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So basicly I have this assignement in C, I have to input numbers until I enter 0, and after I enter 0 I have to print 1st and 2nd min number from all that numbers and I can't use arrays. I get that I have to use do-while loop for input but I can't figure out how to find two smallest from all of them. I think that thing can be done with if loops but don't know how to make it as I have only one variable to enter numbers into it (int a). And in input I have error when I enter 0 I'm able to enter one more number before program quits.
#include <stdio.h>
int main() {
int a;
do {
printf("Enter numbers: ");
scanf("%d\n", &a);
//what to do here
}while(a != 0);
You need to add 2 variables to hold the smallest values detected so far. Like
int smallest = INT_MAX;
int second_smallest = INT_MAX;
Then in the loop you need to test if the new input value is smaller than the values stored so far. Something like:
if (a <= smallest)
{
second_smallest = smallest;
smallest = a;
}
else if (a < second_smallest)
{
second_smallest = a;
}
You can use two variables to do what you need
#include <stdio.h>
#include <limits.h>
int main(void)
{
int a = INT_MAX;
int min_1 = INT_MAX;
int min_2 = INT_MAX;
int valid;
do
{
if (a < min_1)
{
min_2 = min_1;
min_1 = a;
}
else if (a < min_2)
{
min_2 = a;
}
printf("Enter numbers: ");
valid = scanf("%d", &a);
}
while ((a != 0) && (valid == 1));
if (valid == 1)
{
printf("Minimum numbers entered are: %d %d\n", min_1, min_2);
}
else
{
fprintf(stderr, "Error in data input\n");
}
}
So:
use limits.h defines to init min variables to the highest value for int type INT_MAX.
for each loop you must test if entered number is a minimum
you must check that user input is valid: check scanf return value.
remove \n in the format string of scanf.
Related
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I just want a 1 or 0 if it is prime or not.
but I am getting multiple 0's and 1's. How can I solve this.
#include <stdio.h>
int num() {
int a, i;
printf("Enter a number:");
scanf("%d", &a);
for (i = 2; i < a; i++) {
if (a % i == 0)
printf("1");
else
printf("0");
}
}
int main() {
num();
return 0;
}
Based on the naming used and specific combination of function use, I am almost certain OP's code is based off this which is my first google response to c check if number is prime.
I challenged myself to "fix" it with least number of modification to the original code, here is the version that works the way OP expects. It is ugly but the point is to make it clear where it differs from his code.
OP seems to have mixed up the inner if statements with the outer if statements, and completely forgot about the counter. Also OP seems to have got confused in the function num, as it should either print 1 or 0 and be a void function, or return 1 or 0 and take a as input to a function that returns int eg int num(int a) or void num(), whereas OP ended up going halfway int num().
The working(if you can call it that, since fflush(stdout) is not called after printf is called, so the program will not not show the question on mingw without winpty) program would look like this:
#include <stdio.h>
void num() {
// a is the user input number
// c is the count
// i is the iterator
int a, i, c = 0;
printf("Enter a number: ");
scanf("%d", &a);
for (i = 2; i < a; i++) {
if (a % i == 0)
++c;
}
if (c != 0)
printf("0");
else
printf("1");
}
int main() {
num();
return 0;
}
The reason you get multiple 0s and 1s is you print them for every potential factor. You should instead test the factors to determine if none of the factors up to the square root of the number divide the number evenly, printing a 1 in this case and a 0 otherwise.
Function isprime should take an int argument and return 1 or 0, the main() function takes case of getting the number from the user and printing the result.
Here is a modified version:
#include <stdio.h>
int isprime(int a) {
int i;
if (a < 2)
return 0;
for (i = 2; a / i >= i; i++) {
if (a % i == 0)
return 0;
}
return 1;
}
int main() {
int a;
printf("Enter a number:");
if (scanf("%d", &a) == 1) {
printf("%d\n", isprime(a));
}
return 0;
}
Also note how the code is indented and uses spaces to improve readability. Learn how to do this for your next projects.
