write a program that enter a series of integer(stores in array),then sorts the integer by calling the function selection_sort. When given an array with n elements, selection_sort must do the following:
1.search the array to find the largest,then move it to the last position.
2.call itself recursively to sort the first n-1 elements of the array.
following is my code i think the code is errors everywhere i hope some master can help me
#include <stdio.h>
int selection_sort(int a[])//this function have error that "i"and"count"is undeclared
{
int max = 0;
for (i = 1; i <= count; i++)// continuous compare to final
{
if (a[i] > max)
{
max=a[i];
}
}
a[count] = a[i]; //put the max to last position
count--;
}
int main(void)
{
int a[100] = { 0 },i=0,count=0;
while (1)
{
scanf("%d",&a[i]);
if (a[i] = '\n') { break; }//this line have error because '\n' not "int" so when i "enter" it would not break
i++;
count++; //counting how many integer i scanf
}
selection_sort();//call this function (i don't know well about function so i don't known where to put is correct )
return 0;
}
You have to compare. BUt you are assigned..
change this if (a[i] = '\n') { break; } to if (a[i] == '\n') { break; }.
You should change follows in your code.
1) change your function call..
2) declare count and i in function..
3) for taking values in to array,follow other method..
Try yourself...
Here is the complete code.Basically there are many syntax and logical error in your code. Consider this as refer and compare with your original source.
int selection_sort(int a[], int count){
int max = 0, i =0;
for (i = 0; i < count; i++){
if (a[i] > max)
{
max=a[i];
}
}
a[count] = max;
count--;
return 0;
}
int main(void) {
int a[100] = { 0 },i=0,count=0;;
printf ("Enter the total num\n");
scanf("%d",&count);
if ( ( count ) >= (sizeof(a)/sizeof(a[0])) )
return -1;
while (i < count){
scanf("%d",&a[i]);
i++;
}
selection_sort(a, count);
printf ("\nmax:%d\n", a [count]);
return 0;
}
Related
If the user inputs a value that matches with an int in the array continue the program else quit the program.
My issue is that the function loops the whole array and if it finds one of the values doesnt match it will quit.
//Check if the user input matches with one of the ints in the array
void check(int num, int arr[]) {
for (int i = 0; i < 3; i++) {
if (arr[i] != num) {
printf("Invalid input");
exit(0);
}
}
}
void main() {
int arr[3] = { 1, 2, 3 };
int num = 0;
for (int i = 0; i < 3; i++) {
printf("%d ", arr[i]);
}
printf("\nPLease enter a value which matches with the array %d", num);
scanf("%d", &num);
check(num, arr);
}
void check(int num, int arr[]) {
for (int i = 0; i < 3; i++) {
if (arr[i] == num) {
return;
}
}
printf("Invalid input");
exit(0);
}
Your issue is that checks a single element and judges the input on that specific value. If it has run through each value and the function has still not returned, there is not match and we can exit the program.
You have a logic flaw in the check function: you should output the message and quit if none of the values match the input. You instead do this if one of the values does not match. The check always fails.
Here is a modified version:
#include <stdio.h>
//Check if the user input matches with one of the ints in the array
void check(int num, const int arr[], size_t len) {
for (size_t i = 0; i < len; i++) {
if (arr[i] == num) {
// value found, just return to the caller
return;
}
}
// if we get here, none of the values in the array match num
printf("Invalid input: %d\n", num);
exit(1);
}
int main() {
int arr[3] = { 1, 2, 3 };
size_t len = sizeof(arr) / sizeof(*arr); // length of the array
int num;
for (size_t i = 0; i < len; i++) {
printf("%d ", arr[i]);
}
printf("\n");
printf("Please enter a value which matches with the array: ");
if (scanf("%d", &num) == 1) {
check(num, arr, len);
} else {
// input missing or not a number
printf("Invalid input\n");
}
return 0;
}
You are right, you do exit once arr[i] != num (When the value is not the same as the i:th element in the array).
So, you could change it to: arr[i] == num. If it is the same, perhaps print "You got it!", and a return afterwards.
I have the code for finding prime numbers within a range.
The problem is to remove the last comma.
#include<stdio.h>
int main()
{
int a,b,i,x,c,f=1;
scanf("%d%d",&a,&b);
for(x=a;x<=b;(x++,f=0))
{
for(i=2;i<x;i++)
{
if(x%i==0)
{
f=1;
}
}
if(f==0)
printf("%d,",x);
}
}
But the output contains an extra comma in the last.
