So, I'm doing an exercise, and I want to sort a list of float numbers. When I used for loop, it worked perfectly. When I changed to while loop, it shows nothing. I already tried to declare a different variable for each while loop, but remains the same. I'm getting this warning for the file ex005.c:
Please, input one non-integer values: 45.3
Please, input one non-integer values: 32.2
Please, input one non-integer values: 34.5
zsh: segmentation fault ./"ex005"
Here is my code:
#include <stdio.h>
int main(void)
{
float three_numbers[3];
char support_var;
int size_array;
int i;
int z;
size_array = 3;
z = size_array - 1;
i = 0;
while (i < size_array)
{
printf("Please, input one non-integer values: ");
scanf("%f", &three_numbers[i]);
i++;
}
i = 0;
while (i < size_array)
{
while (z != i)
{
if (three_numbers[i] < three_numbers[z])
{
support_var = three_numbers[i];
three_numbers[i] = three_numbers[z];
three_numbers[z] = support_var;
}
z--;
}
i++;
}
i = 0;
while (i < size_array)
{
printf("The %dth place is %.1f. \n", i + 1, three_numbers[i]);
i++;
}
return (0);
}
You probably want to set z in the inner loop otherwise it causes out of bound access of three_numbers[z]. Also, your support_var should be a float. Use the typeof operator if available (with gcc 10.2.1-6 it's still an extension so -std=gnu18):
#include <stdio.h>
#define ARRAY_SIZE 3
int main(void) {
float three_numbers[ARRAY_SIZE];
for(int i = 0; i < ARRAY_SIZE; i++) {
printf("Please, input one non-integer values: ");
scanf("%f", three_numbers + i);
}
for(int i = 0; i < ARRAY_SIZE; i++) {
for(int z = ARRAY_SIZE - 1; z != i; z--) {
if(three_numbers[i] < three_numbers[z]) {
// typeof (*three_numbers) support_var = three_numbers[i];
float support_var = three_numbers[i];
three_numbers[i] = three_numbers[z];
three_numbers[z] = support_var;
}
}
}
for(int i = 0; i < ARRAY_SIZE; i++) {
printf("The %dth place is %.1f. \n", i + 1, three_numbers[i]);
}
}
Not sure why you don't like for-loops but it makes your code easier to read. Also replaced int array_size with a constant ARRAY_SIZE. Minimizing scope of variables makes is usually a good idea.
Related
So here is the problem: Write a program that accept an integer n, print out the largest number but smaller or equal n that is the product of two consecutive even number. Example: Input: 12, Output: 8 ( 2x4 )
Here is my code :
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
for (int i = n; i >= 0; i--)
{
for (int j = 0; j <= n; j = j + 2)
{
if ( i == j * (j+2) )
{
printf("%d ", i);
break;
}
}
}
return 0;
}
So if i input 20, it will print out 8 and 0 instead of 8, if i input 30, it will print out 24,8 and 0 instead of just 24. How do i make it stop after printing out the first number that appropriate ?
You need to stop an outer loop from processing, for example by using a boolean flag (meaning "solution found, we finish work") or a goto statement.
#include <stdio.h>
int main() {
int n;
scanf("%d", &n);
int solutionFound = 0;
for (int i = n; i >= 0; i--) {
// this could also be put into for's condition i.e. "i >= 0 && !solutionFound"
if (solutionFound) {
break;
}
for (int j = 0; j <= n; j = j + 2) {
if ( i == j * (j+2) ) {
printf("%d ", i);
solutionFound = 1;
break;
}
}
}
return 0;
}
EDIT: immediate return as noted in the comments is also a nice idea, if you don't need to do anything later.
Your problem is that you are nested - in a for loop which is inside another for loop - when you want to stop processing.
Some languages would let you code break 2; to indicate that you want to break out of 2 loops. Alas, C i snot such a language.
I would recommend that you code a function. That would serve a few porpoises: 1) your main should be "lean & mean" 2) as your programs get larger, you will learn the benefits of putting individual coding tasks into functions 3) you can use return; instead of break; and it will exit the function immediately.
