C pointer confusion - pointer to char array

Question

This should be simple, but pointers to char arrays still leave me scratching my head sometimes. gcc complains that "subscripted value is neither array nor pointer" in lines 4 and 5, and "invalid type argument of 'unary *'" on lines 8 and 11. Can someone explain what is going wrong here? I changed lines 4 and 5 to dereference the pointer first using brackets, but still can't get what I want.

This should be a pretty simple function:

1  void makesafestr ( const char *unsafe,  const char *safe )
2  {
3    int offset=0;
4    for (; (*safe)[offset] != "\0" ; offset++) {
5      switch ((*unsafe)[offset]) {
6        case "\n":
7        case "\r":
8          *safe[offset] = "~";
9          break;
10       default:
11         *safe[offset] = *unsafe[offset];
12     }
13     offset++;
14   }
}

cnicutar · Accepted Answer · 2012-02-10 18:00:09Z

3

In your code safe is a pointer to a char (not a pointer to an array). So when you say *safe that's a plain char. Instead of (*safe)[offset] try safe[offset]. Same goes for *safe[offset], same goes for unsafe.

A second problem is the way you are comparing characters. "\0" is a string literal. In your code you want a character literal. In short, the for should be:

for (; safe[offset] != '\0' ; offset++)

But you could rewrite it and make it simpler, since \0 is 0:

for (offset=0; safe[offset]; offset++)

edited Feb 10, 2012 at 18:00

answered Feb 10, 2012 at 17:55

cnicutar

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3 Comments

fdk1342 Over a year ago

And this is why you don't need the '*'. A postfix expression followed by an expression in square brackets [] is a subscripted designation of an element of an array object. The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2))). Because of the conversion rules that apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the initial element of an array object) and E2 is an integer, E1[E2] designates the E2-th element of E1 (counting from zero).

Michelle Dupuis Over a year ago

Wow that's complicated. Does this mean that the pointer is ALWAYS dereferenced first, then the subscript added? I was concerned that the compiler might assume it is an array of pointers, and use the address of 'safe' plus [offset], then dereference that address.

John Bode Over a year ago

No. a[i] is equivalent to *(a + i); the result of a + i is dereferenced (IOW, the offset is added before the dereference). Had you declared safe as char (*safe)[] (a pointer to an array of char), then you would need to dereference the pointer before applying the subscript (and the parens around *safe would be mandatory -- subscripting has a higher precedence than dereference, so *safe[i] would be parsed as *(safe[i]), not (*safe)[i], which would cause problems).

Mark Wilkins · Accepted Answer · 2012-02-10 17:55:07Z

3

The character constants need single quotes instead of double quotes: '\0', '\n', '\r', and '~'

Also, the assignment to the safe probably will not be allowed since it is marked as const (after fixing the indirection issues already mentioned in other answers).

answered Feb 10, 2012 at 17:55

Mark Wilkins

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Comments

Ed Heal · Accepted Answer · 2012-02-10 18:03:33Z

The pointer unsafe and safe point to the first item of the character array.

safe[4] will be the fifth item (counting from zero) in the array i.e. the fifth character

So the code should be

void makesafestr ( const char *unsafe,  char *safe )  /* Do not need const for safe, as you are
                                                         constructing it */
{  
  int offset=0;  
  for (; unsafe[offset] != 0 ; offset++) /* Do not need the " - see below. Also would have
                                            thought you need to scan unsafe */
  {  
      switch (unsafe[offset]) {  
        case '\n':  /* Single quotes required as we are dealing with characters and not strings */
        case '\r':  
          safe[offset] = '~';  
          break;  
       default:  
          safe[offset] = unsafe[offset];  
     }  
     offset++; 
   }
   safe[offset] = 0; /* In C and C++ strings end with the null character */  
}

Note, a string ("a string") in C and C++ is an array of characters.

Phonon · Accepted Answer · 2012-02-10 17:55:00Z

1

You don't need asterisks in front of safe and unsafe. They're already pointers. Also, you must use single quotes instead of double quotes. Single quotes are for chars, double quotes are for stings.

answered Feb 10, 2012 at 17:55

Phonon

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