For example
2,3,5,7,
whereas the expected output is
2,3,5,7
Instead of flag you can decide directly what you want to print between numbers
And note that you can break out of the internal loop as soon as f is set to 1
#include<stdio.h>
int main()
{
int a,b,i,x,c,f=1;
const char* delim = "";
scanf("%d%d",&a,&b);
for(x=a; x<=b; (x++,f=0))
{
for(i=2; i<x; i++)
{
if(x%i==0)
{
f=1;
break; //no need to continue the checking
}
}
if(f==0) {
printf("%s%d",delim,x);
delim = ", ";
}
}
putchar('\n');
}
#include<stdio.h>
int main()
{
int a,b,i,x,c,f=1;
char backspace = 8;
scanf("%d%d",&a,&b);
for(x=a;x<=b;(x++,f=0))
{
for(i=2;i<x;i++)
{
if(x%i==0)
{
f=1;
}
}
if(f==0)
printf("%d,",x);
}
printf("\b"); // or printf("%c", backspace);
}
Add another flag, just a simple counter that tells you if you are printing the first time then check the flag to decide what to print, e.g.
#include<stdio.h>
int main()
{
int a,b,i,x,c,first=0,f=1;
scanf("%d%d",&a,&b);
for(x=a;x<=b;(x++,f=0))
{
for(i=2;i<x;i++)
{
if(x%i==0)
{
f=1;
}
}
if(f==0)
{
if(first==0){
printf("%d",x);
}else{
printf(",%d",x);
}
first++
}
}
}
Use a flag to detect the first occurrence of printf() and print the first number as such without any ,. For consecutive number printing precede with ,
#include<stdio.h>
int main()
{
int a,b,i,x,c,f=1,flag=0;//Flag to mark first occurrence
scanf("%d%d",&a,&b);
for(x=a;x<=b;(x++,f=0))
{
for(i=2;i<x;i++)
{
if(x%i==0)
{
f=1;
break;// Once the condition fails can break of the for loop as it fails for the prime number condition at the first case itself
}
}
if(f==0)
{
if(flag==0)
{//Check if it is first time
printf("%d",x);
flag = 1;//If so print without ',' and set the flag
}
else
printf(",%d",x);// On next consecutive prints it prints using ','
}
}
}
This method also avoids the , when only one number is printed.
Eg: When input is 2 and 4. It prints just 3 and not 3,
Simply you need odd number best practice for minimum loop is given below;
#include<stdio.h>
int main()
{
int a,b,i,x,c,f=1;
scanf("%d%d",&a,&b);
while (a < b)
{
if ( (a%2) == 1) {
printf("%d", a);
if ( (a + 1) < b && (a + 2) < b)
printf(",");
}
a = a + 1;
}
}
please check from the site
http://rextester.com/MWNVE38245
Store the result into a buffer and when done print the buffer:
#include <stdio.h>
#include <errno.h>
#define RESULT_MAX (42)
size_t get_primes(int * result, size_t result_size, int a, int b)
{
int i, x, f = 1;
size_t result_index = 0;
if (NULL == result) || (0 == result_size) || ((size_t) -1 == result_size))
{
errno = EINVAL;
return (size_t) -1;
}
for (x = a; x <= b; (x++, f = 0))
{
for (i = 2; i < x; i++)
{
if (x % i == 0)
{
f = 1;
break;
}
}
if (f == 0)
{
result[result_index] = x;
++result_index;
if (result_size <= result_index)
{
fprintf(stderr, "Result buffer full. Aborting ...\n");
break;
}
}
}
return result_index;
}
int main(void)
{
int a = 0, b = 0;
int result[RESULT_MAX];
scanf("%d%d", &a, &b);
{
size_t result_index = get_primes(result, RESULT_MAX, a, b);
if ((size_t) -1 == result_index)
{
perror("get_primes() failed");
}
else if (0 == result_index)
{
fprintf(stderr, "No primes found.\n");
}
else
{
printf("%d", result[0]);
for (size_t i = 1; i < result_index; ++i)
{
printf(", %d", result[i]);
}
}
}
return 0;
}
This example uses a simple fixed-size buffer, if this does not suite your needs replace it by a dynamic one.
