#include <stdio.h>
int main(){
int i,j;
int flag = 0;
int x [3][3] = {{1,2,3},{4,5,6},{7,8,9}};
for (i = 0;i<3;i++){
if (flag ==1){
for (j=2;j<0;j--){
printf(" %d",x[i][j]);
}
flag =0;
}
else {
for (j=0;j<3;j++)
{
printf(" %d ",x[i][j]);
}
flag =1;
}
}
return 0;
}
i'm trying to print the numbers in the array in a zigzag form the expected output should be 123654789 but all i got 123789 for some reason i dont enter the loop in the flag condition i want to know the reason.. thanks in advance
Instead of
for (j=2;j<0;j--){
use
for (j=2;j>-1;j--){
and (only for the nice formatting) instead of
printf(" %d ",x[i][j]);
(near the end, with a space after %d) use
printf(" %d",x[i][j]);
(without that space).
Try this, it will give you the output what you want.
#include <stdio.h>
int main(){
int i,j;
int flag = 0;
int x [3][3] = {{1,2,3},{4,5,6},{7,8,9}};
for (i = 0;i<3;i++){
if (flag ==1){
for (j=2;!(j<0);j--){
printf(" %d",x[i][j]);
}
flag =0;
} else {
for (j=0;j<3;j++) {
printf(" %d",x[i][j]);
}
flag =1;
}
}
return 0;
}
Output:
./a.out
1 2 3 6 5 4 7 8 9
Related
I'm doing a program that check if 5 numbers that the user insert are even or odd, then they will be stored into an array and finally these values will be printed out on screen. In order to do this i've divided this program in two functions just to understand how the functions and the arrays works together, but it doesn't print the values that i've putted in. Why?
int check_even_and_odd(int number, int list[]){
printf("Insert the numbers\n");
scanf("%d", &list[number]);
if (number % 2 == 0) {
printf("Even\n");
}
else{
printf("Odd\n");
}
return 0;
}
int main () {
int k;
int i = 0;
int list2[5] = {0};
while (i < 5) {
i++;
k = check_even_and_odd(i, &list2[i]);
}
i = 0;
while (i < 5) {
i++;
printf("\n%d\n", list2[i]);
}
return 0;
}
Edit: Now that the main issue is gone, I want to add a little improvement to this little project. I want that the program tells to me how many Even or Odd number are in the array, but i don't know how to do it. I was thinking about adding 2 counters into the if statement (one for the even number and one for the odd numbers) but once i do this i don't know how to continue.
The program with the counters is this:
void check_even_and_odd(int number, int list[]){
int even = 0;
int odd = 0;
printf("Insert the numbers\n");
scanf("%d", &list[number]);
if (number % 2 == 0) {
even++;
}
else{
odd++;
}
printf("Even numbers are: %d\n", even);
printf("Odd numbers are: %d\n", odd);
}
int main () {
int i = 0;
int list2[5] = {0};
while (i < 5) {
i++;
check_even_and_odd(i, list2);
}
i = 0;
while (i < 5) {
i++;
printf("\n%d\n", list2[i]);
}
return 0;
}
Obviously it isn't complete, but as i have already said, i don't know how to continue
Your function expects an array argument but you are passing the address of individual elements of the array, so it won't work properly, you'll just need to use the correct argument:
k = check_even_and_odd(i, list2);
Quibble: k is never used so you don't really need it. You can just make your function void and remove the variable:
void check_even_and_odd(int number, int list[]){
printf("Insert the numbers\n");
scanf("%d", &list[number]);
if (number % 2 == 0){
printf("Even\n");
}
else{
printf("Odd\n");
}
}
int main(){
int i = 0;
int list2[5] = {0};
while (i < 5){
i++;
check_even_and_odd(i, list2);
}
i = 0;
while (i < 5){
i++;
printf("\n%d\n", list2[i]);
}
return 0;
}
Your fault is in line scanf("%d", &list[number]); just need to change it to scanf("%d", &list); but i think you are miss understanding whole array and pointer logic. You can't pass list as argument and if you do that, The compiler will changed it to pointer automatically. So if you want to tell a function about your list you just have to pass it your list address in memory (pointer). So you should do it like:
#include <stdio.h>
int How_Many_Odd = 0;
int How_Many_Even = 0;
void Add_To_List(int Number, int *ListIndex){
printf(
"Number %d is %s\n",
Number,
(Number % 2 == 0)? "Even": "Odd" // check if is odd or even
);
if(Number % 2 == 0)
How_Many_Even++;
else
How_Many_Odd++;
// changing value of pointer ListIndex to Number
*ListIndex = Number;
}
int main(){
// first creating integer array with size of 5
int List[5];
for(int i=0; i < 5; i++){
// waiting for user to enter number
int value;
scanf("%d", &value);
// changing value of index 0 to 3
Add_To_List(value, &List[i]);
}
// showing how many odds and how many evens
printf("%d numbers are even and %d numbers are odd\n", How_Many_Even, How_Many_Odd);
// you can show every index value too
for(int i=0; i < 5; i++)
printf("value of index %d is %d\n", i, List[i]);
return 0;
}
I recommend you to learn about pointer that will fix your issues
Here is your code fixed:
#include <stdio.h>
void check_even_and_odd(int number, int list[], unsigned int *even_count)
{
printf("Insert the numbers\n");
scanf("%d", &list[number]);
if (number % 2 == 0) {
printf("Even\n");
*even_count += 1;
}
else
{
printf("Odd\n");
}
}
int main()
{
unsigned int even_count;
int i = 0;
int list2[5] = {0};
while (i < 5)
{
check_even_and_odd(i, list2, &even_count);
i++;
}
i = 0;
while (i < 5)
{
printf("\n%d\n", list2[i]);
i++;
}
printf("There are %d even numbers and %d odd ones.\n", even_count, 5 - even_count);
return 0;
}
Basically, your main problem is in passing the list to the function in k = check_even_and_odd(i, &list2[i]);, as you should be passing the entire list, not a specific number in the list.
