2

我从这里得到代码。

from binascii import hexlify
from ctypes import create_string_buffer, addressof
from socket import socket, AF_PACKET, SOCK_RAW, SOL_SOCKET
from struct import pack, unpack

sniff_interval=120
# A subset of Berkeley Packet Filter constants and macros, as defined in
# linux/filter.h.

# Instruction classes
BPF_LD = 0x00
BPF_JMP = 0x05
BPF_RET = 0x06

# ld/ldx fields
BPF_H = 0x08
BPF_B = 0x10
BPF_ABS = 0x20

# alu/jmp fields
BPF_JEQ = 0x10
BPF_K = 0x00

def bpf_jump(code, k, jt, jf):
    return pack('HBBI', code, jt, jf, k)

def bpf_stmt(code, k):
    return bpf_jump(code, k, 0, 0)


# Ordering of the filters is backwards of what would be intuitive for
# performance reasons: the check that is most likely to fail is first.
filters_list = [
    # Must have dst port 67. Load (BPF_LD) a half word value (BPF_H) in
    # ethernet frame at absolute byte offset 36 (BPF_ABS). If value is equal to
    # 67 then do not jump, else jump 5 statements.
    bpf_stmt(BPF_LD | BPF_H | BPF_ABS, 36),
    bpf_jump(BPF_JMP | BPF_JEQ | BPF_K, 5201, 0, 5),

    # Must be UDP (check protocol field at byte offset 23)
    bpf_stmt(BPF_LD | BPF_B | BPF_ABS, 23),
    bpf_jump(BPF_JMP | BPF_JEQ | BPF_K, 0x06, 0, 3),

    # Must be IPv4 (check ethertype field at byte offset 12)
    bpf_stmt(BPF_LD | BPF_H | BPF_ABS, 12),
    bpf_jump(BPF_JMP | BPF_JEQ | BPF_K, 0x0800, 0, 1),

    bpf_stmt(BPF_RET | BPF_K, 0x0fffffff), # pass
    bpf_stmt(BPF_RET | BPF_K, 0), # reject
]

# Create filters struct and fprog struct to be used by SO_ATTACH_FILTER, as
# defined in linux/filter.h.
filters = ''.join(filters_list)
b = create_string_buffer(filters)
mem_addr_of_filters = addressof(b)
fprog = pack('HL', len(filters_list), mem_addr_of_filters)

# As defined in asm/socket.h
SO_ATTACH_FILTER = 26

# Create listening socket with filters
s = socket(AF_PACKET, SOCK_RAW, 0x0800)
s.setsockopt(SOL_SOCKET, SO_ATTACH_FILTER, fprog)
s.bind(('eth0', 0x0800))

while True:
    data, addr = s.recvfrom(65565)
    #print "*****"
    print 'got data from', addr, ':', hexlify(data) #Have to print data, then only the CPU is 2%

我正在测试iperf3，通过以太网电缆生成从另一台笔记本电脑到我的笔记本电脑的流量。服务器（我的笔记本电脑）在 5021 上列出，客户端（另一台笔记本电脑）发送数据。

  • 如果我注释该行print 'got data from', addr, ':', hexlify(data)并运行脚本，脚本的 CPU 利用率会上升 30%，在 100MB 流量的情况下会上升 40%。
  • 如果我取消注释该行print 'got data from', addr, ':', hexlify(data)并再次运行，CPU 将下降到2%存在相同数量的流量。我登记了 htop

那么，这里发生了什么？

4

1 回答 1

0

我敢打赌hexlify()，或者最有可能print（因为它必须与 STDOUT 同步）正在给你的主线程一个非常需要的休息和一个呼吸的空间，而不是仅仅在无限while循环中敲打套接字读数

尝试添加time.sleep(0.05)（当然是先导入time）而不是 print 语句并再次检查 CPU 使用率。

于 2017-06-14T16:16:03.033 回答