Difference between char* and const char*?

Question

char* is a mutable pointer to a mutable character/string.

const char* is a mutable pointer to an immutable character/string. You cannot change the contents of the location(s) this pointer points to. Also, compilers are required to give error messages when you try to do so. For the same reason, conversion from const char * to char* is deprecated.

char* const is an immutable pointer (it cannot point to any other location) but the contents of location at which it points are mutable.

const char* const is an immutable pointer to an immutable character/string.

char *name

You can change the char to which name points, and also the char at which it points.

const char* name

You can change the char to which name points, but you cannot modify the char at which it points.
correction: You can change the pointer, but not the char to which name points to (https://msdn.microsoft.com/en-us/library/vstudio/whkd4k6a(v=vs.100).aspx, see "Examples"). In this case, the const specifier applies to char, not the asterisk.

According to the MSDN page and http://en.cppreference.com/w/cpp/language/declarations, the const before the * is part of the decl-specifier sequence, while the const after * is part of the declarator.
A declaration specifier sequence can be followed by multiple declarators, which is why const char * c1, c2 declares c1 as const char * and c2 as const char.

EDIT:

From the comments, your question seems to be asking about the difference between the two declarations when the pointer points to a string literal.

In that case, you should not modify the char to which name points, as it could result in Undefined Behavior. String literals may be allocated in read only memory regions (implementation defined) and an user program should not modify it in anyway. Any attempt to do so results in Undefined Behavior.

So the only difference in that case (of usage with string literals) is that the second declaration gives you a slight advantage. Compilers will usually give you a warning in case you attempt to modify the string literal in the second case.

Online Sample Example:

#include <string.h>
int main()
{
    char *str1 = "string Literal";
    const char *str2 = "string Literal";
    char source[] = "Sample string";

    strcpy(str1,source);    //No warning or error, just Undefined Behavior
    strcpy(str2,source);    //Compiler issues a warning

    return 0;
}

Output:

cc1: warnings being treated as errors
prog.c: In function ‘main’:
prog.c:9: error: passing argument 1 of ‘strcpy’ discards qualifiers from pointer target type

Notice the compiler warns for the second case but not for the first.

char mystring[101] = "My sample string";
const char * constcharp = mystring; // (1)
char const * charconstp = mystring; // (2) the same as (1)
char * const charpconst = mystring; // (3)

constcharp++; // ok
charconstp++; // ok
charpconst++; // compile error

constcharp[3] = '\0'; // compile error
charconstp[3] = '\0'; // compile error
charpconst[3] = '\0'; // ok

// String literals
char * lcharp = "My string literal";
const char * lconstcharp = "My string literal";

lcharp[0] = 'X';      // Segmentation fault (crash) during run-time
lconstcharp[0] = 'X'; // compile error

// *not* a string literal
const char astr[101] = "My mutable string";
astr[0] = 'X';          // compile error
((char*)astr)[0] = 'X'; // ok

In neither case can you modify a string literal, regardless of whether the pointer to that string literal is declared as char * or const char *.

However, the difference is that if the pointer is const char * then the compiler must give a diagnostic if you attempt to modify the pointed-to value, but if the pointer is char * then it does not.

CASE 1:

char *str = "Hello";
str[0] = 'M'  //Warning may be issued by compiler, and will cause segmentation fault upon running the programme

The above sets str to point to the literal value "Hello" which is hard-coded in the program's binary image, which is flagged as read-only in memory, means any change in this String literal is illegal and that would throw segmentation faults.

CASE 2:

const char *str = "Hello";
str[0] = 'M'  //Compile time error

CASE 3:

char str[] = "Hello";
str[0] = 'M'; // legal and change the str = "Mello".