Random number generator game in C - c

I have started C recently and am having trouble make the computer think of a random number.
This is the code so far. I need help!
#include <stdio.h>
#include <stdlib.h>
int main ()
{
time_t t;
int userin;
printf("Guess a number from 1 to 10\n");
scanf("%d", userin);
int r = rand() % 11;
if (r == userin)
{
printf ("you are right");
}
else
{
printf("Try again");
}
return 0;
}
Thx a lot guys it worked out!!!!

In your code, r will be a random number from 0 to 10. For a random number between 1 and 10, do this:
int r = rand() % 10 + 1;
Also, you should call
srand(time(NULL));
at the beginning of main to seed the random number generator. If you don't seed the generator, it always generates the same sequence.

There is issue in your scanf statement as well.
You should use
scanf("%d", &userin);
instead of
scanf("%d", userin); /* wrong - you need to use &userin */
scanf needs the address of variables at which it will store the value. For a variable, this is given by the prefexing the variable with &, as in &userin.

There are few issues in your code.
not reading into the address & of your variable using scanf
not considering "legitimate" values of input, result of rand()%11 can also be 0
not checking against "illegal" input values, which can "alias" the result.
not properly initializing seed of the pseudo-random rand() function, so it always returns the same result.
Using printf for debugging your code, as in the following example, based on your code can help a lot:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define DEBG 1
int main (void)
{
time_t t;
int userin;
printf("Guess a number from 1 to 10\n");
if(scanf("%d", &userin) != 1){ // read into the variable's address
printf("Conversion failure or EOF\n");
return 1;
}
if(userin < 1 || userin > 10){ // check against "illegal" input
printf("Offscale, try again\n");
return 1;
}
srand(time(NULL)); // initialize the seed value
int r = 1 + rand() % 10; // revise the formula
if (DEBG) printf("%d\t%d\t", r, userin); //debug print
if (r==userin){
printf ("you are right\n");
}else{
printf("Try again\n");
}
return 0;
}
Please, also consult this SO post.

Problems :
scanf("%d", userin); //you are sending variable
This is not right as you need to send address of the variable as argument to the scanf() not the variable
so instead change it to :
scanf("%d", &userin); //ypu need to send the address instead
and rand()%11 would produce any number from 0 to 10 but not from 1 to 10
as other answer suggests, use :
(rand()%10)+1 //to produce number from 1 to 10
Solution :
And also include time.h function to use srand(time(NULL));
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(void)
{
srand(time(NULL));
int userin;
printf("Guess a number from 1 to 10\n");
scanf("%d", &userin);
int r = (rand() % 10)+1;
if (r==userin)
{
printf ("you are right");
}
else
{
printf("Try again");
}
return 0;
}
Why use srand(time(NULL)) ?
rand() isn't random at all, it's just a function which produces a sequence of numbers which are superficially random and repeat themselves over a period.
The only thing you can change is the seed, which changes your start position in the sequence.
and, srand(time(NULL)) is used for this very reason

This should work
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main ()
{
int userIn = 0; //I like to initialize
printf("Guess a number from 1 to 10\n");
scanf("%d", &userIn);
srand(time(NULL)); //seed your randum number with # of seconds since the Linux Epoch
int r = (rand()%10)+1; //rand%11 gives values 0-10 not 1-10. rand%10 gives 0-9, +1 makes sure it's 1-10
if (r == userIn)
{
printf ("you are right\n");
}
else
{
printf("Try again\n");
}
return 0;
}
Edit: You may want to implement code to verify that the user input is in fact an integer.

Related

Can you Help me correct this simple c program's little mistake?

