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char m = "hello";

char *pm = "hello";

Why can't the *pm be modified?

I see someone say:
char pointer is a copy of address,
char array is a copy of content, which is "hello".

my question is this:

1. why is char m = "hello"; a copy of the content.


2. m itself is also the first address of the array, why does the constant string "hello" assign them to different effects?


3. Is char m = 'h','e','l','l','0'; a Syntactic sugar to char m = "hello";

char m = "hello"; says:

• Define an array of char named m.


• Use the string literal "hello" to initialize the array. This is somewhat different from other initializations; there is a rule for initializing character arrays from string literals that initializes the array elements from the characters in the string literal.


• Set the size of the array from the size of the string literal (including the terminating null character).


• The final result is an array m which you may modify.

char *pm = "hello"; says:

• Define a pointer to char named pm.


• Use the string literal "hello" to initialize the pointer. Since it is not a character array being initialized, the rule for character arrays does not apply. Instead, "hello" is an expression. Since a string literal is itself an array, the rule that arrays are converted to pointers applies. So "hello" is converted to a pointer to its first character. Then pm is initialized to the value of this pointer.


• The final result is that pm points to the array of a string literal, which you may not modify.

In regard to your question:

m itself is also the first address of the array, why does the constant string "hello" assign them to different effects?

First, the name of an array is not the address of the first element of the array. In many circumstances, it effectively acts like that, but there are differences. When an array is used in an expression, it is automatically converted to a pointer to its first element—but not if it is the operand of sizeof or unary &. (And also not if it is a string literal being used to initialize an array.) So, in sizeof m, m is not a pointer to its first element. It is the entire array, and sizeof m produces the size of the entire array, not of the pointer.

Second, using "hello" as the initializer has different effects because of the rule about using string literals to initialize arrays. Specifically, the rule is in C 2018 6.7.9 14, which says:

An array of character type may be initialized by a character string literal or UTF–8 string literal, optionally enclosed in braces. Successive bytes of the string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.

Third, the string literal is not assigned in these definitions. It is used to initialize the object. Assignment and initialization are different and have different rules. Assignments are expressions and behave much like other expressions. Initializations are part of declarations and have some special syntax and semantics that assignments do not. That is why some source code behaves different in initializations than it does in assignments.

In regard to this question:

Is char m = 'h','e','l','l','0'; a Syntactic sugar to char m = "hello";

It is the other way around; we would say char m = "hello"; is the sweeter version, whereas char m = 'h','e','l','l','o',0; is raw. But, yes, they have the same effect. (Note I corrected two errors: There should be an 'o' in the list, and the final element should be 0, not '0'. '0' is the character for the digit zero, while 0 is zero, which serves as the null character.)

why is char m = "hello"; a copy of the content.

This is the array initialization syntax. The compiler will allocate enough new memory for the character array and then copy the contents of the string literal (not the same as a string constant) into it.

m itself is also the first address of the array, why does the constant string >"hello" assign them to different effects?

Using just the array name to get the equivalent of &m[0] is just a compiler convenience. We also say that the array, when used just by it's name, 'decays' into a pointer to the first element.

When you use char *pm = "Hello"; that is not an initialization, but an assignment. You assign the address of the string constant "Hello" to the pointer pm. The string constant is stored in a different memory segment, usually read-only.

Is char m = 'h','e','l','l','0'; a Syntactic sugar to char m = "hello";

No. The first version will NOT include the terminating null character.