英文:
BPF tail call not called
问题
在下面的代码中，tail_prog程序没有从main_prog中调用：
#include <linux/bpf.h>#include <bpf/bpf_helpers.h>struct bpf_map_def SEC("maps") jump_table = {.type = BPF_MAP_TYPE_PROG_ARRAY,.key_size = sizeof(__u32),.value_size = sizeof(__u32),.max_entries = 8,};SEC("xdp")int main_prog(struct xdp_md *ctx) {bpf_printk("Making tail call");bpf_tail_call(ctx, &jump_table, 0);return XDP_PASS;}SEC("xdp_1")int tail_prog(struct xdp_md *ctx) {bpf_printk("Inside tail call");return XDP_PASS;}char _license[] SEC("license") = "GPL";
我观察到只有main_prog中的打印语句被打印出来。
我正在使用Cilium的eBPF Go包加载BPF程序。以下是加载程序和映射的代码：
type BpfObjects struct {MainProg *ebpf.Program `ebpf:"main_prog"`TailProg *ebpf.Program `ebpf:"tail_prog"`JumpTable *ebpf.Map `ebpf:"jump_table"`}var objects BpfObjectsspec, err := ebpf.LoadCollectionSpec("prog.o")if err != nil {log.Fatalln("ebpf.LoadCollectionSpec", err)}if err := spec.LoadAndAssign(&objects, nil); err != nil {log.Fatalln("ebpf.LoadAndAssign", err)}objects.JumpTable.Update(0, objects.TailProg.FD(), ebpf.UpdateAny)
根据这里的说明，跳转表必须从用户空间初始化，这就是我认为上面最后一行代码应该做的事情。然而，无论该行是否存在，我都没有看到任何区别。
英文:
In the following code BPF program tail_prog is not getting tail called from main_prog:
#include <linux/bpf.h>#include <bpf/bpf_helpers.h>struct bpf_map_def SEC("maps") jump_table = {.type = BPF_MAP_TYPE_PROG_ARRAY,.key_size = sizeof(__u32),.value_size = sizeof(__u32),.max_entries = 8,};SEC("xdp")int main_prog(struct xdp_md *ctx) {bpf_printk("Making tail call");bpf_tail_call(ctx, &jump_table, 0);return XDP_PASS;}SEC("xdp_1")int tail_prog(struct xdp_md *ctx) {bpf_printk("Inside tail call");return XDP_PASS;}char _license[] SEC("license") = "GPL";
I observe only the print in main_prog is printed.
I'm loading the BPF programs using Cilium's eBPF Go package. Here's the code for loading the programs and maps:
type BpfObjects struct {MainProg *ebpf.Program `ebpf:"main_prog"`TailProg *ebpf.Program `ebpf:"tail_prog"`JumpTable *ebpf.Map `ebpf:"jump_table"`}var objects BpfObjectsspec, err := ebpf.LoadCollectionSpec("prog.o")if err != nil {log.Fatalln("ebpf.LoadCollectionSpec", err)}if err := spec.LoadAndAssign(&objects, nil); err != nil {log.Fatalln("ebpf.LoadAndAssign", err)}objects.JumpTable.Update(0, objects.TailProg.FD(), ebpf.UpdateAny)
According to this, the jump table has be initialized from user space, which is what I think the last line above is supposed to do. However, I don't see any difference whether that line is there or not.
答案1
得分: 2
我没有注意到Update函数返回的错误：Update can't marshal key: encoding int: binary.Write: invalid type int。因此，程序数组映射没有被更新。我改成了以下方式：
err = objects.JumpTable.Update(uint32(0), uint32(objects.CopyHttpHostnameProg.FD()), ebpf.UpdateAny)if err != nil {println("Update", err.Error())}
如果将0作为键传递，键的大小为8个字节，这就是为什么你需要使用uint32(0)，以匹配映射的定义。现在尾调用成功了。
英文:
I wasn't looking at the error returned from the Update function: Update can't marshal key: encoding int: binary.Write: invalid type int. Therefore, the program array map was not updated. I changed to the following:
err = objects.JumpTable.Update(uint32(0), uint32(objects.CopyHttpHostnameProg.FD()), ebpf.UpdateAny)if err != nil {println("Update", err.Error())}
If you pass 0 as the key, the size of the key is 8 bytes, which is why you have to do uint32(0), which matches the map's definition. Now the tail calls succeed.
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