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
int num, rnum, times = 1;
srand(4383);
rnum=rand() % 300 + 1;
while(times <=8)
{
printf("Guess the numper random number between 1-300: ");
scanf("%d", &num);
if (num<rnum)
{
printf("The random number is biger\n");
}
if (num>rnum)
{
printf("The magic number is smaller\n");
}
if (num == rnum)
{
printf("RIGHT!");
break;
}
times++;
}
printf("FAILURE!");
return 0;
}
The point of the task is to make a program for a user to type and try to guess a numper from 1–300 with 8 attempts. If you find the number it shows RIGHT! and if not it guides you by telling that the number is biger/smaller. If you fail in your 8 atemts then it shows failure. The problem is that it shows failure when you fail to guess in your 8 atempts but when you find the number it prints both RIGHT & FAILURE. What should i correct for the program to print failure only when you cant’t find the number within your 8 tries?
My 2 cents, path of least resistance is to simply return rather than break when the user guesses correctly:
if (num == rnum)
{
printf("RIGHT!");
return 0; // program exits here, no FAILURE print
}
You should also seed the rand function with a changing number, like time. With a constant, you'll find your number to guess is the same every time.
srand(time(NULL)); // randomize seed
You should check if they exceeding the 8 try limit before executing the print statement:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
int num, rnum, times = 1;
srand(4383);
rnum=rand() % 300 + 1;
while(times <=8)
{
printf("Guess the numper random number between 1-300: ");
scanf("%d", &num);
if (num<rnum)
{
printf("The random number is biger\n");
}
if (num>rnum)
{
printf("The magic number is smaller\n");
}
if (num == rnum)
{
printf("RIGHT!");
break;
}
times++;
}
if (times > 8)
{
printf("FAILURE!");
}
return 0;
}
I think you should write return 0; instead of break; after printing RIGHT. because if you guess the rnum it will print RIGHT and breaks out and after that it will print FAILURE too but if you write return 0; it will end the program.
When you break; out after having printed RIGHT! you end up where you print FAILURE! so you need to check this somehow.
Here's my take on it:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main() {
int num = 0, rnum;
// This game will get boring with a static seed.
// I'm assuming you use this seed for testing only.
srand(4383);
rnum = rand() % 300 + 1;
for(int times = 0; times < 8; ++times) { // simpler loop
printf("Guess the random number between 1-300: ");
// if the user fails to enter a number - make it print FAILURE
if(scanf("%d", &num) != 1) break;
if(num < rnum)
puts("The random number is bigger");
else if(num > rnum) // added "else"
puts("The magic number is smaller");
else
break; // neither less nor greater than rnum so it must be correct
}
if(num == rnum) // added check
printf("RIGHT!\n");
else
printf("FAILURE!\n");
}

How to quit loop when requirement is met in C

I have made a while loop and it works partly. I want the code to stop when the values entered are under the parameter, but it keeps going regardless of the output. Here is my code:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
#include <math.h>
#include <time.h>
int main()
{
// defining variables until "till here" comment
int i;
int rollDice;
int firInp;
int secInp;
srand (time(NULL)); // seeding rand so that we get different values every time
// till here
while(rollDice > 0)
{
printf("Enter the amount of faces you want your dice to have (MAX=24, MIN=1): "); // prints the message
scanf("%d", &firInp); // user input stored into firInp
printf("Enter the amount of throws you want(MAX=499, MIN=1): "); // this message is printed after the users first input
scanf("%d", &secInp); // user input stored into secInp
if (((firInp < 25)&&(firInp > 1))&&((secInp < 500)&&(secInp > 1))){ // if statement to check parameters
for(i = 0; i < secInp; i++){
rollDice = (rand()%firInp) + 1;
printf("%d \n", rollDice);
}
}
else{
printf("Sorry, these numbers don't meet the parameters\nPlease enter a number in the right parameters.\n");
}
}
return 0;
}
I'm new to C btw.
edit: I want the loop to continue if the user input is more than 24, 499 respectively.
What you're doing is wrong. Variable rollDice is for storing the values of the outcomes rather than doing a condition check. It will have random values and since the values on the dice can't be negative or zero it may not exit the while loop. I don't know what will rand() will produce so I'm just assuming.
The range for rand() is [0,RAND_MAX), including zero and excluding RAND_MAX. But because of this expression (rand()%firInp) + 1 , you're adding one to it. So it will never become Zero.
You can use a flag variable and set it to 1. When the if conditions are met, you can set the flag to 0. It will exit the while loop.
Corrected code :-
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
#include <math.h>
#include <time.h>
int main()
{
// defining variables until "till here" comment
int i;
int rollDice;
int firInp;
int secInp;
int flag = 1;
srand (time(NULL)); // seeding rand so that we get different values every time
// till here
while(flag)
{
printf("Enter the amount of faces you want your dice to have (MAX=24, MIN=1): "); // prints the message
scanf("%d", &firInp); // user input stored into firInp
printf("Enter the amount of throws you want(MAX=499, MIN=1): "); // this message is printed after the users first input
scanf("%d", &secInp); // user input stored into secInp
if (((firInp < 25)&&(firInp > 1))&&((secInp < 500)&&(secInp > 1))){ // if statement to check parameters
for(i = 0; i < secInp; i++){
rollDice = ((rand() + 1)%firInp);
printf("%d \n", rollDice);
}
flag = 0;
}
else{
printf("Sorry, these numbers don't meet the parameters\nPlease enter a number in the right parameters.\n");
}
}
return 0;
}
EDIT :-
Also, division with 0 is undefined. rand() can attain value 0. You should add 1 to rand() rather than adding to whole modulus. It can create an error if the rand() will give 0 as an output.

