Check if number is a power of two, and if the input is number

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i want to check if the the numbers in array are power of 2,
i wrote the folowing code but it doesn't work it skips the part that chiks if the number is power of two and prints the last sentence.
also if someone can help me in how to check if the input is number and not any other charctar.
thank you!

What I have tried:

#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]==0){
             printf("The number %d is a power of 2:%d=2^%d\n",a,a,count);
             sum+=a;
        }
    }
    printf("Total exponent num is %d\n",sum);
  return 0;
}

Posted 4-Apr-17 13:44pm

Member 13106348

Updated 5-Apr-17 2:40am

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Comments

PIEBALDconsult 4-Apr-17 20:35pm

A spot of debugging should do the trick. I particularly think you should examine numbers[k] after the while loop and before you test for zero.

Member 13106348 4-Apr-17 20:37pm

i tried debugging didn't work :\ i still don't know the problem .. i changed the if statmnet to if (numbers[k]==0) and now it works only for input that is one number.

PIEBALDconsult 4-Apr-17 20:42pm

What was it before?
What set of numbers did you test with? What result did you get?

[no name] 4-Apr-17 21:44pm

Yes I know what you mean. You ran the program and thought you was debugging it but didn't bother using breakpoints to step through your code and examine the variables. So, no you did not debug your program to see what it was doing.

[no name] 4-Apr-17 20:36pm

Learning how to use the debugger is a *really* useful skill that you must learn.

Member 13106348 4-Apr-17 20:40pm

can i learn it in half a day ?

PIEBALDconsult 4-Apr-17 20:41pm

No. But you can start.

nv3 5-Apr-17 3:35am

First, of course, debug your code. Once you have done that, note that there is a much more efficient method of checking whether a number is a power of 2:

int x;

if (x & (x-1) == 0)
// yes, it is a power of two

I leave it up to you to prove why and how this little trick works.

PIEBALDconsult 5-Apr-17 9:34am

But will it tell you the exponent?

nv3 5-Apr-17 10:49am

No, it doesn't. Just whether the number is a power of two or not. The exponent could also be determined in a piece of linear code by using a 12-bit lookup table and three rsp. six shifts.

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Patrice T · Accepted Answer · 2017-04-04T14:51:00

Quote:
i tried debugging didn't work

This quote suggest that you don't know how to use the debugger. the debugger only show you what your code is doing, you have to inspect variables and check if everything is consistent with your expectations.
For a fast starting, you can find tutos on youtube.

This line is in wrong place because it will try to allocate memory whatever is the value of x. You need to make sure that x is positive before memory allocation.

C++

int *numbers=malloc(sizeof(int)*x);

When you don't understand what your code is doing or why it does what it does, the answer is debugger.
Use the debugger to see what your code is doing. Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute, it is an incredible learning tool.

Debugger - Wikipedia, the free encyclopedia[^]
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]

The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.

CPallini · Accepted Answer · 2017-04-05T02:40:00

You know, the C programming language does not prevent you defining functions...
Try

C

#include <stdio.h>
#include <stdlib.h>


int powoftwo(int x)
{
  int count;

  for ( count = 0; x && ((x & 1) == 0); ++count )
    x >>= 1;

  if ( x == 1)
    return count;

  return -1;
}


int main()
{
    int x;
    int i;
    int k;
    printf("Enter size of input:\n");
    scanf("%d",&x);
    int *numbers= NULL;
    if (x<0){
      printf("Invalid size\n");
      return -1;
    }
    else
    {
      numbers = malloc(sizeof(int)*x);
      if ( ! numbers)
      {
        printf("memory allocation failure\n");
      }

       printf("Enter numbers:\n");
       for(i=0;i<x;++i){
         scanf("%d",&numbers[i]);
       }
    }
    for(k=0;k<x;++k)
    {
      int exp = powoftwo(numbers[k]);
      if ( exp > 0)
        printf("The number %d is a power of 2:%d=2^%d\n", numbers[k],numbers[k],exp);

    }
    free(numbers);
    return 0;
}

Check if number is a power of two, and if the input is number

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