Hey Could anyone please debug these C codes for me..!? compiled but wrong output!

1.This code is running but only for a single word input

#include<stdio.h>

#include<ctype.h>

int len[100];

int l=0;

int main(){

char s[1000], words[100][50];

int i=0, j=-1, k=-1, l=-1, countw=0, MAX=0,max_index;

printf("Enter a string : ");

gets(s);

i=0;

while(s[i]!='\0'){

if(s[i]==' '){

i++;

}

else if(isalpha(s[i])){

countw++;

j++;

k=0;

while(!isspace(s[i])){

len[l]++;

words[j][k++]=s[i];

i++;

}

words[j][k]='\0';

if(len[l]>MAX){

max_index=j;

}

l++;

}

} 

printf("The longest word is : ");

for(k=0;words[max_index][k]!='\0'; k++){

printf("%c", words[max_index][k]);

}

printf("\nThe number of words in the string is : %d", countw);

return 0;

}

2. This code displays a negative infinity.

#include<stdio.h>

double f(double x){

return (1/(1+x));

}

int main(){

double x0, xn, fx0, fxn, fxi, h, Y=0.0, result=0.0, i, subint;

printf("Lower Bound, Upper Bound, Subintervals : ");

scanf("%lf, %lf, %d", &x0, &xn, &subint);

h=((xn-x0)/subint);

for (i=1.0;i<subint; i++){

fxi=f(x0+(i*h));

Y=Y+fxi;

}

fx0=f(x0);

fxn=f(xn);

h=h/2.0;

result=h*(fx0+2*Y+fxn);

printf("The result is : %lf", result);

}

3. This code too runs but stops in the middle.

#include<stdio.h>

double f(double x){

return (1/(1+x));

}

int main(){

double a, b, h, result, Yodd=0.0, Yeven=0.0;

int i, subintervals;

printf("Enter the lower bound : ");

scanf("%lf", &a);

printf("Enter the upper bound : ");

scanf("%lf", &b);

printf("Enter the number of subintervals ");

scanf("%d", subintervals);

h=((b-a)/subintervals);

for(i=1;i<subintervals; i++){

if(i%2==1){

Yodd=Yodd+f(a+(i*h));

}

else{

Yeven=Yeven+f(a+(i*h));

}

}

result=(h/3)*((f(a))+(4*Yodd)+(2*Yeven)+(f(b)));

printf("\nThe result is : %lf ", result);

return 0;

}

Reply:

1.

 #include<stdio.h>

double f(double x){
 return (1/(1+x));
}

int main(){
 double a, b, h, result, Yodd=0.0, Yeven=0.0;
 int i, subintervals;
 printf("Enter the lower bound : ");
 scanf("%lf", &a);
 printf("Enter the upper bound : ");
 scanf("%lf", &b);
 printf("Enter the number of subintervals ");

 scanf("%d", &subintervals);                                //put Here "&" Before subinterval.. it's working after that...

  h=((b-a)/subintervals);
 for(i=1;i<subintervals; i++){
  if(i%2==1){
   Yodd=Yodd+f(a+(i*h));
  }
  else{
   Yeven=Yeven+f(a+(i*h));
  }
 }
 result=(h/3)*((f(a))+(4*Yodd)+(2*Yeven)+(f(b)));
 printf("\nThe result is : %lf ", result);
 return 0;
}

 

 

Reply 2:

 

Were you able to crack your first code (i.e based on string) ? 

(or) 

Are you still looking for an answer ?

If yes, I could make it work by adding highlighted statements in the code.

##########################################################

          while((!isspace(s[i])) && (s[i] != '\0')){

                len[l]++;

                words[j][k++]=s[i];

                i++;

            }

            words[j][k]='\0';

            if(len[l]>MAX){

            max_index=j;

            MAX = len[l];

########################################################## 

Console output:

Enter a string : handles msg delay active calls

The longest word is : handles

The number of words in the string is : 5