perl下按位与时&，变量为什么不会自动转为整型？

linux_arm

1. $cat = "168";
   
2. $dog = "255";
   
3. 
   
4. print $cat & $dog; #输出040
   
5. print $cat & $dog + 0; #加0是为了将其转为数值运算，输出168

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这个是不是设计到perl上下文的概念？断断续续自学perl，没事做点小程序玩玩，对perl理解不深入

zhlong8

根据上下文实际上是两个不同的操作符，上面是字符串的位操作符，下面的是数字位操作符。具体见 perldoc perlop 中的 &

linux_arm

谢谢解答，通过查perldoc，答案如下：
Bitstrings of any size may be manipulated by the bitwise operators (~ | & ^).

If the operands to a binary bitwise op are strings of different sizes, | and ^ ops act as though the shorter operand had additional zero bits on the right, while the & op acts as though the longer operand were truncated to the length of the shorter. The granularity for such extension or truncation is one or more bytes.

    # ASCII-based examples
    print "j p \n" ^ " a h";            # prints "JAPH\n"
    print "JA" | "  ph\n";              # prints "japh\n"
    print "japh\nJunk" & '_____';       # prints "JAPH\n";
    print 'p N$' ^ " E<H\n";            # prints "Perl\n";

If you are intending to manipulate bitstrings, be certain that you're supplying bitstrings: If an operand is a number, that will imply a numeric bitwise operation. You may explicitly show which type of operation you intend by using "" or 0+, as in the examples below.

    $foo =  150  |  105;        # yields 255  (0x96 | 0x69 is 0xFF)
    $foo = '150' |  105;        # yields 255
    $foo =  150  | '105';       # yields 255
    $foo = '150' | '105';       # yields string '155' (under ASCII)

    $baz = 0+$foo & 0+$bar;     # both ops explicitly numeric
    $biz = "$foo" ^ "$bar";     # both ops explicitly stringy


还有从FAQ里查到的：
Why doesn't & work the way I want it to?
  The behavior of binary arithmetic operators depends on whether they're
  used on numbers or strings. The operators treat a string as a series of
  bits and work with that (the string "3" is the bit pattern 00110011).
  The operators work with the binary form of a number (the number 3 is
  treated as the bit pattern 00000011).

  So, saying "11 & 3" performs the "and" operation on numbers (yielding
  3). Saying "11" & "3" performs the "and" operation on strings (yielding
  "1").