所以我有一個根據規則格式化文本的任務,并最終列印出字串中的字符數(包括\n和空格,除了\0字串末尾的任何內容)。
基本上,一個有效的輸入以一個點結束,但我認為在最后一個點之后我還有一些空格。我嘗試了幾種方法,例如\0從字串末尾開始替換空格的回圈。似乎沒有任何作業...
編輯要求是:
- 將雙點(..)轉換為新行
- 洗掉多個空格,只留下一個,
- 確保逗號或點之前沒有空格
- 確保逗號或點后有一個空格。
- 不改變撇號之間的原始內容。
- 并驗證僅在正確的位置(新行\段落)有大寫字母。
我們需要在 main 函式中完成所有代碼(不幸的是),而我的代碼通常會將字符數誤認為是 1-2 額外的計數(可能會在最后一個點之后做額外的空格
這是我的代碼在計數字符時失敗的輸入示例
the LANGUAGE "C" is a procedural programming language .It was initially developed by "Dennis Ritchie".. the Main feAtures of "C" language include low-level access to memory, simple set of keywords, and clean style .
int main() {
char ans;
printf("*************** Welcome to the text cleaner ***************\n\n");
do
{
int length, i, j = 0;
int word, sentence, para, space;
char tin[601], tout[601], * dex, * pos;
printf("\nPlease enter text:\n");
gets_s(tin, 600);
length = strlen(tin);
dex = strchr(tin, '.'); //converts double dots to new line
while (dex != NULL)
{
if (tin[dex - tin 1] == '.') {
tin[dex - tin 1] = '\n';
}
dex = strchr(dex 1, '.');
}
length = strlen(tin);
dex = strchr(tin, ' '); //converting multiple spaces to single space
while (dex != NULL)
{
while (dex != NULL && tin[dex - tin 1] == ' ')
{
for (i = dex - tin 1; i < strlen(tin); i )
{
tin[i - 1] = tin[i];
}
dex = strchr(dex, ' ');
j ;
}
dex = strchr(dex 1, ' ');
}
tin[length - j] = '\0';
j = 0;
dex = strchr(tin, '\n');
while (dex != NULL && tin[dex-tin 1] == ' ') //delets spaces in the beggining of new row
{
for (i = dex - tin 1;i < strlen(tin);i ) {
tin[i] = tin[i 1];
}
dex = strchr(dex 1, '\n');
}
dex = strchr(tin, ','); //deletes space before comma
while (dex != NULL && tin[dex - tin - 1] == ' ')
{
for (i = dex - tin - 1; i < strlen(tin); i )
{
tin[i] = tin[i 1];
}
dex = strchr(dex 1, ',');
}
dex = strchr(tin, '.'); //deletes space before dots
while (dex != NULL && tin[dex - tin - 1] == ' ')
{
for (i = dex - tin - 1; i < strlen(tin); i )
{
tin[i] = tin[i 1];
}
dex = strchr(dex 1, '.');
}
dex = strchr(tin, ','); // adds space after comma
while (dex != NULL && tin[dex - tin 1] != ' ')
{
if (tin[dex - tin 1] != '\n')
{
tin[strlen(tin) 1] = '\0';
for (i = strlen(tin); i > dex - tin; i--)
{
if (i == dex - tin 1)
{
tin[i] = ' ';
}
else
{
tin[i] = tin[i - 1];
}
}
dex = strchr(dex 1, ',');
}
}
dex = strchr(tin, '.'); // adds space after dot
while (dex != NULL && tin[dex - tin 1] != ' ')
{
tin[strlen(tin) 1] = '\0';
if (tin[dex - tin 1] == '\n')
{
dex = strchr(dex 1, '.');
}
else
{
for (i = strlen(tin); i > dex - tin; i--)
{
if (i == dex - tin 1)
{
tin[i] = ' ';
}
else
{
tin[i] = tin[i - 1];
}
}
dex = strchr(dex 1, '.');
}
}
strcpy_s(tout, sizeof(tout), tin);
_strlwr_s(tout,sizeof(tout)); //copies and lowercasing the input string
dex = strchr(tin, '"');
if (dex != NULL) {
