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Hello guys!
I am trying to merge a number which is contained in seperate indexes of an array.
So for example I have [0] = 4, [1] = 3, [2] = 7.
That would save into a variable of type int the number "437".
Any number of indexes is possible upto the maximum of int. Thats somewhere around 32000 right?
I am unsure how would be the easier way to do this.
You could either put the number in an array and convert it using atoi() - OR - you could use modular arithmetic to merge it, eg:
437 = (4 * 100) + (3 * 10) + (7 * 1).
When dealing with absurdly large numbers, you're gonna have to implement some kind of a handler for it yourself, or use an external library. You could also just use an array to do it and handle the carry yourself.
So, let's get this right: You have a (potentially large) array of small integers (0-9), that you want to put together into a long number?
The maximum of an int is commonly either 32767 or 2147483648. The latter is the usual case for modern machines, the former was the case on MS-DOS and Windows 2.x machines (and OS/2 1.x).
What have you done so far?
--
Mats
This is what Ive done, but it seems I was way off, and it doesnt work as planned.
Code:#include <stdio.h>
int array[] = {3,5,0,7,4};
int main()
{
int x;
printf("%s", array);
x = my_getnbr(array);
printf("\n%d", x);
}
int my_getnbr(char *str)
{
int x, y, z;
y = 0, x = 0, z = 0;
while (*str != '\0')
{
str++;
y++;
}
while (x <= y)
{
while (z <= *str[x])
{
if (str[x] = 0)
z = z + 0;
z = z + 10000;
}
str++;
while (z <= *str[x])
{
if (str[x] = 0)
z = z + 0;
z = z + 1000;
}
str++;
while (z <= *str[x])
{
if (str[x] = 0)
z = z + 0;
z = z + 100;
}
str++;
while (z <= *str[x])
{
if (str[x] = 0)
z = z + 0;
z = z + 10;
}
str++;
while (z <= *str[x])
{
if (str[x] = 0)
z = z + 0;
z = z + 1;
}
}
}
Your code is decidedly "too complex" (and broken) - I very much doubt that it's even compiled.
This should at the very least give you a compiler warning, as it's passing an integer array to a function that takes a character pointer. Converting arrays to pointers is fine, but don't change the type at the same time, or you'll be getting into trouble.Code:int array[] = {3,5,0,7,4};
...
x = my_getnbr(array);
printf("\n%d", x);
}
int my_getnbr(char *str)
...
This is not right:
It compares a integer array with char, which is OK as such, but in this case isn't the right thing to do, becaue the value of '\0' is the same as 0 - and your array has a zero in the third position [index 2] (and it's valid to have zero's in an array of integers, otherwise the value 1000 couldn't be described in this model, and I don't think that's supposed to be the case).Code:while (*str != '\0')
{
str++;
y++;
}
x and y are unchanged within this loop, so it will never exit the loop if the condition is true when it gets to this point.Code:while (x <= y)
{
What does this do? Absolutely nothing. Adding with zero is meaningless.Code:if (str[x] = 0)
z = z + 0;
I expect this to give a compiler error, as you are taking an indirection (*) on an array that is originally just a pointer to char.Code:while (z <= *str[x])
Edit: And you shouldn't need to duplicate the code multiple times. Imagine how this would work if you had a really large number...
--
Mats
Actually it does do something, just not what was inteneded. It sets str[x] to 0 which then causes the stuff in the if statement to never run, which dident matter beacuase it never did anything anyway. For a comparison it should really have been:Quote:
if (str[x] = 0)
z = z + 0;What does this do? Absolutely nothing. Adding with zero is meaningless.
if (str[x] == 0)
Doh! Of course, you are right. But none of the other code looks particularly right either...Quote:
Originally Posted by mike_g
--
Mats
I'll work on it more, and see if I can come up with something.
Thanks for the help ;)
This is the code I have now:
It doesnt work and the compiler gives no error. I dont understand why.
(So far it should work, if a number has no zeros in it anyway)
Code:#include <stdio.h>
char array[5] = {'3','5','2','7','4'};
int main()
{
int x;
int y;
for (y = 0; y < 5; y++)
{
printf("%c", array[y]);
}
x = my_getnbr(array);
printf("\n%d", x);
system("PAUSE");
return 0;
}
int my_getnbr(char *str)
{
int x, y, z;
y = 0, x = 0, z = 0;
while (*str != '\0')
{
str++;
y++;
}
if (y == 1)
z = z * str[0];
if (y == 2)
z = str[1] + (str[0] * 10);
if (y == 3)
z = str[2] + (str[1] * 10) + (str[0] * 100);
if (y == 4)
z = str[3] + (str[2] * 10) + (str[1] * 100) + (str[0] * 1000);
if (y == 5)
z = str[4] + (str[3] * 10) + (str[2] * 100) + (str[1] * 1000) + (str[0] * 10000);
return z;
}
Use a loop
int result = 0;
Then in a loop
result = result * 10 + ( str[ i ] - '0' );
Because 'array' isn't nul-terminated.
