问题
1.char str[] = "hello"; //legal
2.char str1[];
str1 = "hello"; // illegal
I understand that "hello" returns the address of the string literal from the string literal pool which cannot be directly assigned to an array variable. And in the first case the characters from the "hello" literal are copied one by one into the array with a '\0' added at the end.
Is this because the assignment operator "=" is overloaded here to support this?
I would also like to know other interesting cases wherein initialization is different from assignment.
回答1:
You cannot think of it as overloading (which doesn't exist in C anyway), because the initialization of char arrays with string literals is a special case. The type of a string literal is const char[N], so if it were similar to overloading, you'd be able to initialize a char array with any expression whose type is const char[N]. But you cannot!
const char arr[3];
const char arr1[] = arr; //compiler error. Cannot initialize array with another array.
The language standard simply says that character arrays can be initialized with string literals. Since they say nothing about assignment, the general rules apply, in particular, that an array cannot be assigned to.
As for other cases when initialization is different from assignment: in C++, where there are references and classes, there would be zillions of examples. In C, with no full-fledged classes or references, the only other thing I can think of off the top of my head is const variables:
const int a = 4; //OK;
const int b; //Error;
b = 4; //Error;
Another example: array initialization with braces
int a[3] = {1,2,3}; //OK
int b[3];
b = {1,2,3}; //error
Same with structs
回答2:
If you want to think of it as the operator being overloaded (even though C doesn't use the term), you can of course do that.
Do you also consider this to be overloading:
unsigned char x;
double y;
x = 2;
y = 1.243;
Those are assigning totally different types of data, after all, but using the "same operator", right?
It's just different, to be initializing or to be assigning.
Another big difference is that you used to be able to initialize structures, but there was no corresponding "struct literal" syntax for later assignments. This is no longer true as of C99, where we now have compound literals.
回答3:
char str[] = "hello";
Is array initialization, using syntactic sugar defined in C because string initialization is so common. The compiler allocates some fixed memory in your program an initializes it. The name of the array (str) evaluates to the address of this memory, and it cannot be changed because there is no variable which holds that address.
Grijesh Chauhan explains more details of this.
Other cases depend on what you mean. Extending the current case, you can easily see that other initialized arrays have the same properties, for example
int a[] = { 1, 2, 3, 4 };
回答4:
Array has non modifiable address. You need a pointer as a modifiable lvalue.
By assigning(trying) to a contant string literal, you are taking the address of it. Different address causes that illegality.
"hello" allocates some space in memory and gives and address. Then you take its address to initialize the array.
来源：https://stackoverflow.com/questions/17500960/initialization-different-from-assignment