Answer 1

Line 10
if(isdigit(Password[index]!=0))
Should be
if(isdigit(Password[index])!=0)

Actually, should be
if(isdigit(Password[index]))
But I chose to stick with your style.

And there is one case missing: What happens if the password does not have ANY special characters?

Answer 2

The code I am looking at? Line 10 has not yet been fixed there. If you are using a new code then post theupdated version, please.

Answer 3

You're welcome :) Glad that it is finally solved now 👍 Out of curiosity I was working on it meself in C. The result of which is in my code bits. If you are curious, feel free to check it out :)

Answer 4

Yes, your algorithm is rather convoluted, I must say.

However, which is the current version that is still not fixed?

Answer 5

Manav Roy What is your IsTwoNumbers() function do?
What is the need of it? It always prints weak on a digit encounter at last digits....!

Answer 6

It seems to be working now.

Answer 7

I'll be unsubscribing since there is enough traffic on here. Please mention me if anyone wants to reach me :)

Answer 8

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

int main(){

int i, n=0, m=0;
char s[30];
scanf("%s",s);
if(strlen(s) >= 7){

for(i=0; s[i]!='\0'; i++){
if(s[i]=='#'||s[i]=='#x27;||s[i]=='%'||s[i]=='&'||s[i]=='*'||s[i]=='!'||s[i]=='@')
++n;

if(s[i]=='0'||s[i]=='1'||s[i]=='2'||s[i]=='3'||s[i]=='4'||s[i]=='5'||s[i]=='6'||s[i]=='7'||s[i]=='8'||s[i]=='9')
++m;
}
}

if(n>=2 && m>=2)
printf("Strong");
else
printf("Weak");

return 0;
}

Answer 9

Vaibhav Singh
That's the C version.

Answer 10

Bob_Li i know the task. Also about special characters, am saying same. Seems you not read my all replys.. Read by date sorted way, for clarity..

And also expecting that input don't have special characters other than specified in description. But Manav Roy code is taking all other into count. May it cause problem I thought but now I guess it won't affect.

I just letting @Manav go on his own approach...

Answer 11

Bob_Li
Yes about that also I mentioned more times..
His approach is isalpha(..)==0 && isdigit(..)== 0 means it's a special character... It works but he is having redudent checks..

Answer 12

Bob_Li Yes. That's why I said before

"Task description asked to check special characters !@#$%&* but your also accouting other special character like +-_?.. So if input contains these characters then it may fail otherwise I think it works..."

But I guess now, input don't have special characters other than mentioned in description I guess. So not efficient but this approach works..

I just letting Manav Roy go on his own approach firstfirst from start.

Answer 13

Manav Roy congrats..
Finally.. Hope you try to simplify it..
You're welcome..

Answer 14

Manav Roy
Congratulations!

You powered through it.

In learning, the hard way is often more valuable than the easy way... and you took the hard path.

Respect.

Answer 15

Manav Roy as you saying ..
In IsTwoNumbers function,I basically checked if the password contains **two numbers or not." where are you checking two numbers or not? I can't find it..

It doing like :
If input : 1a2
On '1' it calls function but on 'a' it returns back and breaking loop..

Else for ex: a1234
On 1 it calls function and and comes out loop after 4 and prints weak and exits.. .

Why you need functions here..? Instead why don't you use a counter and just increment it's value.. after last check according to description just...

Answer 16

Oh.. Wait. You want both to return 0..

..
No. Privious one is better..

Reply code works for single character.. But you need for all characters to check..

But why you using functions here instead of just a counter required.. It is more confusing..

Answer 17

Use those in proper way , as it's purpose:

isdigit - Returns 1 if a character is digit.
isalpha - Returns 1 if a character is an alphabet .

Reverse way will
isDigit(..) != 0 => on '9' , condition is true..

isDigit(..) == 0 => on '9' , condition is false.. Also On 'a' condition is false..

So use isdigit(.. ) == 1

Manav Roy
before all actually, are you edited your code to @Ani Jona's suggestion to use
( isDigit(password[index] ) != 0 )

instead of
(isDigit(password[index] != 0 ) ) 👈about braces.. I thought you changed but seems like you are not saved the code after changing.. ?

Answer 18

Task description asked to check special characters !@#$%&* but your also accouting other special character like +-_?.. So if input contains these characters then it may fail otherwise I think it works...

Answer 19

Bob_Li I also saying the same.. Read my replies from start.. may you read only last one.

@Manov code is working but taking count other special characters into count.. So i guess that may causing problem to fail test cases.. So suggestion is to take count those special characters only which described in description..

Answer 20

Jayakrishna🇮🇳
I have read your replies and realize that you too have seen that Manav Roy was taking the long route to the solution.
I just wanted to reinforce that point.
I also admire Ani Jona 🕊 's effort to make the code work. It can be fixed, but that would end up with a janky code that is not efficient at all.