Something like this:
#include <stdio.h>
void FindNeighbouringDivisors(int n)
{
for (int i = n; i >= 0; i--)
{
for (int j = 0; j <= n; j = j + 2)
{
if ( i == j * (j+2) )
{
printf("%d times %d = %d", j, j + 2, i);
return;
}
}
}
printf("There are no two adjacent even numbers which can be multiplied to give %d", n);
}
int main()
{
int n;
scanf("%d", &n); /* could get from comamnd line */
FindNeighbouringDivisors(n);
return 0; /* should be EXIT_SUCCESS */
}
Btw, when you have a problem with your code, ask a question here. When you have it working, consider posting it at our code review site where more experienced programmers can give you advice on how to improve it. It's a great way to learn
Break only breaks you out of immediate loop, so either use flags or just use return to terminate the execution. Or you can even use following code:
#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
for (int j = 0; j <= n; j = j + 2)
{
if ( n < j * (j+2) )
{
printf("%d ", j*(j-2));
break;
}
}
return 0;
}
Sample Input
2
2
5
Sample Output
0 1 1
0 1 1 2 1 2
I already knew how to transfer an integer to its binary and count how many 1 in its binary.
But my code only can input one integer each time. I want it to input many numbers, like the sample input and sample output. To make question more easy to understand, so I drew a picture. Thank you!!
Thanks all of you!! But I need some time to understand those code and my English is pretty basic, so I couldn't reply you guys soon. But I will understand and reply you as soon as possible!!thank you :D
#include <stdio.h>
int main()
{
int n,cnt=0,m;
scanf("%d",&n);
while(n>0){
m=n%2;
if(m==1){
cnt++;
}
n/=2;
}
printf("%d",cnt);
return 0;
}
In practice, the process can be simplified and accelerated by noticing that if we know the result for i < 2^p, then, for all values in [2^p, 2^(p+1)-1], we have
count(j) = 1 + count(j-2^p)
This method is useful here as we have to provide the result for all values less or equal to n.
Moreover, in order to avoid performing the same calculation at different times, we first calculate the maximum n value, and perform the calculation for this maximum value.
#include <stdio.h>
int main() {
int t;
int check = scanf("%d", &t);
if (check != 1) return 1;
int nn[t];
for (int i = 0; i < t; ++i) {
int check = scanf("%d", &nn[i]);
if (check != 1) return 1;
}
int nmax = 0;
for (int i = 0; i < t; ++i) {
if (nn[i] > nmax) nmax = nn[i];
}
int count[nmax+1];
count[0] = 0;
count[1] = 1;
int pow2 = 1;
do {
pow2 *= 2;
for (int j = pow2; (j < 2*pow2) && (j <= nmax); j++) {
count[j] = 1 + count[j-pow2];
}
} while (pow2 <= nmax+1);
for (int i = 0; i < t; ++i) {
for (int j = 0; j <= nn[i]; ++j) {
printf ("%d ", count[j]);
}
printf ("\n");
}
return 0;
}
Let the number of measured values be t. Now if you want to take t number of integers as input, you have to run a loop t times and perform necessary tasks (like taking value of n, calculating cnt etc) inside that loop. Check the following code snippet:
int main()
{
int n,cnt=0,m;
int numberOfMeasuredValues;
scanf("%d", &numberOfMeasuredValues);
while(numberOfMeasuredValues > 0){
cnt = 0;
scanf("%d",&n);
// perform necessary calculations
printf("%d\n",cnt);
numberOfMeasuredValues--;
}
return 0;
}
In the code numberOfMeasuredValues is the t I've used in the explanation.
You can pre-allocate number of cases and work on them with outer and inner loops. I've added the code with comments below.
#include <stdio.h>
int main(void)
{
int cases; // Number of cases
scanf("%d", &cases);
int* nums = malloc(cases * sizeof *nums); // Allocate enough space.
