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once we initialize the array variable, the pointer points base address only & it's fixed
and constant pointer
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What is the easiest way to pass arrays as argument in c?
the simple and efficient way to pass an array is pointer to an array like that int (*p)[30] ; // pointer to an array of integer having 30 element
When does an array behave a pointer?
An array behaves like a pointer when you use its name in an expression without the brackets.int a[10]; /* a array of 10 ints */int *b = a; /* a reference to a as a pointer, making b like a */int c = *(a+3); /* a reference to a[3] using pointer semantics */myfunc(a); /* pass a's address, a pointer to myfunc */Note very carefully that, while an array name and a pointer can almost always be interchanged in context, the are not the same, in that a pointer is an l-value, such as b, above, and can be assigned, whereas a is an r-value and can only be referenced, such as in the same statement, the second statement. Also, an array name does not take up memory, while a pointer does.
What is pointer to a structure?
A pointer is a variable that holds address information. For example, in C++, say you have a Car class and another class that can access Car. Then, declaring Car *car1 =new Car() creates a pointer to a Car object.. The variable "car1" holds an address location.
Why does array index start with zero in C language?
Because of pointers and that all arrays are really pointers. A pointer something like *pointer can also be written as pointer[0] and *(pointer + 1) can also be written as pointer[1]
When does an array behave like a pointer in C plus plus -- provide sample code to support each case?
An array never behaves like a pointer.Understand that a pointer is a variable -- no different to any other variable, other than its type. like any other variable, it is allocated its own memory (4 bytes in a 32-bit system). Those 4 bytes can store any 32-bit value, from 0x00000000 to 0xFFFFFFFF, but because it is declared to be a pointer, that value actually represents a memory location, somewhere within the 4GB address space. 0x00000000 is a reserved memory location -- what we refer to as NULL.An array is not the same as a pointer in any sense. The array name refers to the actual memory (the starting address) where the array resides. But the array name is not a variable -- unlike a pointer, it is not separate from the memory it refers to -- it is merely an alias that refers to the memory allocated to the array itself.That said, pointers and arrays can appear to be the same. For instance, the strlen() function expects you to pass a const char pointer, and yet you can pass an array name instead and it works just fine:char cArray[12];char * p = cArray;strlen( p ); // Pass a pointerstrlen( &cArray[0] ); // Pass pointer to array element via AddressOf operatorstrlen( cArray ); // Pass an array nameAll three of these functions work exactly the same so it would be easy to assume the array is a type of pointer. But it is not. To understand what's really going on here you have to understand how pointers and arrays are actually passed to function.When you pass a pointer to a function, you don't pass the pointer itself, you pass the memory address stored in the memory allocated to the pointer. In other words, pointers are passed by value, not by reference. Similarly, when you pass an array name, you pass the starting address of the array.That's all there is to it! The function's parameter, a pointer, is allocated its own memory. It is a temporary variable, local to the function. All it expects is a memory location so whether you pass a pointer or an array name, that's exactly what it receives.
How does two dimensional array differ from single dimensional array?
A one dimensional array is a scalar value repeated one or more times.A two dimensional array is an array of one dimensional arrays.A three dimensional array is an array of two dimensional arrays, and so forth.The one dimensional array is like a list of things, where the two dimensional array is like an array of things. (Think one row of a spreadsheet versus the whole spreadsheet.)[addendum]Every level of array depth is also a level of pointer depth. For example: A 3 dimensional int array is an int***. So a one dimensional int array is an int*, and a two dimensional int array is an int**. This is only important if you are doing pointer work, but it can become very important.
If an array is declared in user defined function and pointer to this array is returned to main function is it possible to access that array?
In this video, I Cover an array of essential methods for every developer. see these videos on YouTube:- Brain Mentors YouTube Channel.
How are entire arrays passed from one method to another?
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.
Is x plus equals y is post increment or pre increment?
The '+=' operator behaves like a pre increment operator.
When is memory allocated when declaring an array in C plus plus?
It depends how the array is declared (fixed size or variable size) and where it is declared (global scope or local scope). If the array is declared in global scope (outside a function) and is fixed size, it will be allocated in static memory. If it is variable size, the pointer is stored in static memory while the array itself is allocated on the heap. The pointer in static memory points to the start address of the array in heap memory. If the array is declared in local scope (inside a function) and is fixed size, it will be allocated on the stack in whichever thread the function was called. If it is variable size, the local pointer is stored on the stack while the array is allocated on the heap. The pointer will fall from scope when the function returns so the array must not be allowed to outlive the function in which the pointer is declared. If the array must outlive the function that allocates the array, the pointer must be declared at a higher scope in the call stack and must be passed by reference to or returned by value from the function that allocates the array. If you provide your own memory manager, however, an array may be allocated wherever the memory manager's memory pool is allocated, be it in static memory, the stack or the heap. A memory manager essentially allocates an array of bytes which you can then utilise as you see fit (the array of bytes will be allocated as per the previous description for arrays in general).
Why array is a implicit pointer?
An array of pointers is exactly what it sounds like - one or more pointers arranged in order in memory, accessible through a common base name and indexed as needed. Philosophically, there is no difference between an array of pointers and an array of objects...int a[10]; // 10 integers, named a[0], a[1], a[2], ..., a[9]int *b[10]; // 10 pointers to int, named b[0], b[1], b[2], ..., b[9]If you initialize the array of pointers...int i;for (i = 0; i < 10; i++) b[i] = &a[i];... then *b[0] would be the same as a[0], etc.
How do you declare an array of N pointers to functions returning pointers to functions returning pointers to characters?
Please ask just one question at a time!Question 1:How do you declare an array of three pointers to chars?How do you declare an array of three char pointers?Note: both of these questions are merely alternative wordings for the same question.Answer 1:char * a[3];Question 2:How do you declare a pointer to an array of three chars?Answer 2:char a[3]; // an array of three charschar * p = a; // a pointer to an array of three charsQuestion 3:How do you declare a pointer to a function which receives an int pointer?Answer 3:#include // some functions we can point at:void func_1(int * p){}void func_2(int * p){}// note: all functions we wish to point at with the same// pointer must have the same signature.int main(){int* p = NULL; // instantiate an int pointervoid (*pFunc) (int*); // declare a function pointerpFunc = func_1; // point to func_1pFunc(p); // call func_1 via function pointerpFunc = func_2; // point to func_2pFunc(p); // call func_2 via function pointerreturn(0);}Note that the brackets in the function pointer declaration are required. If you omit them, you will end up with a standard function declaration that returns a pointer to void, resulting in a compiler error.
