Array to Int

[taymo2020]

I'm trying to make a algorithm to turn a array of ints to a single int. Here's what I have so far:

Code: Select all

for(int j = 0; j < sizeof(array); j++)
{
     a_int += array[j] * pow(10.0,j);       
}


I'm just not getting the results that I want.

[devil_slayer]

What do you mean an array of ints. What ints are in that array. Aren't you affraid of overflow?

Depending on the declaration of the array, sizeof(array) might not give its number of elements.
Don't forget setting a_int to zero before enetering the loop, too.

Maybe you want:

Code: Select all

int array[]={1,3,4,2,3};
int n=sizeof(array);
int a_int=0;
for(int j=0; j<n; j++)
    a_int+=array[j]*pow(10.0,n-j);


[taymo2020]

What do you mean an array of ints

Uhhhh I mean:

Code: Select all


int array[];


Why should I be afraid of overflow?

[tomcant]

Maybe you want:

Code: Select all

int array[]={1,3,4,2,3};
int n=sizeof(array);
int a_int=0;
for(int j=0; j<n; j++)
    a_int+=array[j]*pow(10.0,n-j);


Close, but `n' will not be the size of the array; it will be the size of the array in bytes. Try this:

Code: Select all

int array[]={1,3,4,2,3};
int n=sizeof(array) / sizeof(int);
int a_int=0;
for(int j=0; j<n; j++)
    a_int+=array[j]*pow(10.0,n-j);


[devil_slayer]

Why should I be afraid of overflow?

Well, what if your array of ints sums up to more then it fits in ine int. Remeber an int can have a max of 2^31 or in other words: 2 147 483 648.

If your array contains more than that the int will overflow.

Can you specify what you are using this for perhaps?

[Emery]

Why should I be afraid of overflow?

Well, what if your array of ints sums up to more then it fits in ine int. Remeber an int can have a max of 2^31 or in other words: 2 147 483 648.

If your array contains more than that the int will overflow.

Can you specify what you are using this for perhaps?

No it wont. It should just truncate to 0.

> "2 147 483 648"
2, 147, 483, 647 acctually

[taymo2020]

Can you specify what you are using this for perhaps?

I'm trying to find more Kaprecar's constants.
http://math.about.com/od/recreationalma ... prekar.htm

[taymo2020]

Ok I tryed this:

Code: Select all

int arrayToInt(int* array)
{
    int a_int = 0;
    int n = sizeof(array)/sizeof(int);
    for(int j = 0; j < n; j++)
    {
        a_int += array[j] * pow(10.0,j);       
    }
    return (a_int);
}

And with a array that looks like this:

Code: Select all


int array[] = {8,9,9,9,9,9,9,9,9,9,9};


All that a_int ends up with is '8';

[Emery]

You're passing a pointer so sizeof is giving you the size of the pointer.

[devil_slayer]

You're passing a pointer so sizeof is giving you the size of the pointer.

Ritz is right. If you send it as an arguemnt through the function it just won't work.

No it wont. It should just truncate to 0.

To me that's an overflow. In any case it's a fault in the program.

One thing. Instead of:
a_int += array[j] * pow(10.0,j);
It should be:
a_int += array[j] * pow(10.0,n-j-1);

The number you specified in the array is too big for a signed int. Try using unsigned long for a_int variable. It takes anything up to 18 446 744 073 709 551 615.

Can someone please explain why:

Code: Select all

int c=4;
int d=(int)pow(10,c-2);

gives 99 instead of 100?

[devil_slayer]

Code: Select all

#include <stdio.h>

int f(int array[],int n)
{
    int a_int=0,j;
    int tens=1;
    for(j=n-1; j>=0; j--)
    {
        a_int += array[j] * tens;
        printf("D:%d\n",a_int);
        tens*=10;
    }
}

int main()
{
    int a[]={4,1,2,3};
   
    printf("%d\n",f(a,4));

    return 0;
}


[tomcant]

Can someone please explain why:

Code: Select all

int c=4;
int d=(int)pow(10,c-2);

gives 99 instead of 100?

[syntax="cpp"]#include <iostream>
#include <cmath>
using namespace std;

int main()
{
int c=4;
int d=(int)pow(10,c-2);

cout<<d;
cin.get();
}[/syntax]

The value of `d' is 100 after the call to pow(...). Must be something your doing slightily different.

[devil_slayer]

Well I compiled it under bloodshed and got this wierd result... Quite strange...

[Emery]

No it wont. It should just truncate to 0.

To me that's an overflow. In any case it's a fault in the program.

By definition that is not an overflow. An overflow is when you write past the end of a buffer into uncharted territories and an overrun is when you read past the end of a buffer. This is the cause of a mathmatical instruction which cannot complete it's routine due to lack of memory. The opposite of a buffer overflow, it avoids a buffer overflow by defaulting to 0.

About that 99 instead of 100 things. I think I remember this coming up before on some other forum but I can't remember for the life of me where it came up or what the solution turned out to be.

[MXP]

Overrun is when you write past the end of a buffer. Overflow is when the value of a calculation is greater than the datatype can store.

As for 99/100, pow() works on floating point numbers and so the value of pow(10.0, 2) might actually be 99.999999 or something. So, when you truncate it the value becomes 99.

If the array is reversed so that the most significant digit is in the first index (since right now it is in the last index) then the algorithm can be changed to not use pow():

Code: Select all

int sum = 0;
for (int i = 0; i < size; ++i) {
   sum = sum * 10 + array[i];
}