I'd like more information on using arrays in C

[zubbaz]

Hi,
could you please give me some more info on how to use arrays in C??

and also, could you tell me whether this statement is correct or not:

void do_child(int data_pipe[][],int i)
close(data_pipe[i][1]);

[randomize]

int a[10];

it's a kind of array from int type..
it has 10 index & it's name is a

[MarshMallow]

Hi,
could you please give me some more info on how to use arrays in C??

and also, could you tell me whether this statement is correct or not:

void do_child(int data_pipe[][],int i)
close(data_pipe[i][1]);

why in the world do you approach advanced topics like InterProcess Communication if you don't even know what an array is???
An array variable stores the first addres of a set of contiguous memory locations where a discrete number of bytes(or words,or double words) is stored.the type of an array variable is therefore a pointer.Please read Ritchie and Kernighan ANSI C book

[weaknessforcats]

Read this and then post again if you still have questions:

First, there are only one-dimensional arrays in C or C++. The number of elements in put between brackets:

int array[5];
 

That is an array of 5 elements each of which is an int.

int array[];
 

won't compile. You need to declare the number of elements.

Second, this array:

int array[5][10];
 

is still an array of 5 elements. Each element is an array of 10 int.

int array[5][10][15];
 

is still an array of 5 elements. Each element is an array of 10 elements where each element is an array of 15 int.

int array[][10];
 

won't compile. You need to declare the number of elements.

Third, the name of an array is the address of element 0

int array[5];
 

Here array is the address of array[0]. Since array[0] is an int, array is the address of an int. You can assign the name array to an int*.

int array[5][10];
 

Here array is the address of array[0]. Since array[0] is an array of 10 int, array is the address of an array of 10 int. You can assign the name array to a pointer to an array of 10 int:

int array[5][10];
 
int (*ptr)[10] = array;
 

Fourth, when the number of elements is not known at compile time, you create the array dynamically:

int* array = new int[value];
int (*ptr)[10] = new int[value][10];
int (*ptr)[10][15] = new int[value][10][15];
 

In each case value is the number of elements. Any other brackets only describe the elements.

Using an int** for an array of arrays is incorrect and produces wrong answers using pointer arithmetic. The compiler knows this so it won't compile this code:

int** ptr = new int[value][10];    //ERROR
 

new returns the address of an array of 10 int and that isn't the same as an int**.

Likewise:

int*** ptr = new int[value][10][15];    //ERROR
 

new returns the address of an array of 10 elements where each element is an array of 15 int and that isn't the same as an int***.

With the above in mind this array:

int array[10] = {0,1,2,3,4,5,6,7,8,9};
 

has a memory layout of

0 1 2 3 4 5 6 7 8 9

Wheras this array:

int array[5][2] = {0,1,2,3,4,5,6,7,8,9};
 

has a memory layout of

0 1 2 3 4 5 6 7 8 9

Kinda the same, right?

So if your disc file contains

0 1 2 3 4 5 6 7 8 9

Does it make a difference wheher you read into a one-dimensional array or a two-dimensional array? No.

Therefore, when you do your read use the address of array[0][0] and read as though you have a
one-dimensional array and the values will be in the correct locations.

[MarshMallow]

Does it make a difference wheher you read into a one-dimensional array or a two-dimensional array? No.

that's because the C language uses a row major order matrix representation.

[zubbaz]

why in the world do you approach advanced topics like InterProcess Communication if you don't even know what an array is???
An array variable stores the first addres of a set of contiguous memory locations where a discrete number of bytes(or words,or double words) is stored.the type of an array variable is therefore a pointer.Please read Ritchie and Kernighan ANSI C book

i know HOW TO DEAL with ARRAYS AND I KNOOOW what arrays are. all this stuff ur giving me.. well i already know it, ive studies C++ and i know all the basics. the problem is that i keep on getting some kind of error on the arrays in my code and i just want to know if theres any difference AT ALL between arrays in C and arrays in C++. u guys keep on giving me very simple info on arrays. as if i dont know what they are!! and besides, u didnt even tell me whether those two code lines were consider correct or not, which was mainly my question :)

[oler1s]

Apparently, you don't know how to do any of the following: write acceptable English, Google, read properly, ask proper detailed questions, show respect to those who actually try to help you, and finally, consider the fact that it's your sloppy coding that might be causing errors.

If you had googled, you would have discovered that C and C++ treat arrays the same.

If you are having trouble dealing with an error, you could post relevant code and the error here.

u didnt even tell me whether those two code lines were consider correct or not, which was mainly my question :)

How are we supposed to know? Outwardly, the syntax looks fine. You might have bungled it up, passing the wrong type of parameter to your defined functions.

