带有特殊字符的字符串的控制台输出

Question

Is there a more or less standard way to output character strings that contain special characters, such ASCII control codes, in such a way that the output is a valid C/C++ literal string, with escape sequences ?

Example of expected output: This\\nis\\na\\\\test\\n\\nShe said, \\"How are you?\\"\\n

Without care, the output would be

This
is
a\test

She said, "How are you?"

not what want.

Answer 1

Printing escaped strings clearly is tricky.

Problems include

1. Control characters and the null character .
2. Escaping the escape character.
3. Negative char .

4. Hexadecimal escape sequence have no specified limit in length. Octal ones are limited to 3 digits.

    void EscapePrint(char ch) { // 1st handle all simple-escape-sequence C11 6.4.4.4 // The important one to detect is the escape character `\\` // Delete or adjust these 2 arrays per code's goals static const char *escapev = "\\a\\b\\t\\n\\v\\f\\r\\"\\'\\?\\\\"; static const char *escapec = "abtnvfr\\"\\'\\?\\\\"; char *p = strchr(escapev, (unsigned char) ch); if (p && *p) { printf("\\\\%c", escapec[p - escapev]); } // if printable, just print it. else if (isprint((unsigned char) ch)) { fputc(ch, stdout); } // else use octal escape else { // Use octal as hex is problematic reading back printf("\\\\%03hho", ch); } }

Pedantic: the octal escape sequence is a problem on rare machines whose char range exceeds 9 bits. This can be handled with a hex escaped sequences at the string level and not on a char by char basis as hex escaped sequences need a limit. Example:

  input 2 `char`:            \1 A
  //                               v----- intervening space
  output text including ":   "\\x1" "A"

Answer 2

Depending on what you might be trying to achieve I can suggest the following solution: Just replace any non-printable character with "\\xnn" (nn being the ascii code of the character. This would boil down to

void myprint (char *s) {
   while (*s) {
      printf(isalnum((unsigned char) *s) ? "%c" : "\\%03hho", *s);
      s++;
   }
}

This will of course not use special abbreviations like \\n (but \\x0a ) instead, but this shouldn't be a problem.

Answer 3

If it helps someone, I have quickly put this together:

void printfe(char* string)
{
    for (char c= *string; c; c= *(++string))
    {
        switch (c)
        {
        case '\a': printf("\\a"); break;
        case '\b': printf("\\b"); break;
        case '\f': printf("\\f"); break;
        case '\n': printf("\\n"); break;
        case '\r': printf("\\r"); break;
        case '\t': printf("\\t"); break;
        case '\v': printf("\\v"); break;
        case '\\': printf("\\\\"); break;
        case '\'': printf("\\\'"); break;
        case '\"': printf("\\\""); break;
        case '\?': printf("\\?"); break;
        default: (c < ' ' || c > '~') ? printf("\\%03o", c) : putchar(c); break;
        }
    }
}

(mind the potential non-portability of c < ' ' || c > '~' , I wanted to avoid any library reference).