why this code doesn’t show a random number , but with different digits??? ( run it till u see same digits in one random number)

Question

https://code.sololearn.com/cX4m50nn9VZz/?ref=app

Answer 1

By random two numbers after another can be the same.
That is random.

Try rolling the dice. At some point you will get the same number again.

Answer 2

You could save the last number and compare it with the new random number. If they are the same get a new random number.

Answer 3

You are right, we must put "a=0" at the start of the main loop, not outside; else, when some replacement is done, "a" will never go back to 0 and infinite loops are executed.

So put at line 20:

a=0;
printf("%d %d %d %d\n", RemainerOfRandom2[0],RemainerOfRandom[1],
RemainerOfRandom[2],RemainerOfRandom[3]);

and you will see how many loops are performed and when some digits are replaced.

Answer 4

Saraadib I'm seeing random number as output they comes randomly no unique value is coming as you mentioned

Answer 5

Abhi run it till u see it ! it will be showed finally ! :(

Answer 6

I want to never see random number with same digits

Answer 7

Dragonxiv what should I do to fix it?

Answer 8

Dragonxiv this code will develope for guessing number game , and for that aim , it must produce random number with totaly different digits!

Answer 9

In your code there are 3 bugs:

BUG 1 -> internal loop at line 23: for(j=i+1;j<3;j++) {
replace 3 with 4, else last digit is never checked for equality with the previous 3 digits

BUG 2 -> line 34: else if(f>=5)
replace with else if(f<=5), else if f<=5 it will be never replaced

BUG 3:
after replacing f you must check again the 4 digits for equality; a not elegant way of doing this is adding a goto to line 21 soon after having replaced f; another way is introducing an outer loop "while (need_further_check)" before line 21, and set need_further_check everytime you replace f, where need_further_check is a bool

Answer 10

Bilbo Baggins really thank u , I tried to do whatever u said , can u check it again? I have a runtime error !

Answer 11

Unfortunately you made again two bugs.

BUG 1 - you are defining a variable "bool a =..." which is not seen by the outer while: the outer while sees the "int a" defined at the top of the program. So please remove the two "bool"

BUG 2 - you are inverting the assignment: replace a=0 with a=1 (which means loop again) and replace a=1 with a=0 (which means stop looping).

But also in this way we could have some cases in which "else a=0" hides a previous a=1, leaving sporadic cases in which there are two equal digits.

So the best solution is:
remove the "else a=0" to avoid hiding
set a=0; before the loop
use a do while instead a while in order to exec the loop at least once:
a = 0;
do { ....; a=1 if some replacement is done, in order to recheck; .... }
while (a);

Happy coding!

Answer 12

Bilbo Baggins thanks for your answer , but I have an error , I think I did everything that u said right :(

Answer 13

line 38: if you have done a replacement, you want to check again; so a must be put to 1, not to 0

Answer 14

Bilbo Baggins but I have error when I put 1
see the code

Answer 15

Bilbo Baggins now it works well😂😻💜really thank u😻😻😻

Answer 16

You welcome! There is yet another little issue, if you really want to emulate a die. You should reenter the main loop even if some digit is equal to 0 or 7 or 8 or 9 :-),

Answer 17

Bilbo Baggins I can‘t get it :( what do u mean?

Answer 18

The four numbers must simulate dices or not?
If yes you have to discard all the numbers outside the range 1-6

Answer 19

Bilbo Baggins no I just want a 4 digit number