This is more of a "language-agnostic" problem: "How do I output a comma-separated list without a final comma?" It is not specifically about prime numbers.
You seem to be thinking of you list as a series of [prime comma] units. It isn't. A better way to think of it is as a single prime as the head of the list, followed by a tail of repeated [comma prime] units.
Some pseudocode to illustrate the general idea:
outputList(theList)
separator = ", "
output(theList.firstItem())
while (theList.hasMoreItems())
output(separator)
output(theList.nextItem())
endwhile
return
/* this is just logic */
for(i=2;i<=n;i++)
{
k=0;
for(j=2;j<=i/2;j++)
{
if(i%j==0)
k=1;
}
if(k==0)
{
c++;
c++;
}
}
System.out.println(c);
for(i=2;i<=n;i++)
{
k=0;
for(j=2;j<=i/2;j++)
{
if(i%j==0)
k=1;
}
if(k==0)
{
System.out.print(i);
b++;
if(b!=c-1)
{
System.out.print(",");
b++;
}
}
}
}
}
//comma separated values
#include <bits/stdc++.h>
using namespace std;
int Prime(int a, int n){
bool prime[n+1];
memset(prime,true,sizeof(prime));
for(int p=2;p*p<=n;p++){
if(prime[p]==true){
for(int i=p*p ; i<=n; i+=p ){
prime[i] = false;
}
}
}
for(int i = 2;i<= n;i++){
if(i==2) cout<<i; // here is the logic first print 2 then for other numbers first print the comma then the values
else if(prime[i]) cout<<","<<i;
}
}
int main(){
int a =2 ;
int n = 30;
Prime(a , n);
}
#include <stdio.h>
int main()
{
int i, j, n, count;
scanf("%d", &n);
for(i=2; i<n; i++)
{
count=0;
for(j=2; j<n; j++)
{
if(i%j==0)
count++;
}
if(count==1)
printf("%d," i);
}
printf("\b \b");
}
\b is a nondestructive backspace. It moves the cursor backward, but doesn't erase what's there, it replaces it. For a a destructive backspace,
use "\b \b" i.e. a backspace, a space, and another backspace.
This Program prints all the prime number up to given number with comma separated
I'm having trouble outputting an invalid statement if the user inputs a letter instead of a number into a 2D array.
I tried using the isalpha function to check if the input is a number or a letter, but it gives me a segmentation fault. Not sure what's wrong any tips?
the following code is just the part that assigns the elements of the matrix.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define MAX 10
void display(int matrix[][MAX], int size);
int main() {
int n, degree;
int matrix[MAX][MAX];
printf("Enter the size of the matrix: "); // assigning size of the matrix
scanf("%d", &n);
if (n <= 1 || n >= 11) { // can't be bigger than a 10x10 matrix
printf("Invalid input.");
return 0;
}
for (int i = 0; i < n; ++i) { // assigning the elements of matrix
printf("Enter the row %d of the matrix: ", i);
for (int j = 0; j < n; ++j) {
scanf("%d", &matrix[i][j]);
if (!isalpha(matrix[i][j])) { // portion I'm having trouble with
continue;
} else {
printf("Invalid input.");
return 0;
}
}
}
...
As the value of n will be number, we can solve it using string instead of int.
char num[10];
int n;
scanf("%s", num);
if(num[0] < '0' || num[0] > '9' || strlen(num) > 2){
printf("invalid\n");
}
if(strlen(num) == 1) n = num[0] - '0';
if(strlen(num) == 2 && num[0] != 1 && num[1] != 0) printf("invalid\n");
else n = 10;
Also we can use strtol() function to convert the input string to number and then check for validity.You can check the following code for it. I have skipped the string input part. Also you have to add #include<stdlib.h> at the start for the strtol() function to work.
char *check;
long val = strtol (num, &check, 10);
if ((next == num) || (*check != '\0')) {
printf ("invalid\n");
}
if(val > 10 || val < 0) printf("invalid\n");
n = (int)val; //typecasting as strtol return long
You must check the return value of scanf(): It will tell you if the input was correctly converted according to the format string. scanf() returns the number of successful conversions, which should be 1 in your case. If the user types a letter, scanf() will return 0 and the target value will be left uninitialized. Detecting this situation and either aborting or restarting input is the callers responsibility.
