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to print in the reverse order, ("Hello World" -> "World Hello")

P: n/a
Hello,

As the subject suggests, I need to print the string in the reverse
order. I made the following program:

# include<stdio.h>

struct llnode
{
char *info;
struct llnode *next;
};

typedef struct llnode NODE;

int main()
{
char msg[50],word[10],*str;
int i=0,length=0,j=0;
NODE *ptr,*front=NULL,*temp,*last=NULL;

printf("Enter the sentence: ");
str=fgets(msg,sizeof(msg),stdin);

while(str[i]!='\0')
{
if((str[i]==' ')||(str[i]=='\n'))
{
word[j]='\0';
j=0;
ptr=(NODE *)malloc(sizeof(NODE));
ptr->info=word;
ptr->next=NULL;

if(front==NULL)
{
front=ptr; // only change the value of
front here
}
else
{
temp=front;
while((temp->next)!=NULL)
{
temp=temp->next;
}
temp->next=ptr;
}
printf("\n##%s\n",front->info); // prints the
words and not //
the first word
}
else
{
word[j]=str[i];
j++;
}
i++;
}

temp=front;
while(temp)
{
length++;
printf("%s ",temp->info);
temp=temp->next;
}
printf("\nLength of Linked List(or, number of words):
%d\n",length);

i=0;
printf("\n************************\n");

while(i<length)
{
temp=front;
while(temp->next!=last)
{
temp=temp->next;
}
last=temp;
printf("%s ",temp->info);
i++;
}

return 0;
}

Here, front is a poiter to the first node. But when I print
front->info, I get different words, I mean "Hello" once and "World",
the second time. I nowhere change the value of fromt except in the
first if statement.

The length of the linked list is printed correctly. And sorry for
pasting such a big code in the post itself, I had no other option.

The output is:

[vijay@vijay ds]$ ./a.out
Enter the sentence: Hello World

##Hello

##World
World World
Length of Linked List(or, number of words): 2

************************
World World [vijay@vijay ds]$
regards,
vijay.

Nov 14 '05 #1
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8 Replies


P: n/a
vijay wrote:
Hello,

As the subject suggests, I need to print the string in the reverse
order.


I came up with this:

#include <stdio.h>
#include <string.h>

void
print_it(const char *str)
{
char *ptr, *lbuf;

if (str == NULL || strlen(str) == 0) return;
lbuf = strdup(str);

printf("printing '%s' in reverse order:\n", str);

while ((ptr = strrchr(lbuf, ' ')) != NULL) {
printf("%s ", ptr + 1);
*ptr = '\0';
}
printf("%s\n", lbuf);
free(lbuf);
}

int
main(void)
{
print_it(NULL);
print_it("");
print_it("world");
print_it("hello world");
print_it("i am lucky i can program c");
}

Nov 14 '05 #2

P: n/a

Peteris Krumins wrote:
vijay wrote:
Hello,

As the subject suggests, I need to print the string in the reverse
order.


I came up with this:


Forgot to #include <stdlib.h>
and return an int from main()
P.Krumins

Nov 14 '05 #3

P: n/a
vijay wrote:
Hello,

As the subject suggests, I need to print the string in the reverse
order. I made the following program:
Please do not use // comments in code posted to usenet as
linebreaks can introduce unintentional bugs which are not
there in your version of the code.
Apart from that, less indentation helps keeping the lines
at a sensible length.
# include<stdio.h>

struct llnode
{
char *info;
struct llnode *next;
};
(*)
Note: info just _points_ to something but does provide
no storage.

typedef struct llnode NODE;

int main()
{
char msg[50],word[10],*str;
int i=0,length=0,j=0;
NODE *ptr,*front=NULL,*temp,*last=NULL;

printf("Enter the sentence: ");
str=fgets(msg,sizeof(msg),stdin);
Check the return value of fgets() -- it may be NULL!

while(str[i]!='\0')
{
If you use a pointer, you might "really" do it.
I.e.
for (str=msg; *str != '\0'; str++)
and all occurrences of str[i] replaced by *str.
Otherwise, I suggest a i = 0 directly before the
loop, so later changes between the initialization
and the loop start do not introduce errors there.
if((str[i]==' ')||(str[i]=='\n'))
{
word[j]='\0';
j=0;
ptr=(NODE *)malloc(sizeof(NODE));
Check the return value of malloc() -- it returns NULL if
no memory could be allocated. You then should do some error
handling and maybe die gracefully.
The cast is completely unnecessary and potentially dangerous.
ptr->info=word;
ptr->next=NULL;
Now, ptr->info points to &word[0].
For every NODE you create, ptr->info contains the
_same_ address. Thus, you get only the content of
word[]. What you want is to make str[i] = 0 and point
ptr->info at &str[i-j].
if(front==NULL)
{
front=ptr; // only change the value of
front here
}
else
{
temp=front;
while((temp->next)!=NULL)
{
temp=temp->next;
}
temp->next=ptr;
}
This is very inefficient.
Create an empty "front" NODE and keep a "curr" NODE* to point at the
last node in the list. Then, you allocate
curr->next;
If this works, you set curr = curr->next; curr->next = NULL;
curr->info = ....

Alternatively, you could _prepend_ the node to the list.
I.e. you allocate and initialise ptr, and then make ptr->next = first;
first = ptr. Then, your list is already sorted the way you want.
printf("\n##%s\n",front->info); // prints the
words and not //
the first word
}
else
{
word[j]=str[i];
j++;
}
i++;
}

temp=front;
while(temp)
{
length++;
printf("%s ",temp->info);
temp=temp->next;
}
printf("\nLength of Linked List(or, number of words):
%d\n",length);

i=0;
printf("\n************************\n");

while(i<length)
{
temp=front;
while(temp->next!=last)
{
temp=temp->next;
}
last=temp;
printf("%s ",temp->info);
i++;
}

return 0;
}

Here, front is a poiter to the first node. But when I print
front->info, I get different words, I mean "Hello" once and "World",
the second time. I nowhere change the value of fromt except in the
first if statement.

