C++ x86, amd64： 为什么 SIGTRAP' ucontext 指令指针没有指向相关的 int3

正如标题所说，ucontext_t中的-rip并不是指int3提高了SIGTRAP.相反，它指向下一条指令.

这与我的(天真的)期望不同，即在信号处理程序返回时将重试每个出错的指令(如果上下文没有更改，并且进程在处理程序中没有显式终止).

另一方面，当获得SIGILL个上下文点时，则指向错误的指令.同样关于ARM和Aarch64-SIGTRAP的上下文也指向了相关的bkpt #0/bpt #0.

测试程序:

/* sigtest.c */ 

#define _GNU_SOURCE

#include <stdio.h>
#include <signal.h>
#include <ucontext.h>

extern void do_int3(void);
extern void do_ud2(void);

void sighandler(int signo, siginfo_t* info, void* context) {
    struct ucontext_t* uctx = context;
    /* Yes, printf() here is bad. I promise to never ever do this in real programs */
    printf("Got signal %d:\n"
           "\tsi_addr: %p\n"
           "\tcontext RIP: %p\n",
           signo,
           info->si_addr,
           (void*)uctx->uc_mcontext.gregs[REG_RIP]);
}

int main(int argc, char** argv) {
    struct sigaction sa = {};
    sa.sa_flags = SA_SIGINFO | SA_ONESHOT;
    sa.sa_sigaction = sighandler;
    sigemptyset(&sa.sa_mask);
    sigaction(SIGTRAP, &sa, NULL);
    sigaction(SIGILL, &sa, NULL);

    void (*fn)(void) = 0;

    if (argv[1][0] == 't') {
        fn = do_int3;
    }
    if (argv[1][0] == 'u') {
        fn = do_ud2;
    }

    printf("call function at %p\n", fn);
    fn();
    printf("call returned\n");
}

; do_int3.S 
    .text
    .globl  do_int3
    .type   do_int3, @function
do_int3:
    int3
    ret
    .size   do_int3, .-do_int3

; do_ud2.S 
    .text
    .globl  do_ud2
    .type   do_ud2, @function
do_ud2:
    ud2
    ret
    .size   do_ud2, .-do_ud2

编译并运行:

$ cc sigtest.c do_int3.S do_ud2.S 
$ ./a.out t
call function at 0x5620b87c6908
Got signal 5:
    si_addr: (nil)
    context RIP: 0x5620b87c6909
call returned
$ ./a.out u
call function at 0x557d1bc1590a
Got signal 4:
    si_addr: 0x557d1bc1590a
    context RIP: 0x557d1bc1590a
Illegal instruction (core dumped)

很容易注意到，对于SIGTRAP，rip值指向下一条指令，而不是int3.

为什么指令指针在SIGTRAP上下文中调整为不重试相关int3？