You can see my code below. I'm trying to check whether the sum of the entered numbers is above 100 or one of the entered numbers is above 20. If one of these happens the program should quit. If I enter numbers so the sum is higher than 100 the program quits. But if I enter a number higher than 20 the program just continues to run.
#include <stdio.h>
#include <string.h>
int main() {
int numbers;
int sum = 0;
printf("Please enter numbers:\n");
for (numbers = 0; sum < 100 || numbers < 20; numbers++) {
scanf("%d", &numbers);
sum += numbers;
}
printf("Sum: %d\n", sum);
return 0;
}
If one of these happens the program should quit
You want && then, not ||. Think about it; if either sum is less than 100 or numbers is less than 20, your program continues. So, when you enter a value for numbers greater than 20, your code will continue to run until and unless sum is also greater than 100.
Related
Something is wrong. I'm trying to make a cade which can count number count of any natural number. Like number count of 2 is 1, 30 is 2, 456 is 3. My code is running for 1 digit numbers but not for two digit numbers.
#include<stdio.h>
void main(void)
{
int num,count,check;
float div;
printf("Enter a natural number\n");
scanf("%d", &num);
while (num<=0)
{
printf("Error\n");
printf("Enter a number\n");
scanf("%d", &num);
}
while(num>1)
{
count=1;
check=10;
div=num/check;
if(div<=1)
{
printf("Number count is\n%d", count);
break;
}
check = check*10;
count = count+1;
}
}
The problem with your solution is that after check and count are modified at the end of the loop, they are re-declared to 1 and 10 respectively at the beginning of the loop at every passage.
You need to move the declaration just before the while loop.
Also div doesn't need to be a float given that the decimal part of this number is irrelevant.
You could also use less variables by replacing check by 10 and
using num directly instead of temporarily storing results in div.
I think this might be a simpler solution
#include <stdio.h>
int main(){
int num = 0, digits = 0;
while (num <= 0)
{
printf("Enter a natural number\n");
scanf("%d", &num);
num == 0 ? printf("Error\n") : 0;
}
for( ; num > 0; digits++)
num /= 10;
printf("number of digits: %d\n", digits);
}
As num is continuously divided by 10, the decimal of the result gets truncated since num is an int while digits steadily increases.
It is time to learn to use a debugger. Using it would have immediately shown the major problem in your code: you reset the value of count and check inside the loop. So if you enter a number greater or equal to 10, you enter an infinite loop because you will consistently divide that number by 10 and find that the result is >= 1!
There is another less important problem: you use if(div<=1) when it should be if(div<1). Because 10/10 is 1 and has 2 digits...
After those fixes you should have:
...
check = 10;
count = 1;
while (num > 1)
{
div = num / check;
if (div < 1)
{
printf("Number count is\n%d", count);
break;
}
check = check * 10;
count = count + 1;
}
return 0; // main shall return an int value to its environment...
}
Which correctly gives the number of decimal digit on positive integers. But as you were said in comments, you should always test the return value of scanf (what is the user inadvertently types a t instead of 5 for example?).
That being said, this answer intends to show you what the problems were, but Keyne's solution is better...
The output should print the statement "Case #(N): I will become a good boy." a certain number of times depending on user input. It should only print prime numbers of N. This is the latest attempt.
#include<stdio.h>
int main(){
int primeNum;
int primeCount;
int primeCheck;
int caseCount=1;
int caseCheck;
scanf("%d", &caseCheck);//gets number of cases
do {
scanf("%d", &primeNum);
primeCheck = 0;
if (primeNum<=1)
{
caseCount++;
}
for (primeCount=2 ; primeCount<=primeNum/2 ; primeCount++)//checks if number is prime
{
if ((primeNum%primeCount) == 0)//checks if number is not prime
{
primeCheck=1;
}
}
if (primeCheck==0)
printf("Case #%d: I will become a good boy.\n", caseCount);//prints if number is prime
} while (caseCount=caseCheck);//while case counter does not exceed number of cases
return(0);}`
The result of this piece of code when the output is 2 [enter] 4 [enter] 2 is "Case #2: I will become a good boy." Why does it not print more than one time and start from number 2?
Just change while (caseCount=caseCheck); to while (caseCount==caseCheck);
The problem with your code is you're not checking the condition. You're just reassigning a value to a variable.
i am really stuck with this been trying to solve it for quite a time now.
i have to write a program where i should input 5 numbers between 1 to 10 and then calculate the average, USING ONLY WHILE LOOP, but it does not have to exit when the number does not meet the requirement. then, i have to write a variation of the same code but this time you can enter all the numbers you want, and when 0 is entered it has to calculate the average and exit
this is where i have gotten so far
#include <stdio.h>
int main(void)
{
int n, i = 1;
float add;
float avg;
do
{
printf("enter the number %d:\n", i++);
scanf("%d", &n);
add = add + n;
} while(n > 0 && n < 11);
avg= (add / 5);
printf("%.1f", avg);
return 0;
}
it will keep asking for numbers after 5 have been entered. and the average is not right anyways
First, you're using nas your while condition variable, but also as the variable to scan the input. If I start your program by scanning 20, for example, your while loop will exit on the first interaction. Use your i variable instead and also increment it every time your loop executes.
do{
...
}while(i <= 5);
Second, if you want only numbers between 1 and 10, then you should write a condition for it. For example:
printf("enter the number %d:\n", i); //do not increment it here!
scanf("%d",&n); //assuming "n" as your variable to scan
if(n > 0 && n < 11){
add += n;
i++; //increment it here instead!
}
Third, initialize your variables in order to not get thrash values
float add = 0;
float avg = 0;
int i = 1;
Finally, assign your result (not mandatory, but since you're using it I'll keep it):
avg = add/5.0f
and display:
printf("%.1f", avg);
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");
}
}
I wanted to print the divisors of given range of numbers. It works alright. But when I try to modify it to put **** at the end of the prime number's divisors it acts like bizarre.
#include <stdio.h>
int main()
{
int a,start,rounds,b,c,k=0;
printf("Please enter a number to start :");
scanf("%d",&start);
printf("Please enter how many numbers you want to print from that number :");
fflush(stdin);
scanf("%d",&rounds);
for(a=start;a<=start+rounds;a++)
{
printf("\n\nThe divisors of number :%d are \n",a);
for(b=1;b<=a;b++)
{
c=a%b;
if(!c)
{
k++;
printf("%d\n",b);
}
}
//printf("%d",k);
if((k==2)||(k==1))
printf("***\n");
}
getchar();
return 0;
}
PS:- The trick I used to find a prime number is counting how many printf statement has been executed before loop ends. Is there any wrong with it? When I remove // from printf statement it prints like below.
start=========>k
1 =========>1
2 =========>3
3 =========>5
4 =========>8
5 =========>10
Why is that?
if((k==2)&&(k==1))
There is no way in todays computers that k can be 2 and 1 at the same time. Maybe you meant to say if k is 2 OR k is 1 ?
You need to reset k back to 0 in your outer loop
printf("\n\nThe divisors of number :%d are \n",a);
k = 0;
(And, as others have pointed out, you need an || operator rather than && for the final test)