Here is a modified version of your code that illustrates both possibilities:
#include <stdio.h>
#define MAX 10
void display(int matrix[][MAX], int size);
int main(void) {
int n, degree;
int matrix[MAX][MAX];
printf("Enter the size of the matrix: "); // assigning size of the matrix
if (scanf("%d", &n) != 1 || n < 2 || n > 10) {
// can't be bigger than a 10x10 matrix nor smaller than 2x2
// aborting on invalid input
printf("Invalid input.");
return 1;
}
for (int i = 0; i < n; i++) { // assigning the elements of matrix
printf("Enter the row %d of the matrix: ", i);
for (int j = 0; j < n; j++) {
if (scanf("%d", &matrix[i][j]) != 1) {
// restarting on invalid input
int c;
while ((c = getchar()) != '\n') {
if (c == EOF) {
printf("unexpected end of file\n");
return 1;
}
}
printf("invalid input, try again.\n");
j--;
}
}
}
...
The isdigit() library function of stdlib in c can be used to check if the condition can be checked.
Try this:
if (isalpha (matrix[i][j])) {
printf ("Invalid input.");
return 0;
}
So if anyone in the future wants to know what I did. here is the code I used to fix the if statement. I am not expecting to put any elements greater than 10000 so if a letter or punctuation is inputted the number generated will be larger than this number. Hence the if (matrix[i][j] > 10000). May not be the fanciest way to do this, but it works and it's simple.
for (int i = 0; i < n; ++i) { // assigning the elements of matrix
printf("Enter the row %d of the matrix: ", i);
for (int j = 0; j < n; ++j) {
scanf("%d", &matrix[i][j]);
if (matrix[i][j] > 10000) { // portion "fixed"
printf("Invlaid input");
return 0;
}
}
}
I used a print statement to check the outputs of several letter and character inputs. The lowest out put is around and above 30000. So 10000 I think is a safe condition.
I would like to know if the number the user enters is number in the array.
Here is my code:
#define ARR_SIZE 10
int main()
{
int my_arr[10];
int secnum = 0;
int i = 0;
for (i = 0; i < ARR_SIZE ; i++)
{
printf("Enter a number: ");
scanf("%d",&my_arr[i]);
}
printf("Enter the another number");
scanf("%d",&secnum);
if(my_arr[i] == secnum)
{
printf("an ex");
}
}
But it doesn't work.
How can I compare a number with another number in array?
Note: I don't know pointers so I need to do it without pointers.
Why it doesn't work and what is wrong with the code?
You are comparing against just one value rather than all the array elements.
The value of i after the scanf loop is 10 so arr[i] would exceed the
array and could cause Illegal memory access.
Check the comments in the program.
#define ARR_SIZE 10
int main()
{
int my_arr[ARR_SIZE]; //Use ARR_SIZE because if ARR_SIZE changes, 10 won't causing unforseen errors.
int secnum = 0;
int i = 0;
for (i = 0; i < ARR_SIZE ; i++)
{
printf("Enter a number: ");
scanf("%d",&my_arr[i]);
}
printf("Enter the another number");
scanf("%d",&secnum);
for (i = 0; i < ARR_SIZE ; i++) // Ensures you are comparing secnum with each array element.
{
if(my_arr[i] == secnum)
{
printf("an ex"); //Do you wish to break here because one you find a match, the goal is attained :)
}
}
}
After the loop, i is equal to ARR_SIZE (10). So you compare my_arr[10] with secnum (0). But my_arr[10], while syntactically correct, points to an undefined value because the size of the array is 10.
#define ARR_SIZE 10
int main()
{
int my_arr[10];
int secnum = 0;
int i = 0;
for (i=0;i<ARR_SIZE ; i++)
{
printf("Enter a number: ");
scanf("%d",&my_arr[i]);
}
printf("Enter the another number");
scanf("%d",&secnum);
for (i=0;i<ARR_SIZE ; i++)
{
if(my_arr[i]==secnum)
{
printf("Given number is in array\n");
break;
}
}
}
As pointed by OP in comments to one of the answers, OP apparently need a routine to check for the key in an array.
So once we have stored an array and have accepted a key to search, we need to pass this array and key to a search function which will return true or false depending upon whether the key is present in the array or not.
