第一次出现，除了C中的转义字符

Question

How can I locate the first unescaped char in a str. In the following code, I get the first char at position 14, but I'm looking the one at position 26.

#include <stdio.h>
#include <string.h>

int main ()
{
  char str[] = "FOO + pHWAx \\\"bar AER/2.1\" BAZ";
  printf ("%s\n",str);
  char * pch;
  pch=strchr(str,'"');
  printf ("found at %d\n",pch-str+1);
  return 0;
}

Answer 1

Use the strpbrk function to look for the first occurence of any one of several characters at once. You must not skip the escape character; you must check whether it is followed by the character that you're really looking for.

Ie suppose we want to look for " which can be escaped as \\" . Actually, this means we must look for either " or \\ . In other words:

char *ptr = strpbrk(string, "\"\\"); /* look for chars in the set { ", \ } */

But we have to do this in a loop, because we are not interested in escaped quotes and have to keep going:

char *quote = 0;
char *string = str; /* initially points to the str array */

while (*string != 0) {
  char *ptr = strpbrk(string, "\"\\");

Next we check whether we found something:

  if (!ptr)
    break;

If we found something is is necessarily a \\ or " :

  if (*ptr == '"') {
    quote = ptr;
    break;
  }

If it is not a quote, then it must be an escape. We increment to the next character. If it is a terminating null it means we have a backslash at the end of a string: an improper escape.

  if (*++ptr == 0)
    break;

Otherwise, we can skip the next character and continue the loop to scan for the next escape or unescaped quote.

  string = ++ptr;
}

If an unescaped quote occurs, then quote points to it after the execution of the while loop. Otherwise quote remains null.

This code assumes that there exist other escapes besides \\" , but that they are all one character long, eg \\b or \\r . It will not work if there are longer escapes like \\xff . Escapes constitute the conventions of a language: you have to know what the language is that you're processing to do it correctly.