How do i print all the odd numbers? - c

Question

I want to print all the odd numbers from 14 to 156 using an infinite loop, break and continue. But, when i run it does not display anything!!
int main()
{
int x;
int y = 14;
while(x) {
if(y % 2 == 0) {
continue;
}
else if(y % 3 == 0) {
printf("%d\n", y);
}
if(y == 156) {
break;
}
y++;
}
return 0;
}

Answer 1

The problem with your code is that it uses an operation with unpredictable results. Specifically, declaring int x; without initializing it, and then using it as your termination condition in while(x) is the problem. On many platforms and compilers, this will retain whatever value was already in the memory occupied by x. In that case, you may see no print statements because x starts with the value zero and the loop never runs.
You should make your loop into an infinite loop:
int main()
{
int x;
for(x = 14; ; x++) {
if(x % 2 == 0) {
continue;
}
if(x >= 156) {
break;
}
printf("%d\n", x);
}
return 0;
}
[IDEOne link]
This will print the values excluding 156 itself. To include 156, put the break condition after the printf call:
printf("%d\n", x);
if(x >= 156) {
break;
}
Alternatively you can change >= to just >.
You do not need else if you have break or continue.
If you have to use a while loop, the situation is a tad more complex because you have to increment before you continue to avoid an infinite loop:
int main() {
int x = 14;
while(1) {
if(x % 2 == 0) {
x++;
continue;
}
if(x >= 156) {
break;
}
printf("%d\n", x++);
}
return 0;
}
[IDEOne Link]
You can avoid the extra increment if you can forgo the continue:
int main() {
int x = 14;
while(1) {
if(x % 2) {
printf("%d\n", x);
}
if(x++ == 156) {
break;
}
}
return 0;
}
[IDEOne Link]
I have also removed your check for x % 3 == 0 since it is not clear what the purpose of that is within the constraints of your problem.

Answer 2

With an infinite loop:
#include <stdio.h>
#include <stdlib.h>
int main() {
int y=14;
while(1){
if(y & 1){
printf("%d\n",y);
}
if(y>=156){
break;
}
y++;
}
return 0;
}
Preferred method, with for loop:
#include <stdio.h>
#include <stdlib.h>
int main() {
int start=14;
start |= 1; //This will increment by one if your start value is even
for (start; start<=146; start+=2) {
printf("%d\n", start);
}
return 0;
}

Answer 3

An infinite loop with a break is not the right way to do this. Use a "for" loop and your code will be much clearer. Start it at 15 and increment by 2 until 156.

Answer 4

There are several problems in your logic.
1. Variable `x` is of no use. You can use variable `y` to terminate the loop.
2. No need to check if the number is a multiple of 3.
3. No need to check for even numbers is either.
I have modified the code and it is giving correct results but I'll urge you to read the above points and try to get the code right.
int main()
{
int y = 14;
while(y) {
if (y==156)
break;
if(y % 2 != 0) {
printf("%d ",y);
}
y++;
}
return 0;
}
A better and faster way would be to start with 15 and increment variable y's value by 2 till its <=156.
int main()
{
int y = 15;
while(y) {
if (y==157)
break;
printf("%d ",y);
y+=2;
}
return 0;
}

Answer 5

Your code has undefined behavior because the variable x is uninitialized and used as the condition in the while statement. You could fix this with while (1) but there would indeed be an infinite loop in your program because the test for evenness occurs before the test for termination on 156.
The problem is solved simply with a standard for loop:
#include <stdio.h>
int main(void) {
for (int y = 14; y < 156; y++) {
if (y % 2 != 0) {
printf("%d\n", y);
}
}
return 0;
}
Which can be simplified as this, only enumerating odd numbers:
#include <stdio.h>
int main(void) {
for (int y = 15; y < 156; y += 2) {
printf("%d\n", y);
}
return 0;
}
If you are required to use break, continue and a infinite loop, you can indeed use the classic forever C loop, a for loop without a termination condition:
#include <stdio.h>
int main(void) {
for (int y = 14;; y++) {
if (y == 156)
break;
if (y % 2 == 0)
continue;
printf("%d\n", y);
}
return 0;
}

Answer 6

Well ...
#include <stdio.h>
#include <stdint.h>
#include <errno.h>
/* Prints even or add numbers between to and from.
Works for negative numbers.
Works up and down.
*/
int print_even_or_odd(long long from, long long to, int even)
{
if (((INT32_MAX < from) || (INT32_MIN > from)) ||
((INT32_MAX < to) || (INT32_MIN > to)))
{
fprintf(stderr, "Invalid input.\n");
errno = EINVAL;
return -1;
}
printf("from=%d to %d\n", (int) from, (int) to);
int sign = (to < from) ?-1 :1;
/* Adjust "from" to next even/odd number. */
if ((!even && !(from & 1)) || (even && (from & 1)))
{
from += sign;
}
/* Adjust "to" to the previous even/odd number. */
if ((!even && !(to & 1)) || (even && (to & 1)))
{
to -= sign;
}
{
size_t steps = (size_t) (sign * ((to - from) / 2)) + 1;
if (0 == steps)
{
printf("Nothing to do.\n");
return 0;
}
while (1)
{
if (0 >= steps)
{
break;
}
--steps;
printf("%d\n", (int) (to - sign * 2 * (int) steps));
continue; /* as having continue is a requirement ... ;-) */
}
}
return 0;
}
int main(void)
{
print_even(14, 156, 0);
print_even(156, 14, 0);
print_even(-14, -156, 0);
print_even(-156, -14, 0);
}