#include <stdbool.h> // We need this for `true` and `false` bool values
bool search(int [], int); // Function declaration
/**** Function Definition *****/
bool search(int numbers[], int key)
{
int i;
for(i = 0; i < ARR_SIZE; i++)
if(numbers[i] == key)
return true;
return false;
}
/** Calling search function from main **/
...
if(search(my_arr, secnum))
printf("Number found in array!\n");
else
printf("Number could NOT be found in array!\n");
To find a value in an array you should iterate through it and compare each value with the wanted number. You should also check the return value of scanf() to control how many item did it actually read. See if this reviewed code is helpfull:
#include <stdio.h>
#define ARR_SIZE 10
int read_numbers(int a[], int size) {
int i = 0;
int r = 0;
while ( i < size ) {
printf("Please, enter a number (%d of %d): ",i+1,size);
r = scanf(" %d",&a[i]); // the space before will ignore any trailing newline
if ( r == EOF ) break; // if scanf fails return
if ( r != 1 ) { // if user don't enter a number, repeat
printf("Wrong input!\n");
scanf("%[^\n]*"); // will read and ignore everything lweft on stdin till newline
continue;
}
++i;
}
return i; // return size, unless scanf fails
}
int find_number(int a[], int size, int x) {
int i = 0;
while ( i < size && a[i] != x ) ++i;
return i; // if x isn't in the array returns size
}
int main()
{
int my_arr[ARR_SIZE];
int secnum = 0;
int i = 0;
int n = 0;
n = read_numbers(my_arr, ARR_SIZE);
if ( n < ARR_SIZE ) {
printf("Warning, only %d numebers read out of %d!\n",n,ARR_SIZE);
} // if happens you have uninitialized elements in array
printf("Now, enter the value you want to find.\n");
if ( read_numbers(&secnum, 1) != 1 ) { // reuse the function
printf("Sorry, an error has occurred.\n");
return -1;
}
i = find_number(my_arr, n, secnum); // If all went right n==ARR_SIZE
if ( i < n ) { // a match has been found
printf("Found!\n");
} else {
printf("Not found.\n");
}
return 0;
}
I am creating a simple "ascending" program. When I find smallest int in array of int I want to replace it with some other value so it won't come as smallest number again in the array. For that, I assigned that int NULL. But now the results are not as expected.
Please tell me if I am doing anything wrong. If so then what I should replace with the value of that int ?
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int a[10],b,c=0,d=0;
printf("Enter number of values you want to enter \n");
scanf("%d",&b);
printf("Enter values \n");
for(int i=0;i<b;i++)
scanf("%d",&a[i]);
while(c<b)
{
for(int k=0;k<b;k++)
{
for(int j=0;j<b;j++)
{
if(a[k] > a[j])
{
d=1;
}
}
if(d!=1 && a[k]!=NULL)
{
c++;
printf("%d ",a[k]);
a[k]='\0' ; //assigning it as NULL
}
if(c >= b)
break;
d=0;
}
}
getch();
}
In C and related languages ints are not "nullable" - you could use a special value instead, e.g. a value that is well outside the expected range of your input data, such as INT_MAX:
#include <limits.h> // required header for INT_MAX et al
...
if(d!=1 && a[k]!=INT_MAX)
{
c++;
printf("%d ",a[k]);
a[k]=INT_MAX
}
However it would probably be a good idea to go back to the drawing board and see if you can come up with a better algorithm that doesn't require special values.
Read the differences between NULL and 0 and '\0' here. There is a type mismatch when you're trying a[k]!=NULL.
#include<stdio.h>
#include<conio.h>
int main()
{
clrscr();
int a[10], b, c = 0, d = 0;
int k, j, i;
printf("Enter number of values you want to enter \n");
scanf("%d",&b);
printf("Enter values \n");
for(i = 0;i < b;i++)
scanf("%d",&a[i]);
while(c < b)
{
for(k = 0;k < b;k++)
{
for(j = 0;j < b;j++)
{
if((a[k] > a[j]) && a[j] != 0)
{
d=1;
}
}
if(d != 1 && a[k] != 0)
{
c++;
printf("%d ",a[k]);
a[k] = 0; //assigning it as NULL
}
if(c >= b)
break;
d=0;
}
}
return 0;
getch();
}
This code fixes the problem.
What you're missing is a[j] != 0 in if((a[k] > a[j]) && a[j] != 0). Also I don't suggest this, as it won't work if there are 0's in the array entered.