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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+