C++

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null terminator same as null and thus falsy value?

johnmerlino (110)
I am looking at one of the functions of an exercise:

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void escape(char * s, char * t) {

    int i, j;

    i = j = 0;

    

    while ( t[i] ) {

        

        /*  Translate the special character, if we have one  */

        

        switch( t[i] ) {

        case '\n':

            s[j++] = '\\';

            s[j] = 'n';

            break;

            

        case '\t':

            s[j++] = '\\';

            s[j] = 't';

            break;

            

        case '\a':

            s[j++] = '\\';

            s[j] = 'a';

            break;

            

        case '\b':

            s[j++] = '\\';

            s[j] = 'b';

            break;

            

        case '\f':

            s[j++] = '\\';

            s[j] = 'f';

            break;

            

        case '\r':

            s[j++] = '\\';

            s[j] = 'r';

            break;

            

        case '\v':

            s[j++] = '\\';

            s[j] = 'v';

            break;

            

        case '\\':

            s[j++] = '\\';

            s[j] = '\\';

            break;

            

        case '\"':

            s[j++] = '\\';

            s[j] = '\"';

            break;

            

        default:

            

            /*  This is not a special character, so just copy it  */

            

            s[j] = t[i];

            break;

        }

        ++i;

        ++j;

    }

    s[j] = t[i];    /*  Don't forget the null character  */

}


Notice the while loop evaluates the current value in t to true or false. When we hit the null terminator, does that get evaluated as 0 and hence evaluates as a falsy value so the while loop exits?
long double main (999)
Yes, a null character will get evaluated to false.
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