what does alignment means in c/c++ malloc

Upon successful completion, the value pointed to by memptr shall be a multiple of alignment.

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syedshan said:

can any one explain what does alignment means when we are using dynamic memory allocation related functions, like

_alligned_maloc()

for example : while searching for posix_memalign() function I get to know

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A memory address a, is said to be n-byte aligned when n is a power of two and a is a multiple of n bytes. In this context a byte is the smallest unit of memory access, i.e. each memory address specifies a different byte. An n-byte aligned address would have log2 least-significant zeros when expressed in binary.

The alternate wording b-bit aligned designates a b/8 byte aligned address (ex. 64-bit aligned is 8 bytes aligned).

A memory access is said to be aligned when the datum being accessed is n bytes long and the datum address is n-byte aligned. When a memory access is not aligned, it is said to be misaligned. Note that by definition byte memory accesses are always aligned.

A memory pointer that refers to primitive data that is n bytes long is said to be aligned if it is only allowed to contain addresses that are n-byte aligned, otherwise it is said to be unaligned. A memory pointer that refers to a data aggregate (a data structure or array) is aligned if (and only if) each primitive datum in the aggregate is aligned.

Note that the definitions above assume that each primitive datum is a power of two bytes long. When this is not the case (as with 80-bit floating-point on x86) the context influences the conditions where the datum is considered aligned or not.

Data structures can be stored in memory on the stack with a static size known as bounded or on the heap with a dynamic size known as unbounded.

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A char (one byte) will be 1-byte aligned.

A short (two bytes) will be 2-byte aligned.

An int (four bytes) will be 4-byte aligned.

A long (four bytes) will be 4-byte aligned.

A float (four bytes) will be 4-byte aligned.

A double (eight bytes) will be 8-byte aligned on Windows and 4-byte aligned on Linux (8-byte with -malign-double compile time option).

A long double (ten bytes with C++Builder and DMC, eight bytes with Visual C++, twelve bytes with GCC) will be 8-byte aligned with C++Builder, 2-byte aligned with DMC, 8-byte aligned with Visual C++ and 4-byte aligned with GCC.

Any pointer (four bytes) will be 4-byte aligned. (e.g.: char*, int*)


The only notable difference in alignment for a 64-bit system when compared to a 32-bit system is:

A long (eight bytes) will be 8-byte aligned.

A double (eight bytes) will be 8-byte aligned.

A long double (eight bytes with Visual C++, sixteen bytes with GCC) will be 8-byte aligned with Visual C++ and 16-byte aligned with GCC.

Any pointer (eight bytes) will be 8-byte aligned.

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#include <stdlib.h>
double *foo(void) 
{
   double *var;//create array of size 10
   int     ok;
 
   ok = posix_memalign((void**)&var, 64, 10*sizeof(double));
 
   if(ok != 0)
     return NULL;
 
   return var;
}

syedshan said:

Can you explain the meaning of this statement used in a sample program I've been trying to understand

Code:

dma_buffer_in = (char *) _alligned_malloc(_MEMORY_SIZE, 4096);

//where this function is used to allocate memory on a specified memory boundary,
// #define _MEMORY_SIZE 4096
// char * dma_buffer_in;//buffer used for DMA transaction

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