unsigned char table length - c

Question

I declared a table of unsigned char as follow:
unsigned char buf[10]={'1','5',0x00,'8'};
in order to know the number of elements of this table i implemented this function
int tablength(unsigned char *buf)
{
int i=0;
for (i=0;buf[i];i++)
;
return i;
}
However this function don't give me the right result when the buffer contains 0 in the middle .Sizeof don't give me the right result since it returns 10 in this case i can't neither use strlen since this is a table of unsigned char.
Do you have any idea to improve this function or any predefined function that help me solve my problem (the result that i 'am waiting for is 4)

Answer 1

Technically, since you declared a statically allocated array of 10 elements, the size of the array is 10. Even though you may not have initialized every element, there is something filling that space. C++ cannot determine whether the value in the array means anything or not.

Answer 2

After
unsigned char x[10] = {1, 2, 3};
the variable x (an array) has 10 elements, the first three initialized to 1, 2 and 3 and all the others initialized to 0. In other words that definition is absolutely identical to
unsigned char x[10] = {1, 2, 3, 0, 0, 0, 0, 0, 0, 0};
An array in C and C++ is just a fixed area of memory and doesn't include a counter of how many "interesting" elements are there.
If you are looking for a container with a variable number of elements consider instead std::vector (C++ only). With that std::vector::size() returns the current number of elements in the container.

Answer 3

If you need the array to contain exactly the number of elements you've specified, just declare it without a specific size:
unsigned char buf[]={'1','5',0x00,'8'};
cout << sizeof(buf); // should be 4
If you want to store a variable amount of data (in C++) use std::vector instead of an array.
Otherwise you'll need to keep track of the number of valid elements yourself. There's nothing in the language that will do it for you.

Answer 4

Compilers cannot know how you would like to use an array instance.
Therefore you must follow the language's semantics. By declaring your array globally or locally, but with the storage class specifier static you are initializing every element to 0 on default and your function will work.
0x00 is false. 0x00 (which same as 0x0) is a hex number representing 0 (false). This is where your counting loop will stop at - the 3rd element.
Another thing you can do is declare your array with non-fixed size.
unsigned char buf[]={'1','5',0x00,'8'};
In that case, the sizeof operator works as expected.
Because that way, you will have an array of 4 elements.
strlen() obviously won't work as it is designed to work with strings, not a buffer.
As for a function that counts on a smarter way:
size_t arrcnt (unsigned char source[], size_t size)
{
size_t i;
for(i = size; i >= 0 && !source[i]; i--);
return i + 1;
}
Usage:
printf("size of buf: %u", arrcnt(buf, sizeof(buf));

Answer 5

buf[i] evaluates to false when buf[i] contains 0.
You cannot do what you want unless you know one value which can never occur in your array between 0 to UCHAR_MAX (255). Say the value is 255, then you first preinitialize the full array to 255 before you start filling it up.
memset(buf, 255, sizeof(char) * sizeof(buf));
Then you fill other elements like you want and then you can use the following
for(i = 0; buf[i] != 255, ++i)

Answer 6

try putting :
for(int i=0;buf[i]!= 0;i++)
count++;
return count;