Why does allocated memory is different than the size of the string?

问题

Please consider the following code:

char    **ptr;

str = malloc(sizeof(char *) * 3); // Allocates enough memory for 3 char pointers
str[0] = malloc(sizeof(char) * 24);
str[1] = malloc(sizeof(char) * 25);
str[2] = malloc(sizeof(char) * 25);

When I use some printf to print the memory adresses of each pointer:

printf("str[0] = '%p'\nstr[1] = '%p'\nstr[2] = '%p'\n", str[0], str[1], str[2]);

I get this output:

str[0] = '0x1254030'
str[1] = '0x1254050'
str[2] = '0x1254080'

I expected the number corresponding to the second adress to be the sum of the number corresponding to the first one, and 24, which corresponds to the size of the string str[0] in bytes (since a char has a size of 1 byte). I expected the number corresponding to the second adress to be 0x1254047, considering that this number is expressed in base 16 (0123456789abcdef).

It seems to me that I spotted a pattern: from 24 characters in a string, for every 16 more characters contained in it, the memory used is 16 bytes larger. For example, a 45 characters long string uses 64 bytes of memory, a 77 characters long string uses 80 bytes of memory, and a 150 characters long string uses 160 bytes of memory.

Here is an illustration of the pattern:

I would like to understand why the memory allocated isn't equal to the size of the string. Why does it follow this pattern?

回答1:

Three reasons:

(1) The string "ABC" contains 4 characters, because every string in C has to have a terminating '\0'.

(2) Many processors have memory address alignment issues that make it most efficient when any block allocated by malloc() starts on an address that is a multiple of 4, or 8, or whatever the "natural" memory size is.

(3) The malloc() function itself requires some memory to store information about what has been allocated where, so that free() knows what to do.

回答2:

There at least two reasons malloc() may return memory in the manner you've noted.

Efficiency and peformance. By returning blocks of memory in only a small set of actual sizes, a request for memory is much more likely to find an already-existing block of memory that can be used, or wind up producing a block of memory that can be easily reused in the future. This will make the request return faster and has the side effect of limiting memory fragmentation.
Implications arising from the memory alignment requirements 7.22.3 Memory management functions of the C Standard states "The order and contiguity of storage allocated by successive calls to the aligned_alloc, calloc, malloc, and realloc functions is unspecified. The pointer returned if the allocation succeeds is suitably aligned so that it may be assigned to a pointer to any type of object with a fundamental alignment requirement ..." Note the italicized part. Since the malloc() implementation has no knowledge of what the memory is to be used for, the memory returned has to be suitably aligned for any possible use. This usually means 8- or 16-byte alignment, depending on the platform.

来源：https://stackoverflow.com/questions/44189651/why-does-allocated-memory-is-different-than-the-size-of-the-string

标签

memory