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Write a c program to find, no of occurance of a given word
in a file. The word is case sensitive.

Answers were Sorted based on User's Feedback



Write a c program to find, no of occurance of a given word in a file. The word is case sensitive...

Answer / vadivelt

First provide an input string, which will be saved in a file.
Then provide a word to be searched. So that it will return
the occurance of a given word in the file.

Here am creating a file and writting the data, and performing
search operation. But if you want to find a word in already
existing file please comment the functions fopen(), gets() and
fclose() which are used in write mode ie., "w+". And give the
file path in fopen() which is used in read mode.

Code here.,

#include<stdio.h>
#include<conio.h>

main()
{
FILE *fp;
/*Here ch = 'a' assignment is done only just to enter the
while loop for the 1st time*/
char ch = 'a', *p1, *dupP1, *dupP2, *src, *dupSrc, *pmain;
int count = 0;

pmain = (char *)malloc(1000);
fp = fopen("vel.txt", "w+");

if(pmain != '\0' && fp != '\0')
{
printf("ENTER THE STRING TO BE WRITTEN IN FILE\n");
pmain = gets(pmain);
fputs(pmain, fp);
}

src = (char *)malloc(20);
if(src != '\0')
{
printf("\nENTER THE STRING TO BE SEARCHED IN THE FILE\n");
gets(src);
}

p1 = (char *)malloc(100);
dupP2 = dupP1 = p1;
dupSrc = src;

fclose(fp);

fp = fopen("vel.txt", "r");
while(ch != EOF)
{
ch = fgetc(fp);
if(ch != ' ' && ch != '\n' && ch != EOF)
{
*(dupP1++) = ch;
}
else
{
*dupP1 = '\0';
while(*dupSrc != '\0' && *dupP2 != '\0')
{
if((*dupSrc == *dupP2) )
{
if((*(dupSrc + 1) == '\0'
&& *(dupP2 + 1) == '\0'))
{
count++;
}
}
else
{
break;
}
dupP2++;
dupSrc++;
}
dupP2 = dupP1 = p1;
dupSrc = src;
}
}
printf("\nIN '%d' PLACE(S) THE STRING '%s' AVAILABLE IN \
THE FILE\n", count, src);

fclose(fp);

getch();
}

Is This Answer Correct ?    4 Yes 3 No

Write a c program to find, no of occurance of a given word in a file. The word is case sensitive...

Answer / ashutosh shashi

//function return occurance of word
int findoc(char* str,char* str1)
{
char* temp = str1;
int count =0;
while(*str != NULL)
{
while((*str == *temp)&&(*temp != NULL))
{
str++;
temp++;
if(*temp == NULL)
count++;
}
temp = str1;
if(*str != *temp)
str++;
}
return count;
}

///if function is called like

char* str = "ashashushuaashu";
char* str1 = "ashu";
int count = findoc(str,str1);
printf("%d",count);
///it prints 2, because str1 is appears 2 times in str

Is This Answer Correct ?    1 Yes 5 No

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