#include<conio.h>
#include<stdio.h>
#include<iostream.h>
#define NULL 0
int main()
{
char name[20],c;
int nw=0;
int j=0;
int t=0;
char s[] = "newas"; // find the frequency of this word in abc.txt
char p[5];
FILE *fpt;
//printf("Enter the name of file to be checked:- ");
//gets(name);
fpt=fopen("abc.txt","r");
if (fpt==NULL)
{
printf("ERROR - can/'t open file %s",name);
getch();
exit(0);
}
else
{
while ((c=getc(fpt))!=EOF)
{
switch(1)
{
case 1:
if (c==' ')
{
point:
while((c=getc(fpt))==' ');
if (c!=' ')
nw=nw+1;
// if(c==' ')
// nw--;
if(j < 5)
p[j++] = c;
printf("\n %c ",p[j]);
if(j == 5)
{
if(p == s)
{
t++;
j = 0;
}
}
}
if(c==' ')
{
j = 0;
goto point;
}
}
}
}
printf("\n The no. of words is %d. ",nw);
printf("\n Freq of words %s is %d. ",s,t);
getch();
}
上面的代码给出了单词总数的正确答案，但没有给出特定单词的频率值[给定代码中的s]，请对此进行评论，如何计算特定单词的频率在文本文件中。
答案 0 :(得分：2)
当你包括iostream.h时，我猜这应该是某种形式的C ++，而不是C.如果是这样，这就是你的字频率：
#include <iostream>
#include <map>
#include <string>
#include <fstream>
using namespace std;
typedef map <string, int> FreqMap;
int main() {
FreqMap frequencies;
ifstream ifs( "words.txt" );
string word;
while( ifs >> word ) {
frequencies[ word ] += 1;
}
for ( FreqMap::const_iterator it = frequencies.begin();
it != frequencies.end(); ++it ) {
cout << it->first << " " << it->second << "\n";
}
}
答案 1 :(得分：2)
此代码查找有趣的单词：
if(p == s)
{
t++;
j = 0;
}
错了。您无法比较C中的字符串，这只会比较指针值，而不是指向的字符（字符串的内容）。
假设代码的其余部分设置正确，以便p真正指向一个真正的字符串，您可以这样做：
if(strcmp(p, s) == 0)
{
t++;
j = 0;
}
这要求p指向一个完全以0结尾的字符串，如果它指向一行中间的某个字符，则上述字符串将不起作用。
答案 2 :(得分：1)
我并没有完全回答这个问题，但这是一些可能会帮助你的反馈......
#include<conio.h>
#include<stdio.h>
#include<iostream.h>
#define NULL 0
int main()
{
/*
* GIVE YOUR VARIABLES NAMES THAT MAKE SENSE
* j, t, c, s, nw are meaningless to anybody picking up the code
*/
char name[20],c;
int nw=0;
int j=0;
int t=0;
char s[] = "newas"; // find the frequency of this word in abc.txt
/*
* Personally, I'd tend to have p as an array of 6, so that it's the same size as
* s and I'd initialize it to "", so that it's got a null terminator.
*/
char p[5];
FILE *fpt;
fpt=fopen("abc.txt","r");
if (fpt==NULL)
{
printf("ERROR - can/'t open file %s",name);
getch();
exit(0);
}
/*
* you don't need an else here... the other flow has already terminated */
*/
else
{
while ((c=getc(fpt))!=EOF)
{
/*
* What is the point of this switch statement? It may as well say if(true)
*/
switch(1)
{
case 1:
if (c==' ')
{
/*
* If you start using goto's in your code, it's usually a good sign that there's
* something wrong
*/
point:
/*
* It's hard to follow what you're doing because your variables don't have names
* and your code has no clear intent. If the while loop was in a function
* 'SkipToNextWord', the intent would be clearer, which would make it easier to find
* issues. What happens if there is a space at the end of your file?
*/
while((c=getc(fpt))==' ');
/*
* 'c' is never going to equal ' ', if it did, you'd still be in the while loop
*/
if (c!=' ')
nw=nw+1;
// if(c==' ')
// nw--;
if(j < 5)
p[j++] = c;
printf("\n %c ",p[j]);
/*
* This as written, could be a compound if statement...
* if(j == 5 && p == s)
*/
if(j == 5)
{
/*
* However, it looks like you're trying to do a string comparison?
* if(strncmp(p, s, sizeof(s)-1)==0)
*/
if(p == s)
{
t++;
/*
* This 'j=0' should be outside of the inner if, otherwise if there isn't a match
* you don't reset j to 0
*/
j = 0;
}
}
}
/*
* If you have a six letter word in your file, j is never reset to
* 0 and next time round the loop, you're not going to collect the
* letters correctly
*/
if(c==' ')
{
j = 0;
goto point;
}
}
}
}
printf("\n The no. of words is %d. ",nw);
printf("\n Freq of words %s is %d. ",s,t);
getch();
}
答案 3 :(得分：0)
我认为以下代码将回答您的问题：
#include <stdio.h>
#include <conio.h>
int main(int argc, char* argv[])
{
char* name = "abc.txt";
char* word = "newas";
FILE* fpt = fopen(name, "rt");
int c;
int nw = 0;
int t = 0;
int i;
if (fpt == NULL)
{
printf("ERROR - can't open file %s\n", name);
getch();
return 0;
}
while ((c = getc(fpt)) != EOF)
{
// Skip spaces
if (c == ' ')
continue;
// Increase num of words
nw++;
// Check match
i = 0;
while ((c != EOF) && (c != ' ') && ((char)c == word[i]) && (word[i] != '\0'))
{
c = getc(fpt);
i++;
}
if (((c == ' ') || (c == EOF)) && (word[i] == '\0'))
t++;
// skip this word
while ((c != EOF) && (c != ' '))
c = getc(fpt);
}
fclose(fpt);
printf("\n The no. of words is %d.\n", nw);
printf("\n Freq of words %s is %d.\n", word, t);
getch();
return 0;
}