C中的摩尔斯电码转换器
是的功课
我们假设有char二维数组保持字符，大小限制为255 char char string[100][255];
程序neede：用户将输入的莫尔斯代码更改为字母/英文字母（大写字母）
示例输入
2
.... . .-.. .-.. --- / .-- --- .-. .-.. -..
.--- --- -.- .
Sample OutPut
-
情况＃1：
HELLO WORLD -
情况＃2：
JOKE
我唯一的想法是让一个单词的第一个字符由用户输入..要检查它是否是'。'或' - '然后仔细并手动分配..嵌套的if(string[i][c]=='.')和每个嵌套的最后一个if将是if(string[i][c]==' ')然后打印出字母“E”示例
if(string[i][c]=='.') {
isspace(string[i][c+1])
printf("E");
}
现在我的问题是..有没有更容易解决这个问题的方法？我不需要输入相同的' - '和'。'如果声明..和东西？我的思绪爆炸了吗？因为我用if或case语句丢失了对应下一个char的跟踪？
9 个答案:
答案 0 :(得分：11)
您已经发现可以对每个莫尔斯信号进行分支，并且将所有这些信号硬编码为if - else语句是很烦人的。当您这样做时，您将注意到具有更深嵌套条件的特定结构。您可以将此结构表示为树：
*
/ \
E T
/ \ / \
I A N M
/ \ / \ / \ / \
S U R W D K G O
/ \ / \ / \ / \ / \ / \ / \ / \
H V F * L * P J B X C Y Z Q * *
同一棵树可以在middle sections of the Wikipedia entry on Morse code中以（稍微）更漂亮的形式找到。 （最下面一行中的星号表示不是英文字母26个字母之一的编码。）
你从顶部开始。分支留在dit上，分支在dah上，并在完成后读取值。
实现树的方法有很多种。在这种情况下，树的分支都具有相同的深度，至少如果我们考虑星号也是如此。您可以通过逐行索引节点将树表示为线性数组。当toe top node为1时，你得到：
1
/ \
2 3
/ \ / \
4 5 6 7
/ \ / \ / \ / \
8 9 10 11 12 13 14 15
/ \ / \ / \ / \ / \ / \ / \ / \
16 ... ... 31
您可以看到从节点n左侧的分支将您带到节点2*n，并且右侧分支将您带到索引为2*n + 1的右邻居。从1开始构建索引，然后在数组中查找您的字母：
const char *letter = "**ETIANMSURWDKGOHVF?L?PJBXCYZQ??";
（前面的两个星号表示非法指数。）
答案 1 :(得分：2)
我的简短版本::
#include <stdio.h>
#include <string.h>
typedef struct
{
char* morse;
char* ascii;
} morse_table_t;
int main(void) {
char input[] = ".- -... -.-.";
morse_table_t table[] = { {".-", "A"},
{"-...", "B"},
{"-.-.", "C"}
/* TODO: Fill in the rest of the Morse Table Here */
};
char* segment;
int i;
segment = strtok(input, " ");
while(segment)
{
for(i=0; i<ARRAY_SIZE(table); ++i)
{
if (!strcmp(segment, table[i].morse)) puts(table[i].ascii);
}
segment = strtok(NULL, " ");
}
return 0;
}
答案 2 :(得分：1)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
static const char *alpha[] = {
".-", //A
"-...", //B
"-.-.", //C
"-..", //D
".", //E
"..-.", //F
"--.", //G
"....", //H
"..", //I
".---", //J
"-.-", //K
".-..", //L
"--", //M
"-.", //N
"---", //O
".--.", //P
"--.-", //Q
".-.", //R
"...", //S
"-", //T
"..-", //U
"...-", //V
".--", //W
"-..-", //X
"-.--", //Y
"--..", //Z
};
static const char *num[] = {
"-----", //0
".----", //1
"..---", //2
"...--", //3
"....-", //4
".....", //5
"-....", //6
"--...", //7
"---..", //8
"----.", //9
};
static const char **table[] = { alpha, num };
typedef enum kind {
ALPHA, NUM
} Kind;
typedef struct mtree {
char value;
struct mtree *dot;
struct mtree *bar;
} MTree;
MTree *root;
void make_tree(void);
void drop_tree(void);
void encode_out(const char *s);
void decode_out(const char *s);
int main(void){
make_tree();
encode_out("HELLO WORLD");
encode_out("JOKE");
decode_out(".... . .-.. .-.. --- / .-- --- .-. .-.. -..");
decode_out(".--- --- -.- .");
drop_tree();
return 0;
}
void encode_out(const char *s){
for(;;++s){
char ch = *s;
if(ch == '\0')
break;
if(isalpha(ch)){
ch = toupper(ch);
fputs(table[ALPHA][ch - 'A'], stdout);//`-'A'` depend on the sequence of character code
} else if(isdigit(ch))
fputs(table[NUM][ch - '0'], stdout);
else if(ch == ' ')
fputc('/', stdout);//need rest space skip ?
