What happens if you use the 32-bit int 0x80 Linux

Question

int 0x80 on Linux always invokes the 32-bit ABI, regardless of what mode it's called from: args in ebx, ecx, ... and syscall numbers from /usr/include/asm/unistd_32.h. (Or crashes on 64-bit kernels compiled without CONFIG_IA32_EMULATION).

64-bit code should use syscall, with call numbers from /usr/include/asm/unistd_64.h, and args in rdi, rsi, etc. See What are the calling conventions for UNIX & Linux system calls on i386 and x86-64. If your question was marked a duplicate of this, see that link for details on how you should make system calls in 32 or 64-bit code. If you want to understand what exactly happened, keep reading.

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syscall system calls are faster than int 0x80 system calls, so use native 64-bit syscall unless you're writing polyglot machine code that runs the same when executed as 32 or 64 bit. (sysenter always returns in 32-bit mode, so it's not useful from 64-bit userspace, although it is a valid x86-64 instruction.)

Related: The Definitive Guide to Linux System Calls (on x86) for how to make int 0x80 or sysenter 32-bit system calls, or syscall 64-bit system calls, or calling the vDSO for "virtual" system calls like gettimeofday. Plus background on what system calls are all about.

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Using int 0x80 makes it possible to write something that will assemble in 32 or 64-bit mode, so it's handy for an exit_group() at the end of a microbenchmark or something.

Current PDFs of the official i386 and x86-64 System V psABI documents that standardize function and syscall calling conventions are linked from https://github.com/hjl-tools/x86-psABI/wiki/X86-psABI.

See the x86 tag wiki for beginner guides, x86 manuals, official documentation, and performance optimization guides / resources.

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But since people keep posting questions with code that uses int 0x80 in 64-bit code, or accidentally building 64-bit binaries from source written for 32-bit, I wonder what exactly does happen on current Linux?

Does int 0x80 save/restore all the 64-bit registers? Does it truncate any registers to 32-bit? What happens if you pass pointer args that have non-zero upper halves?

Does it work if you pass it 32-bit pointers?

Answer 1

TL:DR: int 0x80 works when used correctly, as long as any pointers fit in 32 bits (stack pointers don't fit). Also, strace decodes it wrong, decoding register contents as if it was the 64-bit syscall ABI.

int 0x80 zeros r8-r11, and preserves everything else. Use it exactly like you would in 32-bit code, with the 32-bit call numbers. (Or better, don't use it!)

Not all systems even support int 0x80: The Windows Ubuntu subsystem is strictly 64-bit only: int 0x80 doesn't work at all. It's also possible to build Linux kernels without IA-32 emulation either. (No support for 32-bit executables, no support for 32-bit system calls).

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The details: what's saved/restored, which parts of which regs the kernel uses

int 0x80 uses eax (not the full rax) as the system-call number, dispatching to the same table of function-pointers that 32-bit user-space int 0x80 uses. (These pointers are to sys_whatever implementations or wrappers for the native 64-bit implementation inside the kernel. System calls are really function calls across the user/kernel boundary.)

Only the low 32 bits of arg registers are passed. The upper halves of rbx-rbp are preserved, but ignored by int 0x80 system calls. Note that passing a bad pointer to a system call doesn't result in SIGSEGV; instead the system call returns -EFAULT. If you don't check error return values (with a debugger or tracing tool), it will appear to silently fail.

All registers (except eax of course) are saved/restored (including RFLAGS, and the upper 32 of integer regs), except that r8-r11 are zeroed. r12-r15 are call-preserved in the x86-64 SysV ABI's function calling convention, so the registers that get zeroed by int 0x80 in 64-bit are the call-clobbered subset of the "new" registers that AMD64 added.

This behaviour has been preserved over some internal changes to how register-saving was implemented inside the kernel, and comments in the kernel mention that it's usable from 64-bit, so this ABI is probably stable. (I.e. you can count on r8-r11 being zeroed, and everything else being preserved.)

