You don't pass the entire array. You pass a single value, the address of one of its elements, usually the first element.
int a[] = {1, 3, 5, 7, 9, 2, 4, 6, 8, 10};
This creates an int array of length 10, with initial values as stated. The first element is a[0], having value of 1, and the last element is a[9], having value of 10.
somefunction (int * b);
somefunction(a);
This declares a function that receives an address to an int array. The call passes the address of a[0] to the function. You could just as easily pass some other element, such as a[3], by saying somefunction (&a[3]); or somefunction (a+3); Of course, in this case, you would not want to access b[9], because that would violate the addressability limit of the pointer, but you could certainly access b[3], which would be the same as a[6].
Note, in this example, that I mix array syntax with pointer syntax. In fact, the two are the same, and you can easily inter-mix them. You could just as easily say...
somefunction (int b[]);
somefunction(a);
An array is still an array, regardless of how you pass it into functions. It is still an array-type variable at the beginning of a function. However, the function itself may manipulate the array by "imploding" it into a string with a delimiter, using array values with only specific keys, and such.
Reference:cprogramming-bd.com/c_page1.aspx# array programming
In C programming, a string doesn't have a specific return type as it's essentially an array of characters. So, if a function is returning a string, it should be declared to return a pointer to a char (char*), since a string in C is represented as an array of characters terminated by a null character ('\0').
Basically in c++ passing an array as an argument only provides a pointer to the first value and that function won't know how many values it has.If you read beyond the size you will just get garbage from memory.
AnswerUnfortunately your question is to broad. All progamming languages vary in the way things are done. I will just give a general way of doing it.You have to pass the multidimensional array into the function by including it with calling the function. On the receiving end you have to declare another multidimensional array so the information can be passed into it. Depending on the language, you may not be passing in a multidimensional array, instead that array may be stored in an object which you can pass instead.Hope this helps some.-Ashat-in C, when passing two dimensional arrays the compiler needs to know the width so it can calculate memory offsets.passing a 2d array of width 4:voidFunc(type array[][4]);
If the identifier you want to pass is an ordinary identifier, pass it as the address of... function(&identifier); If the identifier you want to pass is an array identifier, pass its name... function(arrayname);
An array is still an array, regardless of how you pass it into functions. It is still an array-type variable at the beginning of a function. However, the function itself may manipulate the array by "imploding" it into a string with a delimiter, using array values with only specific keys, and such.
In some programming languages, like C, you can pass the new method (or function) an address pointer to the first element in the array. As long as you don't leave the scope of the method the array was created in, the array will remain valid. In other languages that don't support memory addresses, like FORTRAN, it must be done by making the array global.
Reference:cprogramming-bd.com/c_page1.aspx# array programming
In C programming, a string doesn't have a specific return type as it's essentially an array of characters. So, if a function is returning a string, it should be declared to return a pointer to a char (char*), since a string in C is represented as an array of characters terminated by a null character ('\0').
Basically in c++ passing an array as an argument only provides a pointer to the first value and that function won't know how many values it has.If you read beyond the size you will just get garbage from memory.
AnswerUnfortunately your question is to broad. All progamming languages vary in the way things are done. I will just give a general way of doing it.You have to pass the multidimensional array into the function by including it with calling the function. On the receiving end you have to declare another multidimensional array so the information can be passed into it. Depending on the language, you may not be passing in a multidimensional array, instead that array may be stored in an object which you can pass instead.Hope this helps some.-Ashat-in C, when passing two dimensional arrays the compiler needs to know the width so it can calculate memory offsets.passing a 2d array of width 4:voidFunc(type array[][4]);
In C++ you would pass a std::array if the array is fixed-length, otherwise you'd use a std::vector. Most object oriented languages will provide some method of passing a self-contained array object to a function. In C and other non-object oriented languages you would pass a reference or pointer to the start address of the array along with a variable indicating the number of valid elements within the array. The array type will determine the size of each element.
It 's address is received by the function . that any changes in the value of array elements in the function will result in actual change.
There are no commands in C-programming, you should use function sqrt from math.h
A type construction: one or more values with the same type and name.
Sizeof is an example.