Some problems in coding a "guessing random number in C" under some conditions such as using input(), output()

I tried going beyond just guessing random numbers. The conditions were these:
use input() numbers used from 1 to100 and if inserted numbers that are out of this range, to show a line to re-enter a number
use output() to show the output(but show the last line```You got it right on your Nth try!" on the main())
make the inserted number keep showing on the next line.
Basically, the program should be made to show like this :
insert a number : 70
bigger than 0 smaller than 70.
insert a number : 35
bigger than 35 smaller than 70.
insert a number : 55
bigger than 55 smaller than 70.
insert a number : 60
bigger than 55 smaller than 60.
insert a number : 57
You got it right on your 5th try!
I've been working on this already for 6 hours now...(since I'm a beginner)... and thankfully I've been able to manage to get the basic structure so that the program would at least be able to show whether the number is bigger than the inserted number of smaller than the inserted number.
The problem is, I am unable to get the numbers to be keep showing on the line. For example, I can't the inserted number 70 keep showing on smaller than 70.
Also, I am unable to find out how to get the number of how many tries have been made. I first tried to put it in the input() as count = 0 ... count++; but failed in the output. Then I tried to put in in the output(), but the output wouldn't return the count so I failed again.
I hope to get advice on this problem.
The following is the code that I wrote that has no errors, but problems in that it doesn't match the conditions of the final outcome.
(By the way, I'm currently using Visual Studio 2017 which is why there is a line of #pragma warning (disable : 4996), and myflush instead of fflush.)
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#pragma warning (disable : 4996)
int input();
int random(int);
void myflush();
void output(int, int);
int main()
{
int num;
int i;
int ran;
srand((unsigned int)time(NULL));
i = 0;
while (i < 1) {
ran = 1 + random(101);
++i;
}
num = input();
output(ran, num);
printf("You got it right on your th try!");a
return 0;
}
int input()
{
int num;
printf("insert a number : ");
scanf("%d", &num);
while (num < 1 || num > 100 || getchar() != '\n') {
myflush();
printf("insert a number : ");
scanf("%d", &num);
}
return num;
}
int random(int n)
{
int res;
res = rand() % n;
return res;
}
void myflush()
{
while (getchar() != '\n') {
;
}
return;
}
void output(int ran, int num) {
while (1) {
if (num != ran){
if (num < ran) {
printf("bigger than %d \n", num); //
}
else if (num > ran) {
printf("smaller than %d.\n", num);
}
printf("insert a number : ");
scanf("%d", &num);
}
else {
break;
}
}
return;
}
There are many problem and possible simplifications in this code.
use fgets to read a line then scanf the line content. This avoids the need of myflush which doesn’t work properly.
the function random is not needed since picking a random number is a simple expression.
if the range of the random number is [1,100], you should use 1+rand()%100.
there is no real need for the function output since it’s the core of the main program. The input function is however good to keep to encapsulate input.
you should test the return value of scanf because the input may not always contain a number.
Here is a simplified code that provides the desired output.
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#pragma warning (disable : 4996)
int input() {
char line[100];
int num, nVal;
printf("insert a number : ");
fgets(line, sizeof line, stdin);
nVal = sscanf(line, "%d", &num);
while (nVal != 1 || num < 1 || num > 100) {
printf("insert a number : ");
fgets(line, sizeof line, stdin);
nVal = sscanf(line, "%d", &num);
}
return num;
}
int main()
{
int cnt = 0, lowerLimit = 0, upperLimit = 101;
srand((unsigned int)time(NULL));
// pick a random number in the range [1,100]
int ran = 1 + rand()%100;
while(1) {
cnt++;
int num = input();
if (num == ran)
break;
if (num > lowerLimit && num < upperLimit) {
if (num < ran)
lowerLimit = num;
else
upperLimit = num;
}
printf("bigger than %d and smaller than %d\n", lowerLimit, upperLimit);
}
printf("You got it right on your %dth try!\n", cnt);
return 0;
