pointer to char vs pointer to arrays

Question

I thought I had a decent knowledge of pointers until I decided to review some so called trivial examples.

One thing I know is that when declaring an array say:

int arr[2] {3, 5};

arr will hold the value of the first element in the array so attempting to print that (cout << arr) gives obviously the address of arr[0]. Even thought my program uses pointers it's still similar.

My question is why can I print h and have bonjour as output, but I can't do the same with p?

It also looks like when I increment h++ and print it again I get onjour. How are pointers different with char?

#include <iostream>
#include <string>

int main()
{
    char* h = "bonjour";
    int k[4]{3, 4, 5, 6};
    int * p = k;

    std::cout << "Hello, "<< h << "!\n";
}

Outputting arrays of characters as strings is so common that there is a << overload for const char * that does it for you. Practically no one ever wants to output an array of int without some special formatting, and even then how would << know when to stop? With a char array, 0 terminates the string, but 0 is just another number in an int array. — user4581301
– user4581301, Commented Oct 27, 2015 at 20:02

Barry · Accepted Answer · 2015-10-27 20:01:06Z

9

When you stream h, you are using this overload:

template< class Traits >
basic_ostream<char,Traits>& operator<<( basic_ostream<char,Traits>& os,  
                                        const char* s );

But when you stream p, you are using this one:

basic_ostream& operator<<( const void* value );

The former will print every byte until it gets to a \0, the latter will simply print the address. That is - there's simply a special case for const char* that you're taking advantage of.

answered Oct 27, 2015 at 20:01

Barry

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Comments

R Sahu · Accepted Answer · 2015-10-28 15:23:48Z

why can i print h and have bonjour as output and i can't do the same with p? it also looks like when i increment h++ and print it again i get onjour. how are pointers different with char?

The standard library provides two overloads of the operator<< function for dealing with pointers.

A member function overload:

basic_ostream& operator<<( const void* value );

and a non-member function overload:

template< class Traits >
basic_ostream<char,Traits>& operator<<( basic_ostream<char,Traits>& os,  
                                        const char* s );

The first one is invoked when cout << ptr is used for all pointers that are not of type char* or char const*. The second is used for all pointers that are of tpe char* or char const*. The second one outputs everything up to the terminating null characters while the first one outputs just the pointer value.

That explains the output.

It's instructive to understand why the standard library treats them differently.

The language and the standard libraries make exceptions to the null terminated strings that are represented by char* and char const*. When dealing with strings, the null character ('\0') acts as a sentinel value that marks the end of the string. There is no such value for other types. Hence, an int* cannot be treated the same was a char*.

znkr · Accepted Answer · 2015-10-27 20:41:17Z

1

char* can be special in some circumstances, because C decided to use it for representing strings. In that case a char* refers to a null terminated array of characters, aka C-string. The reason you can print h but not p is because there exists a function that does the special handling for char*:

 std::ostream::operator<<(std::ostream& os, const char*);

You could provide your own overload to print null terminated integer arrays:

std::ostream& operator<<(std::ostream& os, const int* arr) {
   while (int i = *arr++) {
      os << i;
      if (*arr) os << ", ";
   }
   return os;
}

int main()
{
  const int arr[] = {1, 2 ,3, 4, 0};
  std::cout << arr << "\n";
}

However, this is extremely dangerous because most integer arrays are not terminated by a zero and not all pointer to integers are arrays.

edited Oct 27, 2015 at 20:41

answered Oct 27, 2015 at 20:27

znkr

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5 Comments

Keith Thompson Over a year ago

C doesn't use char* for storing strings. It uses char* to point to a string.

Bobby Over a year ago

so if (*arr) evaluates to true if there is a value and false otherwise? interesting. So instead of the usual for loop to print (or manupulate idividual element) the array for (int i=0, i<sizeof(array);), i can just use a while loop then: while((*array)) do somestuff and array++!(?).

znkr Over a year ago

@Bobby: That only works if the last element is 0. :-)

znkr Over a year ago

@KeithThompson, you're absolutely right. I changed it.

Keith Thompson Over a year ago

@Bobby: if (*arr) is essentially equivalent to if (*arr != 0). (In C it's exactly equivalent; C++ operator overloading might create other opportunities.)

Ziezi · Accepted Answer · 2015-10-27 20:08:47Z

0

I think this has to do with operator << for char* and int* more than anything else. C programmers are used to null terminated strings printing out so a method was created to print them. ints can be printed directly. A char will not be able to print out a whole string.

edited Oct 27, 2015 at 20:08

Ziezi

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answered Oct 27, 2015 at 20:02

Robert Jacobs

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