The length of the linked list is printed correctly. And sorry for
pasting such a big code in the post itself, I had no other option.

The output is:

[vijay@vijay ds]$ ./a.out
Enter the sentence: Hello World

##Hello

##World
World World
Length of Linked List(or, number of words): 2

************************
World World [vijay@vijay ds]$
regards,
vijay.

--
E-Mail: Mine is an /at/ gmx /dot/ de address.
Nov 14 '05 #4

P: n/a
Peteris Krumins wrote:
vijay wrote:
Hello,

As the subject suggests, I need to print the string in the reverse
order.

I came up with this:

#include <stdio.h>
#include <string.h>

void
print_it(const char *str)
{
char *ptr, *lbuf;

if (str == NULL || strlen(str) == 0) return;
lbuf = strdup(str);


strdup() is not a standard library function, so you should
grace us with a prototype and description or do an example
implementation.

printf("printing '%s' in reverse order:\n", str);

while ((ptr = strrchr(lbuf, ' ')) != NULL) {
printf("%s ", ptr + 1);
*ptr = '\0';
}
printf("%s\n", lbuf);
free(lbuf);
}

int
main(void)
{
print_it(NULL);
print_it("");
print_it("world");
print_it("hello world");
print_it("i am lucky i can program c");
}

--
E-Mail: Mine is an /at/ gmx /dot/ de address.
Nov 14 '05 #5

P: n/a
Michael Mair wrote:

strdup() is not a standard library function, so you should
grace us with a prototype and description or do an example
implementation.


Excuse me, I forgot that strdupn was not a stdlib. function.

My implementation of strdup is as follows:

#include <string.h>
#include <stdlib.h>

char
*strdup(const char *s)
{
int size;
char *newthing;

if (s == NULL) return NULL;
size = strlen(s) + 1;
if ((newthing = malloc(size)) == NULL) return NULL;
return memcpy(newthing, s, size);
}
P.Krumins

Nov 14 '05 #6

P: n/a
"vijay" <Ta***********@yahoo.com> wrote:
# Hello,
#
# As the subject suggests, I need to print the string in the reverse
# order. I made the following program:

If that's all you want, you can write a function say f

f(string) =
if string is not empty,
let w = last entity of string
let p = first part of string (may be empty)
f(p)
print w

For example to reverse characters

...
void f(char *s,int n) {
if (n>0) {
char w = s[n-1];
f(s,n-1);
fputc(w,stdout);
}
}
...
f(string,strlen(string));
...

--
SM Ryan http://www.rawbw.com/~wyrmwif/
We found a loophole; they can't keep us out anymore.
Nov 14 '05 #7

P: n/a
<snip>
ptr->info=word;
ptr->next=NULL;
Now, ptr->info points to &word[0].
For every NODE you create, ptr->info contains the
_same_ address. Thus, you get only the content of
word[]. What you want is to make str[i] = 0 and point
ptr->info at &str[i-j].


This was the problem. It always pointed to the last word, and therefore
"front" always printed the last word.
if(front==NULL)
{
front=ptr; // only change the value of front here
}
else
{
temp=front;
while((temp->next)!=NULL)
{
temp=temp->next;
}
temp->next=ptr;
}
This is very inefficient.
Create an empty "front" NODE and keep a "curr" NODE* to point at the
last node in the list. Then, you allocate
curr->next;
If this works, you set curr = curr->next; curr->next = NULL;
curr->info = ....


Is this how linked lists are created? "front" poiting to the empty node
which in turns points to the first node and "curr" pointing to the
first node.

Alternatively, you could _prepend_ the node to the list.
I.e. you allocate and initialise ptr, and then make ptr->next = first; first = ptr. Then, your list is already sorted the way you want.


Yes, even this is a good work-around.

Thanks to Peteris too. Your solution was very simple. This is where I
guess is the diference between a good programmer and a not-so-good
programmer, the design of the solution. I thought of making it with the
help of linked list, pinch of complexity in itself.

regards,
vijay.

Nov 14 '05 #8

P: n/a
vijay wrote:
Hello,

As the subject suggests, I need to print the string in the reverse
order. I made the following program:

Well, this is something that you should work on, because this type of
programming is found in alot of C books. I 'm surprised your here with
this post, generally this excercise is suppose to train you how to use
the debugger with arrays and loops and so on. It also makes you think a
little on the problem, and to be creative with programming in C or any
language for that matter, but here is what the code looks like, but I
could of made a syntax error, so don't take my word for it

there are many answers to this........:)

#include <stdlib>

#define MAX 80 /* the constant which defines the array(limit).*/

int main(void)
{
char ch, letters[MAX];
int count = 0;

printf("Enter a sentence and hit Enter -> ");
while ((ch = getchar()) != '\n'))/* Searching for the end of chars*/
{
letters[count++] = ch;
}

letters[count] = '\0'; /* insert NULL last in the string array*/

for (--count, count >= 0, --count) /* counts back each letter*/
{
putchar(letters[count]); /* at the last count in the array */

} /* then prints each one*/

}
Simply enough, you could of used the strlen() function to count the
characters if you want to, but this is by far much easier. I have seen
it done with recursion, that makes the code very small, but hard to
understand.

OK
-neil

Nov 14 '05 #9

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