Question

#include<stdio.h>
#include<string.h>
int main()
{
        char a[]="aaa";
        char *b="bbb";
        strcpy(a,"cc");
        printf("%s",a);
        strcpy(b,"dd");
        printf("%s",b);
        return 0;
}

We could not modify the contents of the array but the above program does not show any compile time error.When run it printed cc and terminated. The contents of the array i think will get stored in the read only section of the data segment and so its not possible to change the value of array as its a const.But here in the above program the value got changed to cc and the program terminated.The value got changed here why is it so.Please help me understand.

Answer 1

Here’s a hypothetical memory map showing how the string literals, array, and pointer all relate to each other:

Item              Address            0x00  0x01  0x02  0x03
----              -------            ----  ----  ----  ----
"aaa"             0x00040000          'a'   'a'   'a'  0x00
"bbb"             0x00040004          'b'   'b'   'b'  0x00
"cc"              0x00040008          'c'   'c'   0x00 ???
"dd"              0x0004000C          'd'   'd'   0x00 ???
...
  a               0x08000000          'a'   'a'   'a'  0x00
  b               0x08000004          0x00  0x04  0x00 0x00

This is the situation at line 6 in your code, after a and b have been declared and initialized. The string literals "aaa", "bbb", "cc", and "dd" all reside somewhere in memory such that they exist over the lifetime of the program. They are stored as arrays of char (const char in C++). Attempting to modify the contents of a string literal (in the case of this hypothetical, attempting to write to any memory location starting with 0x0004) invokes undefined behavior. Some platforms store string literals in read-only memory, some store them in writable memory, but in all cases, they should be treated as though they are unwritable.

The object a is an array of char, and it has been initialized with the contents of the string literal "aaa". The object b is a pointer to char, and it has been initialized with the address of the string literal "bbb". In the line

strcpy(a, "cc");

you’re copying the contents of the string literal "cc" to a; after the line is executed, your memory map looks like this:

Item              Address            0x00  0x01  0x02  0x03
----              -------            ----  ----  ----  ----
"aaa"             0x00040000          'a'   'a'   'a'  0x00
"bbb"             0x00040004          'b'   'b'   'b'  0x00
"cc"              0x00040008          'c'   'c'   0x00 ???
"dd"              0x0004000C          'd'   'd'   0x00 ???
...
  a               0x08000000          'c'   'c'   0x00 0x00
  b               0x08000004          0x00  0x04  0x00 0x00

So when you print a to standard output, you should see the string cc. Note: printf is buffered, so it’s possible that output may not be written to the terminal immediately – either add a newline character to the format string (printf("%s\n", a);) or call fflush(stdout); after the printf to make sure all your output shows up.

In line 9, you attempt to copy the contents of the string literal "dd" to the location pointed to by b; unfortunately, b points to another string literal, which as mentioned above invokes undefined behavior. At this point, your program could literally do anything from run as expected to crash outright to anything in between. This could be part of the reason you only see the output for cc.