带有全局变量的简单 ebpf 代码：分配失败

Question

我有一个简单的 ebpf 测试代码，其中很少有全局变量（据我所知，全局变量是通过我的版本（5.13 Linux）中的映射机制访问的）。这是代码。

int global_var = 911;
int * global_ptr = &global_var;

int __attribute__ ((noinline)) assign_value_via_var(void) {
    global_var = 119;
    return global_var;
}
int __attribute__ ((noinline)) assign_value_via_ptr1(void) {
    global_var = 118; 
    return *global_ptr;
}
int __attribute__ ((noinline)) assign_value_via_ptr2(void) {
    *global_ptr = 912; 
    return *global_ptr;
}
SEC("tracepoint/syscalls/sys_enter_execve")
int tracepoint__syscalls__sys_enter_execve_(struct trace_event_raw_sys_enter *ctx)
{
    bpf_printk("--before assignment-- global_var = %d \n",global_var);
    assign_value_via_var();
    bpf_printk("--after assignment-- global_var = %d \n",global_var);
    assign_value_via_ptr1();
    bpf_printk("--after  assignment-- global_var = %d \n",global_var);
    assign_value_via_ptr2();test
    bpf_printk("--after  assignment-- global_var = %d \n",global_var);
    return 0;
}

上述代码触发验证器在 allocate_value_via_ptr1 和 allocate_value_via_ptr2 函数中发出“R1 invalid mem access 'inv'”错误。以下是 allocate_value_via_ptr1 的验证程序日志：

Validating assign_value_via_ptr1() func#2...
; global_var = 118;
32: (18) r1 = 0xffffc181c0224000
34: (b4) w2 = 118
35: (63) *(u32 *)(r1 +0) = r2
 R1_w=map_value(id=0,off=0,ks=4,vs=5040,imm=0) R2_w=invP118 R10=fp0
; return *global_ptr;
36: (18) r1 = 0xffffc181c0224008
38: (79) r1 = *(u64 *)(r1 +0)
 R1_w=map_value(id=0,off=8,ks=4,vs=5040,imm=0) R2_w=invP118 R10=fp0
; return *global_ptr;
39: (61) r0 = *(u32 *)(r1 +0)
R1 invalid mem access 'inv'

以下是 allocate_value_via_ptr2 的验证程序日志：

Validating assign_value_via_ptr2() func#2...
; *global_ptr = 912;
32: (18) r1 = 0xffffc181c0224008
34: (79) r1 = *(u64 *)(r1 +0)
 R1_w=map_value(id=0,off=8,ks=4,vs=5040,imm=0) R10=fp0
35: (b4) w2 = 912
; *global_ptr = 912;
36: (63) *(u32 *)(r1 +0) = r2
R1 invalid mem access 'inv'

如果我将上面的两个函数更改为以下内容。一切开始运转良好。

int __attribute__ ((noinline)) assign_value_via_ptr1(void) {
    return *global_ptr;
}
int __attribute__ ((noinline)) assign_value_via_ptr2(void) {
    return *global_ptr;
}

谁能帮忙解释一下背后的原理？谢谢。

仅供参考，以下是这些函数的字节码：

0000000000000000 assign_value_via_var:
;       global_var = 119;
       0:       18 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 r1 = 0 ll
                0000000000000000:  R_BPF_64_64  global_var
       2:       b4 02 00 00 77 00 00 00 w2 = 119
       3:       63 21 00 00 00 00 00 00 *(u32 *)(r1 + 0) = r2
;       return global_var;
       4:       b4 00 00 00 77 00 00 00 w0 = 119
       5:       95 00 00 00 00 00 00 00 exit

0000000000000030 assign_value_via_ptr1:
;       global_var = 118;
       6:       18 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 r1 = 0 ll
                0000000000000030:  R_BPF_64_64  global_var
       8:       b4 02 00 00 76 00 00 00 w2 = 118
       9:       63 21 00 00 00 00 00 00 *(u32 *)(r1 + 0) = r2
;       return *global_ptr;
      10:       18 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 r1 = 0 ll
                0000000000000050:  R_BPF_64_64  global_ptr
      12:       79 11 00 00 00 00 00 00 r1 = *(u64 *)(r1 + 0)
      13:       61 10 00 00 00 00 00 00 r0 = *(u32 *)(r1 + 0)
      14:       95 00 00 00 00 00 00 00 exit

0000000000000078 assign_value_via_ptr2:
;       *global_ptr = 912;
      15:       18 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 r1 = 0 ll
                0000000000000078:  R_BPF_64_64  global_ptr
      17:       79 11 00 00 00 00 00 00 r1 = *(u64 *)(r1 + 0)
      18:       b4 02 00 00 90 03 00 00 w2 = 912
      19:       63 21 00 00 00 00 00 00 *(u32 *)(r1 + 0) = r2
;       return *global_ptr;
      20:       b4 00 00 00 90 03 00 00 w0 = 912
      21:       95 00 00 00 00 00 00 00 exit