The return value is sign-extended to fill 64-bit rax. (Linux declares 32-bit sys_ functions as returning signed long.) This means that pointer return values (like from void *mmap()) need to be zero-extended before use in 64-bit addressing modes

Unlike sysenter, it preserves the original value of cs, so it returns to user-space in the same mode that it was called in. (Using sysenter results in the kernel setting cs to $__USER32_CS, which selects a descriptor for a 32-bit code segment.)

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strace decodes int 0x80 incorrectly for 64-bit processes. It decodes as if the process had used syscall instead of int 0x80. This can be very confusing. e.g. since strace prints write(0, NULL, 12 <unfinished ... exit status 1> for eax=1 / int $0x80, which is actually _exit(ebx), not write(rdi, rsi, rdx).

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int 0x80 works as long as all arguments (including pointers) fit in the low 32 of a register. This is the case for static code and data in the default code model ("small") in the x86-64 SysV ABI. (Section 3.5.1 : all symbols are known to be located in the virtual addresses in the range 0x00000000 to 0x7effffff, so you can do stuff like mov edi, hello (AT&T mov $hello, %edi) to get a pointer into a register with a 5 byte instruction).

But this is not the case for position-independent executables, which many Linux distros now configure gcc to make by default (and they enable ASLR for executables). For example, I compiled a hello.c on Arch Linux, and set a breakpoint at the start of main. The string constant passed to puts was at 0x555555554724, so a 32-bit ABI write system call would not work. (GDB disables ASLR by default, so you always see the same address from run to run, if you run from within GDB.)

Linux puts the stack near the "gap" between the upper and lower ranges of canonical addresses, i.e. with the top of the stack at 2^48-1. (Or somewhere random, with ASLR enabled). So rsp on entry to _start in a typical statically-linked executable is something like 0x7fffffffe550, depending on size of env vars and args. Truncating this pointer to esp does not point to any valid memory, so system calls with pointer inputs will typically return -EFAULT if you try to pass a truncated stack pointer. (And your program will crash if you truncate rsp to esp and then do anything with the stack, e.g. if you built 32-bit asm source as a 64-bit executable.)

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How it works in the kernel:

In the Linux source code, arch/x86/entry/entry_64_compat.S defines ENTRY(entry_INT80_compat). Both 32 and 64-bit processes use the same entry point when they execute int 0x80.

entry_64.S is defines native entry points for a 64-bit kernel, which includes interrupt / fault handlers and syscall native system calls from long mode (aka 64-bit mode) processes.

entry_64_compat.S defines system-call entry-points from compat mode into a 64-bit kernel, plus the special case of int 0x80 in a 64-bit process. (sysenter in a 64-bit process may go to that entry point as well, but it pushes $__USER32_CS, so it will always return in 32-bit mode.) There's a 32-bit version of the syscall instruction, supported on AMD CPUs, and Linux supports it too for fast 32-bit system calls from 32-bit processes.

I guess a possible use-case for int 0x80 in 64-bit mode is if you wanted to use a custom code-segment descriptor that you installed with modify_ldt. int 0x80 pushes segment registers itself for use with iret, and Linux always returns from int 0x80 system calls via iret. The 64-bit syscall entry point sets pt_regs->cs and ->ss to constants, __USER_CS and __USER_DS. (It's normal that SS and DS use the same segment descriptors. Permission differences are done with paging, not segmentation.)

entry_32.S defines entry points into a 32-bit kernel, and is not involved at all.

