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assign values to malloc array

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macOS   C: How do I assign a value to a pointer following malloc?

  • Thread starter moonman239
  • Start date Dec 3, 2015
  • Sort by reaction score
  • Apple Programming
  • Dec 3, 2015

I'm an Obj-C developer who's starting to pick up ANSI C, and I've been wondering how to give a pointer a value to point to following a call to malloc. In Obj-C, all variables are initialized at some point, like this: Code: NSObject *obj = [[NSObject alloc] init] As I understand, this is functionally equivalent to doing something like this in ANSI C: Code: NSObject *obj = malloc(sizeof(NSObject)); *obj = // some value I tried running the following ANSI C code, and the program outputted 6: Code: int main(int argc, const char * argv[]) { int *p = malloc(sizeof(int)); *p = 6; printf("%d/n",*p); return 0; }  

chown33

A malloc'ed pointer is not an NSObject pointer. Do not assign the pointer returned by malloc() to any kind of Objective-C object pointer or id type. This code is wrong, in both C and Obj-C: Code: NSObject *obj = malloc(sizeof(NSObject)); *obj = // some value The reason a malloc'ed pointer isn't an object is because it hasn't been properly filled in. malloc() doesn't guarantee anything about the contents of the memory it allots. It could contain anything. If the Obj-C message-sending code receives such a block of memory, it may fail badly at the first message-send. If you want to look at the details, look into the underlying mechanics of Obj-C's isa (is-a) pointer, and how the compiler and runtime create objects. Your plain-C code is correct: Code: int *p = malloc(sizeof(int)); If you want more examples, look in a C tutorial or book. malloc() and calloc() are both standard C. Any reference explaining standard C should cover them. An Obj-C tutorial or book will cover Obj-C, which is a superset of plain C. To learn plain standard C, use something that explains plain standard C.  

macrumors 68040

In most cases you will not get away with a simple assignment to initialize a C object, usually it's something more complex than a primitive type like an int.  

chown33 said: A malloc'ed pointer is not an NSObject pointer. Do not assign the pointer returned by malloc() to any kind of Objective-C object pointer or id type. This code is wrong, in both C and Obj-C: Code: NSObject *obj = malloc(sizeof(NSObject)); *obj = // some value The reason a malloc'ed pointer isn't an object is because it hasn't been properly filled in. malloc() doesn't guarantee anything about the contents of the memory it allots. It could contain anything. If the Obj-C message-sending code receives such a block of memory, it may fail badly at the first message-send. If you want to look at the details, look into the underlying mechanics of Obj-C's isa (is-a) pointer, and how the compiler and runtime create objects. Your plain-C code is correct: Code: int *p = malloc(sizeof(int)); If you want more examples, look in a C tutorial or book. malloc() and calloc() are both standard C. Any reference explaining standard C should cover them. An Obj-C tutorial or book will cover Obj-C, which is a superset of plain C. To learn plain standard C, use something that explains plain standard C. Click to expand...
moonman239 said: But the plain-C code I posted works regardless of what C type p belongs to, right? Click to expand...
moonman239 said: No, I think you misunderstand. I'm not trying to use Obj-C in C code; what I was trying to do is figure out what to do to a plain-C variable after an malloc call. In my mind, the first two blocks of code I posted were simply analogous to each other. Click to expand...
In the tutorial I was following, I didn't see what happens after malloc, thus my question. Click to expand...
But the plain-C code I posted works regardless of what C type p belongs to, right? Click to expand...
chown33 said: Scalars or structs/unions, or typedefs of them, will work as shown. Click to expand...
subsonix said: Provided that you have a copy around in the desired init state, which isn't usually the case. Click to expand...

