如何使用sed/awk在十六进制数字之间添加零?

问题描述

我的文件包含该字符串

    abc = <0x12345678>;
    abc = <0x01234 0x56789>;
    abc = <0x123 0x456 0x789>;
    abc = <0x0 0x01234 0x0 0x56789>;
    abc = <0x012 0x345>, <0x678 0x901>;
    def = <0x12345 0x67890>;

我需要将其转换为文件包含

I need to convert it to file contains

    abc = <0 0x12345678>;
    abc = <0 0x01234 0 0x56789>;
    abc = <0x123 0x456 0x789>;
    abc = <0x0 0x01234 0x0 0x56789>;
    abc = <0 0x012 0 0x345>, <0 0x678 0 0x901>;
    def = <0x12345 0x67890>;

因此，如果字符串以'abc ='开头，则我需要在十六进制数字前添加零，即在两个三角括号之间不超过2个十六进制数字，并且没有 0x0之间的十六进制数字.如何使用 sed ， awk 或其他bash工具做到这一点?

So I need to add zeros before HEX numbers if strings starts with 'abc = ', that are no more than 2 HEX numbers between couple of triangular brackets and there isn't 0x0 between that HEX numbers. How I can do it with sed, awk or another bash tools?

推荐答案

sed '/abc = </{/<[^ >]* [^ >]* /!s/0x/0 &/g}'

说明:

/abc = </{/<[^ >]* [^ >]* /!s/0x/0 &/g}
/abc = </                                Operate on lines containing "abc = <"
         {                            }  Group the commands together
                           !             Negate the condition
          /<[^ >]* [^ >]* /              Match if lines contain more than two elements after <
                            s/0x/0 &/    Prepend "0 " to "0x"
                                     g   Globally, i.e. replace as many times as possible

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