else
;//invalid character => ignore
fputc(' ', stdout);
}
fputc('\n', stdout);
}
static void decode_out_aux(MTree *tree, const char *s){
if(tree == NULL) return;
if(*s == '\0')
fputc(tree->value, stdout);
else if(*s == '/')
fputc(' ', stdout);
else if(*s == '.')
decode_out_aux(tree->dot, ++s);
else if(*s == '-')
decode_out_aux(tree->bar, ++s);
}
void decode_out(const char *s){
char *p;
while(*s){
p = strchr(s, ' ');
if(p){
if(p-s != 0){
char code[p-s+1];
memcpy(code, s, p-s);
code[p-s]='\0';
decode_out_aux(root, code);
}
s = p + 1;
} else {
decode_out_aux(root, s);
break;
}
}
fputc('\n', stdout);
}
static void insert_aux(MTree **tree, char ch, const char *s){
if(*tree == NULL)
*tree = calloc(1, sizeof(**tree));
if(*s == '\0')
(*tree)->value = ch;
else if(*s == '.')
insert_aux(&(*tree)->dot, ch, ++s);
else if(*s == '-')
insert_aux(&(*tree)->bar, ch, ++s);
}
static inline void insert(char ch, const char *s){
if(*s == '.')
insert_aux(&root->dot, ch, ++s);
else if(*s == '-')
insert_aux(&root->bar, ch, ++s);
}
void make_tree(void){
root = calloc(1, sizeof(*root));
//root->value = '/';//anything
int i;
for(i = 0; i < 26; ++i)
insert('A'+i, table[ALPHA][i]);
for(i = 0; i < 10; ++i)
insert('0'+i, table[NUM][i]);
}
static void drop_tree_aux(MTree *root){
if(root){
drop_tree_aux(root->dot);
drop_tree_aux(root->bar);
free(root);
}
}
void drop_tree(void){
drop_tree_aux(root);
}
答案 3 :(得分：0)
蛮力方法是最简单的 - 不像你提议的那样粗暴。
- 创建一个包含摩尔斯电码的输入字符串数组。
- 创建一个输出字符串数组，其中包含＃1中的字符串所代表的字符串（大多数字符串将是单个字符）。确保它与＃1中的数组完全相同。 （您可以使用二维数组或结构同时执行这两项操作，也可以使用更高级的方法执行这两种操作。使用您最了解的内容。）
- 外部循环开始：将目标字符串初始化为空。
- 从输入字符串一次读取一个字符，并且：
一个。如果它是短划线或点，请将其添加到目标字符串;
湾如果没有，请结束此循环。 - 重复＃4，直到遇到没有破折号或点的东西。
一个。将新字符串与数组＃1中的每个莫尔斯代码进行比较。找到后，从阵列＃2中写入相应的输出代码
湾跳过输入字符串中的空格;
C。如果你遇到斜线，写一个空格;和
d。如果遇到输入字符串的结尾，那么就完成了。 - 在3处重复循环，直到遇到输入结束。
答案 4 :(得分：0)
这是一个评论代码，可以回答您的问题！
#include <stdio.h>
#include <string.h>
int main()
{
/* string array will contain the whole line morse code */
char string[256]="";
/* T is the number of test c ases */
int T;
scanf("%d ",&T);
/* morse array contains all the letters from A to Z in */
/* morse code */
const char morse[][10]={
".-", //morse code of letter A
"-...", //morse code of letter B
"-.-." , //morse code of letter C
"-.." , //morse code of letter D
"." , //morse code of letter E
"..-." , //morse code of letter F
"--." , //morse code of letter G
"...." , //morse code of letter H
".." , //morse code of letter I
".---" , //morse code of letter J
"-.-" , //morse code of letter K
".-.." , //morse code of letter L
"--" , //morse code of letter M
"-." , //morse code of letter N
"---" , //morse code of letter O
".--." , //morse code of letter P
"--.-" , //morse code of letter Q
".-." , //morse code of letter R
"..." , //morse code of letter S
"-" , //morse code of letter T
"..-" , //morse code of letter U
"...-" , //morse code of letter V
".--" , //morse code of letter W
"-..-" , //morse code of letter X
"-.--" , //morse code of letter Y
"--.." //morse code of letter Z
};
/* i will be used to print the number of test case */
int i=1;
/* while loop to do every case */
while(T--)
{
printf("Case#%d:\n",i);
/* read the line of more code via f gets */
fgets(string,sizeof(string),stdin);
/* strtok is for extracting every word from the line */
char *p=strtok(string," ");
while(p!=NULL)
{
/* check if the word is / print space and go to the next word */
if(p[0]=='/')
{
printf(" ");
goto next;
}
int j=0;
for(j=0; j<26;j++)
{
//check the correspondant string in morse array
if(!strcmp(p,morse[j]))
{
/* print the equivalent letter after finding the subscript */
/* A=65+0 .... Z=90=65+25 */
printf("%c",(j+65));
}
}
next:
/* fetch the next word by ignoring space tab newline*/
p=strtok(NULL,"\t \n");
}
printf("\n");
i++;
}
return 0;
}
请记住，这不是最佳解决方案，因为例如搜索模式是线性的，而您可以在对数组进行排序后使用二进制搜索！
简单地说，上面的代码可以改进!!