pos = strchr(dex 1, '"');
while (dex != NULL)
{
for (i = dex - tin; i < pos - tin; i ) {
tout[i] = tin[i];
}
dex = strchr(pos 1, '"');
if (dex)
{
pos = strchr(dex 1, '"');
}
} //making sure that the letters in the quotes have't been lowercased
}
_strupr_s(tin, sizeof(tin));
dex = strchr(tout, '.');
pos = strchr(tin, '.');
while (dex != NULL && pos != NULL)
{
tout[dex - tout 2] = tin[pos - tin 2];
dex = strchr(dex 1, '.');
pos = strchr(pos 1, '.');
}
//CAPSLOCK
dex = strchr(tout, '.'); //deletes space before dots
while (dex != NULL)
{
if (tout[dex - tout - 1] == ' ')
{
for (i = dex - tout - 1; i < strlen(tout); i )
{
tout[i] = tout[i 1];
}
}
dex = strchr(dex 1, '.');
}
if (tout[0] == ' ') {
for (i = 0 ;i < strlen(tout); i ) {
tout[i] = tout[i 1];
}
}//handeling single space in the beggining of the string
if (tout[0] >= 'a' && tout[0] <= 'z') {
tout[0] -= 32;
} //First letter always capital
word = 0;
sentence = 0;
para = 1;
space = 0;
length = strlen(tout);
for (i = 0; tout[i] != '\0';i )
{
if (tout[i] == ' ' && tout[i 1] != ' ')
word ;
}
dex = strchr(tout, '.');
while (dex != NULL)
{
sentence ;
dex = strchr(dex 1, '.');
}
dex = strchr(tout, '\n');
while (dex != NULL)
{
space ;
para ;
word ;
dex = strchr(dex 1, '\n');
}
//dex = strchr(tout, '-');
//while (dex != NULL)
//{
// word ;
// dex = strchr(dex 1, '-');
//}
printf_s("\nText after cleaning:\n------------------------------------------------------------------------------------------------\n");
printf_s("%s\n\n", tout);
printf_s("characters: %d | words: %d | sentences: %d | paragraphs: %d\n------------------------------------------------------------------------------------------------\n",length, word, sentence, para);
printf_s("\nIf you want to clean another string press (y): ");
scanf_s(" %c", &ans, 1);
if (ans == 'y')
{
gets_s(tin, 600);
}
} while (ans =='y');
uj5u.com熱心網友回復:
正如我在頂部評論中提到的,這可以在帶有狀態變數的單個回圈中完成。
幾個假設:
- 每當我們看到
..(轉換為換行符)時,它都會開始一個新段落 - 你所說的“撇號”,我稱之為雙引號(因為這是唯一有意義的事情)。
- 在引號內沒有任何轉換
- 引號 [它們自己] 被復制(即不被剝離)
不幸的是,我不得不完全重構代碼。它是注釋的。我意識到你必須只使用main. 額外的功能僅用于除錯,因此它們“不算數”:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int opt_d; // debug
int opt_quo; // preserve quote
#if DEBUG
#define dbgprt(_fmt...) \
do { \
if (opt_d) \
printf(_fmt); \
} while (0)
#else
#define dbgprt(_fmt...) do { } while (0)
#endif
#define COPYX(_chr) \
do { \
*dst = _chr; \
dbgprt("COPY %2.2X/%s\n",_chr,showchr(_chr)); \
} while (0)
#define COPY \
COPYX(chr)
#define WHITEOUT \
do { \
if (! white) \
break; \
COPYX(' '); \
white = 0; \
ctr_word = 1; \
} while (0)
const char *
showchr(int chr)
{
static char buf[10];
if ((chr >= 0x20) && (chr <= 0x7E))
sprintf(buf,"%c",chr);
else
sprintf(buf,"{%2.2X}",chr);
return buf;
}
void