OK, so what do I do when a simple assignment won't suffice?  

chown33 said: I took the question to refer only to the malloc(). That is, this line of code: Click to expand...
moonman239 said: OK, so what do I do when a simple assignment won't suffice? Click to expand...

lee1210

It sounds like this isn't actually a question about using pointers, but about initialization of C types. What you're interested in is initialization of C types, and maybe scratching at hiding implementation details of a type (making it opaque). For one, we have to be clear that you're really dealing with structs (or arrays) when you're malloc'ing normally. There are a few options, but it really comes down to a few approaches: Code: foo_t *createFoo(int, char, double) void initFoo(foo_t *, int, char, double) The first would both allocate and initialize the struct (and make the caller implicitly in charge of free'ing the pointer). The second takes an allocated struct and initializes it. This gives the caller more flexibility, because they can choose how and where to allocate the memory (and free it). One way you can get real value out of patterns like this vs directly accessing members is in making opaque structures that can change (field names, types, order, etc.) and still stay compatible with existing code, even binaries. It forces consumers to use a more verbose API, calling methods to do anything with this data instead of just manipulating it directly. It's a tradeoff. Transparent data types vs. ease of future upgrades (and better control of data state). -Lee  

  • Dec 9, 2015
lee1210 said: It sounds like this isn't actually a question about using pointers, but about initialization of C types. What you're interested in is initialization of C types, and maybe scratching at hiding implementation details of a type (making it opaque). For one, we have to be clear that you're really dealing with structs (or arrays) when you're malloc'ing normally. There are a few options, but it really comes down to a few approaches: Code: foo_t *createFoo(int, char, double) void initFoo(foo_t *, int, char, double) Click to expand...
moonman239 said: Well, I'm not necessarily thinking about initializing my own types per se. What I'm kind of leaning towards now is how to initialize any type following an malloc. It seems there isn't exactly a one-size-fits-all approach. Coming from an Obj-C background, I find this fact awkward because in Obj-C, we alloc & init in the same line of code. Click to expand...

Toutou

macrumors 65816

  • Dec 10, 2015

There's no need to allocate memory for a simple int, you just declare and initialize at once: Code: int number = 6; If you, for whatever reason, wanted to allocate memory for an array of integers (maybe to be able to resize it later), you would have to use malloc, then assign values to them manually: Code: int * array = (int *)malloc(5*sizeof(int)); *array = 1; *(array+1) = 2; *(array+2) = 3; *(array+3) = 4; *(array+4) = 5; (or by a function, often a loop that goes through the memory by incrementing a pointer (in this case "int * pointer") and assigns a default value to every position) An allocated chunk of memory is just a chunk of memory. It doesn't care what's stored in it. Malloc allocates the memory and gives you a pointer to its beginning and doesn't care anymore. It is possible to malloc a memory, store a hundred integers, then - in the middle - switch to a char * pointer and start storing chars and then put a structure at the end. Nobody would care, but you'd have to be very careful with your pointers and functions handling the data. In other words - allocated memory isn't typed, it's an empty box and whatever you'd like to have inside you have to put there first manually.  

Toutou said: There's no need to allocate memory for a simple int Click to expand...
Toutou said: Code: int * array = (int *)malloc(5*sizeof(int)); *array = 1; *(array+1) = 2; *(array+2) = 3; *(array+3) = 4; *(array+4) = 5; Click to expand...
lee1210 said: This is a somewhat puzzling example. It's certainly correct and does what's advertised, but [] alleviates a lot of the manual pointer arithmetic and dereference. Code: array[1] = 2; Seems a lot easier to follow. I guess my point is that knowledge of pointer arithmetic is worthwhile, but might over-complicate things for a beginner. -Lee Click to expand...

How to use "malloc" in C

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Memory allocation (malloc), is an in-built function in C. This function is used to assign a specified amount of memory for an array to be created. It also returns a pointer to the space allocated in memory using this function.

The need for malloc

In the world of programming where every space counts, there are numerous times when we only want an array to have a specific amount of space at run time. That is, we want to create an array occupying a particular amount of space, dynamically. We do this using malloc .

We know what malloc returns and we know what it requires as an input, but how does the syntax of the function work. The illustration below shows that:

Note: malloc will return NULL if the memory specified is not available and hence, the allocation has failed

Now that we know how malloc is used and why it is needed, let’s look at a few code examples to see how it is used in the code.

