当您将字符串传递给 C 中的函数时，幕后会发生什么

Question

Does the compiler convert it to a string literal?

Is there any difference between these two?

printf("Hello world\n");

and

char* c = "Hello world\n";
printf(c); //I know you shouldn't do this

Answer 1

A string is just an array of char (where content is terminated by the presence of a nul character).

char s[] = "Hi";                // These two declarations
char s[] = { 'H', 'i', '\0' };  // are equivalent

Arrays are not first class citizens in C. They cannot be passed to or returned from functions.

Pointers , on the other hand, are scalar types and can be passed to and returned from functions just as any other scalar type: by value. That is, the value of the pointer (the memory address it holds) is passed to or returned from a function.

So when you type:

puts( "Hello world" );

what happens is that the address of the first character of the literal array is passed to the function.

As you suspect, this is no different than:

const char * s = "Hello world";  // s is a pointer to read-only memory
puts( s );

Again, s is a pointer containing the address of the first character in the literal array. Its value (the address) is passed to the function.

If s were an array:

char s[] = "Hello world";  // s is an array, content copied from read-only memory
puts( s );

then the address of the first element of the array would be passed to the function.

You find this same thing true of any kind of array in C:

int xs[] = { 2, 3, 5, 7, 11, 13 };
some_function( xs );

The array itself is not passed to the function, but the value of the pointer &(xs[0]) is passed , where the value is the address of the first element of the array.

In C we call this “array-to-pointer decay”.

(An array is an address, element type, and count of elements. A pointer is just an address and element type.)