并排打印两个二维数组
Printing two 2D arrays side by side
我刚开始用 C 编写一个单元,我的第一个任务是制作经典战舰游戏的简单控制台版本。
其中的一部分——也是我目前遇到麻烦的部分——是在每个回合向玩家显示信息。
这就是我们被要求的显示效果:
并且,经过最少的大惊小怪,我的看起来像这样:
(我的 Player
网格目前是空的,因为它的二维数组是空的,但如果我使用一个包含一些东西的数组,就没问题。)
我的头文件:
/* 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]);
我的 C 源文件:
#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.*/
}
}
}
最终,PlayerHidden
数组将覆盖在 PlayerReveal
数组上,当玩家的船被击沉且其身份 'revealed' 时,但现在我只想让电脑显示器的一半工作。
只需在打印内容的第二个循环之后将其添加到您的代码中即可:
printf(" %c |", row);
for(j=0; j<SIZE; j++)
{
/*Print the char at [i][j] with a '|' afterwards.*/
printf("%c|",computReveal[i][j]);
}
因此您的代码如下所示:
//...
/*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]);
}
//...
您的问题是您只打印了播放器部分。看一下代码:
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]);
}
// ...
}
您需要有另一个循环来打印 computerReveal 数组
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]);
}
// ...
}
应该可以了。您正在打印全宽行分隔符,而不是创建单元格分隔符的数组部分
我刚开始用 C 编写一个单元,我的第一个任务是制作经典战舰游戏的简单控制台版本。
其中的一部分——也是我目前遇到麻烦的部分——是在每个回合向玩家显示信息。
这就是我们被要求的显示效果:
并且,经过最少的大惊小怪,我的看起来像这样:
(我的 Player
网格目前是空的,因为它的二维数组是空的,但如果我使用一个包含一些东西的数组,就没问题。)
我的头文件:
/* 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]);
我的 C 源文件:
#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.*/
}
}
}
最终,PlayerHidden
数组将覆盖在 PlayerReveal
数组上,当玩家的船被击沉且其身份 'revealed' 时,但现在我只想让电脑显示器的一半工作。
只需在打印内容的第二个循环之后将其添加到您的代码中即可:
printf(" %c |", row);
for(j=0; j<SIZE; j++)
{
/*Print the char at [i][j] with a '|' afterwards.*/
printf("%c|",computReveal[i][j]);
}
因此您的代码如下所示:
//...
/*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]);
}
//...
您的问题是您只打印了播放器部分。看一下代码:
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]);
}
// ...
}
您需要有另一个循环来打印 computerReveal 数组
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]);
}
// ...
}
应该可以了。您正在打印全宽行分隔符,而不是创建单元格分隔符的数组部分