Thread: Help with inverse...

I want to enter 10 words in a vector and print the inverse.

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

void reverse(char str[]);

Code:

 
main()
{
clrscr();
char str[200],str1[200];
int len;
int i;
int j;

for(int i=1; i<=10; i++){
printf("\nEnter your nickname:");
gets( str[i]);
len=strlen(str);
for ( i=0,j=len-1;i<len;i++,j--)
  str1[i]=str[j];
str1[i]='\0';
puts(str1);
getch(); }
}

Use strrev() and print as many reverse numbers as you want :d
Also don't you think it would be good if you indent your code properly?Here you are only writing only 10 lines so it's not a big deal but do you think you would be able to debug your code if you write more than 1000 lines of code?which braces are for which that itself you wouldn't be able to distinguish

If I see someone using gets() with like a 4k buffer or something, I usually do not utter a peep. But with a 20 byte buffer I do have to ask what the writer is expecting to happen. You do realize that upset users are prone to typing stuff like

Code:

asdfkl;wjerq0981341=3-4908a;fkljadfjk;l1234l;jkadf;jkl

At least that is what I do when a command line program keeps complaining about every little thing that its crazy input demands. So keep in mind that you shouldn't really ever use gets() but if you feel you must, don't use it with such rediculously small spans of memory.

Ehh whatever I was bored, saw this topic, had my editor opened, and so I coded an implementation. I don't think there are any bugs, so this should help you out.

Code:

#include <stdlib.h>
#include <stdio.h>

int revstr(char * orig, char ** new_rev);

int main(int argc, char * argv[])
{
	char * org = "Google would more than likely reveal the answer, but I was bored...";
	char * new_s = malloc(200 * sizeof(char));
	if (!new_s) return 0;
	
	revstr(org, &new_s);
	
	printf("&#37;s\n", new_s);
	
	free(new_s);
	return 0;
}

int revstr(char * orig, char ** new_rev)
{
	if (orig == NULL || *new_rev == NULL) return -1;
	int i = 0;
	int j = 0;
	
	while (orig[i] != '\0') i++;
	
	while (--i >= 0) {
		(*new_rev)[j] = orig[i];
		j++;
	}
	(*new_rev)[j] = '\0';
	
	return 0;
}

How about something even more simplistic?

Code:

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

int revstr(char * orig, char ** new_rev);

int main(int argc, char * argv[])
{
	char * org = "Google would more than likely reveal the answer, but I was bored...";
	char * new_s = 0;

	revstr(org, &new_s);
	
	printf("&#37;s\n", new_s);
	
	free(new_s);
	return 0;
}

int revstr(char * orig, char ** new_rev)
{
	char *tmp;

	if(!*new_rev)
	{
		if(!(tmp = malloc(strlen(orig) + 1)))
			return 1;
	} else
		if(!(tmp = realloc(*new_rev, strlen(orig) + 1)))
			return 1;
	strcpy(orig, tmp);
	strrev(tmp);

	*new_rev = tmp;
	return 0;
}

Yours is probably faster, but mine copes with stuff a little nicer... unless you don't initialize the pointer you pass to the function.

Nothing wrong with multiple solutions.

I hadn't even thought about doing it that way actually.

Spanish... lol...... Got my third and final year of that done last year, hopefully I want have to take that crap in College.

Also, if I get bored enough I may time the implementations and post the results. I'd much rather do that then coding something useful, lol.

>>hopefully I want have to take that crap in College

I am not so fortunate.

I speak spanish damn near fluently and had four years of it back in high school.

I guess speaking it fluently would be pretty cool. Although I just don't have the desire to learn a foreign language.