}
I am unable to find out how to get the number of how many tries have been made.
Change the output function from void to int so it can return a value for count, and note comments for other changes:
int output(int ran, int num) {//changed from void to int
int count = 0;//create a variable to track tries
while (1) {
if (num != ran){
count++;//increment tries here and...
if (num < ran) {
printf("bigger than %d \n", num); //
}
else if (num > ran) {
printf("smaller than %d.\n", num);
}
printf("insert a number : ");
scanf("%d", &num);
}
else {
count++;//... here
break;
}
}
return count;//return value for accumulated tries
}
Then in main:
//declare count
int count = 0;
...
count = output(ran, num);
printf("You got it right on your %dth try!", count);
With these modifications, your code ran as you described above.
(However, th doesn't work so well though for the 1st, 2nd or 3rd tries)
If you want the program to always display the highest entered number that is lower than the random number ("bigger than") and the lowest entered number that is higher then the random number ("smaller than"), then your program must remember these two numbers so it can update and print them as necessary.
In the function main, you could declare the following two ints:
int bigger_than, smaller_than;
These variables must go into the function main, because these numbers must be remembered for the entire duration of the program. The function main is the only function which runs for the entire program, all other functions only run for a short time. An alternative would be to declare these two variables as global. However, that is considered bad programming style.
These variables will of course have to be updated when the user enters a new number.
These two ints would have to be passed to the function output every time it is called, increasing the number of parameters of this function from 2 to 4.
If you want a counter to count the number of numbers entered, you will also have to remember this value in the function main (or as a global variable) and pass it to the function output. This will increase the number of parameters for the function to 5.
If you don't want to pass so many parameters to output, you could merge the contents of the functions output and input into the function main.
However, either way, you will have to move most of the "smaller than" and "bigger than" logic from the function output into the function main, because that logic is required for changing the new "bigger_than" and "smaller_than" int variables which belong to the function main. The function output should only contain the actual printing logic.
Although it is technically possible to change these two variables that belong to the function main from inside the function output, I don't recommend it, because that would get messy. It would require you to pass several pointers to the function output, which would allow that function to change the variables that belong to the function main.
I have now written my own solution and I found that it is much easier to write by merging the function output into main. I also merged all the other functions into main, but that wasn't as important as merging the function output.
Here is my code:
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#pragma warning (disable : 4996)
int main()
{
const char *ordinals[4] = { "st", "nd", "rd", "th" };
int num_tries = 0;
int bigger_than = 0, smaller_than = 101;
int input_num;
int random_num;
srand( (unsigned int)time( NULL ) );
random_num = 1 + rand() % 101;
for (;;) //infinite loop, equivalent to while(1)
{
printf( "Bigger than: %d, Smaller than: %d\n", bigger_than, smaller_than );
printf( "enter a number: " );
scanf( "%d", &input_num );
printf( "You entered: %d\n", input_num );
num_tries++;
if ( input_num == random_num ) break;
if ( input_num < random_num )
{
if ( bigger_than < input_num )
{
bigger_than = input_num;
}
}
else
{
if ( smaller_than > input_num )
{
smaller_than = input_num;
}
}
}
printf( "You got it right on your %d%s try!", num_tries, ordinals[num_tries<3?num_tries:3] );
return 0;
}
Also, I made sure that the program would print "1st", "2nd" and "3rd", whereas all the other solutions simply print "1th", "2th", "3th". I used the c++ conditional operator for this.