regarding:
if (number % 2 == 0) {
This is checking the passed in parameter rather than the value entered by the user. Suggest:
...
if( (list[number] % 2) == 0 )
{
printf( "%s\n", "number is even" );
....
I wrote the following code, but it is not giving the correct output. For input n = 5, the output should be 4 6 7 9 10, but my output is 4 5 6 7 8. Can anyone help me out? Where am I going wrong?
#include<stdio.h>
int isFibo(int n){
if(n<=0)
return 1;
if(n==1)
return 1;
if(n==2)
return 1;
int t1=0;
int t2=1,t3;
//int i=1;
while(t3<n){
t3=t1+t2;
t1=t2;
t2=t3;
if(t3==n)
return 1;
}
return 0;
}
void printNonFibo(int n){
int i=0,count=0;
while(count<n){
if(!isFibo(i))
{
printf("\nNon Fibo: %d ",i);
count++;
}
i++;
}
}
int main(){
int n;
printf("Enter n: ");
scanf("%d",&n);
printNonFibo(n);
return 0;
}
I'm a beginner at C programming. I'm making a program that will input numbers and delete the last input even number from the array using stack or the push-pop method.
The problem is I can't pop the last even number and I don't know what is wrong.
#include <stdio.h>
#include <stdlib.h>
#define MAX 100
int top = -1;
int stack[MAX];
void deleteEven(int num[], int i);
int main() {
int num[100];
int i, size;
printf("\n-----------------\n\n");
printf("Enter size of array: ");
scanf("%d", &size);
for (i = 0; i < size; i++) {
printf("Enter number: ");
scanf("%d", &num[i]);
top++;
stack[top] = num[i];
}
printf("\nList: ");
for (i = 0; i < size; i++) {
printf("%d, ", num[i]);
}
printf("\n");
printf("Even: ");
for (i = 0; i < size; i++) {
if (num[i] % 2 == 0) {
printf("%d, ", num[i]);
}
}
deleteEven(num, i);
return 0;
}
void deleteEven(int num[], int i) {
printf("\nAnswer: ");
if (num[i] % 2 == 0) {
stack[top--];
}
for (int j = top; j >= 0; --j) {
printf("%d, ", stack[j]);
}
}
I have implement the working one in C with implementing on your code, you can see below. I added int checkEven(int stack[], int stackSize) function which control the array if there is any even number or not. If not, so end the problem with returning 0 or whatever your error code is, other side if there is even number it returns the index of it and deleteEven function swipe the array (stack). It working for size of 5 array but you can fix it. I use 5 for easy testing.
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#define MAX 5
int top = -1;
int stack[MAX];
void deleteEven(int num[], int indexOfEven);
int checkEven(int stack[], int stackSize);
int main() {
int num[5];
int i, size;
printf("\n-----------------\n\n");
printf("Enter size of array: ");
scanf("%d", &size);
for (i = 0; i < size; i++) {
printf("Enter number: ");
scanf("%d", &num[i]);
top++;
stack[top] = num[i];
}
printf("\nList: ");
for (i = 0; i < size; i++) {
printf("%d, ", num[i]);
}
printf("\n===stack===");
for( i = 0; i <size; i++){
printf("%d ", stack[i]);
}
int indexOfEven = checkEven(stack,5);
if(indexOfEven >= 0){
printf("This sequence has even number");
printf("the index => %d ",indexOfEven);
deleteEven(stack, indexOfEven);
}else{
printf("this sequence has no even number");
/*
no even number
exit
*/
return 0;
}
return 0;
}
int checkEven(int stack[], int stackSize){
for(int i = stackSize - 1; i >= 0; i--){
if(stack[i] % 2 == 0){
return i;
}
}
return -1;
}
void deleteEven(int num[], int indexOfEven) {
int simpleArray[5];
for(int t = 0; t < 5; t++){
simpleArray[t] = num[t];
}
int c;
for (c = indexOfEven; c < 4; c++)
simpleArray[c] = num[c+1];
for (c = 0; c < 4; c++){
printf("\n%d\n", simpleArray[c]);
}
}
So far you see the O(n) implementation of it with array but you describe that you want to implement it with push() - pop() - peek() stack mechanism. I want to write sudo code for fully Stack implementation.
