The difference between char * and char[] [duplicate]

The most straightforward answer is:

The difference here is that

char *s = "Hello world";

will place Hello world in the read-only parts of the memory and making s a pointer to that, making any writing operation on this memory illegal. While doing:

char s[] = "Hello world";

puts the literal string in read-only memory and copies the string to newly allocated memory on the stack. Thus making

s[0] = 'J';

legal.

A more lengthy explanation would include what segments the memory is stored in, and how much memory is allocated:

Example:                       Allocation Type:     Read/Write:    Storage Location:   Memory Used (Bytes):
===========================================================================================================
const char* str = "Stack";     Static               Read-only      Code segment        6 (5 chars plus '\0')
char* str = "Stack";           Static               Read-only      Code segment        6 (5 chars plus '\0')
char* str = malloc(...);       Dynamic              Read-write     Heap                Amount passed to malloc
char str[] = "Stack";          Static               Read-write     Stack               6 (5 chars plus '\0')
char strGlobal[10] = "Global"; Static               Read-write     Data Segment (R/W)  10

References

  1. What is the difference between char s[] and char *s in C?, Accessed 2014-09-03, <https://stackoverflow.com/questions/1704407/what-is-the-difference-between-char-s-and-char-s-in-c>
  2. Difference between declared string and allocated string, Accessed 2014-09-03, <https://stackoverflow.com/questions/16021454/difference-between-declared-string-and-allocated-string>

Edit

To address the edit in the question and the comment issued with it, I’ve added notes to your solution:

#include <stdio.h>

int main() {
   char ch[] = "Hello"; /* OK; Creating an array of 6 bytes to store
                         * 'H', 'e', 'l', 'l', 'o', '\0'
                         */
   char *p1 = ch;       /* OK; Creating a pointer that points to the
                         * "Hello" string.
                         */
   char *p2 = p1;       /* OK; Creating a second pointer that also
                         * points to the "Hello" string.
                         */
   char *p3 = *p1;      /* BAD; You are assigning an actual character
                         * (*p1) to a pointer-to-char variable (p3);
                         * It might be more intuitive if written in
                         * 2 lines:
                         * char* p3;
                         * p3 = *p1; //BAD
                         */
   printf("ch : %s\n", ch);   /* OK */
   printf("p1 address [%d] value is %s\n", p1, *p1);  /* Bad format specifiers */
   printf("p2 address [%d] value is %s\n", p2, *p2);  /* Bad format specifiers */
   printf("p3 address [%d] value is %s\n", p3, *p3);  /* Bad format specifiers */
   return 0;
}

So, three major bugs.

  1. You are assigning a char value to a pointer-to-char variable. Your compiler should be warning you about this. (char *p3 = *p1).
  2. Depending on your compiler, you may have to use the pointer %p format specifier to print out an address rather than the %d (integer) format specifier.
  3. You are using the string %s specifier with a char data type (ie: printf("%s", 'c') is wrong). If you are printing a single character, you use the %c format specifier, and the matching argument should be a character (ie: ‘c’, char b, etc). If you are printing an entire string, you use the %s format specifier, and the argument is a pointer-to-char.

Examples

#include <stdio.h>

int main(void) {
   char c = 'H';                    // A character
   char* pC = &c;                   // A pointer to a single character; IS NOT A STRING
   char cArray[] = { 'H', 'e', 'l', 'l', 'o' };   // An array of characters; IS NOT A STRING
   char cString[] = { 'H', 'e', 'l', 'l', 'o', '\0' };   // An array of characters with a trailing NULL charcter; THIS IS A C-STYLE STRING
   // You could also replace the '\0' with 0 or NULL, ie:
   //char cString[] = { 'H', 'e', 'l', 'l', 'o', (char)0 };
   //char cString[] = { 'H', 'e', 'l', 'l', 'o', NULL };
   const char* myString = "Hello world!"; // A C-style string; the '\0' is added automatically for you

   printf("%s\n", myString);        // OK; Prints a string stored in a variable
   printf("%s\n", "Ducks rock!");   // OK; Prints a string LITERAL; Notice the use of DOUBLE quotes, " "
   printf("%s\n", cString);         // OK; Prints a string stored in a variable

   printf("%c\n", c);               // OK; Prints a character
   printf("%c\n", *pC);             // OK; Prints a character stored in the location that pC points to
   printf("%c\n", 'J');             // OK; Prints a character LITERAL; Notice the use of SINGLE quotes, ' '

   /* The following are wrong, and your compiler should be spitting out warnings or even not allowing the
    * code to compile. They will almost certainly cause a segmentation fault. Uncomment them if you
    * want to see for yourself by removing the "#if 0" and "#endif" statements.
    */
#if 0
   printf("%s\n", c);               // WRONG; Is attempting to print a character as a string, similar
                                    // to what you are doing.
   printf("%s\n", *pC);             // WRONG; Is attempting to print a character as a string. This is
                                    // EXACTLY what you are doing.
   printf("%s\n", cArray);          // WRONG; cArray is a character ARRAY, not a C-style string, which is just
                                    // a character array with the '\0' character at the end; printf
                                    // will continue printing whatever follows the end of the string (ie:
                                    // random memory, junk, etc) until it encounters a zero stored in memory.
#endif
   return 0;
}

Code Listing – Proposed Solution

#include <stdio.h>

int main() {
   char ch[] = "Hello"; /* OK; Creating an array of 6 bytes to store
                         * 'H', 'e', 'l', 'l', 'o', '\0'
                         */
   char *p1 = ch;       /* OK; Creating a pointer that points to the
                         * "Hello" string.
                         */
   char *p2 = p1;       /* OK; Creating a second pointer that also
                         * points to the "Hello" string.
                         */
   char *p3 = p1;       /* OK; Assigning a pointer-to-char to a 
                         * pointer-to-char variables.
                         */
   printf("ch : %s\n", ch);   /* OK */
   printf("p1 address [%p] value is %s\n", p1, p1);  /* Fixed format specifiers */
   printf("p2 address [%p] value is %s\n", p2, p2);  /* Fixed format specifiers */
   printf("p3 address [%p] value is %s\n", p3, p3);  /* Fixed format specifiers */
   return 0;
}

Sample Output

ch : Hello
p1 address [0x7fff58e45666] value is Hello
p2 address [0x7fff58e45666] value is Hello
p3 address [0x7fff58e45666] value is Hello

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