How do I get the size of an array from a pointer to the array?

**jeffcobb** · 06-02-2010

Actually if you know that much (seeing posts waaaay up there) your sizeof() attempt almost works; the basic tactic for seeing the number of bytes consumed by an array is something like so:

Code:

double array[10]; // for example
int size = sizeof(array[0]) * 10;

If *all* you have is a ptr then you are screwed unless you get fancy and have debugging turned on, find the bounding array flags (how the debugger knows when you have overwritten the ends of an array), dereference the original pointer and start walking memory till you hit the marker.

The only thing more fun than debugging with the proper tools is debugging without them.

**robwhit** · 06-02-2010

Originally Posted by jeffcobb

Code:

double array[10]; // for example
int size = sizeof(array[0]) * 10;

better yet,

Code:

double array[10]; // for example
size_t size = sizeof(array);

sizeof evaluates to a value of type size_t, which is an unsigned integer (int != integer, int is a kind of integer) type.

**Programmer_P** · 06-03-2010

Originally Posted by robwhit

better yet,

Code:

double array[10]; // for example
size_t size = sizeof(array);

sizeof evaluates to a value of type size_t, which is an unsigned integer (int != integer, int is a kind of integer) type.

The problem with that approach is the array is a function parameter, so it defaults to a pointer. And you can't use the sizeof operator on a pointer and get the size.
That is why I get the size of the array by iterating through it and incrementing a size variable so long as the current character is not a null-terminated character.

Have either of you actually read the code I linked to a couple posts back? You'll see what I'm talking about.

EDIT: I see jeffcobb probably only read the first post and responded to it...

**syzygy** · 06-03-2010

You already know the length of the string so just pass it in as a parameter if you are so intent on using cstrings in your function. There is no point wasting cycles to figure out something you already know.

**MK27** · 06-03-2010

Originally Posted by Programmer_P

That is why I get the size of the array by iterating through it and incrementing a size variable so long as the current character is not a null-terminated character.

Ah, so you re-wrote strlen() as well?

Not such a big deal.

**Programmer_P** · 06-03-2010

Originally Posted by syzygy

You already know the length of the string so just pass it in as a parameter if you are so intent on using cstrings in your function. There is no point wasting cycles to figure out something you already know.

True, but that's only in my test code. In my real program which uses that function, I'm passing to it a c-style buffer string which stores characters read from a line which will be a case-by-case basis, and hence the length of the actual string(s) is not known (though I suppose I could always call ifstream::count() to get the number of characters read from a line...). Yes, I suppose I could do that.

**Programmer_P** · 06-03-2010

Originally Posted by MK27

Ah, so you re-wrote strlen() as well?

Not such a big deal.

Yeah, I suppose I did...oops.

Oh well, it was a good learning practice.

**Programmer_P** · 06-03-2010

And the cool thing about my containsString() function is, it will handle any null-terminated strings without any extra work in the caller function. Whereas if I passed in the size of a string to the function, I would need to do that for every function call, and there would be more code in each caller function which calls it. This way, I can just pass in any null-terminated string, and rest assured that my function is able to figure out the size of the strings on its own.

**MK27** · 06-03-2010

Originally Posted by Programmer_P

Whereas if I passed in the size of a string to the function, I would need to do that for every function call, and there would be more code in each caller function which calls it. This way, I can just pass in any null-terminated string, and rest assured that my function is able to figure out the size of the strings on its own.

Yeah, that would be a normal way to deal with C-strings -- use strlen() inside the called function, rather than use it first and pass an extra parameter. With most other kinds of arrays, though, you would have to supply the length since there is no way to tell the length. However, you can null terminate pointer arrays and then iterate thru them without knowing the length.

**Elysia** · 06-03-2010

That is why you, in C++, never use built-in native arrays. If you really need a static array for some reason (and not std::string), then you use std::array. It interfaces neatly with the STL and even keeps track of its size.
But I still fail to see why you need an array in the first place. Doesn't getline work for you?

**Programmer_P** · 06-03-2010

Originally Posted by Elysia

That is why you, in C++, never use built-in native arrays. If you really need a static array for some reason (and not std::string), then you use std::array. It interfaces neatly with the STL and even keeps track of its size.
But I still fail to see why you need an array in the first place. Doesn't getline work for you?

Of course it works for me. The function expects a c-style string buffer though. I know I can use string::c_str() to get the cstring representation of the std::string, but...
I don't know. I started out with a char array buffer before I learned of that function, and I'm too lazy to change it back now.

**Elysia** · 06-03-2010

...So why did it expect a c-style buffer in the first place?

**Programmer_P** · 06-03-2010

Originally Posted by Elysia

...So why did it expect a c-style buffer in the first place?

I don't know. Ask the developers of the ifstream class.

**Elysia** · 06-03-2010

So you are saying you were trying to call a function in the ifstream class that accepted a c-style string?

**Programmer_P** · 06-03-2010

Originally Posted by Elysia

So you are saying you were trying to call a function in the ifstream class that accepted a c-style string?

Yeah, ifstream::getline() does.

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