showbuf(const char *buf,const char *who)
{
const char *sep = "'";
printf("%s: %s",who,sep);
for (int chr = *buf ; chr != 0; chr = *buf )
printf("%s",showchr(chr));
printf("%s\n",sep);
}
int
main(int argc,char **argv)
{
char inp[1000];
char buf[1000];
char *src;
char *dst;
--argc;
argv;
for (; argc > 0; --argc, argv) {
char *cp = *argv;
if (*cp != '-')
break;
cp = 2;
switch (cp[-1]) {
case 'd':
opt_d = ! opt_d;
break;
case 'q':
opt_quo = ! opt_quo;
break;
}
}
opt_quo = ! opt_quo;
const char *file;
if (argc > 0)
file = *argv;
else
file = "inp.txt";
FILE *xfsrc = fopen(file,"r");
if (xfsrc == NULL) {
perror(file);
exit(1);
}
while (fgets(inp,sizeof(inp),xfsrc) != NULL) {
strcpy(buf,inp);
src = buf;
dst = buf;
int quo = 0;
int white = 0;
int dot = 1;
int ctr_sent = 0;
int ctr_word = 0;
int ctr_para = 1;
for (int chr = *src ; chr != 0; chr = *src ) {
dbgprt("LOOP %2.2X/%s quo=%d white=%d dot=%d word=%d sent=%d para=%d\n",
chr,showchr(chr),quo,white,dot,
ctr_word,ctr_sent,ctr_para);
// got a quote
if (chr == '"') {
if (! quo)
WHITEOUT;
if (opt_quo)
COPY;
quo = ! quo;
continue;
}
// non-quote
else {
if (quo) {
COPY;
continue;
}
}
// got a dot
if (chr == '.') {
dot = 1;
// double dot --> newline (new paragraph)
if (*src == '.') {
COPYX('\n');
src;
ctr_para = 1;
continue;
}
COPY;
white = 1;
continue;
}
// from fgets, this can _only_ occur at the end of the buffer
if (chr == '\n') {
dot = 1;
white = 1;
COPY;
break;
}
// accumulate/skip over whitespace
if (chr == ' ') {
white = 1;
continue;
}
// output accumulated whitespace
WHITEOUT;
// got uppercase -- convert to lowercase if we're not at the start
// of a sentence
if (isupper(chr)) {
if (! dot)
chr = tolower(chr);
}
// got lowercase -- capitalize if we're just starting a sentence
else {
if (islower(chr)) {
if (dot)
chr = toupper(chr);
}
}
COPY;
// count sentences
if (dot)
ctr_sent = 1;
dot = 0;
}
*dst = 0;
showbuf(inp,"inp");
showbuf(buf,"buf");
#if 0
if (dot)
ctr_word = 1;
#endif
printf("TOTAL: length=%zu sentences=%d paragraphs=%d words=%d\n",
strlen(buf),ctr_sent,ctr_para,ctr_word);
}
fclose(xfsrc);
return 0;
}
這是程式輸出:
inp: 'the LANGUAGE "C" is a procedural programming language .It was initially developed by "Dennis Ritchie".. the Main feAtures of "C" language include low-level access to memory, simple set of keywords, and clean style .{0A}'
buf: 'The language "C" is a procedural programming language. It was initially developed by "Dennis Ritchie"{0A} The main features of "C" language include low-level access to memory, simple set of keywords, and clean style.{0A}'
TOTAL: length=214 sentences=3 paragraphs=2 words=31
更新:
太好了謝謝!我已經采取了你所做的并在我的代碼上實作了它,基本上,教授將我們限制在 stdio.h 和 string.h 庫中,所以我不能使用任何其他函式...... – Nitai Dan
別客氣!