/* Take the inputs. */
for (int i = 0; i < cases; i++)
{
scanf("%d", (nums + i));
}
int cnt = 0; // Counter for number of 1s.
// Outer loop for going through cases.
for (int i = 0; i < cases; i++)
{
// Inner loop for couting towards inputted number.
for (int j = 0; j < *(nums + i) + 1; j++)
{
/*
* You can Implement this part by yourself if you wish.
*/
for (int k = 0; k < 32; k++)
{
if ((j >> (31 - k)) & 0x01) cnt++; // Increment counter if digit is 1.
}
printf("%d ", cnt);
cnt = 0;
}
printf("\n");
}
return 0;
}
Here's the following code to your problem
#include<iostream>
using namespace std;
int main()
{
int tcs;
cin >> tcs;
while (tcs--) {
int n;
cin >> n;
for (int i = 0 ; i <= n ; i++) {
cout << __builtin_popcount(i) << " ";
}
cout << "\n";
}
return 0;
}
I'm pretty much done with my code which calculates all primes between 2 and 1000. However, I need make a code which scans my prime array and prints whether its a prime or not.
For those who dont understand what I want: After calculating all primes between 2 and 1000, I want the user to give a random number which then tells the user whether its a prime or not. With this code below, it doesnt work for some reason... can someone tell me what I did wrong?
#include <stdio.h>
#include <stdbool.h>
int main () {
int durch [1000] = {0};
for (int i = 2; i <= 1000; i++)
{
if (!durch[i-2])
{
for (int j = 2*i; j <= 1000; j+= i)
{
durch[j-2] = 1;
}
}
}
int primzahlen [1000];
int anzahl = 0;
int n;
for (int i = 2; i <= 1000; i++)
{
if (!durch [i-2])
{
primzahlen[anzahl] = i;
anzahl++;
// printf ("%i : %i\n", anzahl, i);
scanf ("%i\n", &n);
if (n = primzahlen[anzahl]){
printf ("Yes it is a prime!\n");
break;
}
else {
printf ("No it isnt a prime!\n");
break;
}
}
}
return 0;
}
I really cant figure that part out.. I feel like I miss something? Can somebody help me out please?
you are discarding value of n variable here:
if (n = primzahlen[anzahl]){
you probably intended to compare numbers instead of assignment
if (n == primzahlen[anzahl]){
The whole second part of your program isn't necessary. You can just simplify it to:
#include <stdio.h>
#include <stdbool.h>
int main () {
int i;
int durch [1000] = {0};
int n;
for (i = 2; i <= 1000; i++)
{
if (!durch[i-2])
{
for (int j = 2*i; j <= 1000; j+= i)
{
durch[j-2] = 1;
}
}
}
scanf ("%i", &n);
if ((n > 1) && !durch [n-2]) {
printf ("Yes it is a prime!\n");
}
else {
printf ("No it isnt a prime!\n");
}
return 0;
}
And note scanf ("%i\n", &n); will ask for 2 numbers. You need to use scanf ("%i", &n);
See: scanf() curious behaviour!
I have an array of numbers ex.[5,1,2,4,6,8,12], and I want to find the length of longest arithmetic progression within the sequence and to print it. Longest arithmetic progression means an increasing sequence with common difference, in this case [2,4,6,8].
