Question

我试图在2d数组中找到一个1d数组（找到一个带有给定2d数组的单词）并且我得到的一切都是正确的，除非它打印出find_ten数组，它打印＆＃39;，0＆＃39 ;而不是＆＃39; ten＆＃39;然后打印错误的位置＆＃39; -1，-1＆＃39;。找到我的问题所在的任何帮助都会很棒。我的程序继续打印：

set found at: 0,0

sat found at: 0,0

ere found at: 0,1

,0  found at: -1,-1

axe found at: -1,-1

而不是这个，这是我想要达到的目的：

set found at: 0,0

sat found at: 0,0

ere found at: 0,1

ten  found at: 2,0

axe found at: -1,-1

以下是main的代码：

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include "find_word.h"

int main(){

    char find_set[3] = {'s', 'e', 't'};
    char find_sat[3] = {'s', 'a', 't'};
    char find_ere[3] = {'e', 'r', 'e'};
    char find_ten[3] = {'t', 'e', 'n'};
    char find_axe[3] = {'a', 'x', 'e'};
    char two_d[3][3] = {{'s', 'e', 't'}, {'a', 'r', 'e'}, {'t', 'e', 'n'}};

    word_find(two_d, find_set);
    word_find(two_d, find_sat);
    word_find(two_d, find_ere);
    word_find(two_d, find_ten);
    word_find(two_d, find_axe);

    return (0);
}

功能：

#include <stdio.h>
#include <stdlib.h>
#include "find_word.h"
#include <string.h>

void word_find(char stringof_3 [3][3], char the_word[3]){

    char *row_col;
    int print_letter = 0;
    int word_count = 0;

    if(stringof_3[0][0] == the_word[0] && stringof_3[0][1] == the_word[1] && stringof_3[0][2] == the_word[2])
            strcpy(row_col, "0,0  ");
    else if(stringof_3[0][0] == the_word[0] && stringof_3[1][0] == the_word[1] && stringof_3[2][0] == the_word[2])
            row_col = "0,0  ";
    else if(stringof_3[1][0] == the_word[0] && stringof_3[1][1] == the_word[1] && stringof_3[1][2] == the_word[2])
            row_col = "1,0  ";
    else if(stringof_3[2][0] == the_word[0] && stringof_3[2][1] == the_word[1] && stringof_3[2][2] == the_word[2])
            row_col = "2,0  ";
    else if(stringof_3[0][1] == the_word[0] && stringof_3[1][1] == the_word[1] && stringof_3[2][1] == the_word[2])
            row_col = "0,1  ";
    else if(stringof_3[0][2] == the_word[0] && stringof_3[1][2] == the_word[1] && stringof_3[2][2] == the_word[2])
            row_col = "0,2 ";
    else
            row_col = "-1,-1";

    for(print_letter = 0; print_letter < 3; print_letter++){
            printf("%c", the_word[print_letter]);
    }
    printf(" found at: %s", row_col);
    printf("\n");
}

标题：

extern void word_find(char stringof_3[3][3], char the_word[3]);  /*Inputs a 2d char array and finds the given 3 letter  word */

Answer 1

将strcpy(row_col, "0,0 ");替换为row_col= "0,0 ";

Answer 2

您没有为row_col缓冲区分配空间。你刚刚宣布它就是为什么它会给你奇怪的输出。所以请为row_col缓冲区分配空间。其次请使用strcpy而不是直接将值赋给row_col缓冲区。下面是word_find函数的代码，纠正了这两个错误。

void word_find(char stringof_3 [3][3], char the_word[3]){

    char *row_col;
    row_col = malloc(6);   // allocate space to buffer
    int print_letter = 0;
    int word_count = 0;

    if(stringof_3[0][0] == the_word[0] && stringof_3[0][1] == the_word[1] && stringof_3[0][2] == the_word[2])
            strcpy(row_col, "0,0  ");
    else if(stringof_3[0][0] == the_word[0] && stringof_3[1][0] == the_word[1] && stringof_3[2][0] == the_word[2])
            strcpy(row_col, "0,0  ");
    else if(stringof_3[1][0] == the_word[0] && stringof_3[1][1] == the_word[1] && stringof_3[1][2] == the_word[2])
            strcpy(row_col, "1,0  ");
    else if(stringof_3[2][0] == the_word[0] && stringof_3[2][1] == the_word[1] && stringof_3[2][2] == the_word[2])
            strcpy(row_col, "2,0  ");
    else if(stringof_3[0][1] == the_word[0] && stringof_3[1][1] == the_word[1] && stringof_3[2][1] == the_word[2])
            strcpy(row_col, "0,1  ");
    else if(stringof_3[0][2] == the_word[0] && stringof_3[1][2] == the_word[1] && stringof_3[2][2] == the_word[2])
            strcpy(row_col, "0,2 ");
    else
            strcpy(row_col, "-1,-1");

    for(print_letter = 0; print_letter < 3; print_letter++){
            printf("%c", the_word[print_letter]);
    }
    printf(" found at: %s", row_col);
    printf("\n");
}

C程序 - 修复打印字符数组的错误

2 个答案: