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:
Don't use sizeof (char), since it's always 1 it doesn't add any value.
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.
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.
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.
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.
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.
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
In your case, you want to alloc an array of length characters. You will store in sample a pointer to an array of length times the size of what you point to. The sizeof(char*) is the size of a pointer to char. Not the size of a char.
A good practice is
sample = malloc(length * sizeof(*sample));
Using that, you will reserve length time the size of what you want to point to. This gives you the ability to change the data type anytime, simply declaring sample to be another kind of data.
int *sample;
sample = malloc(length * sizeof(*sample)); // length * 4
char *sample;
sample = malloc(length * sizeof(*sample)); // length * 1
Provided the length already accounts for the nul terminator, I would write either:
sample = malloc(length);
or:
sample = malloc(length * sizeof(*sample));
sizeof(char*) is the size of the pointer, and it is completely irrelevant to the the size that the allocated buffer needs to be. So definitely don't use that.
My first snippet is IMO good enough for string-manipulation code. C programmers know that memory and string lengths in C are both measured in multiples of sizeof(char). There's no real need to put a conversion factor in there that everybody knows is always 1.
My second snippet is the One True Way to write allocations in general. So if you want all your allocations to look consistent, then string allocations should use it too. I can think of two possible reasons to make all your allocations look consistent (both fairly weak IMO, but not actually wrong):
some people will find it easier to read them that way, only one visual pattern to recognise.
you might want to use the code in future as the basis for code that handles wide strings, and a consistent form would remind you to get the allocation right when the length is no longer measured in bytes but in wide chars. Using sizeof(*sample) as the consistent form means you don't need to change that line of code at all, assuming that you update the type of sample at the same time as the units in which length is measured.
Other options include:
sample = calloc(length, 1);
sample = calloc(length, sizeof(char));
sample = calloc(length, sizeof(*sample));
They're probably fairly pointless here, but as well as the trifling secondary effect of zeroing the memory, calloc has an interesting difference from malloc that it explicitly separates the number and size of objects that you're planning to use, whereas malloc just wants the total size.
For any type T, the usual form is
T *p = malloc(N * sizeof *p);
or
T *p;
...
p = malloc(N * sizeof *p);
where N is the number of elements of type T you wish to allocate. The expression *p has type T, so sizeof *p is equivalent to sizeof (T).
Note that sizeof is an operator like & or *, not a library function; parentheses are only necessary if the operand is a type name like int or char *.
Please visit this Linkhttps://www.codesdope.com/c-dynamic-memory/for understand how it allocat the memory dynamically at run time. It might be helpful to understand the concept of malloc and how it allocate the amount of memory to the variable.
In your example;
char *sample;
sample = malloc ( length * sizeof(char) );
here, you are declare a pointer to character for sample without declaring how much memory it required. In the next line, length * sizeof(char) bytes memory is assigned for the address of sample and (char*) is to typecast the pointer returned by the malloc to character.
Related
This question already has answers here:
Closed 13 years ago.
Possible Duplicate:
newbie questions about malloc and sizeof
I am trying to read strings into a program. When I noticed that the strings were sometimes being corrupted, I tried the following code:
void *mallocated = malloc(100);
printf("sizeof(mallocated) = %d\n", sizeof(mallocated));
According to my program, the size of mallocated was 8, even though I allocated 100 bytes for it. Because of this, whenever I try to store a string longer than 8 bytes, everything after the 8th byte will sometimes disappear. Why is this happening, and how can I prevent it?
Because the size of the "string" pointer is 8 bytes. Here are some examples of using sizeof() with their appropriate "size". The term size_of() is sometimes deceiving for people not used to using it. In your case, the size of the pointer is 8 bytes.. below is a representation on a typical 32-bit system.
sizeof (char) = 1
sizeof (double) = 8
sizeof (float) = 4
sizeof (int) = 4
sizeof (long) = 4
sizeof (long long) = 8
sizeof (short) = 2
sizeof (void *) = 4
sizeof (clock_t) = 4
sizeof (pid_t) = 4
sizeof (size_t) = 4
sizeof (ssize_t) = 4
sizeof (time_t) = 4
Source
You are leaving out how you are determining your string is disappearing (char array). It is probably being passed to a function, which you need to pass the explicit length as a variable or track it somewhere. Using sizeof() won't tell you this.
