char * variable address vs. char [] variable address

Question

I am printing out addresses and strings from the following two declarations and initializations:

  char * strPtr = (char *) "This is a string, made on the fly.";
  char charArray [] = "Chars in a char array variable.";

When printed, the following output occurs with wildly different addresses for the variables charArray and strPtr. The question is, "Why?"

Printing:

  printf( "%10s%40s%20p\n", "strPtr", strPtr, &(*strPtr));    
  printf( "%10s%40s%20p\n", "charArray", charArray, charArray);

Output:

    strPtr      This is a string, made on the fly.            0x400880    
 charArray         Chars in a char array variable.      0x7fff12d5ed30

The different addresses, as you see, are: 0x400880 vs. 0x7fff12d5ed30

The rest of the variable declared before this have addresses like that of charArray.

Again, the question is, "Why are the addresses so different?" Thanks for any assistance.

Just some advice; it is more advantageous to declare pointers to string literals as for example, const char *str = "string";. This helps prevent the undefined behaviour of attempting to modify string literals. — dreamlax
– dreamlax, Commented Dec 5, 2010 at 3:33

Alex Budovski · Accepted Answer · 2010-12-05 03:14:31Z

4

Because string literals, e.g. "foo bar" get allocated in a "different place" than your char array.

This is implementation dependent, but a typical implementation will put string literals in your .rdata ("read-only data") section of your executable, and your char array is declared locally, and hence goes on the stack.

And different sections of your image will get mapped to vastly different addresses when they get loaded in RAM.

answered Dec 5, 2010 at 3:14

Alex Budovski

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Tim · Accepted Answer · 2010-12-05 03:13:58Z

3

I am guessing the compiler/linker puts the char array on the stack, whereas the the other string is put into a static string table.

answered Dec 5, 2010 at 3:13

Tim

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Karl Knechtel · Accepted Answer · 2010-12-05 03:46:39Z

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The text "Chars in a char array variable." and "This is a string, made on the fly." are probably quite near each other. However, char charArray[] = ... requests space on the stack into which the corresponding bit of text is copied. The stack is practically in a different universe from the original hard-coded text, once the OS is done with its virtualization etc.

answered Dec 5, 2010 at 3:46

Karl Knechtel

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Srikar Appalaraju · Accepted Answer · 2010-12-05 07:39:08Z

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This is how it goes - I remember reading about this in [Unix: Systems Programming]

1 alt text

As you can see Initialized static data gets stored in a different location on the heap as opposed to uninitialized static data.

answered Dec 5, 2010 at 7:39

Srikar Appalaraju

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caf · Accepted Answer · 2010-12-06 01:05:48Z

The crucial thing to realise here is that in the case of strPtr, you are dealing with two different objects, whereas in the case of charArray, you are dealing with only one.

charArray is a single array object, filled with the characters of the "Chars in a char array variable." string.

strPtr itself is a single pointer object. Its value is the address of a second anonymous, unmodifiable array object that in turn contains the characters of the "This is a string, made on the fly." string.

When you print out charArray using %p, you are printing the address of charArray[0] (due to a special rule for arrays). When you print out &(*strPtr) (which is exactly the same as just strPtr), you are printing the address of the anonymous, unmodifiable array object mentioned earlier - and this is why it appears so different to the addresses of the other variables involved.

If you print out &strPtr using %p, you will see that the address of the variable strPtr itself is in a similar range to the other local variables.

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