C memory allocating - char* and char sizeof

Related searches

What form is correct in allocating string in C?

char *sample;

sample = malloc ( length * sizeof(char) );

or

sample = malloc ( length * sizeof(char*) );

Why does char* take 4 bytes when char takes 1 byte?

Assuming the goal is to store a string of length characters, the correct allocation is:

sample = malloc(length + 1);

Notes:

  1. Don't use sizeof (char), since it's always 1 it doesn't add any value.
  2. Remember the terminator, I assumed (based on name) that length is the length in visible characters of the string, i.e. the return of strlen() will be length.
  3. I know you didn't, but it's worth pointing out that there should be no cast of the return value from malloc(), either.

The reason char * is larger is that it's a pointer type, and pointers are almost always larger than a single character. On many systems (such as yours, it seems) they are 32 bit, while characters are just 8 bits. The larger size is needed since the pointer needs to be able to represent any address in the machine's memory. On 64-bit computers, pointers are often 64 bits, i.e. 8 characters.

Dynamic memory allocation in C : calloc, malloc, free, realloc in C, So, the size of address will be 50 * sizeof(char). char *address → Here we declared a pointer to character for address without specifying how much memory is� char *address → Here we declared a pointer to character for address without specifying how much memory is required. address = (char*)malloc (50 * sizeof (char)) → By writing this, we assigned a memory of '50 * sizeof (char)' bytes for address. We used (char*) to typecast the pointer returned by malloc to character.

Why does char* take 4 bytes when char takes 1 byte?

Because you are on a 32-bit systems, meaning that pointers take four bytes; char* is a pointer.

char always takes exactly one byte, so you do not need to multiply by sizeof(char):

sample = malloc (length);

I am assuming that length is already padded for null termination.

Assignment #6 Answers, Introductory C Programming How much memory does the call malloc(10) allocate? What if you want char *cp = malloc(10); int *ip = malloc(sizeof(int)); It allocates 10 bytes of space for pointer ‘ch’ of type char and base address is stored in ch. Note: the storage space allocated dynamically has no name and therefore its contents can be accessed only through a pointer. We should check whether the allocation is successful before using the memory pointer.

sample = malloc ( length * sizeof(char) );

First is the correct one if you want to allocate memory for length number of characters.

char* is of type pointer which happens to be 4 bytes on your platform. So sizeof(char*) returns 4.

But sizeof(char) is always 1 and smae is guaranteed by the C standard.

[PDF] Character Array (i.e. string) example, char arr[4]; strncpy(arr, ptr, sizeof("abc")); // Copy the string "abc" into the array arr printf ("\n %s The function malloc() will allocate a block of memory that is size bytes large. If http://www.codingunit.com/c-reference-stdlib-h-function- malloc. Or alternatively, use an additional level of pointers to emulate pass by reference : Looks complex, but you're basically just replacing every instance of game in allocate_memory with *game.

In the given cases you are doing two different things:

In the first case : sample = malloc ( length * sizeof(char) );

You are allocating length multiplied by the size of type char which is 1 byte

While in the second case : sample = malloc ( length * sizeof(char*) );

You are allocating length multiplied by the size of pointer to char which is 4 byte on your machine.

Consider that while case 1 remains immutable, on the second case the size is variable.

Intro to Dynamic Allocation, With static allocation, if we wanted a char array label to be able to hold various strings, What would the size of the array needed to hold these things be? C provide a set of routines for allocating space for any type of object at run-time. You tell malloc() how many bytes of memory to allocate and it returns a pointer to the� Dynamic Memory Allocation Examples using C programs 1) C program to create memory for int, char and float variable at run time. In this program we will create memory for int, char and float variables at run time using malloc() function and before exiting the program we will release the memory allocated at run time by using free() function.

sample = malloc(length);

is the right one

char* is a pointer, a pointer uses 4 bytes (say on a 32-bit platform)

char is a char, a char uses 1 byte

Introduction to C, A pointer usually has an associated type, e.g., an int pointer vs. a char pointer. Declare an array variable without a size, and allocate memory dynamically. A2 = (int*) malloc (sizeof(int) * 10); for (i=0; i < 10; i++) { A2[i] = i * 100; // Fill the� This works fine in C but writing in this form is a bad idea in C++. That’s why compiler shows warning of “deprecated conversion from string constant to ‘char*'” because in C string literals are arrays of char but in C++ they are constant array of char. Therefore use const keyword before char*. const char* str = "This is GeeksForGeeks";

char arrays and memory allocation . char arrays and memory allocation dhaneku (The sizeof a reference or a reference type is the size of the referenced type).

The main use of the concept of dynamic memory allocation is for allocating arrays when we have to declare an array by specifying its size but are not sure about the size. Consider a situation when we want the user to enter the name but are not sure about the number of characters in the name that the user will enter.

You won’t get too far programming C if you don’t read how pointers and memory allocation and such work. Anyways, wc is some variable, containing a word count. This code is allocating an array of wc char pointers. sizeof() tells it how large a char pointer is. There’s ‘wc’ char pointers.

Comments
  • Though sizeof(char) doesn't add anything, it makes the intent explicit about what's going on. I guess it's a matter of taste.
  • @KingsIndian: it's a matter of how you read the malloc call (and I agree that in turn is a matter of taste). unwind and I would read malloc(length + 1) as "allocate length+1 bytes" or "allocate length+1 chars", and know that the two mean exactly the same thing in C. Some might read it as "allocate length+1 of something, but of what? The author doesn't say, maybe they don't know!". To those readers, throwing an explicit sizeof(char) in there might help convince them that I meant what I said.
  • @SteveJessop It may confuse when using length+sizeof(int) for allocating length number of integers or other data type. Why not calculate the same way length*4 bytes (if sizeof(int)=4) and same way for other data types. This will obviously lead to problems when sizeof(int) changes. Requires to have a complete understanding of subtleties involved that sizeof(char) is always 1, sizeof(int) may vary etc. So length+sizeof(char) reads "length chars" and "length ints", "length floats" etc. IMO, this is very explicit and don't have to think about chars as some kind of a special case.
  • @KingIndian: I don't understand who you're saying it would confuse. You? Some hypothetical newbie who will quickly get over their ignorance? Or actual colleagues? Like I say, I don't think it's wrong to put the size in there, just a matter of taste and redundant for real C programmers that I have encountered. And I might write the code differently in a tutorial than in production, because it's for a different audience (newbies), but by the end of the tutorial I would reveal the production way(s).
  • And btw I'm not thinking of chars as a special case of allocation, I'm thinking of chars as the fundamental unit of memory in C. So malloc(1) means, "allocate 1 char". malloc(sizeof(int)) means, "allocate a number of chars equal to the size of int in chars". It's not my choice that they're a special case, but they are. You can choose to ignore this, and write your allocations as though the fundamental unit might be something other than char. I don't see any huge advantage, just the minor ones in my answer. It's actively harmful if it prevents people learning what that unit is in C.
  • length * sizeof(char) +1 is somewhat bizarre. I can understand (length + 1) * sizeof(char), and length + 1 is fine too, but why add 1 after multiplication?