C++ Pointer question - c++

Question

I'm new to pointers in C++. I'm not sure why I need pointers like char * something[20] as oppose to just char something[20][100]. I realize that the second method would mean that 100 block of memory will be allocated for each element in the array, but wouldn't the first method introduce memory leak issues.
If someone could explain to me how char * something[20] locates memory, that would be great.
Edit:
My C++ Primer Plus book is doing:
const char * cities[5] = {
"City 1",
"City 2",
"City 3",
"City 4",
"City 5"
}
Isn't this the opposite of what people just said?

Answer 1

You allocate 20 pointers in the memory, then you will need to go through each and every one of them to allocate memory dynamically:
something[0] = new char[100];
something[1] = new char[20]; // they can differ in size
And delete them all separately:
delete [] something[0];
delete [] something[1];
EDIT:
const char* text[] = {"These", "are", "string", "literals"};
Strings specified directly in the source code ("string literals", which are always const char *) are quite different to char *, mainly because you don't have to worry about alloc/dealloc of them. They are also generally handled very different in memory, but this depends on the implementation of your compiler.

Answer 2

You're right.
You'd need to go through each element of that array and allocate a character buffer for each one.
Then, later, you'd need to go through each element of that array and free the memory again.
Why you would want to faff about with this in C++ is anyone's guess.
What's wrong with std::vector<std::string> myStrings(20)?

Answer 3

It will allocate space for twenty char-pointers.
They will not be initialized, so typical usage looks like
char * something[20];
for (int i=0; i<20; i++)
something[i] = strdup("something of a content");
and later
for (int i=0; i<20; i++)
if (something[i])
free(something[i]);

Answer 4

You're right - the first method may introduce memory leak issues and the overhead of doing dynamic allocations, plus more reads. I think the second method is usually preferable, unless it wastes too much RAM or you may need the strings to grow longer than 99 chars.
How the first method works:
char* something[20]; // Stores 20 pointers.
something[0] = malloc(100); // Make something[0] point to a new buffer of 100 bytes.
sprintf(something[0], "hai"); // Make the new buffer contain "hai", going through the pointer in something[0]
free(something[0]); // Release the buffer.

Answer 5

char* smth[20] does not allocate any memeory on heap. It allocates just enough space on the stack to store 20 pointers. The value of those pointers is undefined, so before using them, you have to initialize them, like this:
char* smth[20];
smth[0] = new char[100]; // allocate memory for 100 chars, store the address of the first one in smth[0]
//..some code..
delete[] smth[0];

Answer 6

First of all, this almost inapplicable in C++. The normal equivalent in C++ would be something like: std::vector<std::string> something;
In C, the primary difference is that you can allocate each string separately from the others. With char something[M][N], you always allocate exactly the same number of strings, and the same space for each string. This will frequently waste space (when the strings are shorter than you've made space for), and won't allow you to deal with any more strings or longer of strings than you've made space for initially.
char *something[20] let's you deal with longer/shorter strings more efficiently, but still only makes space for 20 strings.
The next step (if you're feeling adventurous) is to use something like:
char **something;
and allocate the strings individually, and allocate space for the pointers dynamically as well, so if you get more than 20 strings you can deal with that as well.
I'll repeat, however, that for most practical purposes, this is restricted to C. In C++, the standard library already has data structures for situations like these.

Answer 7

C++ has pointers because C has pointers.
Why do we use pointers?
To track dynamically-allocated memory. The memory allocation functions in C (malloc, calloc, realloc) and the new operator in C++ all return pointer values.
To mimic pass-by-reference semantics (C only). In C, all function arguments are passed by value; the formal parameter and the actual parameter are distinct objects, and modifying a formal parameter doesn't affect the actual parameter. We get around this by passing pointers to the function. C++ introduced reference types, which serve the same purpose, but are a bit cleaner and safer than using pointers.
To build dynamic, self-referential data structures. A struct cannot contain an instance of itself, but it can contain a pointer to an instance. For example, the following code
struct node
{
data_t data;
struct node *next;
};
creates a data type for a simple linked-list node; the next member explicitly points to the next element in the list. Note that in C++, the STL containers for stacks and queues and vectors all use pointers under the hood, isolating you from the bookkeeping.
There are literally dozens of other places where pointers come up, but those are the main reasons you use them.
Your array of pointers could be used to store strings of varying length by allocating just enough memory for each, rather than relying on some maximum size (which will eventually be exceeded, leading to a buffer overflow error, and in any case will lead to internal memory fragmentation). Naturally, in C++ you'd use the string data type (which hides all the pointer and memory management behind the class API) instead of pointers to char, but someone has decided to confuse you by starting with low-level details instead of the big picture.

Answer 8

I'm not
sure why I need pointers like char *
something[20] as oppose to just char
something[20][100]. I realize that the
second method would mean that 100
block of memory will be allocated for
each element in the array, but
wouldn't the first method introduce
memory leak issues.
The second method will suffice if you're only referencing your buffer(s) locally.
The problem comes when you pass the array name to another function. When you pass char something[10] to another function, you're actually passing char* something because the array length doesn't go along for the ride.
For multidimensional arrays, you can declare a function that takes in an array of determinate length in all but one direction, e.g. foo(char* something[10]).
So why use the first form rather than the second? I can think of a few reasons:
You don't want to have the restriction that the entire buffer must reside in continuous memory.
You don't know at compile-time that you'll need each buffer, or that the length of each buffer will need to be the same size, and you want the flexibility to determine that at run-time.
This is a function declaration.

Answer 9

char * something[20]
Assuming this is 32Bit, this allocates 80 bytes of data on the stack.
4 Bytes for each pointer address, 20 pointers total = 4 x 20 = 80 bytes.
The pointers are all uninitialized, so you need to write additional code to allocate/free
the buffers for doing this.
It roughly looks like:
[0] [4 Bytes of Uninitialized data to hold a pointer/memory address...]
[1] [4 Bytes of ... ]
...
[19]
char something[20][100]
Allocates 2000 bytes on the stack.
100 Bytes for each something, 20 somethings total = 100 x 20 = 2000 bytes.
[0] [100 bytes to hold characters]
[1] [100 bytes to hold characters]
...
[19]
The char *, has a smaller memory overhead, but you have to manage the memory.
The char[][] approach, has bigger memory overhead, but you don't have additional memory management.
With either approach, you have to be careful when writing to the buffer allocated not to exceed/overwrite the memory alloc'd for it.