How can I fix the problem with '\' character?

Question

In the code I am converting every character to '(' if the appears only once in the whole string, or to ')' if it appears more than once. I pass almost all tests, except the test with input "$$\". It gives "missing terminating character" error. I see that problem is the '\' char. and if I add a second '\' it is good, but is there a quick way to fix it, or I should somehow add 1 to the size of the pointer and then add the second '\'?

PS The input is fixed. I cannot change it.

char text[] = "$$\";
char *res = malloc(strlen(text));
int counter = 0;
for(int i = 0; i < strlen(text); i++) {
    for(int j = 0; j < strlen(text); j++) {
        if( tolower(text[i]) == tolower(text[j]) )  {
            counter++;
        }
    }
    if(counter == 1) {
        res[i] = '(';
    } else {
        res[i] = ')';
    }
    printf("%c", res[i]);
    counter = 0;
}
return 0;

Answer 1

Since we are already proposing trigraphs, you can alternatively also use a magic number instead (x86 solution):

#include <stdio.h>
int main(void) 
{ 
  char* text = (char*)&(int) <%6038564%>;
  puts(text);
}

Output:

$$\

Or if you like the magic number 027022044 better, then go for that one instead.

(Not really a serious answer, but it's Friday, so...)

Answer 2

Nothing can escape from an escape character, even an escape character....

In C, all escape sequences consist of two or more characters

The first of which is the backslash, \ , called the Escape character.

The remaining characters determine the interpretation of the escape sequence.

Escape sequence of \\ equals 5C hex value in ascii which represents character of Backslash . Therefore in order to use backslash, you have to use it twice.

Answer 3

You can use trigraph sequence

char text[] = "$$??/"; // { dollar-sign, dollar-sign, backslash, zero }