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Not doing anything special. Just taking a user's input between 1 - 500, and then printing the number using for loop for each iteration. It crashes at the for loop. It does not print anything at all.
#include <stdio.h>
int forCounter() {
int num;
int count = 0;
printf("Pick a positive number (1 - 500): ");
scanf("%d", &num);
while (num < 1 || num > 500) {
printf("Out of range, try again (1 - 500): ");
scanf("%d", &num);
}
int i = num;
for (i; count <= i; count++) {
printf(count);
}
getchar();
return 0;
}
I take first input applied to 'num' and check if it's above or below allowed amount with the while loop.
When that's done it leaves and should start the for loop with i taking over for num. I tried using num in the place of i but it didn't work so I tried using a separate variable to see if it'd work.
I get two warnings seen in the image providedTwo warnings
You have to specify the format of output in printf.
int printf(const char *format, ...)
Your code:
#include <stdio.h>
int forCounter(void) {
int num;
printf("Pick a positive number (1 - 500): ");
scanf("%d", &num);
while (num < 1 || num > 500) {
printf("Out of range, try again (1 - 500): ");
scanf("%d", &num);
}
for (int i = 1; i <= num; i++) {
// printf(count); --> Bad
printf("Value = %d\n", i);
}
getchar(); // this will return immediately
return 0;
}
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I want to check if the numbers in an array are a power of two.
I wrote the following code, but it doesn't work it skips the part that checks if the number is the power of two and prints the last sentence.
Also, if someone can help me in how to check if the input is a number and not any other character.
Thank you!
update the power of two thing is working but i still haven't figure out how to check if the input is a number and not any other characher
#include <stdio.h>
#include <stdlib.h>
int main()
{
int x;
int i;
int k;
int count=0;
int a;
int sum=0;
printf("Enter size of input:\n");
scanf("%d",&x);
int *numbers=malloc(sizeof(int)*x);
if (x<0){
printf("Invalid size\n");
}
else {
printf("Enter numbers:\n");
for(i=0;i<x;++i){
scanf("%d",&numbers[i]);
}
}
for(k=0;k<x;++k)
{
count=0;
a=numbers[k];
while (((numbers[k] % 2) == 0) && numbers[k] > 1){ /* While x is even and > 1 */
numbers[k]/= 2;
++count;
}
if (numbers[k]==1&&a!=1){
printf("The number %d is a power of 2:%d=2^%d\n",a,a,count);
sum+=count;
}
}
printf("Total exponent num is %d\n",sum);
return 0;
}
Your check for the power of two is wrong: you divide out two all the way down to 1, but the following if incorrectly checks numbers[k]==0.
The check should be numbers[k]==1 instead, because when you divide out all twos from a power of two you end up with 20, which is 1.
Note: You can check if a number is a power of two without a loop by using a bit trick described in this Q&A.
There's much in your example that's incidental to the problem. For example, allocating an array and reading user input is just a distraction from finding the solution. Concentrate first on debugging your algorithm:
#include <stdbool.h>
bool is_power_of_two(int n)
{
while (n % 2 == 0 && n > 1){ /* While x is even and > 1 */
n/= 2;
}
return n == 0;
}
int main()
{
return !is_power_of_two(2);
}
Now, you can refine that function until it gives the correct result. The simple fix is to replace n == 0 with n == 1. Now you can add more tests, running the program as you add each one:
int main()
{
return is_power_of_two(0)
+ !is_power_of_two(1)
+ !is_power_of_two(2)
+ is_power_of_two(3)
+ !is_power_of_two(4)
/* negative numbers can never be an exact power of a positive */
+ is_power_of_two(-1)
+ is_power_of_two(-2)
+ is_power_of_two(-3);
}
Once you have some confidence in your function, you can then use it in your program to process arrays.
