我正在编写一个代码,我应该在其中找到字符串中的单词数,知道每个单词可以被除了AZ(或az)。我写的代码只有在句子开头没有标点的情况下才能正常工作。但是,如果用引号等标点符号来启动句子(即“仅连接”,将显示3个单词而不是2),麻烦就来了。我使用Dev-C++以C++编程。您的帮助将不胜感激。我的代码如下:无法在我的字符串中找到准确的字数,以标点符号开头,例如引号
#include <cstring>
#include <iostream>
#include <conio.h>
#include <ctype.h>
using namespace std;
int getline(); //scope
int all_words(char prose[]); //scope
int main()
{
getline();
system ("Pause");
return 0;
}
int getline()
{
char prose[600];
cout << "Enter a sentence: ";
cin.getline (prose, 600);
all_words(prose);
return 0;
}
int all_words(char prose[])
{ int y, i=0, k=0; int count_words=0; char array_words[600], c;
do
{
y=prose[i++];
if (ispunct(y)) //skeep the punctuation
i++;
if ((y<65 && isspace(y)) || (y<65 && ispunct(y))) //Case where we meet spacial character or punctuation follwed by space
count_words++; //count words
if (ispunct(y)) //skeep the punctuation
i++;
}while (y); //till we have a character
cout <<endl<<" here is the number of words "<< count_words <<endl;
return 0;
}
***********************************Output******************************
Enter a sentence: "Only connect!"
here is the number of words 3
Press any key to continue . . .
对你的问题回答[1]。 [1]:http://stackoverflow.com/questions/53849/how-do-i-tokenize-a-string-in-c –