How to convert a c string into its escaped version in c?

Question

Is there any buildin function or a alternative simple and fast way of escape a C character array that if used with e.g printf should yield original character array again.

char* str = "\tHello World\n";
char* escaped_str = escape(str); //should contain "\tHello World\n" with char \ ,t.
printf(escaped_str); //should print out [TAB]Hello World[nextline] similar to if str was printed.

Is there a simple way in c to escape a string with c escape characters.

Update

I have buffer containing a string with escape character. And i want to include in a C file. For that i need to escape it so it can be complied. I just need to know if there is simple way of doing it instead of scanning the buffer for \n \t etc and generating there c escape char.

for(int i=0; i< strlen(buffer);i++)
    if(buffer[i]=='\n')
      sprintf(dest,"\\n")
    else ....

Update 2

I wrote this function. It work fine.

char* escape(char* buffer){
    int i,j;
    int l = strlen(buffer) + 1;
    char esc_char[]= { '\a','\b','\f','\n','\r','\t','\v','\\'};
    char essc_str[]= {  'a', 'b', 'f', 'n', 'r', 't', 'v','\\'};
  char* dest  =  (char*)calloc( l*2,sizeof(char));
    char* ptr=dest;
    for(i=0;i<l;i++){
        for(j=0; j< 8 ;j++){
            if( buffer[i]==esc_char[j] ){
              *ptr++ = '\\';
              *ptr++ = essc_str[j];
                 break;
            }
        }
        if(j == 8 )
      *ptr++ = buffer[i];
    }
  *ptr='\0';
    return dest;
}

Answer 1

No, there isn't any standard function for creating the source code version of the string. But you could use the iscntrl function to write one, or just use the switch keyword.

But, unless your program writes out a C source file intended to be run through the compiler, you don't need to work with escaped strings. printf doesn't process character escape sequences, only variable insertions (%d, %s, etc)

Specifically, the following produce the same output:

printf("\tHello World\n");

and

const char* str = "\tHello World\n";
printf(str);

and

const char* str = "\tHello World\n";
printf("%s", str);

The second one isn't a good idea, because if str contained % your program would produce bad output and could crash.

EDIT: For producing the source code version, there are a couple of approaches:

Simpler, but less readable output:

if (iscntrl(ch) || ch == '\\' || ch == '\"' || ch == '\'') {
   fprintf(outf, "\\%03o", ch);
}
else
   fputc(ch, outf);

More readable results:

switch (ch) {
  case '\"':
    fputs("\\\"", outf);
    break;
  case '\'':
    fputs("\\\'", outf);
    break;
  case '\\':
    fputs("\\\\", outf);
    break;
  case '\a':
    fputs("\\a", outf);
    break;
  case '\b':
    fputs("\\b", outf);
    break;
  case '\n':
    fputs("\\n", outf);
    break;
  case '\t':
    fputs("\\t", outf);
    break;
  // and so on
  default:
    if (iscntrl(ch)) fprintf(outf, "\\%03o", ch);
    else fputc(ch, outf);
}

Answer 2

If you don't require the resulting string to be human-readable, and your compile-time character set is the same as your execution character set, then the simplest way is to use code point escapes for everything:

int print_string_literal(char *s)
{
    putchar('\"');

    while (*s)
    {
        unsigned cp = (unsigned char)*s++;
        printf("\\x%.2x", cp);
    }

    putchar('\"');
}

You could finess this some to produce nicer looking strings, but you did ask for something simple...