希望它有所帮助!!
答案 5 :(得分：0)
#include<stdio.h>
#include<string.h>
#include<ctype.h>
#define MAX 100
#define SIZE 255
int main(){
char string[MAX][SIZE];
char destination[MAX][5];
char *input[37]={".-","-...","-.-.","-..",".","..-.","--.",
"....","..",".---","-.-",".-..","--","-.",
"---",".--.","--.-",".-.","...","-","..-",
"...-",".--","-..-","-.--","--..","-----",
".----","..---","...--","....-",".....",
"-....","--...","---..","----.","/"};
char *output[37]= {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O",
"P","Q","R","S","T","U","V","W","X","Y","Z","0","1","2","3",
"4","5","6","7","8","9"," "};
int i, c, x, m, j;
printf("Enter the number of Cases:");
scanf("%d", &x);
for(i=0;i<x;i++){
printf("Case#%d: ", i+1);
if (i==0){
gets(string[0]); }
gets(string[i]);
}
for(i=0;i<x;i++){
printf("Case#%d: ",i+1);
for(c=0,m=0;string[i][c]!='\0';c++,m++){
if(isspace(string[i][c])){
m++;}
else{
destination[m][c]=string[i][c]; }
}
for(j=0,m=0;j<37;j++){
if(destination[m]==input[j]){
printf("%d %s \n", i+1, output[j]); m++;
}
}
}
return 0;
}
I might have done something stupid here... ._. i'm just trying though.. does this not work?
答案 6 :(得分：0)
使用strtok
分割字符串的示例#include <stdio.h>
#include <string.h>
#define MAX 100
#define SIZE 255
int main(){
char string[MAX][SIZE] = {
".... . .-.. .-.. --- / .-- --- .-. .-.. -..",
".--- --- -.- ."
};
char destination[MAX][8];
int x = 2;//number of input
int i, j, m;
char *code, *separator = " ";//" " --> " \t\n"
for(i=0;i<x;++i){
j = 0;
for(code = strtok(string[i], separator);
code != NULL;
code = strtok(NULL, separator)){
printf("'%s'\n", code);
strcpy(destination[j++], code);
}
m = j;
if(strcmp(destination[0], "....")==0)
puts("yes, It's 'H'.");
}
return 0;
}
答案 7 :(得分：0)
我找到了解决方案！ ：D学分为BLUEPIXY
for(i=0,j=0;i<x;++i){
for(code = strtok(string[i], separator);code != NULL;code = strtok(NULL,separator)){
strcpy(destination[i][j++], code);} }
谢谢你们
#include<stdio.h>
#include<string.h>
#define MAX 100
#define SIZE 255
int main(){
char string[MAX][SIZE];
char destination[MAX] [MAX][8];
char *input[38]={".-","-...","-.-.","-..",".","..-.","--.",
"....","..",".---","-.-",".-..","--","-.",
"---",".--.","--.-",".-.","...","-","..-",
"...-",".--","-..-","-.--","--..","-----",
".----","..---","...--","....-",".....",
"-....","--...","---..","----.","/"};
char *output[38]={"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O",
"P","Q","R","S","T","U","V","W","X","Y","Z","0","1","2","3",
"4","5","6","7","8","9"," "};
char *code, *separator = " ";
int i, c, x, j;
int m[MAX];
printf("Enter the number of Cases:");
scanf("%d", &x);
getchar();
for(i=0;i<x;i++){
printf("Case#%d: ", i+1);
gets(string[i]);
}
for(i=0,j=0;i<x;++i){
for(code = strtok(string[i], separator);code != NULL;code = strtok(NULL, separator)){
strcpy(destination[i][j++], code);
}
m[i] = j;
}
for(i=0;i<x;i++){
printf("Case#%d: ", i+1);
for(j=0;j<m[i];j++){
for(c=0;c<37;c++){
if(strcmp(destination[i][j], input[c])==0){
printf("%s",output[c]);}
}
}
printf("\n");
}
return 0;
}
答案 8 :(得分：0)
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