Use a loop.
Im sorry I couldn't quite grasp that.Quote:
Use a loop
int result = 0;
Then in a loop
result = result * 10 + ( str[ i ] - '0' );
Use a loop where? One for the whole function?
Yes, something like:
Code:/* it's not a string since it's not nul-terminated */
int my_getnbr(const char * arr, size_t n)
{
int result = 0;
size_t i = 0;
for(i = 0; i < n; i++)
result = result * 10 + (arr[i] - '0');
return result;
}
Again Im sorry but I dont understand.
If result = 0 then wont result = result * 10 always give you zero?
No, the + (arr[i] - '0'); part will start by adding something to the 0 (at some point) so it will eventually be non-zero.Quote:
Originally Posted by +Azazel+
bah, I think I am destined to never get this. I spent some time dissecting that loop and trying to understand it but I dont get it :/
I'm sure someone here can explain exactly every part of it, but please do explain what you think it does first - because otherwise we'll be spending a lot of time explaining what you already know, and perhaps not enough to explain what you need to know.
--
Mats
I dont understand the "dash - " and I dont really understand how it solves my problem.
Looking through what the atoi() function does, I think I need exactly that (not sure though). Where can I see the source code for it?
In the end this is just a curious question, and it its too hard for me I might as well dump it :/
You don't understand the minus symbol?Quote:
Originally Posted by +Azazel+
Say you input "123" as the string.
- result = result * 10 + ('1' - '0')
- result = 0 * 10 + 1
- result = 1
- result = result * 10 + ('2' - '0')
- result = 1 * 10 + 2
- result = 12
- result = 12 * 10 + ('3' - '0')
- result = 12 * 10 + 3
- result = 123
In the end, result = 123 which is the converted integer value of the string "123".
Ohhh! I see, yes. Thank you! But what if I dont know the size of the inputted number? It could be 2,3,4,5 digits long (Its an int so 5 would be max). How would I determine that?
You have to realize that every base-10 number(the numbering system that we use) is composed of ones, tens, hundres, thousands, etc.Quote:
Originally Posted by +Azazel+
Eg: 1234 = (1 * 1000) + (2 * 100) + (3 * 10) + (4 * 1).
Just try that in the calculator, or by hand
Yes?Code:
1000
+ 200
+ 30
+ 4
-----------
=1234
If you put that number into the function posted:
There will be 4 steps, as there are 4 digits:Code:result = 0;
for(i = 0; i < n; i++)
result = result * 10 + (arr[i] - '0');
If you understand all this, but don't understand (arr[i] - '0'), read on:Code:result = 0 * 10 + 1 = 0 + 1 = 1
result = 1 * 10 + 2 = 10 + 2 = 12
result = 12 * 10 + 3 = 120 + 3 = 123
result = 123 * 10 + 4 = 1230 + 4 = 1234
result = 1234
In essence, all things in a computer are numbers. Your compiler and C programs 'convert' chars to integers, as integers are native to the computer(Which is why, when possible, always use ints instead of char). With this in mind, you can look at an ascii chart and find the integer value for for 0(zero), which is 48.
Now, if you look at the ASCII chart at take a random number on it, 1 for example, you will see that its integer value is 49, right after 0. If you subtract 48 from 49, you get 1, which is why the algorithm he posted works.
FYI, ASCII stands for American Standard Code (for) Information Interchange. Basically, this standard simply controls which integers correspond to which letters. The ASCII table lists the characters and what their integer, hex and octal equivalents are.
There are numerous other standards though, such as EBCDIC for example, and there are also other ASCII sets(Often called Extended ASCII sets) which define the sets for various foreign alphabets.
Any further questions? - Google, then ask.
The glibc source code is available freely through the CVS web interface, and here's the atoi code (you have to click "download" on this page):
http://sourceware.org/cgi-bin/cvsweb...?cvsroot=glibc
It's not going to help you much, as it's calling strtol(), which calls a more generic internal version:
http://sourceware.org/cgi-bin/cvsweb...&cvsroot=glibc
Now, if you think that little loop was hard to understand, you won't have much chance to understand the above code... It's quite a lot more complex.
--
Mats
Alright. Well Im almost done but one little problem.
char array[5] = {'3','5','0','7','0'};
If I do sizeof(array) it gives me "5". (Correct)
But if I do
It gives me 1. Why and how would I find the sizeof array through a pointer?Code:int my_getnbr(char *str)
{
int z = sizeof(*str);
printf("%d", z);
}
That's correct. Sizeof only tells you the size of whatever you asked for. In this case, what's the size of (*str) - which is one. The size of the array, on the other hand, is 5, because the array has 5 elements of one byte each.
You need to pass the size in, or use some other method to know the size (e.g.which would make the a string that is zero-terminated, and you can go back to checking for a zero at the end of the string).Code:char array[] = "12345";
--
Mats