C language address for array name

Author: Qing Dujun

Have you ever thought about getting an address for a one-dimensional array name and then using this address for addition and subtraction. What will happen?

Example:

Int a [5] = {1, 2, 3, 4, 5 };

Int * p = (int *) (& a + 1 );

Printf ("% d \ n", * (p-1 ));

What is the output?

Why? Why do we need to forcibly convert the type in the second line?

A: a is the name of a one-dimensional array, and a corresponds to a pointer pointing to a one-dimensional array. How do you feel so familiar? Is that a row pointer to an array? Int (* p) [].

The row pointer + 1 is used to add or subtract the pointer. The result is still a row pointer ~~~ Originally, it was required to be forcibly converted to a single pointer.

So what is the output result?

A: Of course, it is 5.

Why?

A: I didn't mean that & a is a row pointer, that is, a row pointer. The row pointer + 1 does not point to the next row. Here the row is [5], so int * p = (int *) (& a + 1) it is necessary to deviate from five positions on the basis of a: (the last line indicates the number of digits, and the next line indicates the corresponding data in)

0 1 2 3 4 5

1 2 3 4 5 *

Isn't that out of bounds? Moving the five digits all ran out of the [] array in the '*' position. Oh, I see the output of printf. What you output is P-1). Here, p is a single pointer-1, which is just a shift to the left. Isn't it the return to the position 5? It turns out that the output result is 5.

Next we will use a piece of code to show the displacement ~~~

# Include
 
  
Int main () {int a [5] = {1, 2, 3, 4, 5}; int * p = (int *) (& a + 1 ); // + 1 is equivalent to moving five digits of printf ("% p \ n % p \ nbit = % d \ n", a, p, p-); printf ("% d \ n", * (p-1); return 0 ;}