我很高興您能夠將我的代碼合并到您的代碼中。這是所有可能的學習場景中最好的。
我不確定我是否添加了足夠的注釋以使我的演算法清晰,所以我對其進行了一些清理,并在考慮是否發布它。它張貼在下面。
根據您剛才所說,我 [仍然] 不確定是否允許創建您自己的函式。正如我所說,如果除錯函式不改變演算法,它們 [可能] 就可以了。
無論如何,這是更新的代碼:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int opt_d; // debug
int opt_quo; // preserve quote
int opt_x; // show space as hex
#define _dbgprt(_fmt...) \
printf(_fmt)
#if DEBUG
#define dbgprt(_fmt...) \
do { \
if (opt_d) \
_dbgprt(_fmt); \
} while (0)
#else
#define dbgprt(_fmt...) do { } while (0)
#endif
#if DEBUG
#define COPYX(_chr) \
do { \
*dst = _chr; \
copy = 1; \
if (_chr != src[-1]) \
_dbgprt(" COPY %s\n",_showchr(_chr,1)); \
} while (0)
#else
#define COPYX(_chr) \
do { \
*dst = _chr; \
} while (0)
#endif
#define COPY \
COPYX(chr)
// output accumulated white space and increment count of number of words
#define SPACEOUT \
do { \
if (! spc) \
break; \
spc = 0; \
COPYX(' '); \
ctr_word = 1; \
} while (0)
const char *
_showchr(int chr,int xflg)
{
static char buf[10];
int lo = xflg ? 0x21 : 0x20;
if ((chr >= lo) && (chr <= 0x7E))
sprintf(buf,"%c",chr);
else
sprintf(buf,"{%2.2X}",chr);
return buf;
}
const char *
showchr(int chr)
{
return _showchr(chr,opt_x);
}
void
showbuf(const char *buf,const char *who)
{
const char *sep = "'";
printf("%s: %s",who,sep);
for (int chr = *buf ; chr != 0; chr = *buf )
printf("%s",showchr(chr));
printf("%s\n",sep);
}
void
dbgint(const char *sym,int val)
{
do {
int ctr = isupper(sym[0]);
if (! ctr) {
if (! val)
break;
}
_dbgprt(" %s",sym);
if (ctr)
_dbgprt("%d",val);
} while (0);
}
int
main(int argc,char **argv)
{
char inp[1000];
char buf[1000];
char *src;
char *dst;
--argc;
argv;
for (; argc > 0; --argc, argv) {
char *cp = *argv;
if (*cp != '-')
break;
cp = 2;
switch (cp[-1]) {
case 'd':
opt_d = ! opt_d;
break;
case 'q':
opt_quo = ! opt_quo;
break;
case 'x':
opt_x = (*cp != 0) ? atoi(cp) : 1;
break;
}
}
opt_quo = ! opt_quo;
const char *file;
if (argc > 0)
file = *argv;
else
file = "inp.txt";
FILE *xfsrc = fopen(file,"r");
if (xfsrc == NULL) {
perror(file);
exit(1);
}
while (fgets(inp,sizeof(inp),xfsrc) != NULL) {
strcpy(buf,inp);
src = buf;
dst = buf;
// state variables
int quo = 0; // 1=within quoted string
int spc = 0; // 1=space seen
int dot = 1; // 1=period/newline seen
// counters
int ctr_sent = 0; // number of sentences
int ctr_word = 0; // number of words
int ctr_para = 1; // number of paragraphs
#if DEBUG
int copy = 0;
int ochr = 0;
#endif
for (int chr = *src ; chr != 0; chr = *src ) {
#if DEBUG
if (opt_d) {
// show if we skipped the prior char (and it was _not_ a space)
if ((! copy) && (ochr != ' '))
_dbgprt("SKIP\n");
copy = 0;
ochr = chr;
_dbgprt("LOOP %s",showchr(chr));
dbgint("W:",ctr_word);
dbgint("S:",ctr_sent);
dbgint("P:",ctr_para);
dbgint("quo",quo);
dbgint("spc",spc);
dbgint("dot",dot);
_dbgprt("\n");
}
#endif
// got a quote
if (chr == '"') {
// flush whitespace if starting a quoted string
if (! quo)
SPACEOUT;
// copy the quote
if (opt_quo)
COPY;
// flip the quote mode
quo = ! quo;
continue;
}
// non-quote
else {
// if inside a quoted string, just copy out the char verbatim
if (quo) {
COPY;
continue;
}
}
// got a dot
if (chr == '.') {
dot = 1;
// double dot --> newline (new paragraph)
if (*src == '.') {
COPYX('\n');
src;
ctr_para = 1;
continue;
}
COPY;
// force whitespace mode (ensure space after dot)
// (e.g.) change:
// i go.he goes.