We're here to help, but you have no excuse to act offended if you get the wrong answers because you can't write proper English and you can't ask the right question.

[weaknessforcats]

that's because the C language uses a row major order matrix representation.

That has nothing to do with it. row-major is a database table concept.

In C and C++ the array elements are required to be contiguous in memory and the size of the element must be known at compile time. Otherwise, the compiler cannot perform the necessary pointer aritmetic needed to navigate the array.

and i just want to know if theres any difference AT ALL between arrays in C and arrays in C++.

There is absolutely no difference between arrays in C and arrays in C++ just like oler1s said.

[MarshMallow]

That has nothing to do with it. row-major is a database table concept.

Please,open Patterson Hennessy,Computer Architecture:A Quantitative Approach
Chapter 5:"Compiler Optimizations to ReduceMiss Rate" and read it carefully,very carefully........

[bonzo]

That has nothing to do with it. row-major is a database table concept.

Row major order is a methods for storing multidimensional arrays. Rows in a matrix are stored consecutively (a[0][0], a[0][1], a[0][2],...a[0][n], a[1][0], a[1][1],...a[m][n]).

So, just you have already said, you can access elements increasing a pointer.

Row Major Order <=> Column Major Order

Swap the concepts... :)

[weaknessforcats]

There is a disconnect here since C and C++ only have one-dimensional arrays.

Read this:

First, there are only one-dimensional arrays in C or C++. The number of elements in put between brackets:

int array[5];
 

That is an array of 5 elements each of which is an int.

int array[];
 

won't compile. You need to declare the number of elements.

Second, this array:

int array[5][10];
 

is still an array of 5 elements. Each element is an array of 10 int.

int array[5][10][15];
 

is still an array of 5 elements. Each element is an array of 10 elements where each element is an array of 15 int.

int array[][10];
 

won't compile. You need to declare the number of elements.

Third, the name of an array is the address of element 0

int array[5];
 

Here array is the address of array[0]. Since array[0] is an int, array is the address of an int. You can assign the name array to an int*.

int array[5][10];
 

Here array is the address of array[0]. Since array[0] is an array of 10 int, array is the address of an array of 10 int. You can assign the name array to a pointer to an array of 10 int:

int array[5][10];
 
int (*ptr)[10] = array;
 

Fourth, when the number of elements is not known at compile time, you create the array dynamically:

int* array = new int[value];
int (*ptr)[10] = new int[value][10];
int (*ptr)[10][15] = new int[value][10][15];
 

In each case value is the number of elements. Any other brackets only describe the elements.

Using an int** for an array of arrays is incorrect and produces wrong answers using pointer arithmetic. The compiler knows this so it won't compile this code:

int** ptr = new int[value][10];    //ERROR
 

new returns the address of an array of 10 int and that isn't the same as an int**.

Likewise:

int*** ptr = new int[value][10][15];    //ERROR
 

new returns the address of an array of 10 elements where each element is an array of 15 int and that isn't the same as an int***.

With the above in mind this array:

int array[10] = {0,1,2,3,4,5,6,7,8,9};
 

has a memory layout of

0 1 2 3 4 5 6 7 8 9

Wheras this array:

int array[5][2] = {0,1,2,3,4,5,6,7,8,9};
 

has a memory layout of

0 1 2 3 4 5 6 7 8 9

Kinda the same, right?

So if your disc file contains

0 1 2 3 4 5 6 7 8 9

Does it make a difference wheher you read into a one-dimensional array or a two-dimensional array? No.

Therefore, when you do your read use the address of array[0][0] and read as though you have a
one-dimensional array and the values will be in the correct locations.

[MarshMallow]

There is a disconnect here since C and C++ only have one-dimensional arrays.

Read this:

'Father, forgive them, for they do not know what they are doing' " (23:33-34)

[bonzo]

There is a disconnect here since C and C++ only have one-dimensional arrays.
Read this:

this is blasphemy this is madness...

So array of integer doesn't exist because they are four-dimensional array of byte, isn't it?

This is not a C/C++ approach: this is assembly.

Yes, all memory is a huge one-dimensional array of byte, but in a high level language we talk about n-dimensional arrays. In C they are not the same. For compilers it makes difference (type checking).

Only brutal cast can break the rule, but this is discouraged.

P.S: Sorry for my English.

[MarshMallow]

the truth must be visible to everyone in this forum,my dear weakness,do not try again to hide the truth;you can make it once,maybe twice...but how long are you supposed to do it?Accept the fact that you are wrong,and PLEASE do not try again to make a thread float down just to make it not visible to the others.
Say hi to phanepip and begin to study,seriously.