The int 0x80 entry point in Linux 4.12's entry_64_compat.S:

/*
 * 32-bit legacy system call entry.
 *
 * 32-bit x86 Linux system calls traditionally used the INT $0x80
 * instruction.  INT $0x80 lands here.
 *
 * This entry point can be used by 32-bit and 64-bit programs to perform
 * 32-bit system calls.  Instances of INT $0x80 can be found inline in
 * various programs and libraries.  It is also used by the vDSO's
 * __kernel_vsyscall fallback for hardware that doesn't support a faster
 * entry method.  Restarted 32-bit system calls also fall back to INT
 * $0x80 regardless of what instruction was originally used to do the
 * system call.
 *
 * This is considered a slow path.  It is not used by most libc
 * implementations on modern hardware except during process startup.
 ...
 */
 ENTRY(entry_INT80_compat)
 ...  (see the github URL for the full source)

The code zero-extends eax into rax, then pushes all the registers onto the kernel stack to form a struct pt_regs. This is where it will restore from when the system call returns. It's in a standard layout for saved user-space registers (for any entry point), so ptrace from other process (like gdb or strace) will read and/or write that memory if they use ptrace while this process is inside a system call. (ptrace modification of registers is one thing that makes return paths complicated for the other entry points. See comments.)

But it pushes $0 instead of r8/r9/r10/r11. (sysenter and AMD syscall32 entry points store zeros for r8-r15.)

I think this zeroing of r8-r11 is to match historical behaviour. Before the Set up full pt_regs for all compat syscalls commit, the entry point only saved the C call-clobbered registers. It dispatched directly from asm with call *ia32_sys_call_table(, %rax, 8), and those functions follow the calling convention, so they preserve rbx, rbp, rsp, and r12-r15. Zeroing r8-r11 instead of leaving them undefined was probably a way to avoid info-leaks from the kernel. IDK how it handled ptrace if the only copy of user-space's call-preserved registers was on the kernel stack where a C function saved them. I doubt it used stack-unwinding metadata to find them there.

The current implementation (Linux 4.12) dispatches 32-bit-ABI system calls from C, reloading the saved ebx, ecx, etc. from pt_regs. (64-bit native system calls dispatch directly from asm, with only a mov %r10, %rcx needed to account for the small difference in calling convention between functions and syscall. Unfortunately it can't always use sysret, because CPU bugs make it unsafe with non-canonical addresses. It does try to, so the fast-path is pretty damn fast, although syscall itself still takes tens of cycles.)

Anyway, in current Linux, 32-bit syscalls (including int 0x80 from 64-bit) eventually end up indo_syscall_32_irqs_on(struct pt_regs *regs). It dispatches to a function pointer ia32_sys_call_table, with 6 zero-extended args. This maybe avoids needing a wrapper around the 64-bit native syscall function in more cases to preserve that behaviour, so more of the ia32 table entries can be the native system call implementation directly.

Linux 4.12 arch/x86/entry/common.c

if (likely(nr < IA32_NR_syscalls)) {
  /*
   * It's possible that a 32-bit syscall implementation
   * takes a 64-bit parameter but nonetheless assumes that
   * the high bits are zero.  Make sure we zero-extend all
   * of the args.
   */
  regs->ax = ia32_sys_call_table[nr](
      (unsigned int)regs->bx, (unsigned int)regs->cx,
      (unsigned int)regs->dx, (unsigned int)regs->si,
      (unsigned int)regs->di, (unsigned int)regs->bp);
}

syscall_return_slowpath(regs);

In older versions of Linux that dispatch 32-bit system calls from asm (like 64-bit still does), the int80 entry point itself puts args in the right registers with mov and xchg instructions, using 32-bit registers. It even uses mov %edx,%edx to zero-extend EDX into RDX (because arg3 happen to use the same register in both conventions). code here. This code is duplicated in the sysenter and syscall32 entry points.

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Simple example / test program:

I wrote a simple Hello World (in NASM syntax) which sets all registers to have non-zero upper halves, then makes two write() system calls with int 0x80, one with a pointer to a string in .rodata (succeeds), the second with a pointer to the stack (fails with -EFAULT).

Then it uses the native 64-bit syscall ABI to write() the chars from the stack (64-bit pointer), and again to exit.

So all of these examples are using the ABIs correctly, except for the 2nd int 0x80 which tries to pass a 64-bit pointer and has it truncated.

