What does it mean to be 32 bit aligned?

Data structure alignment
Data structure alignment is the way data is arranged and accessed in computer memory. For instance, in a 32-bit architecture, the data may be aligned if the data is stored in four consecutive bytes and the first byte lies on a 4-byte boundary.

Is malloc 16 byte aligned?

On OS X / iOS all calls to malloc/calloc/etc. are always 16 byte aligned. If you needed 32 byte aligned for AVX, for example, then you can use posix_memalign: void *buf = NULL; int err = posix_memalign( &buf, 32 /*alignment*/, 1024 /*size*/); if( err ) RunInCirclesWaivingArmsWildly(); free(buf);

What does it mean to be 8 byte aligned?

An object that is “8 bytes aligned” is stored at a memory address that is a multiple of 8. Many CPUs will only load some data types from aligned locations; on other CPUs such access is just faster.

What happens if memory address is not 32 bit aligned?

Small amounts of data (say 4 bytes, for example) fit nicely in a 32-bit word if it is 4-byte aligned. If it is not aligned, it can cross a 32-bit boundary and require additional memory fetches. Modern CPUs have other optimizations as well that improve performance for address aligned data.

What does it mean when an address is 4 byte aligned?

Sequential addresses refer to sequential bytes in memory. An address that is “4-byte aligned” is a multiple of 4 bytes. In other words, the binary representation of the address ends in two zeros ( 00 ), since in binary, it’s a multiple of the binary value of 4 ( 100b ). The test for 4-byte aligned address is, therefore:

Is the base address of an array 32 byte or 32 byte?

In each case, the variable must be 32-byte aligned. In the array, the base address of the array, not each array member, is 32-byte aligned. The sizeof value for each array member is unaffected when you use __declspec (align (#)).

How to align 32-byte aligned memory in C-stack overflow?

So it would be ~0x1F. To align a pointer to a 32 byte boundary, you want the last 5 bits of the address to be 0. (Since 100000 is 32 in binary, any multiple of 32 will end in 00000.) 0x1F is 11111 in binary. Since it’s a pointer, it’s actually some number of 0’s followed by 11111 – for example, with 64-bit pointers, it would be 59 0’s and 5 1’s.