C Program to find prime number

Hey guys so I need to make a program which asks the user to enter a number as a argument and then let them know if it is a prime number or 0 otherwise. So the code I have so far is as follows but I am a little confused on how to make it run through all the possible values of the and make sure that it isn't a non-prime number. Right now what happens is that the program opens, I enter a value and nothing happens. Note: I have math in the header as I am unsure if it is needed or not at this stage.
EDIT: SO I MADE THE CHANGES SUGGESTED AND ALSO ADDED A FOR LOOP HOWEVER WHEN I GO TO COMPILE MY PROGRAM I GET AN WARNING SOMETHING ALONG THE LINES OF 'CONTROL MAY REACH END OF NON-VOID FUNCTION'. HOWEVER THE PROGRAM DOES COMPILE WHEN I GO TO ENTER A NUMBER AND HIT ENTER IRRELEVANT OT WHETHER OR NOT IT IS A PRIME NUMBER I GET AN ERROR BACK SAYING 'FLOATING POINT EXCEPTION: 8'.
EDIT 2: THE FLOATING POINT ERROR HAS BEEN FIXED HOWEVER NOW THE PROGRAM SEEMS TO THINK THAT EVERY NUMBER IS NON - PRIME AND OUTPUTS IT THIS WAY. I CAN'T SEEM TO SEE WHY IT WOULD DO THIS. I AM ALSO STILL GETTING THE 'CONTROL MAY REACH END OF NON-VOID FUNCTION' WARNING
#include <stdio.h>
#include <math.h>
int prime(int a){
int b;
for(b=1; b<=a; b++){
if (a%b==0)
return(0);
}
if(b==a){
return(1);
}
}
int main(void){
int c, answer;
printf("Please enter the number you would like to find is prime or not= ");
scanf("%d",&c);
answer = prime(c);
if(answer==1){
printf("%d is a prime number \n",c);
}
else
printf("%d is not a prime number\n",c);
}
1. You never initialized i (it has indeterminate value - local variable).
2. You never call function is_prime.
And using a loop will be good idea .Comparing to what you have right now.
I just modified your function a little. Here is the code
#include <stdio.h>
#include <math.h>
int prime(int a)
{
int b=2,n=0;
for(b=2; b<a; b++)
{
if (a%b==0)
{
n++;
break;
}
}
return(n);
}
int main(void)
{
int c, answer;
printf("Please enter the number you would like to find is prime or not= ");
scanf("%d",&c);
answer = prime(c);
if(answer==1)
{
printf("%d is not a prime number \n",c);
}
else
{
printf("%d is a prime number\n",c);
}
return 0;
}
Explanation-
In the for loop, I am starting from 2 because, I want to see if the given number is divisible by 2 or the number higher than 2. And I have used break, because once the number is divisible, I don't want to check anymore. So, it will exit the loop.
In your main function, you had not assigned properly for the printf() statement. If answer==1, it is not a prime number. (Because this implies that a number is divisible by some other number). You had written, it is a prime number(which was wrong).
If you have any doubts, let me hear them.
I suggest you start with trial division. What is the minimal set of numbers you need to divide by to decide whether a is prime? When can you prove that, if a has a factor q, it must have a smaller factor p? (Hint: it has a prime decomposition.)
Some errors your program had in your prime finding algorithm:
You start the loop with number 1 - this will make all numbers you test to be not prime, because when you test if the modulo of a division by 1 is zero, it's true (all numbers are divisible by 1).
You go through the loop until a, which modulo will also be zero (all number are divisible by themselves).
The condition for a number to be prime is that it must be divisible by 1 and itself. That's it. So you must not test that in that loop.
On main, the error you're getting (control reaches end of non-void function) is because you declare main to return an int.
int main(void)
And to solve that, you should put a return 0; statement on the end of your main function. Bellow, a working code.
#include <stdio.h>
#include <math.h>
int prime(int a)
{
int b;
for (b = 2; b < a; b++) {
if (a % b == 0)
return (0);
}
return 1;
}
int main(void)
{
int c, answer;
printf
("Please enter the number you would like to find is prime or not= ");
scanf("%d", &c);
answer = prime(c);
if (answer == 1) {
printf("%d is a prime number \n", c);
} else {
printf("%d is not a prime number\n", c);
}
return 0;
}
On a side note, don't use the CAPSLOCK to write full sentences. Seems like you're yelling.
Mathematically the maximum divisor of a number can be as a large as the square of it, so we just need to loop until sqrt(number).
A valid function would be:
//Function that returns 1 if number is prime and 0 if it's not
int prime(number) {
int i;
for (i = 2; i < sqrt(number); i++) {
if (a % i == 0)
return (0);
}
return 1;
}
#include<stdio.h>
int main()
{
int n , a, c = 0;
printf ("enter the value of number you want to check");
scanf ("%d", &n);
//Stopping user to enter 1 as an input.
if(n==1)
{
printf("%d cannot be entered as an input",n);
}
for(a = 2;a < n; a++)
{
if(n%a==0)
{
c=1;
break;
}
}
if(c==0 && n!=1)
{
printf("%d is a prime number \n",n);
}
else
{
if(c!=0 && n!=1)
{
printf("%d is not a prime number \n",n);
}
}
return 0;
}
#include <stdio.h>
#include <stdlib.h>
int main()
{
int x,i;
printf("enter the number : ");
scanf("%d",&x);
for ( i=2; i<x;i++){
if ( x % i == 0){
printf("%d",x);
printf(" is not prime number ");
printf("it can be divided by : ");
printf("%d",i);
break;
}[this is best solution ][1]
}
if( i>=x) {
printf("%d",x);
printf(" is prime number");
}
}

