Is malloc 16 byte aligned?
malloc is required by the C standard to return a pointer that’s properly aligned for any data type. glibc malloc on x86-64 returns 16-byte-aligned pointers. Alignment requirements specify what address offsets can be assigned to what types. malloc is required to return an address suitable for any alignment requirement.
How do you align 16 bytes?
Each byte is 8 bits, so to align on a 16 byte boundary, you need to align to each set of two bytes. Similarly, memory aligned on a 32 bit (4 byte) boundary would have a memory address that’s a multiple of four, because you group four bytes together to form a 32 bit word.
Why is stack 16 byte aligned?
The compiler is maintaining a 16-byte alignment of the stack pointer when a function is called, adding padding to the stack as necessary. The compiler knows that the stack will always be aligned correctly, so it can emit instructions with alignment requirements without risk of triggering their fault conditions.
What does 4-byte aligned mean?
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. Data alignment is the aligning of elements according to their natural alignment.
What does it mean to align the stack?
That’s because the four bytes are stored so that the 101001 leading bits are significant. (The processor really accesses these through a “don’t care” by fetching 101001XX.) So to align something in memory means to rearrange data (usually through padding) so that the desired item’s address will have enough zero bytes.
What is byte alignment?
In this context, a byte is the smallest unit of memory access, i.e. each memory address specifies a different byte. A memory access is said to be aligned when the data 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.
What is aligned memory?
Alignment refers to the arrangement of data in memory, and specifically deals with the issue of accessing data as proper units of information from main memory. Example: A 32bit memory that is byte addressable. Each row denotes a location with a fixed size of eight bits (1byte) labeled zero through seven.
What’s the maximum alignment size for GCC compiler?
(For some linkers, the maximum supported alignment may be very very small.) If your linker is only able to align variables up to a maximum of 8 byte alignment, then specifying aligned (16) in an __attribute__ will still only provide you with 8 byte alignment.
When to leave out the alignment factor in GCC?
Whenever you leave out the alignment factor in an aligned attribute specification, the compiler automatically sets the alignment for the declared variable or field to the largest alignment which is ever used for any data type on the target machine you are compiling for.
Why is the stack aligned 16 byte in x86-64?
The x86-64 CPU actually tries to keep the stack 16-byte aligned, e.g., it’ll do so when an IRQ comes in. So the reason it doesn’t work in the kernel mostly comes down to the fact that the struct pt_regs which lives near the top of the stack is 168 bytes which is not a multiple of 16.
How to set addressability of an object in GCC?
Use this attribute on the M32R/D to set the addressability of an object. The identifier model-name is one of small, medium , or large, representing each of the code models. Small model objects live in the lower 16MB of memory (so that their addresses can be loaded with the ld24 instruction).