// into:
// i go. he goes.
spc = 1;
continue;
}
// from fgets, this can _only_ occur at the end of the buffer
if (chr == '\n') {
dot = 1;
spc = 1;
COPY;
break;
}
// accumulate/skip over whitespace
if (chr == ' ') {
spc = 1;
continue;
}
// output accumulated whitespace
SPACEOUT;
// convert case
if (dot)
chr = toupper(chr);
else
chr = tolower(chr);
// output the current character -- it's _not_ special
COPY;
// count sentences
if (dot)
ctr_sent = 1;
// we're no longer at the start of a sentence
dot = 0;
}
*dst = 0;
if (opt_x == 1)
opt_x = 0;
showbuf(inp,"inp");
showbuf(buf,"buf");
#if 0
if (dot)
ctr_word = 1;
#endif
printf("TOTAL: length=%zu sentences=%d paragraphs=%d words=%d\n",
strlen(buf),ctr_sent,ctr_para,ctr_word);
}
fclose(xfsrc);
return 0;
}
輸出-d:
SKIP
LOOP t W:0 S:0 P:1 dot
LOOP h W:0 S:1 P:1
LOOP e W:0 S:1 P:1
LOOP W:0 S:1 P:1
LOOP L W:0 S:1 P:1 spc
COPY {20}
LOOP A W:1 S:1 P:1
LOOP N W:1 S:1 P:1
LOOP G W:1 S:1 P:1
LOOP U W:1 S:1 P:1
LOOP A W:1 S:1 P:1
LOOP G W:1 S:1 P:1
LOOP E W:1 S:1 P:1
LOOP W:1 S:1 P:1
LOOP W:1 S:1 P:1 spc
LOOP " W:1 S:1 P:1 spc
COPY {20}
LOOP C W:2 S:1 P:1 quo
LOOP " W:2 S:1 P:1 quo
LOOP W:2 S:1 P:1
LOOP i W:2 S:1 P:1 spc
COPY {20}
LOOP s W:3 S:1 P:1
LOOP W:3 S:1 P:1
LOOP a W:3 S:1 P:1 spc
COPY {20}
LOOP W:4 S:1 P:1
LOOP p W:4 S:1 P:1 spc
COPY {20}
LOOP r W:5 S:1 P:1
LOOP o W:5 S:1 P:1
LOOP c W:5 S:1 P:1
LOOP e W:5 S:1 P:1
LOOP d W:5 S:1 P:1
LOOP u W:5 S:1 P:1
LOOP r W:5 S:1 P:1
LOOP a W:5 S:1 P:1
LOOP l W:5 S:1 P:1
LOOP W:5 S:1 P:1
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP W:5 S:1 P:1 spc
LOOP p W:5 S:1 P:1 spc
COPY {20}
LOOP r W:6 S:1 P:1
LOOP o W:6 S:1 P:1
LOOP g W:6 S:1 P:1
LOOP r W:6 S:1 P:1
LOOP a W:6 S:1 P:1
LOOP m W:6 S:1 P:1
LOOP m W:6 S:1 P:1
LOOP i W:6 S:1 P:1
LOOP n W:6 S:1 P:1
LOOP g W:6 S:1 P:1
LOOP W:6 S:1 P:1
LOOP l W:6 S:1 P:1 spc
COPY {20}
LOOP a W:7 S:1 P:1
LOOP n W:7 S:1 P:1
LOOP g W:7 S:1 P:1
LOOP u W:7 S:1 P:1
LOOP a W:7 S:1 P:1
LOOP g W:7 S:1 P:1
LOOP e W:7 S:1 P:1
LOOP W:7 S:1 P:1
LOOP W:7 S:1 P:1 spc
LOOP W:7 S:1 P:1 spc
LOOP W:7 S:1 P:1 spc
LOOP W:7 S:1 P:1 spc
LOOP . W:7 S:1 P:1 spc
LOOP I W:7 S:1 P:1 spc dot
COPY {20}
LOOP t W:8 S:2 P:1
LOOP W:8 S:2 P:1
LOOP w W:8 S:2 P:1 spc
COPY {20}
LOOP a W:9 S:2 P:1
LOOP s W:9 S:2 P:1
LOOP W:9 S:2 P:1
LOOP i W:9 S:2 P:1 spc
COPY {20}
LOOP n W:10 S:2 P:1
LOOP i W:10 S:2 P:1
LOOP t W:10 S:2 P:1
