搜索expect_out(缓冲区)的十六进制值

Question

我有下面的expect脚本，它应该将预期值存储到2个变量hex1和十六进制2中。

for {set chan 0} {$chan < 6} {incr chan 1} { 
  incr $chan_enable_value 1
  puts $chan_enable_value\r
  send -- "i2c_write -b 0  -r reg -w chan_enable\r"
  expect "barebox: /"
  send -- "i2c_probe\r"
  expect -re "barebox: .* (0x[[:xdigit:]]{2}) (0x[[:xdigit:]]{2})\r\n"
  set hex1 $expect_out(1, string)
  set hex1 $expect_out(2, string)

在运行脚本时，我会得到以下错误

invalid command name ":xdigit:"

Answer 1

你可能想：

  send -- "i2c_probe\r"
  # .....^ note the space
  
  # expect uses \r\n line endings: find 2 space-separated hex numbers at the end of line
  expect -re "barebox: .* (0x[[:xdigit:]]{2}) (0x[[:xdigit:]]{2})\r\n"

  # extract the values captured by the regex parentheses
  set hex1 $expect_out(1,string)
  set hex2 $expect_out(2,string)
  
  puts [list $hex1 $hex2]

大脑屁:方括号是Tcl的命令替换语法，双引号允许内插。

选其中一个

  expect -re "barebox: .* (0x\[\[:xdigit:]]{2}) (0x\[\[:xdigit:]]{2})\r\n"
  # .........................^.^...................^.^

  set regex {barebox: .* (0x[[:xdigit:]]{2}) (0x[[:xdigit:]]{2})}
  expect -re "$regex\r\n"

我们需要防止将[...]解释为命令替换，但允许解释\r\n，因此有点尴尬。