![在 shell 中解析 C 字符串文字](https://linux22.com/image/8739/%E5%9C%A8%20shell%20%E4%B8%AD%E8%A7%A3%E6%9E%90%20C%20%E5%AD%97%E7%AC%A6%E4%B8%B2%E6%96%87%E5%AD%97.png)
我有一个有效的 C 字符串文字,我想将其转换为其实际表示形式。遗憾的是 printf 的 %b 转换说明符没有定义 ? 的转义符' 和 ",它们在 C 字符串中都是有效的。
现在,我有一个小型 C 程序,用于执行转换,但我更希望能够使用标准 POSIX 工具完成所有操作。
我希望它只与 POSIX 工具一起工作(所以没有 GNUisms),如果需要的话也可能与 bash 一起工作,尽管 sh 是首选。
答案1
我想到了一种在 awk 中做到这一点的方法;虽然它有点难看,所以如果有人有更好的解决方案,我很感兴趣。这适用于所有有效的 C 字符串,但不能处理所有无效字符串的情况(例如 \q,其中 q 是跟在反斜杠后面的无效字符,只会消耗该字符):
function hexnumber(str, ret, n, k)
{
n = length(str)
ret = 0
for (i = 1; i <= n; i++) {
c = substr(str, i, 1)
c = tolower(c)
# index() returns 0 if c not in string,
# includes c == "0"
k = index("123456789abcdef", c)
ret = ret * 16 + k
}
return ret;
}
function octnumber(str, ret, n, k)
{
n = length(str)
ret = 0
for (i = 1; i <= n; i++) {
c = substr(str, i, 1)
# index() returns 0 if c not in string,
# includes c == "0"
k = index("1234567", c)
ret = ret * 8 + k
}
return ret;
}
BEGIN { RS="\\" ; notQuoted = 0}
#{ print $0 ; next}
NR == 1 {
if(match($0,/^"/)) {
split($0,matches,"\"")
printf("%s",matches[2]);
if(match($0,/^"[^"]*"/)) {
exit 0;
}
next
} else {
exit 1;
}
}
#This will happen on the 2nd (and subsequent) consecutive backslash
/^$/ {
if (notQuoted) {
notQuoted = 0
} else {
notQuoted = 1
printf("\\")
next
}
}
notQuoted == 1 {
if(match($0,/([^"]*)"/))
{
split($0,matches,"\"")
printf("%s",matches[1])
exit 0
}
printf("%s",$0);
next
}
# If we reach here, then the first character is part of a backslash escape
{
first = substr($0,1,1);
rest = substr($0,2);
if(first == "a") printf("\a")
else if(first == "b") printf("\b")
else if(first == "f") printf("\f")
else if(first == "n") printf("\n")
else if(first == "r") printf("\r")
else if(first == "t") printf("\t")
else if(first == "v") printf("\v")
else if(first == "\"") printf("\"")
else if (first == "x") {
second = substr(rest,1,1);
third = substr(rest,2,1);
if(match(third,/[1234567890ABCDEFabcdef]/)) {
printf("%c",hexnumber(second third)+0);
rest = substr(rest,3);
}
else {
printf("%c",hexnumber(second)+0);
rest = substr(rest,2);
}
}
else if (match(first,/[01234567]/)) {
second = substr(rest,1,1);
third = substr(rest,2,1);
fourth = substr(rest,3,1);
if(match(second third fourth,/^[01234567]*$/)) {
printf("%c",octnumber("0" first second third fourth)+0);
rest = substr(rest,length(first second third fourth));
}
else if(match(second third,/^[01234567]*$/)) {
printf("%c",octnumber(first second third)+0);
rest = substr(rest,length(first second third));
}
else if(match(second,/^[01234567]*$/)) {
printf("%c",octnumber(first second)+0);
rest = substr(rest,length(first second));
}
else {
printf("%c",octnumber(first)+0);
}
}
if(match(rest,/([^"]*)"/)) # this comment fixes vim's syntax matching "
{
split(rest,matches,"\"")
printf("%s",matches[1])
exit 0
}
printf("%s",rest);
}