LOOP i W:10 S:2 P:1
LOOP a W:10 S:2 P:1
LOOP l W:10 S:2 P:1
LOOP l W:10 S:2 P:1
LOOP y W:10 S:2 P:1
LOOP W:10 S:2 P:1
LOOP d W:10 S:2 P:1 spc
COPY {20}
LOOP e W:11 S:2 P:1
LOOP v W:11 S:2 P:1
LOOP e W:11 S:2 P:1
LOOP l W:11 S:2 P:1
LOOP o W:11 S:2 P:1
LOOP p W:11 S:2 P:1
LOOP e W:11 S:2 P:1
LOOP d W:11 S:2 P:1
LOOP W:11 S:2 P:1
LOOP b W:11 S:2 P:1 spc
COPY {20}
LOOP y W:12 S:2 P:1
LOOP W:12 S:2 P:1
LOOP " W:12 S:2 P:1 spc
COPY {20}
LOOP D W:13 S:2 P:1 quo
LOOP e W:13 S:2 P:1 quo
LOOP n W:13 S:2 P:1 quo
LOOP n W:13 S:2 P:1 quo
LOOP i W:13 S:2 P:1 quo
LOOP s W:13 S:2 P:1 quo
LOOP W:13 S:2 P:1 quo
LOOP R W:13 S:2 P:1 quo
LOOP i W:13 S:2 P:1 quo
LOOP t W:13 S:2 P:1 quo
LOOP c W:13 S:2 P:1 quo
LOOP h W:13 S:2 P:1 quo
LOOP i W:13 S:2 P:1 quo
LOOP e W:13 S:2 P:1 quo
LOOP " W:13 S:2 P:1 quo
LOOP . W:13 S:2 P:1
COPY {0A}
LOOP W:13 S:2 P:2 dot
LOOP W:13 S:2 P:2 spc dot
LOOP W:13 S:2 P:2 spc dot
LOOP W:13 S:2 P:2 spc dot
LOOP W:13 S:2 P:2 spc dot
LOOP W:13 S:2 P:2 spc dot
LOOP W:13 S:2 P:2 spc dot
LOOP W:13 S:2 P:2 spc dot
LOOP W:13 S:2 P:2 spc dot
LOOP W:13 S:2 P:2 spc dot
LOOP W:13 S:2 P:2 spc dot
LOOP W:13 S:2 P:2 spc dot
LOOP t W:13 S:2 P:2 spc dot
COPY {20}
COPY T
LOOP h W:14 S:3 P:2
LOOP e W:14 S:3 P:2
LOOP W:14 S:3 P:2
LOOP M W:14 S:3 P:2 spc
COPY {20}
COPY m
LOOP a W:15 S:3 P:2
LOOP i W:15 S:3 P:2
LOOP n W:15 S:3 P:2
LOOP W:15 S:3 P:2
LOOP f W:15 S:3 P:2 spc
COPY {20}
LOOP e W:16 S:3 P:2
LOOP A W:16 S:3 P:2
COPY a
LOOP t W:16 S:3 P:2
LOOP u W:16 S:3 P:2
LOOP r W:16 S:3 P:2
LOOP e W:16 S:3 P:2
LOOP s W:16 S:3 P:2
LOOP W:16 S:3 P:2
LOOP o W:16 S:3 P:2 spc
COPY {20}
LOOP f W:17 S:3 P:2
LOOP W:17 S:3 P:2
LOOP " W:17 S:3 P:2 spc
COPY {20}
LOOP C W:18 S:3 P:2 quo
LOOP " W:18 S:3 P:2 quo
LOOP W:18 S:3 P:2
LOOP l W:18 S:3 P:2 spc
COPY {20}
LOOP a W:19 S:3 P:2
LOOP n W:19 S:3 P:2
LOOP g W:19 S:3 P:2
LOOP u W:19 S:3 P:2
LOOP a W:19 S:3 P:2
LOOP g W:19 S:3 P:2
LOOP e W:19 S:3 P:2
LOOP W:19 S:3 P:2
LOOP i W:19 S:3 P:2 spc
COPY {20}
LOOP n W:20 S:3 P:2
LOOP c W:20 S:3 P:2
LOOP l W:20 S:3 P:2
LOOP u W:20 S:3 P:2
LOOP d W:20 S:3 P:2
LOOP e W:20 S:3 P:2
LOOP W:20 S:3 P:2
LOOP l W:20 S:3 P:2 spc
COPY {20}
LOOP o W:21 S:3 P:2
LOOP w W:21 S:3 P:2
LOOP - W:21 S:3 P:2
LOOP l W:21 S:3 P:2
LOOP e W:21 S:3 P:2
LOOP v W:21 S:3 P:2
LOOP e W:21 S:3 P:2
LOOP l W:21 S:3 P:2
LOOP W:21 S:3 P:2
LOOP a W:21 S:3 P:2 spc
COPY {20}
LOOP c W:22 S:3 P:2
