I need help with this code for the Deja Vu task, there output says ‘no output’ so I don’t know exactly where’s the mistake.

Question

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

int main() {

char str [20];
fgets (str, 20, stdin);

char first, con, ch;
int i, len, t;
t = 0;
ch = 'n';
len = strlen (str);

for (i = 0; i == len; i++){
first = str [t];
con = str [i + 1];
if (first == con){
ch = 'y';
break;
}
else if (i != len - 1){
i++;}
else if (i == len - 1){
t++;
i = t;}
else if (i == len){
break;}
}
if (ch == 'n')
printf ("Unique");
else
printf ("Deja Vu");
return 0;
}

Answer 1

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

int main()
{
char mash[128];
scanf("%s", mash);

for(int i = 0; i < strlen(mash); i++)
{
for(int j = i + 1; j < strlen (mash); j++)
{
if(mash[i] == mash[j])
{
printf("Deja Vu");
return 1;
}
}
}

characters
printf("Unique");
return 0;
}
luis gonzalez
You can just try this..... Hope this helps..

Answer 2

How do I break a loop whitout using the switch function?
Also why is i < len false when i is brought back to 0?

Answer 3

Thanks, it helped

Answer 4

luis gonzalez welcome

Answer 5

Input : "ABC"
'A' == 'B' false
I != len-1 true I++,
I++ now I=2
'A' == '\0' false
I != len-1 false
t!= len-1 true so t++, I=0,
next i++, I=1, t=1,

'B' == 'C' false
I!=len-1 true, i++
next I++, I=3 <len false. so comes out loop. No output.

In first case, no output

break the loop when you print once "deja vu"..

Answer 6

if( first == con)
{
printf("Deja vu") ;
break; // causes loop break;
}
And
Oh. Yes, I did not see I=0, but still same output.. I edited..

edit: oh that example is not good enough example but it is outputing more than dublicate number of time.
output either "Deja vu" or "Unique" , only once.

Answer 7

Ok so I placed break after printf (“Deja Vu”) but it still doesnt have any output, secondly, how do I fix the ‘Unique’ part of the loop?

Answer 8

Take a check variable like a char or bool.
char ch = 'n';

On printing "Deja vu", set it to ch='y';

At end, print result by
if( ch == 'n')
printf("Unique") ; // on change in ch, it won't print.

In your loop, also instead of else if, use if and tak e I, not t : like
if( i == len-1)
printf("Unique") ; also works...

Answer 9

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

int main() {

char str [20];
fgets (str, 20, stdin);

char first, con, ch;
int i, len, t;
t = 0;
ch = "n";
len = strlen (str);

for (i = 0; i == len; i++){
first = str [t];
con = str [i + 1];
if (first == con){
ch = "y";
break;
}
else if (i != len - 1){
i++;}
else if (i == len - 1){
t++;
i = t;}
else if (t == len - 1){
break;}
}
if (ch = "n")
printf ("Unique");
else
printf ("Deja Vu");
return 0;
}

// So I fixed some things that were wrong and added the ch = “n”.
Now it only outputs ‘Unique’ regardless of the imput. Also I dont understand the alternative for ‘else if’ that you explained.

Answer 10

ch = 'n' is assignment. Use operator == for comparison. => ch == 'n'
Also note the for character, use single quotes. Using double quotes means it's a string form( "n"), not character.

You have
else if (t == len-1)
printf("Unique") ; in loop.
Instead of it, use
if( i == len-1 )
printf("Unique");

So i mean it's also work..

Answer 11

Ok so I changed those things but the output is the same, also I replaced the t on the last else but this doesn’t seem to affect the output.

I edited the post to show the latest code

Answer 12

Don't edit original post, next time at least.
Save your update code in playground and share link here.. That helps debug problem easily..

ch == 'n'; it's comparison, return false but you are not saving result so no effect. You need ch = 'n'; here..
i=0; i==len; is false so loop won't starts now.. You need i != len; instead better to use, i < len;

again
ch = 'n'; initializing 'n' into ch variable.
ch == 'n' is comparing ch value with 'n' character..