How malloc works (in assembly level)

Question

Rajanikant Hawaldar · Answer

There are some syscalls involved for enlarging the heap, beside those the whole allocation/deallocation system runs in user mode. There is nothing special assemblywise There are a bunch of different ways for your code to legally claim some memory, including: 1) Call C++ "new" operator, like "int *p=new int[10];". But "new" isn't easy to call from assembly language, because it's a compiler-builtin operator, not just a function. 2) Call the C function "malloc", which takes a byte count and returns a pointer, like "int *p=(int *)malloc(40);". This is a perfectly ordinary function, so it's easy to call from assembly language. 3) Allocate space on the stack It's a pretty straightforward function: pass the number of *BYTES* you want as the only parameter, in rdi. "call malloc." You'll get back a pointer to the allocated bytes returned in rax. To clean up the space afterwards, copy the pointer over to rdi, and "call free" (I'm leaving off the free below, because you need the stack to do that properly). Here's a complete example of assembly memory access. I call malloc to get 40 bytes of space. malloc returns the starting address of this space in rax (the 64-bit version of eax). That is, the rax register is acting like a pointer. I can then read and write from the pointed-to memory using the usual assembly bracket syntax: mov edi, 40; malloc's first (and only) parameter: number of bytes to allocate extern malloc call malloc ; on return , rax points to our newly-allocated memory mov ecx,7; set up a constant mov [rax],ecx; write it into memory mov edx,[rax]; read it back from memory mov eax,edx; copy into return value register ret Rather than copy via the ecx register, you can specify you want a 32-bit memory write and read using "DWORD" in front of the brackets, like this: mov edi, 40; malloc's first (and only) parameter: number of bytes to allocate extern malloc call malloc ; on return , rax points to our newly-allocated memory mov DWORD [rax],7; write constant into memory mov eax, DWORD [rax]; read it back from memory ret

Suri · Answer

Thanks