用于将EBCDIC可打印内容转换为ASCII的C代码 [英] C code to convert EBCDIC printables to ASCII in-place

问题描述

用C语言将EBCDIC编码的字符串就地转换为等效的ASCII的最简单方法是什么.

What is the simplest way in C to convert an EBCDIC-encoded string to its ASCII equivalent in-place.

唯一需要转换的字符是空格，字母数字和来自集合<=>()+-*/&|!$#@.,;%_?"的字符.所有其他字符都可以简单地替换为..

The only characters that need to be converted are the space, alphanumerics, and from the set <=>()+-*/&|!$#@.,;%_?". All other characters can simply be replaced with ..

函数签名基本上是:

void ebcdicToAscii (char *s);

目前，我倾向于针对EBCDIC的各个部分使用一系列查找表和多个if语句，但是我想知道是否有更好的方法.

At the moment, I'm leaning towards a series of lookup tables and multiple if statements for the various EBCDIC sections, but I wonder if there's a better way.

推荐答案

在此处，从我的头顶开始:

static const unsigned char e2a[256] = {
          0,  1,  2,  3,156,  9,134,127,151,141,142, 11, 12, 13, 14, 15,
         16, 17, 18, 19,157,133,  8,135, 24, 25,146,143, 28, 29, 30, 31,
        128,129,130,131,132, 10, 23, 27,136,137,138,139,140,  5,  6,  7,
        144,145, 22,147,148,149,150,  4,152,153,154,155, 20, 21,158, 26,
         32,160,161,162,163,164,165,166,167,168, 91, 46, 60, 40, 43, 33,
         38,169,170,171,172,173,174,175,176,177, 93, 36, 42, 41, 59, 94,
         45, 47,178,179,180,181,182,183,184,185,124, 44, 37, 95, 62, 63,
        186,187,188,189,190,191,192,193,194, 96, 58, 35, 64, 39, 61, 34,
        195, 97, 98, 99,100,101,102,103,104,105,196,197,198,199,200,201,
        202,106,107,108,109,110,111,112,113,114,203,204,205,206,207,208,
        209,126,115,116,117,118,119,120,121,122,210,211,212,213,214,215,
        216,217,218,219,220,221,222,223,224,225,226,227,228,229,230,231,
        123, 65, 66, 67, 68, 69, 70, 71, 72, 73,232,233,234,235,236,237,
        125, 74, 75, 76, 77, 78, 79, 80, 81, 82,238,239,240,241,242,243,
         92,159, 83, 84, 85, 86, 87, 88, 89, 90,244,245,246,247,248,249,
         48, 49, 50, 51, 52, 53, 54, 55, 56, 57,250,251,252,253,254,255
};

void ebcdicToAscii (unsigned char *s)
{
    while (*s)
    {
        *s = e2a[(int) (*s)];
        s++;
    }
}

对于您的特定要求，我建议如下:

For your specific requirements, I would suggest something like:

#include <stdio.h>

void inSituEbcdicToAscii (char *s) {
    static char etoa[] =
        "                                "
        "                                "
        "           .<(+|&         !$*); "  // first char here is real space
        "-/         ,%_>?         `:#@'=\""
        " abcdefghi       jklmnopqr      "
        "  stuvwxyz                      "
        " ABCDEFGHI       JKLMNOPQR      "
        "  STUVWXYZ      0123456789      ";

    while (*s != '\0') {
        *s = etoa[(unsigned char)*s];
        s++;
    }
}

int main (void) {
    char str[] = "\xc8\x85\x93\x93\x96\x40\xa3\x88\x85\x99\x85\x5a";
    inSituEbcdicToAscii (str);
    printf ("%s\n", str);
    return 0;
}

从等效的EBCDIC字符输出Hello there!.尽管您可以将其更改为其他字符，但除您感兴趣的字符以外的所有其他字符都将转换为空格(请确保不要修改EBCDIC代码0x40，即 real 空格)

which outputs Hello there! from the equivalent EBCDIC characters. All other characters beyond those you showed an interest in are converted to a space, though you can change that to something else (make sure you don't modify EBCDIC code 0x40 which is the real space).

这篇关于用于将EBCDIC可打印内容转换为ASCII的C代码的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！