05-06-2011
Quote:
Originally Posted by
rupeshkp728
I want to find the size of the total memory allocated on the heap for the following statement:
int* a = new int[1000];How can I use the sizeof operator for this?
I used:
printf("\t===> %d\n",sizeof(*a));
Is this statement correct?
I have asked the question because when I checked the memory of heap allocated in windbg it shows me the size as fc4 i.e. 4036 which is more then 4000 and not as desired. Any hint as to what may be the cause?
Hi ,
You can use sizeof() operator to find the size of any data structure by providing the base address of the structure. the implementation of memory allocation are different on different architectures and they may allocate slightly more space than you requested for in order to have alignment of the memory allocated which can increase performance benefits.
Thanks and Regards,
Gaurav.
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LEARN ABOUT NETBSD
posix_memalign
POSIX_MEMALIGN(3) BSD Library Functions Manual POSIX_MEMALIGN(3)
NAME
posix_memalign -- aligned memory allocation
LIBRARY
Standard C Library (libc, -lc)
SYNOPSIS
#include <stdlib.h>
int
posix_memalign(void **ptr, size_t alignment, size_t size);
DESCRIPTION
The posix_memalign() function allocates size bytes of memory such that the allocation's base address is an even multiple of alignment, and
returns the allocation in the value pointed to by ptr.
The requested alignment must be a power of 2 at least as large as sizeof(void *).
Memory that is allocated via posix_memalign() can be used as an argument in subsequent calls to realloc(3) and free(3).
RETURN VALUES
The posix_memalign() function returns the value 0 if successful; otherwise it returns an error value.
ERRORS
The posix_memalign() function will fail if:
[EINVAL] The alignment parameter is not a power of 2 at least as large as sizeof(void *).
[ENOMEM] Memory allocation error.
SEE ALSO
free(3), malloc(3), realloc(3), valloc(3)
STANDARDS
The posix_memalign() function conforms to IEEE Std 1003.1-2001 (``POSIX.1'').
BSD
November 19, 2007 BSD