C语言中基于用户输入的数组打印

我正在尝试编写一个程序，其中用户输入文本，控制台以7×5的星号网格打印文本

e、 g

但是，我在识别字符并让程序知道时遇到了困难

我迄今为止的努力：

#include <stdio.h>
#include <string.h>


char a[7][6] = { 
"     ",
"     ",
"**** ",
"    *",
"*****",
"*   *",
"*****"
};

...other letters... 

char word[20];
int i;
int j;
char letter[2];

int main() {//main

fgets(word, sizeof(word), stdin);

for (j = 0; j < strlen(word); j++) {
strcpy(letter, word[j]);
    for (i = 0; i < 7 ; i++) {
        printf("%s %d\n", letter[i]);
    }
}

printf("Total number of characters processed: %d\n", strlen(word) - 1);

return (0);
}//*main

#包括
#包括
字符a[7][6]={
"     ",
"     ",
"**** ",
"    *",
"*****",
"*   *",
"*****"
};
……其他信件。。。
字符字[20]；
int i；
int j；
字符字母[2]；
int main（）{//main
fgets（word、sizeof（word）、stdin）；
对于（j=0；j

字母以单独的阵列制作，并通过回路逐行打印，以便它们在水平方向上一个接一个地打印

---

我最好的想法是使用变量

字母

，它会将值更改为从

word

读取的当前字符，但我知道它不正确。

我将所有字母放入一个数组：

char letters[26][7][6] = {{ 
    "     ",
    "     ",
    "**** ",
    "    *",
    "*****",
    "*   *",
    "*****"
    },
    {
    "*    ",
    "*    ",
    "*    ",
    "*    ",
    "*****",
    "*   *",
    "*****"
    },
    ...
};

然后你可以得到这样的正确答案：

int letterIdx;
for (j = 0; j < strlen(word); j++) {
    // make sure this character is a letter, if not go to the next one
    if (!isalpha(word[i]) continue;
    // subtract the ASCII value of 'a' from the ASCII value of
    // the lower case version of the current letter to get a number between 0 and 25
    letterIdx = tolower(word[i]) - 'a';  
    for (i = 0; i < 7 ; i++) {
        printf("%s\n", letters[letterIdx][i]);
    }
}

int-letterIdx；
对于（j=0；j

这是一个以星号形式打印整个单词的工作示例（代码注释）：

---

你想只使用C还是使用C++特征？BTW，我认为这是一个很好的方法，将整个字母映射到星号网格中，然后将输入解析为一个函数，该函数将字符串/字符转换成网格（多维数组）中转换的星号。你可以用C水平打印。你介意扩展一下你的意思吗？我想把所有的字母都放在一个3d数组中，但我不知道最终如何逐行调用它们。你介意进一步解释一下吗？特别是第三行和第四行，以及程序如何使用输入来查找正确的co数组中字母的组合。

int letterIdx;
for (j = 0; j < strlen(word); j++) {
    // make sure this character is a letter, if not go to the next one
    if (!isalpha(word[i]) continue;
    // subtract the ASCII value of 'a' from the ASCII value of
    // the lower case version of the current letter to get a number between 0 and 25
    letterIdx = tolower(word[i]) - 'a';  
    for (i = 0; i < 7 ; i++) {
        printf("%s\n", letters[letterIdx][i]);
    }
}

#include <stdio.h>
#include <stdlib.h>

#define maxchar 10      // number of characters in asterisk form
#define maxrows 6               // vertical length 
#define maxcols 5               // horizontal length
#define imax    5               // number of chars to be displayed in asterisk form

int main()
{
  // testing word with spaces
  char number[imax] = "92 02";

  // array of numbers
  char letter[maxrows][maxchar][maxcols] = 
    { //  j=0 ,    j=1 ,    j=2 ,    j=3 ,    j=4 ,    j=5 ,    j=6 ,    j=7 ,    j=8 ,    j=9
      { " ** ",  "*** ",  "*** ",  "*** ",  "*  *",  "****",  "*   ",  "****",  " ** ",  " ***" },  // row=0
      { "*  *",  "  * ",  "   *",  "   *",  "*  *",  "*   ",  "*   ",  "   *",  "*  *",  "*  *" },  // row=1
      { "*  *",  "  * ",  "   *",  " ** ",  "****",  "*** ",  "*** ",  "   *",  " ** ",  "*  *" },  // row=2
      { "*  *",  "  * ",  "*** ",  "   *",  "   *",  "   *",  "*  *",  "   *",  "*  *",  " ***" },  // row=3
      { "*  *",  "  * ",  "*   ",  "   *",  "   *",  "   *",  "*  *",  "   *",  "*  *",  "   *" },  // row=4
      { " ** ",  "****",  "****",  "*** ",  "   *",  "*** ",  "*** ",  "   *",  " ** ",  "   *" }   // row=5
    };


  // "iterators"
  int row, col, i;

  for(row=0; row < maxrows; ++row) // print from up to down
    {
      for(i=0; i < imax; ++i)      // for each character in array "number"
        {
          int j = number[i] - '0'; // get position in "letter". For characters, change to  j = number[i] - 'a';

          if(j<0) 
            printf("    ");        // if number[i] is not a valid character, print a space   
          else
            for(col = 0; col < maxcols; ++col)    
              printf("%c", letter[row][j][col]);   // print each * at given row

          printf("  ");            // print a small space between characters
        }
      printf("\n");                // proceed to next row
    }

  return 0;
}

  ***  ***          **   ***   
 *  *     *        *  *     *  
 *  *     *        *  *     *  
  ***  ***         *  *  ***   
    *  *           *  *  *     
    *  ****         **   ****