Some difficult things about the "extended Knowledge 3" array

Some difficult things about the "extended Knowledge 3" array

Extended Directory

1. &array+ 1

2. array+1

3. &array[0]+ 1

Questions about &array+, array+ 1 and &array[0]+ 1 are particularly difficult to understand ~-~. So I'll explain it today.

Since the storage unit for each element in the array is continuously allocated, you can access the array as a pointer, which is the first address of the number.

such as: intarray[100];

Array is the first address of the array with values equal to & Array[0]

PS: Through the address, you can quickly and easily access the array of other element values, the method is as follows:

First address + offset

The value of the element I can be A * (Array+i), * (&array[0]+i), array[i]

Array is the first address of the first element of the array , with the &array[0] Same, while &array is to take the first address of the array.

[ program 1]

#include <stdio.h> int main (void) {         int array[5]= {1, 2, 3, 4, 5};                 printf ("        array[1]=%d\n", array[1]);         printf ("        * (array+1) =%d\n", * (array+ 1));         printf ("* (&array[0]+ 1) =%d\n", * (&array[0]+ 1));                 return 0;}

Operation Result:

array[1]= 2

* (array+1) = 2

* (&array[0]+ 1) = 2

[ program 2]

#include <stdio.h> int main (void) {         int array[5]= {1, 2, 3, 4, 5};                 printf ("        array[1]=%d\n", array[1]);         printf ("        * (array+1) =%d\n", * (array+ 1));        printf ("     * (&array+ 1) =%d\n", * (&array+ 1));//cross-border!!!                  return 0;}

Operation Result:

array[1]= 2

* (array+1) = 2

* (&array[0]+ 1) = 2293440

Analysis:

Array: Is the first address of the first element of the arrays, that is, the first address of array[0]

&array: The first address of the array is taken.

&array+ 1:

Take the first address of the array, the value of this address plus sizeof (array), which is &array+ 5* sizeof (int), and the first address of the next array .

array+ 1://equivalent to &array[0]+ 1;

is to take the first address of the next element of the array, which is array[1].

：

The address of &array+ 1 was found to be out of bounds, the address of which is array[5], i.e. &array+ 1= &array+ 5* sizeof (int). The address of array+1 and &array[0] is the address of array[1].

"Smile at the Fingertips" error is unavoidable, hope to get everyone's correction ^ ^

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Some difficult things about the "extended Knowledge 3" array