LOOP c W:22 S:3 P:2
LOOP e W:22 S:3 P:2
LOOP s W:22 S:3 P:2
LOOP s W:22 S:3 P:2
LOOP W:22 S:3 P:2
LOOP t W:22 S:3 P:2 spc
COPY {20}
LOOP o W:23 S:3 P:2
LOOP W:23 S:3 P:2
LOOP m W:23 S:3 P:2 spc
COPY {20}
LOOP e W:24 S:3 P:2
LOOP m W:24 S:3 P:2
LOOP o W:24 S:3 P:2
LOOP r W:24 S:3 P:2
LOOP y W:24 S:3 P:2
LOOP , W:24 S:3 P:2
LOOP W:24 S:3 P:2
LOOP s W:24 S:3 P:2 spc
COPY {20}
LOOP i W:25 S:3 P:2
LOOP m W:25 S:3 P:2
LOOP p W:25 S:3 P:2
LOOP l W:25 S:3 P:2
LOOP e W:25 S:3 P:2
LOOP W:25 S:3 P:2
LOOP s W:25 S:3 P:2 spc
COPY {20}
LOOP e W:26 S:3 P:2
LOOP t W:26 S:3 P:2
LOOP W:26 S:3 P:2
LOOP o W:26 S:3 P:2 spc
COPY {20}
LOOP f W:27 S:3 P:2
LOOP W:27 S:3 P:2
LOOP k W:27 S:3 P:2 spc
COPY {20}
LOOP e W:28 S:3 P:2
LOOP y W:28 S:3 P:2
LOOP w W:28 S:3 P:2
LOOP o W:28 S:3 P:2
LOOP r W:28 S:3 P:2
LOOP d W:28 S:3 P:2
LOOP s W:28 S:3 P:2
LOOP , W:28 S:3 P:2
LOOP W:28 S:3 P:2
LOOP a W:28 S:3 P:2 spc
COPY {20}
LOOP n W:29 S:3 P:2
LOOP d W:29 S:3 P:2
LOOP W:29 S:3 P:2
LOOP c W:29 S:3 P:2 spc
COPY {20}
LOOP l W:30 S:3 P:2
LOOP e W:30 S:3 P:2
LOOP a W:30 S:3 P:2
LOOP n W:30 S:3 P:2
LOOP W:30 S:3 P:2
LOOP s W:30 S:3 P:2 spc
COPY {20}
LOOP t W:31 S:3 P:2
LOOP y W:31 S:3 P:2
LOOP l W:31 S:3 P:2
LOOP e W:31 S:3 P:2
LOOP W:31 S:3 P:2
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP W:31 S:3 P:2 spc
LOOP . W:31 S:3 P:2 spc
LOOP {0A} W:31 S:3 P:2 spc dot
inp: 'the LANGUAGE "C" is a procedural programming language .It was initially developed by "Dennis Ritchie".. the Main feAtures of "C" language include low-level access to memory, simple set of keywords, and clean style .{0A}'
buf: 'The language "C" is a procedural programming language. It was initially developed by "Dennis Ritchie"{0A} The main features of "C" language include low-level access to memory, simple set of keywords, and clean style.{0A}'
TOTAL: length=214 sentences=3 paragraphs=2 words=31
uj5u.com熱心網友回復:
這是一個可能的解決方案。
希望這將表明您不需要所有的重復。
我只用給出的例子對其進行了測驗,很可能仍然存在它可能會中斷的邊緣情況。您可能希望分配buffer而不是使用特定值,但您可能需要檢查是否有可能導致擴展的輸入。
關于正確位置的大寫字母,原文中除了一行之外沒有任何范圍,因此沒有段落的概念。因此,我在句子的開頭使用了大寫字母。
注意: OP 沒有指定正確的輸出是什么,因為該帖子的標題是“錯誤的字符數”,所以這是基于要求和 OP 代碼的一些見解的最佳猜測(可能如上所述沒有產生正確的結果)。
我不認為這里的重點是修復 OP 的錯誤,而是說明更接近或實作解決方案的替代方法。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
char original[] = "the LANGUAGE \"C\" is a procedural programming language .It was initially developed by \"Dennis Ritchie\".. the Main feAtures of \"C\" language include low-level access to memory, simple set of keywords, and clean style .";
