Okay im a complete newbie when it comes to C. I need to write a function that takes a 2D array as an input and then counts the number of rows in the 2D array. I have played around a bit and did some random testing but i keep hitting a wall(C not keeping track of array boundaries!). The general idea behind the code is simple and i just find it hard to believe something so simple is impossible to implement. HELP!
The arrays size, including it's number of rows, are fixed, when you create the array.
You need to post some code, because what you're saying, doesn't agree with da C.
Code:size_t element_count = sizeof(array) / sizeof(*array);
Please show us what you've tried.
My homepage
Advice: Take only as directed - If symptoms persist, please see your debugger
Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"
But he's asking for the number of rows, instead of elements.Originally Posted by msh
Ambiguous variable naming on my part. In an array of arrays (rows), rows are the elements. At least that's how I understand it.
True ^^^, but the OP specified a 2D array.
Wriggle room is very limited here!
Fair enough. But now I can't edit because you already quoted it.True ^^^, but the OP specified a 2D array.
Wriggle room is very limited here!
alright thanks guys for the quick response. Below is some code iv'e tried.
char str[3][3] ={"1","2","3"};
int i = 0;
while(str[i] != NULL || *str[i] = '\0')
{
i++
}
this however doesn't work at all. As apparently the out of boundary areas posses values.
Help?
Code for what you want was already posted; read the bloody thread.
Oh!.... Sorry there!
I couldn't tell if what you posted, was exactly what you wanted( you stated that you couldn't edit it?).
either way thanks a lot!
Testing it out right now.
Last edited by Kyoukoku; 09-22-2010 at 08:10 AM.
Use code tags [code][/code] when posting code samples.
str[i] will never be null since (in this context) it is an address that always exists regardless of how many actual elements exist. If str[0] exists at address 1000 and str[1] exists at 1003 and str[2] exists at 1006, then str[3] exists at 1009 regardless of whether there is a str[3] at all. This is because of simple pointer arithmetic. An example (granted this one is C++) would easily show this:
Output is below, of note just look at the address listed. It simply increases by 3 even though the elements beyond str[2] do not exist.Code:char str[3][3] = {"1","2","3"}; for( int i = 0; i < 5; ++i ) { std::cout << "str[" << i << "] is : " << str[i]; std::cout << " the address is: " << reinterpret_cast<void*>(str[i]) << std::endl; }
Last edited by hk_mp5kpdw; 09-22-2010 at 08:32 AM. Reason: Can't count 1,2,4?
"Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
-Christopher Hitchens
I wonderCode:char str[3][3] ={"1","2","3"}; int i = 0; while(str[i] != NULL || *str[i] = '\0') { i++ }char str[3][3]
if youchar str[3][3]
could tell mechar str[3][3]
how manychar str[3][3]
rows are inchar str[3][3]
my char array?
char str[3][3]
Help?