Random guessing game stuck

I am practicing the basics of C programming using the book SAMS Teach yourself C in 21 days.
On one of the type and run sections, they have the find the number (or guess the number) program, I typed it and run it, however the program on the console gets stuck showing the following:
Getting a random number
I waited for some time but nothing happens, even pushing some keys it doesn't do anything.
I am not familiar yet with the srand(), time() and rand() routines so I don't know how to fix it and make it run properly.
below is the code:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define NO 0
#define YES 1
int main (void)
{
int guess_value = -1;
int number;
int nbr_of_guesses;
int done = NO;
printf("\nGetting a random number\n");
/*use the time to seed the random number generator*/
srand( (unsigned) time(NULL));
number = rand();
nbr_of_guesses = 0;
while (done == NO);
{
printf("\nPick a number between 0 and %d>", RAND_MAX);
scanf("%d", &guess_value); /*get a number*/
nbr_of_guesses++;
if (number == guess_value)
{
done = YES;
}
else
if (number < guess_value)
{
printf("\nYou guessed high!");
}
else
{
printf("\nYou guessed low!");
}
}
printf("\nCongratulations! you guessed right in %d Guesses!", nbr_of_guesses);
printf("\n\nThe number was %d", number);
return 0;
}
I can see 2 problems.
There is a semicolon on the while that's causing the program to hang.
There is nothing that ensures that the output buffer is flushed before you read the guess.
I have put comments to indicate the code changes.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define NO 0
#define YES 1
int main (void)
{
int guess_value = -1;
int number;
int nbr_of_guesses;
int done = NO;
printf("\nGetting a random number\n");
/*use the time to seed the random number generator*/
srand( (unsigned) time(NULL));
number = rand();
nbr_of_guesses = 0;
while (done == NO) // Removed the ;
{
printf("\nPick a number between 0 and %d>", RAND_MAX);
fflush(stdout); // stdout is line buffered, and since there is no \n in the printf we need an explicit call to fflush,
scanf("%d", &guess_value); /*get a number*/
nbr_of_guesses++;
if (number == guess_value)
{
done = YES;
}
else
if (number < guess_value)
{
printf("\nYou guessed high!");
}
else
{
printf("\nYou guessed low!");
}
}
printf("\nCongratulations! you guessed right in %d Guesses!", nbr_of_guesses);
printf("\n\nThe number was %d", number);
return 0;
}

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