When you do introduce a function to read input, you'll want to check that x isn't negative before using in the argument to malloc(). Better would be to ensure it's not negative, by using an unsigned type:
unsigned int x;
printf("Enter size of input:\n");
if (scanf("%u", &x) != 1) {
fprintf(stderr, "That's not a valid size!\n");
return EXIT_FAILURE;
}
int *numbers = malloc(x * sizeof *numbers);
if (!numbers) {
fprintf(stderr, "Couldn't allocate memory for %u numbers!\n", x);
return EXIT_FAILURE;
}
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#include <stdio.h>
#include <conio.h>
void main() {
int a, n, x;
clrscr();
printf("enter a number");
scanf("%d", &a);
n > 1;
a != n && n < a;
if (a / n == x)
printf("a is not a prime no");
else
printf("a is a prime no");
}
If I run this and put a composite number, it still shows it as prime.
your if statement is never true duo to n and x are not initialized. Therefore you only get your else as return. Moreover your expression n>1; and a != n && n < a; return a bool which is not compered to anything. In that case you need to use a for loop.
Here is a link About for loops
int main()
{
int a,n,x = 0;
printf("enter a number: ");
scanf("%d",&a);
for(n=2; n<=a/2; ++n)
{
if(a%n==0)
{
x=1;
break;
}
}
if (x==0)
printf("",n);
else
printf("a is not a prime no");
return 0;
}
#include<stdio.h>
#include<conio.h>
int main()
{
clrscr();//clearing the screen
int n,x=2,count=0;//Here count is initialised to 0,if it is not prime it remains the same,else it will be equal to 1.You will understand this as you go down
//A number is a prime number if it is not divisible by any other number from 2 and the number before it.
printf("Enter a number : ");
scanf("%d",&n);
while(x<n)//As this checking process should continue till the number just preceding it
{
if(n%x==0)//checking if the number n is divisible by x or not
{
count++;//IF divisible,there is no meaning in continuing,So we are coming out of the loop by incrementing the variable "count"
break;
}
else
x++;
}
if(count==0)
{
printf("%d is a prime number",n);
return 0;//Here if number is prime,There is no need to go further and execute till end,To reduce time complexity ,We will write a return statement to stop executing the code.
}
printf("%d is not a prime number",n);
return 0;
}
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I've been asked to write a program that accepts a list of numbers until a non-numeric is encountered (up to 30 numbers), putting the numbers into an array, and keeping track of how many numbers were inserted. Then it should scan through the array to find the largest number, and print the largest.
This is what I've come up with:
#include<stdio.h>
int main()
{
const int INPUT = 30 ;
int size [INPUT];
int i, big;
printf("Type integer numbers, followed by q to quit: ");
while (scanf("%d", &size[INPUT]) != 'q')
{
for(i=0;i<size;i++)
scanf("%d",&INPUT[i]);
big = INPUT[0];
for(i=1;i<size;i++)
{
if(big<INPUT[i])
big=INPUT[i];
}
printf("The largest number is %d",big);
return 0;
}
Besides the problems, I listed in the comments. You seems to be comfused by the varaible names~ Anyway, I made some code for you.
#include<stdio.h>
int main()
{
const int MAX_INPUT = 30 ;
int input[MAX_INPUT];
int size=0, big;
printf("Type integer numbers, followed by q to quit: ");
while(size < MAX_INPUT){
if(scanf("%d", &input[size]) != 1){
break;
}
++size;
}
if(size ==0){
return 0;
}
big = input[size-1];
while( size-- > 0)
{
if(big<input[size]){
big=input[size];
}
}
printf("The largest number is %d\n",big);
return 0;
}
Tested with GCC 4.1.2 and Linux.
Return value of scanf:
Upon successful completion, these functions return the
number of successfully matched and assigned input items
further, you are mixing the size and input, you actually want the size to be a constant and input to be an array:
const int SIZE = 30 ;
int input[SIZE];
So the while loop should look like:
while (scanf("%d", &input[some_index]) == 1)
and of course this is wrong:
scanf("%d",&INPUT[i]); // should be ==> &input[i]