If you built it as a position-independent executable, the first one would fail too. (You'd have to use a RIP-relative lea instead of mov to get the address of hello: into a register.)

I used gdb, but use whatever debugger you prefer. Use one that highlights changed registers since the last single-step. gdbgui works well for debugging asm source, but is not great for disassembly. Still, it does have a register pane that works well for integer regs at least, and it worked great on this example.

See the inline ;;; comments describing how register are changed by system calls

global _start
_start:
    mov  rax, 0x123456789abcdef
    mov  rbx, rax
    mov  rcx, rax
    mov  rdx, rax
    mov  rsi, rax
    mov  rdi, rax
    mov  rbp, rax
    mov  r8, rax
    mov  r9, rax
    mov  r10, rax
    mov  r11, rax
    mov  r12, rax
    mov  r13, rax
    mov  r14, rax
    mov  r15, rax

    ;; 32-bit ABI
    mov  rax, 0xffffffff00000004          ; high garbage + __NR_write (unistd_32.h)
    mov  rbx, 0xffffffff00000001          ; high garbage + fd=1
    mov  rcx, 0xffffffff00000000 + .hello
    mov  rdx, 0xffffffff00000000 + .hellolen
    ;std
after_setup:       ; set a breakpoint here
    int  0x80                   ; write(1, hello, hellolen);   32-bit ABI
    ;; succeeds, writing to stdout
;;; changes to registers:   r8-r11 = 0.  rax=14 = return value

    ; ebx still = 1 = STDOUT_FILENO
    push 'bye' + (0xa<<(3*8))
    mov  rcx, rsp               ; rcx = 64-bit pointer that won't work if truncated
    mov  edx, 4
    mov  eax, 4                 ; __NR_write (unistd_32.h)
    int  0x80                   ; write(ebx=1, ecx=truncated pointer,  edx=4);  32-bit
    ;; fails, nothing printed
;;; changes to registers: rax=-14 = -EFAULT  (from /usr/include/asm-generic/errno-base.h)

    mov  r10, rax               ; save return value as exit status
    mov  r8, r15
    mov  r9, r15
    mov  r11, r15               ; make these regs non-zero again

    ;; 64-bit ABI
    mov  eax, 1                 ; __NR_write (unistd_64.h)
    mov  edi, 1
    mov  rsi, rsp
    mov  edx, 4
    syscall                     ; write(edi=1, rsi='bye\n' on the stack,  rdx=4);  64-bit
    ;; succeeds: writes to stdout and returns 4 in rax
;;; changes to registers: rax=4 = length return value
;;; rcx = 0x400112 = RIP.   r11 = 0x302 = eflags with an extra bit set.
;;; (This is not a coincidence, it's how sysret works.  But don't depend on it, since iret could leave something else)

    mov  edi, r10d
    ;xor  edi,edi
    mov  eax, 60                ; __NR_exit (unistd_64.h)
    syscall                     ; _exit(edi = first int 0x80 result);  64-bit
    ;; succeeds, exit status = low byte of first int 0x80 result = 14

section .rodata
_start.hello:    db "Hello World!", 0xa, 0
_start.hellolen  equ   $ - _start.hello

Build it into a 64-bit static binary with

yasm -felf64 -Worphan-labels -gdwarf2 abi32-from-64.asm
ld -o abi32-from-64 abi32-from-64.o

Run gdb ./abi32-from-64. In gdb, run set disassembly-flavor intel and layout reg if you don't have that in your ~/.gdbinit already. (GAS .intel_syntax is like MASM, not NASM, but they're close enough that it's easy to read if you like NASM syntax.)

(gdb)  set disassembly-flavor intel
(gdb)  layout reg
(gdb)  b  after_setup
(gdb)  r
(gdb)  si                     # step instruction
    press return to repeat the last command, keep stepping

Press control-L when gdb's TUI mode gets messed up. This happens easily, even when programs don't print to stdout themselves.