See my previous question about this and you'll see even my lack of initial understanding.
In C89, sizeof operator only finds the size of a variable in bytes at compile time (in this case a void pointer of 8 bytes). It works the way you'd expect it to work on plain arrays, because their size is known at compile time.
char arr[100]; // sizeof arr == 100
char *p = arr; // sizeof p == 4 (or 8 on 64-bit architectures)
char *p = malloc(100); // sizeof p == 4 (or 8). Still!
To know the size of heap-allocated memory, you need to keep track of it manually, sizeof won't help you.
sizeof returns the size of the pointer (void *) that you gave it, not the size of the memory you allocated. You would have to store the size of the memory in a separate variable if you want to use it later.
You cannot. As pointed out, you can get the size of the void * mallocated, but that does not tell you much.
You cannot get the size of the malloed *mallocated. That is to say, there is no function call to return 100 (in your example) - unless you write your own memory management routines.
Simplest is just to remember it somewhere ... maybe ....
stuct {
void *data;
unsigned int size
} myStructureWhichRemebersItsSize;
myStructureWhichRemebersItsSize *someData;
someData.data = malloc(100); // and check for NULL
someData.size = 100;
void* is the type of a location in memory if you don't know what it contains. It should be avoided.
char* is the type of a value which points to some location in memory which holds a char. Identifying a location in memory takes eight bytes.
sizeof tells you how many bytes a particular type takes. Not how many were allocated with malloc but just how much memory compiler knows the type should take. Applying sizeof to values is often considered bad style, and as someone else mentioned here, sometimes invokes smart behavior in C99.
char[100] the type of a value which holds 100 chars. char[100] a; is a string of 100 chars on the stack.
char(*)[100] is the type of a pointer to a value that holds 100 chars. char(*b)[100]; makes b point to 100 chars, possibly on the heap. What you probably want is
char (*s)[100] = malloc( sizeof( char[100] ) );
printf( "%u byte pointer to %u bytes of size %u elements\n",
sizeof s, sizeof *s, sizeof **s );
for (int a=0; a<10; ++a) {
printf ("%d", a);
}
char *foo;
foo = (char*)malloc(a);
I want to store more than one char value in foo variable.
Should I change it to an array, since the buffer is only allocating 1 char length?
Is 1 the longest length that can be stored in this buffer?
Well, foo now points to some useable address of a bytes, because this is how malloc() works. It doesn't matter if its type is char *, void * or anything else, you can only use a bytes.
Here, you increment a to 10. That means you can store 10 bytes, being 10 chars, (because in the context of C, 1 char = 1 byte), starting at the address where foo points to. Using a pointer or an array is strictly equivalent.
Since the buffer is only allocating 1 char length...
No, it is not the case here.
Quoting from the C11 standard, chapter ยง7.22.3.4, The malloc function
void *malloc(size_t size);
The malloc function allocates space for an object whose size is specified by size and
whose value is indeterminate.
So, in case of
foo = malloc(a); //yes, the cast is not required
a memory of size same as the value of a will be allocated, considering malloc() is successful.
Simply put, if I write a snippet like
int * p = malloc(10 * sizeof*p);
then, I can also write
for (int i = 0; i < 10, i++)
p[i] = i;
because, I have allocated the required memory for 10 ints.
That said, please see this discussion on why not to cast the return value of malloc() and family in C..
There are a couple of things you could do in a case like this.
If you know at compile time how many chars you want to store you could make it an array char foo[10]; If you know that there is always going to be 10 (or less) characters you want to store.
If you are not sure how many chars it needs to hold at compile time you would typically do dynamic allocation of memory using malloc. Now when using malloc you specify how many bytes of memory you want so for 12 chars you would do malloc(12) or malloc(12 * sizeof(char)). When using malloc you need to manually free the memory when you are done using it so the benefit of being able to ask for arbitrary (within limits) sizes of memory comes at the cost of making memory management harder.
