Let's understand: setjmp()/longjmp()

Feb 9, 2016 · 2039 words · 10 minute read assembly linux lowlevel

Pretty recently I learned about setjmp() and longjmp(). They’re a neat pair of libc functions which allow you to save your program’s current execution context and resume it at an arbitrary point in the future (with some caveats1). If you’re wondering why this is particularly useful, to quote the manpage, one of their main use cases is “…for dealing with errors and interrupts encountered in a low-level subroutine of a program.” These functions can be used for more sophisticated error handling than simple error code return values.

I was curious how these functions worked, so I decided to take a look at musl libc’s implementation for x86. First, I’ll explain their interfaces and show an example usage program. Next, since this post isn’t aimed at the assembly wizard, I’ll cover some basics of x86 and Linux calling convention to provide some required background knowledge. Lastly, I’ll walk through the source, line by line.


int setjmp(jmp_buf env);

setjmp() takes a single jmp_buf opaque type, returns 0, and continues execution afterward normally. A jmp_buf is the structure that setjmp() will save the calling execution context in. We’ll examine it more closely later on.

void longjmp(jmp_buf env, int val);

longjmp() takes a jmp_buf and an int, simply returning back the given int value (unless it was 0, in which case it returns 1). The unusual aspect is that when it returns, the program’s execution resumes as if setjmp() had just been called. This allows the user to jump back an arbitrary amount of frames on the current call stack (presumably out of some deep routine which had an error). The return value allows the code following the setjmp() call to differentiate if setjmp() or longjmp() had just been called, and proceed accordingly.

Here’s a simple example.

#include <setjmp.h>
#include <stdio.h>

void fancy_func(jmp_buf env);

int main() {
    jmp_buf env;
    int ret = setjmp(env);
    if (ret == 0) {
        puts("just returning from setjmp!");
    } else {
        puts("now returning from longjmp and exiting!");


void fancy_func(jmp_buf env) {
    puts("doing fancy stuff");
    longjmp(env, 1);


$ ./main
just returning from setjmp!
doing fancy stuff
now returning from longjmp and exiting!

The above code creates a jmp_buf and calls setjmp(), saving the current execution context. Since setjmp() returns 0, the code follows the first branch, calling fancy_func() and forwarding on the jmp_buf. fancy_func() does some fancy stuff, then calls longjmp(), passing in the jmp_buf and 1. Execution returns to the if statement on line 9, except this time, ret is 1 instead of 0, because we’re returning from longjmp(). Now the code follows the else path which prints and exits. 2

Background Knowledge

I’ve mentioned “execution context” a few times, but let’s make that a little more concrete. In this case, a program’s execution context can be defined by the state of the processor’s registers.

On x86, the relevant registers are the general purpose, index, and pointer registers.

General Purpose: eax, ebx, ecx, edx
Index:           esi, edi
Pointer:         ebp, esp, eip

ebx, ecx, edx, esi, and edi don’t have particularly special meaning here and can be thought of as arbitrary 32 bit storage locations. However eax, ebp, and eip are a little different.

  • eax is used for function return values (specified by the cdecl calling convention)
  • ebp, the frame pointer, contains a pointer to the start of the current stack frame.
  • eip, the instruction pointer, contains a pointer to the next instruction to execute.

With this in mind, I initially thought that a jmp_buf would be an array of 9 ints or something, in order to hold each register.

As it happens, jmp_buf is instead declared as (link):

typedef struct __jmp_buf_tag {
    __jmp_buf __jb;
    unsigned long __fl;
    unsigned long __ss[128/sizeof(long)];
} jmp_buf[1];

And for x86, __jmp_buf is declared as (link):

typedef unsigned long __jmp_buf[6];

I had never seen this syntax of using bracket operators at the end of a typedef but searched and found out that the __jmp_buf declaration declares a fixed size array of 6 unsigned longs, and the jmp_buf declaration declares an array of 1 struct __jmp_buf_tag. The reason for the array of 1 is so the pointer semantics of arrays kick in and the struct __jmp_buf_tag is actually passed by reference in calls to setjmp()/longjmp() (as opposed to being copied).

Anyway, apparently my guess of 9 ints was incorrect, and it’s actually 6 (longs).

Before we can dig into the source to understand why this is, we need to understand what the state of the program stack is at the point setjmp() is called, and to do that, we need to understand which calling convention is being used. Since we assume x86 Linux, this will be cdecl. The relevant parts of cdecl for this case are:

  • arguments passed on the stack
  • integer values and memory addresses returned in eax (as mentioned above)
  • eax, ecx, edx are caller saved, the rest are callee saved

setjmp()’s code executes immediately after setjmp()’s call instruction, so at the point the first instruction of setjmp() executes, the stack looks something like this.

> high memory <
| ...                       |
| caller's caller saved eip |
| caller's caller saved ebp | < ebp
| caller stack var 1        | // caller's stack frame
| caller stack var 2        |
| caller stack var ...      |
| caller stack var n        |
| pointer to jmp_buf        | // argument to setjmp
| caller saved eip          | < esp
+---------------------------+ // setjmp's stack frame
> low memory <

(In this illustration, the stack grows down.)

At the top of the stack is the eip value that the call instruction pushed, or where to return to after setjmp() finishes. Above that is the first, and only argument, the pointer to the given jmp_buf. Lastly, above that is the caller’s stack frame. esp points to the top of the stack as usual, and ebp is still pointing to the start of the caller’s stack frame. Usually the first thing a function does is push ebp on the stack, and set ebp to esp to now point to the current stack frame (a.k.a the prologue), but since setjmp() is such a minimal function, it doesn’t do this. Furthermore, since ebp is one of the registers that needs to be saved, setjmp() needs it to be unperturbed.

