To count Word and Blank

[alpesh]

Quote:

#include<stdio.h>
#include<conio.h>
void main()
{
char *s;
int i,b=0,w=0;
clrscr();
printf("Enter here your string:");
gets(s);
for(i=0;s[i];i++)
{
if(s[i]==' ')
b=b+1;
if(s[i]>0)
w=w+1;
}
w=w-b;
printf("\nWord=%d\n",w);
printf("Blank=%d\n",b);
getch();
}

[jamonathin]

Cool, maybe now you mite want to hit up the ctype.h predefs and apply some of those to your program.

Oh and for future reference use [ syntax="language"][ /syntax] tags.

[nireksh]

probably u calculated
blanks and letters not words and also in ur prog there r some technical mistakes

may be u dont need w
>>w is calculating nothing but stringlength
>>u can calculate that as printf("no. of letters =%d",strlen(s)-b) dirctly to get letters which u say as words
>>u dont even require s[i]>0, putting alone s[i] in if() will work


see my program

/* ...Count letters and words of a string where user can give any
no. of spaces at any place and blank string is also allowed */

#include<stdio.h>

void main()
{
char str[100];
int words=0,letters=0,i=0;

clrscr();
printf("\n"); //ff... for formatting
gets(str);

if(*str==' ')
{
while(str[i++]==' '){ } // `{}' and `;' both will work
i--;
}

if( *(str+strlen(str)-1)==' ')
words--;

if(*str=='\0') // ...3
words--;

while(str[i])
{
if (str[i++]==' ')
{
while(str[i++]==' ');
i--;
words++;
}

else
letters++;
}

printf("\nwords=%d",++words);
printf("\nletters=%d",letters);

getch();

}




/* 3. may be books are right... testing is the most
difficult phase of development life cycles */

/* TEST CASES ...
1. NULL STRING
2. BLANK SPACES
3. SPACES AFTER END OF THE STRING */

[GiffordBruno]

This program is really not working properly....I am trying to do it correct and i will be back soon. Thanks