As a side note: You typically do not want to cast the return value of malloc since it can hide some types of bugs and void *, that malloc returns can be implicitly cast to any pointer type anyway.
I have always used the malloc function as, for exemple,
int size = 10000;
int *a;
a = malloc(size * sizeof(int));
I recently run into a piece of code that discards the sizeof(int) part, i.e.
int size = 10000;
int *a;
a = malloc(size);
This second code seems to be working fine.
My question is then, which form is correct? If the second form is, am I allocating needless space with the first form.
The argument to malloc is the number of bytes to be allocated. If you need space for an array of n elements of type T, call malloc(n * sizeof(T)). malloc does not know about types, it only cares about bytes.
The only exception is that when you allocate space for (byte/char) strings, the sizeof can be omitted because sizeof(char) == 1 per definition in C. Doing something like
int *a = malloc(10000);
a[9000] = 0;
may seem to work now, but actually exploits undefined behavior.
malloc allocates a given number of bytes worth of memory, suitably aligned for any type. If you want to store N elements of type T, you need N * sizeof(T) bytes of aligned storage. Typically, T * p = malloc(N * sizeof(T)) provides that and lets you index the elements as p[i] for i in [0, N).
From the man page:
The malloc() function allocates size bytes and returns a pointer to the allocated memory.
The first form is correct.
Even if the sizeof(int) on the machine you are targeting is one (which is sometimes true on 8-bit microcontrollers) you still want your code to be readable.
The reason the "second code seems to be working fine" is that you are lucky.
The version of malloc you are using might be returning a pointer to an area of memory that is larger than what you requested. No matter what is happening behind the scenes, the behavior may change if you switch to a different compiler, so you do not want to rely on it.
I'm trying to write a function that converts a base 10 integer to a binary number.
I am constructing the function like this:
void dec2bin(int dec, char * bin);
The function will store the binary result by means of the pointer in the argument. It is passed a variable declared elsewhere as follows:
char *r1 = (char *)malloc(24);
The first thing the function has to do is fill the "24 bit result" with all zeros. It works great now since I've coded it based on me knowing the "size," but I can't figure out how to determine the "size" of an arbitrary "r1". Every way I can think to use the sizeof function does not give me 24.
I am confused at this point, and not connecting the relationship between the respective sizes of a char, char*, etc and how I can use those sizes to get what I'm looking for with respect to "r1".
Would someone mind giving me a pointer? Pun intended.
Thanks!
You may just pass the size as another parameter:
void dec2bin(int dec, char * bin, size_t max_chars);
I can't figure out how to determine the "size" of an arbitrary "r1".
You cannot do that: the information about the size that you passed to malloc is irreversibly lost after the call. All you have at that point is a pointer to the chunk of at least 24 bytes of memory. If you need the size, you need to store it somewhere else - for example, in a separate variable that you carry around.
Two typical work-arounds to this issue include (1) storing a zero-terminated string in the allocated space, and using the length of that string as the proxy to the size of the allocated chunk, and (2) defining your own structure that contains both the size and the allocated pointer. Neither solution is ideal, but at least the choice is with you.
The type of r1 is a char* (char pointer), so, like all pointers, its size is 4 (or 8, depending on your situation). No matter the size of the memory block you're allocating to it.
You can't retrieve the size of the allocated block from a pointer, so if you need the size later, you will have to remember the size yoursef (e.g. by storing it in a different variable).
r1 is a variable of pointer type, and its size is fixed and always the same (just like the size of any variable of a given type), which is no larger than (and usually equal to) sizeof(void *). What you really want to know is the size of the array that's sitting at *r1 (note the asterisk), and that is something you cannot know. You have to keep track of it yourself, typically like this:
size_t const n = get_some_number();
foo * p = calloc(n, sizeof(foo));
Now you use the pair (p, n) to describe your dynamic array.
(Note that C99 and C11 have the notion of a variable-length array, in whose context sizeof does actually (sometimes) give you the actual array size, but that's a somewhat murky affair, and in any case it isn't what you asked for.)