After setjmp() returns, the stack will look something like this:

> high memory <
| ...                       |
| caller's caller saved eip |
| caller's caller saved ebp | < ebp
| caller stack var 1        | // caller's stack frame
| caller stack var 2        |
| caller stack var ...      |
| caller stack var n        |
| pointer to jmp_buf        | < esp
> low memory <

It’s nearly identical, except eip has been popped off the stack, and is now executing the next instruction after the caller’s call setjmp. esp has also been updated accordingly. This is the state of the program that setjmp() will need to record, and that longjmp() will restore.

Before reading the source I tried to reason about what I expected would happen. I presume:

  • General purpose and index registers (eax, ebx, ecx, edx, esi, edi) which don’t have any effect on control flow can be trivially saved and restored
  • ebp can similarly be saved “as is”, since its value when setjmp() executes is exactly what it needs to be restored to in longjmp()
  • esp cannot be saved “as is” because when setjmp() executes, there is the extra eip on the stack that is not there after the function returns. Therefore, the value for esp that should be saved is esp+4 to match the expected state of the stack after return
  • The eip that should be saved is the address of the instruction after the call setjmp instruction, which can be retrieved from the top of the stack by dereferencing esp

With all that out of the way, let’s read the source (all annotations by me) (link). Since this type of low level register manipulation isn’t available from C (modulo compiler intrinsics), both of these functions are necessarily written in assembly.

    mov 4(%esp), %eax     ; get pointer to jmp_buf, passed as argument on stack
    mov    %ebx, (%eax)   ; jmp_buf[0] = ebx
    mov    %esi, 4(%eax)  ; jmp_buf[1] = esi
    mov    %edi, 8(%eax)  ; jmp_buf[2] = edi
    mov    %ebp, 12(%eax) ; jmp_buf[3] = ebp
    lea 4(%esp), %ecx     ; get previous value of esp, before call
    mov    %ecx, 16(%eax) ; jmp_buf[4] = esp before call
    mov  (%esp), %ecx     ; get saved caller eip from top of stack
    mov    %ecx, 20(%eax) ; jmp_buf[5] = saved eip
    xor    %eax, %eax     ; eax = 0
    ret                   ; pop stack into eip

The first line retrieves the argument off the stack, placing a pointer to the jmp_buf (remember, an array of 6 unsigned longs) in eax. It then moves ebx, esi, edi, and ebp “as is” into the int array. It adds 4 to esp with a lea and stores that next. Next, it dereferences esp and stores that in the last slot in the array. Lastly, it zeroes out eax and returns.

The final state of the jmp_buf after setjmp() returns looks like:

   0    1    2    3    4    5
[ ebx, esi, edi, ebp, esp, eip ]

Now let’s look at longjmp() (link).

    mov  4(%esp),%edx ; get pointer to jmp_buf, passed as argument 1 on stack
    mov  8(%esp),%eax ; get int val in eax, passed as argument 2 on stack
    test    %eax,%eax ; is int val == 0?
    jnz 1f
    inc     %eax      ; if so, eax++
    mov   (%edx),%ebx ; ebx = jmp_buf[0]
    mov  4(%edx),%esi ; esi = jmp_buf[1]
    mov  8(%edx),%edi ; edi = jmp_buf[2]
    mov 12(%edx),%ebp ; ebp = jmp_buf[3]
    mov 16(%edx),%ecx ; ecx = jmp_buf[4]
    mov     %ecx,%esp ; esp = ecx
    mov 20(%edx),%ecx ; ecx = jmp_buf[5]
    jmp *%ecx         ; eip = ecx

The first two lines retrieve the arguments (pointer to jmp_buf, int return val) from the stack into edx and eax, respectively. The int val is incremented to 1 if it is 0, according to the spec. Next, ebx, esi, edi, and ebp are reset to their saved state, stored in the jmp_buf, in a straightforward manner. As you can see, both setjmp() and longjmp() need to precisely agree on where each particular register is saved in the jmp_buf. esp is mysteriously restored in an indirect manner, via ecx3 and finally, eip is reset to the saved state via an indirect jump.

So I was mostly correct, but it seems like eax, ecx, and edx were not saved in the jmp_buf! If we look back on the details of cdecl, it becomes clear why.

  • eax doesn’t need to be saved, because it is reserved for the return value
  • ecx and edx are caller saved. This means that a callee subroutine is free to trash these registers, and it is the responsibility for the caller to save and restore them after the subroutine returns. Because of this, if the function that calls setjmp() needs to use ecx or edx after the call, it will already have code to save and restore those registers before and after the function call. Since longjmp() resumes execution as if setjmp() had immediately returned, execution will automatically hit the code that restores ecx and edx, making it unnecessary to save them in the jmp_buf.

This is just one example of how great musl libc is at providing an understandable resource for learning the internals of systems software. I often find myself referencing it when I’m curious about libc internals, and if you’re not familiar with it, I highly recommend checking it out!

  1. Mainly, that the function which called setjmp() cannot have returned before the corresponding longjmp() call. [return]
  2. A slight aside: This situation where a function returns different values based on the execution context also appears in fork(), in which the caller uses the return value to differentiate whether it is now executing in the child or parent process.
  3. I actually have no good explanation for this and am pretty curious why it’s done this way. Something like mov $esp, [$edx+16] is a perfectly valid instruction (tested with rasm2)! I asked on the musl mailing list, but no one responded :(. If you have an explanation, please let me know! [return]
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