(), you want to put for each element in the matrix a string. And this is impossible.
You have to use one of the following definition:
1)
char p[5][5][MAX_SIZE_OF_INPUT_STRINGS];
and the scanf should be
scanf("%s", p[i][j]);
2)
char *p[5][5];
and the scanf should be
scanf("%ms", p[i][j]);
p[i][j] here is pointed to a memory allocated dynamically by scanf() and you have to free it with free(p[i][j]) when the p[i][j] become useless in your program
The reason is because strings are just arrays of chars(terminated by \0 or null. Right now, you're indexing chars, and scanf is just taking the char reference you pass it, and filling the rest of the structure with it until it's done filling with the string.
what you're doing is setting p to
* * * * *
* h e l l
o\0 * * *
* * * * *
* * * * *
and then
* * * * *
* h h i\0
o\0 * * *
* * * * *
* * * * *
If I were you, I'd populate closer to something like this
char p[5][6];
int i,j;
for(i=0;i<2;i++)
{
printf("\nInput Product code of day %d: ", i);
scanf("%s", &p[i]);
}
for(i=0;i<2;i++)
{
printf("%s\n", &p[i]);
}

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