When you allocate memory, you get back a pointer, which has a fixed size, so there's no way to get the size of the allocated memory with sizeof.
As others have said, if it's just a pointer, you won't be able to tell anything (unless it's zero-terminated, in which case you can just iterate until you hit an element equal to zero (not the character '0')).
However, if you set it up as follows, you can actually see the buffer size after calling sizeof()... at least until you send it to a function, where the array decays to a plain char* pointer.
void test( char* s )
{
int i2 = sizeof( s ) / sizeof( char );
}
char c1[] = "Testing!";
int i1 = sizeof( c1 ) / sizeof( char ); // i1 is 9 (size of buffer)
test( c1 ); // i2 is 4 (size of pointer)
This question already has answers here:
Closed 13 years ago.
Possible Duplicate:
newbie questions about malloc and sizeof
I am trying to read strings into a program. When I noticed that the strings were sometimes being corrupted, I tried the following code:
void *mallocated = malloc(100);
printf("sizeof(mallocated) = %d\n", sizeof(mallocated));
According to my program, the size of mallocated was 8, even though I allocated 100 bytes for it. Because of this, whenever I try to store a string longer than 8 bytes, everything after the 8th byte will sometimes disappear. Why is this happening, and how can I prevent it?
Because the size of the "string" pointer is 8 bytes. Here are some examples of using sizeof() with their appropriate "size". The term size_of() is sometimes deceiving for people not used to using it. In your case, the size of the pointer is 8 bytes.. below is a representation on a typical 32-bit system.
sizeof (char) = 1
sizeof (double) = 8
sizeof (float) = 4
sizeof (int) = 4
sizeof (long) = 4
sizeof (long long) = 8
sizeof (short) = 2
sizeof (void *) = 4
sizeof (clock_t) = 4
sizeof (pid_t) = 4
sizeof (size_t) = 4
sizeof (ssize_t) = 4
sizeof (time_t) = 4
Source
You are leaving out how you are determining your string is disappearing (char array). It is probably being passed to a function, which you need to pass the explicit length as a variable or track it somewhere. Using sizeof() won't tell you this.
See my previous question about this and you'll see even my lack of initial understanding.
In C89, sizeof operator only finds the size of a variable in bytes at compile time (in this case a void pointer of 8 bytes). It works the way you'd expect it to work on plain arrays, because their size is known at compile time.
char arr[100]; // sizeof arr == 100
char *p = arr; // sizeof p == 4 (or 8 on 64-bit architectures)
char *p = malloc(100); // sizeof p == 4 (or 8). Still!
To know the size of heap-allocated memory, you need to keep track of it manually, sizeof won't help you.
sizeof returns the size of the pointer (void *) that you gave it, not the size of the memory you allocated. You would have to store the size of the memory in a separate variable if you want to use it later.
You cannot. As pointed out, you can get the size of the void * mallocated, but that does not tell you much.
You cannot get the size of the malloed *mallocated. That is to say, there is no function call to return 100 (in your example) - unless you write your own memory management routines.
Simplest is just to remember it somewhere ... maybe ....
stuct {
void *data;
unsigned int size
} myStructureWhichRemebersItsSize;
myStructureWhichRemebersItsSize *someData;
someData.data = malloc(100); // and check for NULL
someData.size = 100;
void* is the type of a location in memory if you don't know what it contains. It should be avoided.
char* is the type of a value which points to some location in memory which holds a char. Identifying a location in memory takes eight bytes.
sizeof tells you how many bytes a particular type takes. Not how many were allocated with malloc but just how much memory compiler knows the type should take. Applying sizeof to values is often considered bad style, and as someone else mentioned here, sometimes invokes smart behavior in C99.
char[100] the type of a value which holds 100 chars. char[100] a; is a string of 100 chars on the stack.
char(*)[100] is the type of a pointer to a value that holds 100 chars. char(*b)[100]; makes b point to 100 chars, possibly on the heap. What you probably want is
char (*s)[100] = malloc( sizeof( char[100] ) );
printf( "%u byte pointer to %u bytes of size %u elements\n",
sizeof s, sizeof *s, sizeof **s );