#include <stdio.h>
void main()
{
int array[100], i, num,diff;
printf("Enter the size of an array \n");
scanf("%d", &num);
printf("Enter the elements of the array \n");
for (i = 0; i < num; i++) {
scanf("%d", &array[i]);
}
printf("\n Numbers in a.p: ");
for (i = 0; i < num; i++) {
diff = array[i+1]-array[i];
if (array[i]-diff == array[i+1]-diff);
{
printf("%d, ", array[i]);
}
}
printf("\n Common difference:%d", diff);
}
like this
#include <stdio.h>
int main(void){
#if DEBUG
int array[] = {5,1,2,4,6,8,12};
int num = sizeof(array)/sizeof(*array);
int i, diff;
#else
int array[100], i, num,diff;
printf("Enter the size of an array \n");
scanf("%d", &num);
printf("Enter the elements of the array \n");
for (i = 0; i < num; i++) {
scanf("%d", &array[i]);
}
#endif
int j, len, longest_len = 2, longest_i = 0;
for (i = 0; i < num - longest_len; i += len-1) {
diff = array[i+1]-array[i];
for(j = i+2; j < num && array[j-1]+diff == array[j]; ++j);
len = j - i;
if(longest_len < len){
longest_len = len;
longest_i = i;
}
}
printf("\n Numbers in a.p: ");
diff = array[longest_i+1] - array[longest_i];
printf("[ ");
for(i = 0; i < longest_len; ++i){
printf("%d", array[longest_i + i]);
if(i == longest_len-1)
printf(" ]\n");
else
printf(", ");
}
printf("\n Common difference:%d", diff);
}
Since this seems to be a homework or challenge I will only help by solving your immediate problems, caused by very strange code.
Here is code which at leat detects the progressions correctly.
You can yourself count the length, store the longest length and its index, then print that sequence after parsing all the array.
There are two assumptions here, which you might want to avoid for challenge/homework:
no identical numbers entered
no overlapping progressions,
i.e. no number is the last of one and the first of next progression
The representation of more than one sequence in the output is a little jittery (missing ")"), but that is not relevant for finding and exclusively printing the longest one.
I did not bother about your ratio output, no idea what that is supposed to be.
Code:
#include <stdio.h>
// avoid a warning
int main() {
int array[100], i, num, diff=0, lastdiff=0, first=1;
printf("Enter the size of an array \n");
// for good code, this should check the result
scanf("%d", &num);
// in case of scaf-failure, cleanup here for good code
// asking for the number of elements
// and then relying on that number to be entered
// is less elegant than checking for and end condition
// like EOF or negative input
printf("Enter the elements of the array \n");
for (i = 0; i < num; i++) {
// for good code, this should check the result
scanf("%d", &array[i]);
// in case of scaf-failure, cleanup here for good code
}
printf("\n Numbers in a.p: ");
for (i = 1; i < num; i++) {
lastdiff=diff;
diff = array[i]-array[i-1];
if (diff==lastdiff)
{ if(first==1)
{ first=0;
printf("(%d, %d",array[i-2], array[i-1]);
}
printf(", %d", array[i]);
} else
{ first=1;
}
}
printf(")\n Ratio:%d", diff);
// avoid a warning
return 0;
}
There are a number of ways to approach this challenge. You can either check each diff against each value in the array or work the other way around. Given you are not sorting the values, you may benefit by nesting the check of values within your loop over all possible diffs. Something similar to the following works:
#include <stdio.h>
int main (void) {
int a[] = {5, 1, 2, 4, 6, 8, 12},
n = sizeof a / sizeof *a,
startidx = 0,
maxlen = 0;
for (int d = 1; d < n; d++) { /* loop over diffs */
int idx = -1,
len = 1;
for (int i = 1; i < n; i++) /* loop over values */
if (a[i - 1] + d == a[i]) {
if (idx < 0) /* if index not set */
idx = i - 1; /* set to 1st index */
len++; /* increment length */
}
if (idx >= 0 && len > maxlen) { /* if progression found */
maxlen = len; /* save length */
startidx = idx; /* save start index */
}
}
printf ("longest progression: '%d' elements.\n", maxlen);
for (int i = 0; i < maxlen; i++)
printf (i ? ", %d" : "%d", a[startidx + i]);
putchar ('\n');
return 0;
}
Example Use/Output
$ ./bin/maxprogression
longest progression: '4' elements.
2, 4, 6, 8
Investigate several ways to approach it, and finally settle on the one that makes the most sense to you. You can work on optimizing later.
As far as the code you posted goes, always validate all user input by, at minimum, checking that the number of expected conversions to type took place, e.g.
if (scanf("%d", &num) != 1) {
fprintf (stderr, "error: input conversion failed for 'num'.\n");
return 1;
}
You would do the same in your values loop. Let me know if you have any questions. Good luck with your coding.