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C Pointers and Memory Allocation

Other resources, declarations, creating a valid pointer value.

int *ip; int x; x = 42; ip = &x;

Using a pointer

printf("%d %d\n", x, *ip);

Memory Allocation and Deallocation

(void *) malloc(size_t numbytes);
int *ip; ip = (int *) malloc(sizeof(int));

Pointers and arrays

int *ip; ip = (int *) malloc( sizeof(int)*10 ); // allocate 10 ints ip[6] = 42; // set the 7th element to 42 ip[10] = 99; // WRONG: array only has 10 elements (this would corrupted memory!)

Freeing memory, and memory leaks

int *ip; ip = (int *) malloc( sizeof(int)*10 ); // allocate 10 ints ... (use the array, etc.) ip = (int *) malloc( sizeof(int)*100 ); // allocate 100 ints ...
int *ip; ip = (int *) malloc( sizeof(int)*10 ); // allocate 10 ints ... (use the array, etc.) free(ip); // de-allocate old array ip = (int *) malloc( sizeof(int)*100 ); // allocate 100 ints ...

Pointers and arrays as arguments

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Since C is a structured language, it has some fixed rules for programming. One of them includes changing the size of an array. An array is a collection of items stored at contiguous memory locations. 

arrays

As can be seen, the length (size) of the array above is 9. But what if there is a requirement to change this length (size)? For example, 

  • If there is a situation where only 5 elements are needed to be entered in this array. In this case, the remaining 4 indices are just wasting memory in this array. So there is a requirement to lessen the length (size) of the array from 9 to 5.
  • Take another situation. In this, there is an array of 9 elements with all 9 indices filled. But there is a need to enter 3 more elements in this array. In this case, 3 indices more are required. So the length (size) of the array needs to be changed from 9 to 12.

This procedure is referred to as Dynamic Memory Allocation in C . Therefore, C Dynamic Memory Allocation can be defined as a procedure in which the size of a data structure (like Array) is changed during the runtime. C provides some functions to achieve these tasks. There are 4 library functions provided by C defined under <stdlib.h> header file to facilitate dynamic memory allocation in C programming. They are: 

Let’s look at each of them in greater detail.

C malloc() method

The “malloc” or “memory allocation” method in C is used to dynamically allocate a single large block of memory with the specified size. It returns a pointer of type void which can be cast into a pointer of any form. It doesn’t Initialize memory at execution time so that it has initialized each block with the default garbage value initially. 

Syntax of malloc() in C

ptr = (int*) malloc(100 * sizeof(int)); Since the size of int is 4 bytes, this statement will allocate 400 bytes of memory. And, the pointer ptr holds the address of the first byte in the allocated memory.  

assign values to malloc array

If space is insufficient, allocation fails and returns a NULL pointer.

Example of malloc() in C

C calloc() method.

  • “calloc” or “contiguous allocation” method in C is used to dynamically allocate the specified number of blocks of memory of the specified type. it is very much similar to malloc() but has two different points and these are:
  • It initializes each block with a default value ‘0’.
  • It has two parameters or arguments as compare to malloc().

Syntax of calloc() in C

For Example:  

ptr = (float*) calloc(25, sizeof(float)); This statement allocates contiguous space in memory for 25 elements each with the size of the float.  

assign values to malloc array

Example of calloc() in C

C free() method.

“free” method in C is used to dynamically de-allocate the memory. The memory allocated using functions malloc() and calloc() is not de-allocated on their own. Hence the free() method is used, whenever the dynamic memory allocation takes place. It helps to reduce wastage of memory by freeing it.

Syntax of free() in C

assign values to malloc array

Example of free() in C

C realloc() method.

“realloc” or “re-allocation” method in C is used to dynamically change the memory allocation of a previously allocated memory. In other words, if the memory previously allocated with the help of malloc or calloc is insufficient, realloc can be used to dynamically re-allocate memory . re-allocation of memory maintains the already present value and new blocks will be initialized with the default garbage value.

Syntax of realloc() in C

assign values to malloc array

Example of realloc() in C

One another example for realloc() method is:

assign values to malloc array

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