let it inputs be 1 - 2 - 3 - 5 - 7
describe inputSize
describe mainStack
describe helperStack
read inputs to mainStack
show stacks
mainStack -> [1-2-3-5-7]
helperStack -> []
while mainStack.peek() != NULL :
if mainStack.peek() % 2 == 0: // even number
mainStack.pop()
break the loop
else:
describe popValue = mainStack.pop()
helperStack.push( popValue )
if inputSize == helperStack:
// no even number
// so nothing break the loop, every value is odd so, all there is another stack
// finish program with error code or return main array / inputs
show stacks
mainStack -> [ 1 ]
helperStack -> [ 3 5 7 ]
now pop() the all helperStack and push it to mainStack
while helperStack.peek() != NULL:
mainStack.push( helperStack.pop() )
show stacks
mainStack -> [ 1 3 5 7 ]
helperStack -> [ ]
Return mainStack as array format.
It seems that the last loop before the call to deleteEven will increment i until the end of the stack array regardless the last number is even or not, because all you do is checking if the current number is even and then printing it, and right after that going to the next one. that will iterate through all the numbers which will result in calling deleteEven with the last index of the array.
how about going from the last element of the array to index 0 (backwards) and printing the first encounter with even number?
Also, not really sure why you're using two different arrays and copying elements one by one after using scanf.
In the below code I am taking an array to read 4 numbers and I need only to display the distinct number.
#include <stdio.h>
int main(){
int ch[3];
int s[3];
int count = 0;
int i;
int j;
for(i = 0; i < 4; i++){
scanf("%d", &s[i]);
for(i = 0; i < 4; i++){
ch[i] = s[i];
printf("ch= %d", ch[j]);
}
if(ch[i] == s[i]){
count = count + 1;
printf("%d =", count);
}
}
}
Somewhat I'm not getting the output, it is giving me a weird output.
Question:
What could be the cause of the strange output?
int ch[3];
int s[3];
This mean valid references to these arrays can be made from 0 to 2 , whereas your for loop is looping 4 times.
printf("ch= %d",ch[j]);
you mean ch[i] ?
what do you use "int j" for ?
Cant understand the question but i wrote a code for taking 4 inputs and giving the distinct number among them as an output(or the numbers which have only occurred once).
#include <stdio.h>
int main(){
int ch[4];
int s[4];
int i;
int j;
for(j=0;j<4;j++){
ch[j]=0;
}
for( i=0;i<4;i++){
scanf("%d",&s[i]);
}
for(i=0;i<4;i++){
for(j=0;j<4;j++){
if(s[i]==s[j]){
ch[i]++;
}
}
}
for(i=0;i<4;i++){
if(ch[i]==1){
printf("\n%d is the distinct number.", s[i]);
}
}
}
i have this code
#include <math.h>
#include <stdio.h>
const int n = 3;
const int s = 3;
int getm(int mat[n][s]);
int printm(int mat[n][s]);
int main()
{
int m[n][s];
getm(m);
printm(m);
return 0;
}
int getm(int mat[n][s])
{
for(int x = 0;x < n;x++)
{
for (int y = 0;y<s;y++)
{
scanf("%i ", &mat[x][y]);
}
}
return 0;
}
int printm(int mat[n][s])
{
for(int x = 0;x<n;x++)
{
for(int y = 0;y<s;y++)
{
printf("%i ", mat[x][y]);
if(y==(s-1))
{
printf("\n");
}
}
}
}
which shoud ask for 9 numbers to make a 3x3 matrix array, but it actually asks for 10 numbers, printm is working well - printing only 9 numbers. Where is error?
I think the problem is the space after %i: you do not need the tenth number, but your code is asking for it anyway, because it waits to get a space after the ninth number.
Separately, your printing code can be optimized a little by dropping the if:
for(int x = 0;x<n;x++)
{
for(int y = 0;y<s;y++)
{
printf("%i ", mat[x][y]);
}
printf("\n");
}
scanf("%i ", &mat[x][y]);
Get rid of the space after %i, so it only reads a number:
scanf("%i", &mat[x][y]);