Following is the complete working code. You can see it working here:
#include <stdio.h>
int main()
{
int array[100], i, num,diff, resDiff, startIndx, resStartIndx, countCur, countPre;
printf("Enter the size of an array \n");
scanf("%d", &num);
printf("Enter the elements of the array \n");
for (i = 0; i < num; i++) {
scanf("%d", &array[i]);
}
//Now code changes
startIndx =0, resStartIndx=0, countPre=0, resDiff=0;
for (i = 0; i < num; /*No increment required here*/) {
countCur =0;
startIndx=i;
countCur++;
if(++i < num)
{
diff = array[i] - array[startIndx];
countCur++;
i++;
while((i < num) && (diff == (array[i] - array[i-1])))
{
countCur++;
i++;
}
if(countCur > countPre)
{
resStartIndx = startIndx;
countPre = countCur;
resDiff = diff;
}
}
}
countPre += resStartIndx;
printf("\n Numbers in a.p: ");
for (i = resStartIndx; i < countPre; i++) {
printf("%d, ", array[i]);
}
printf("\n Common difference:%d", resDiff);
return 0;
}
int main(void)
{
int i,j=0,k; //initialization
char equation[100]; //input is a string (I think?)
int data[3]; //want only 3 numbers to be harvested
printf("Enter an equation: ");
fgets(equation, 100, stdin); //not so sure about fgets()
for (i = 0; i < equation[100]+1; i++) { //main loop which combs through
//"equation" array and attempts
//to find int values and store
while (j <= 2) { //them in "data" array
if (isdigit(equation[i])) {
data[j] = equation[i]
j++;
}
}
if (j == 2) break;
}
for (k = 0; k <= 2; k++) { //this is just to print the results
printf("%d\n", data[k]);
}
return 0;
}
Hello! This is my program for my introductory class in C, I am trying to comb through an array and pluck out the numbers and assign them to another array, which I can then access and manipulate.
However, whenever I run this I get 0 0 0 as my three elements in my "data" array.
I am not sure whether I made an error with my logic or with the array syntax, as I am new to arrays.
Thanks in advance!!! :)
There are a few problems in your code:
for (i = 0; i < equation[100]+1; i++) { should be something like
size_t equ_len = strlen(equation);
for (i = 0; i < equ_len; i++) {
Whatever the input is, the value of equation[100] is uncertain, because char equation[100];, equation only has 100 element, and the last of them is equation[99].
equation[i] = data[j]; should be
data[j] = equation[i];
I suppose you want to store digit in equation to data.
break; should be deleted.
this break; statement will jump out of the while loop, the result is you will store the last digit in equation to data[0] (suppose you have switched data and equation, as pointed out in #2).
If you want the first three digits in equation, you should do something like
equ_len = strlen(equation);
j = 0;
for (i = 0; i < equ_len; i++) {
if (j <= 2 && isdigit(equation[i])) {
data[j] = equation[i];
j++;
}
if (j > 2) break;
}
printf("%d\n", data[k]); should be printf("%c\n", data[k]);
%d will give the ASCII code of data[k], for example, if the value of data[k] is character '1', %d will print 50 (the ASCII code of '1') instead of 1.
Here is my final code, based on the OP code:
#include <ctype.h>
#include <string.h>
#include <stdio.h>
int main(void)
{
int i,j,k;
char equation[100];
int data[3];
int equ_len;
printf("Enter an equation: ");
fgets(equation, 100, stdin);
equ_len = strlen(equation);
j = 0;
for (i = 0; i < equ_len; i++) {
if (j <= 2 && isdigit(equation[i])) {
data[j] = equation[i];
j++;
}
if (j > 2) break;
}
for (k = 0; k <= 2; k++) {
printf("%c\n", data[k]);
}
return 0;
}
Tested with:
$ ./a.out
Enter an equation: 1 + 2 + 3
1
2
3