int main() {
char buffer[256];
strcpy(buffer, original);
char *src, *dest;
bool quoted = false;
bool sentence = false;
int periods = 0;
bool space = false;
bool paragraph = true;
bool comma = false;
int letters = 0;
int words = 0;
int sentences = 0;
int paragraphs = 0;
src = dest = buffer;
for ( ; *src ; src ) {
if (quoted) {
switch (*src) {
case '"':
quoted = false;
break;
}
if ((*src >= 'a' && *src <= 'z') || (*src >= 'a' && *src <= 'z')) {
letters ;
} else {
if (letters) {
words ;
letters = 0;
}
}
*dest = *src;
} else {
switch (*src) {
case '"':
quoted = true;
break;
case ',':
comma = true;
continue;
case ' ':
if (letters) {
words ;
letters = 0;
}
space = true;
continue;
case '.':
if ( periods == 2) {
*dest = '.';
*dest = '\n';
periods = 0;
paragraph = true;
} else {
sentence = true;
}
continue;
}
if (comma) {
*dest = ',';
*dest = ' ';
comma = space = false;
}
if (periods) {
*dest = '.';
periods = 0;
}
if (space) {
if (!paragraph) {
*dest = ' ';
}
space = false;
}
if ((*src >= 'a' && *src <= 'z') || (*src >= 'a' && *src <= 'z')) {
letters ;
} else {
if (letters) {
words ;
letters = 0;
}
}
*dest = sentence || paragraph ? toupper(*src) : tolower(*src);
if (sentence || paragraph) {
if (letters) {
words ;
}
letters = 0;
}
if (sentence) {
sentences ;
}
if (paragraph) {
paragraphs ;
}
sentence = paragraph = false;
}
}
if (sentence) {
sentences ;
}
if (paragraph) {
paragraphs ;
}
if (periods) {
*dest = '.';
}
*dest = '\n';
*dest = '\0';
printf("\nInput Chars=%d\n\n\"%s\"\n", (int)strlen(original), original);
printf("\nOutput Chars=%d, Words=%d, Sentences=%d, Paragraphs=%d\n\n\"%s\"\n", (int)strlen(buffer), words, sentences, paragraphs, buffer);
return 0;
}
這會產生:
Input Chars=259
"the LANGUAGE "C" is a procedural programming language .It was initially developed by "Dennis Ritchie".. the Main feAtures of "C" language include low-level access to memory, simple set of keywords, and clean style ."
Output Chars=214, Words=34, Sentences=3, Paragraphs=2
"The language "C" is a procedural programming language. It was initially developed by "Dennis Ritchie".
The main features of "C" language include low-level access to memory, simple set of keywords, and clean style.
"
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