Array elements to Integer

[flavius doroftei]

Hi guys, i am looking for a function to fuse the elements of an array into a single "int" variable, to make it simple i am looking for the opposite of this function, couldn`t really figure it out

	for(i = 9 ; i >= 0 ; i--)
	{
		array[i] = N%10;
		N = N / 10;
	}

[Athar]

The opposite of division is multiplication.

[flavius doroftei]

I could not figure out what to do with the modulus operator ...

[Athar]

In this case the reverse operation would be addition.

[flavius doroftei]

for(i = 0 ; i<=9 ; i++)
{
	a *= 10;
	tab[i] = a+10;
}

always a=0 ...

[Athar]

0 * 10 is 0? How very shocking.

[flavius doroftei]

look pal, if you don`t wanna help, don`t do it, ok ?!

and i wonder where did you got that 0*10 from ?

[Athar]

If you don't want to use your brain, then why on earth are you trying to learn programming?
When you have the digits 1, 2 and 3, how do you get to 123? This is first grade maths.

and i wonder where did you got that 0*10 from ?

What value does a have? What happens when you multiply it by 10? Or even better, if a is 0 after you multiplied the previous value by 10, what was the previous value?

[D80DSL3A]

I say stop trying to figure out how to "invert" that function and just write the calculation. If int digits[5]; holds the digits of a 5 digit number then the N you are looking for is:
int N = digits[0] + 10*digits[1] + 100*digits[2] + ...
Just put that in a loop.

@Athar: Hope you don't mind me stepping in here.
Question for you. The meaning of big-endian vs. little-endian in a number representation.
I have chosen big-endian above because I am storing the most significant digit last, correct?

[Athar]

I have chosen big-endian above because I am storing the most significant digit last, correct?

That's little-endian, actually. How you store them is up to you, but with OP's code it would be big-endian.

[D80DSL3A]

Thanks, so I was wrong.

The endianness is named for what is in element 0 then, not the last element?
I store the least significant digit (the little one) in digits[0], so it is little-endian?
I see, by refering to OP's OP (original post), that he is indeed assigning the ones digit to array[9] on the 1st iteration, so it would be big-endian.
I was confused because I thought the "end" in "endian" referred the "end" of the array, not the beginning.

The term should be beginianness then, I say. LOL

[flavius doroftei]

I am sorry Arthar, my bad ...

always a=0 ...

i wanted to point at the fact that "&a" is the address in which the 12345 number has to be stored, but no matter how hard i tried it kept outputing "0"

[flavius doroftei]

And about using my brain, yes, i used my brain doing this, so far it worked ok, but i can only insert 5 digits, no more, no less

#include<conio.h>
#include<stdio.h>

int i;
int tab[5];
int tab1[5]={10000, 1000, 100, 10, 1};
int tab2[5];

void main()
{
	printf("\n-----------------------------------------------------");
	printf("\nInsert 5 digits!!!");
	printf("\n-----------------------------------------------------");
	printf("\n");
	for(i=0 ; i<5 ; i++)
	{
		printf("Digit [%d] = ", i+1);
		scanf("%d", &tab[i]);
	}
	printf("\n-----------------------------------------------------\n");
	for(i=0 ; i<5 ; i++)
	tab2[i]=tab1[i]*tab[i];
	int a=tab2[0]+tab2[1]+tab2[2]+tab2[3]+tab2[4];
	printf("A = %d", a);
	printf("\n-----------------------------------------------------");
	getch();
}