C language FAQ 4.5

One-dimensional arrays and pointers relationship:

1. The one-dimensional array name simplify operation

. 1 #include <stdio.h>
 2  void main ()
 . 3  {
 . 4      int I, A [ . 5 ], B [ . 5 ];
 . 5      int * P;
 . 6      for (I = 0 ; I < . 5 ; I ++ )
 . 7              Scanf ( " % D% D " , a + i, & b [i]); // a i + directly at the address, and remove the standard & compared to the address b i of the array, the two equivalent, but The former method is no longer an ampersand.

Then there is a problem: that a [i] = i [a] it?

According to C99, the compiler uses the pointer arithmetic internally to access array elements. For example: a [8] = * (a + 8), then in accordance with exchangeable addition, * (a + 8) = * (8 + a) = 8 [a]

2. The one-dimensional array of numbers pointer operation

Can be done by any array subscripting, pointers can borrow be implemented by means of the pointer from the Operational simplify operation, using a pointer to an array storage helps produce occupies a small space, fast, high-quality code , but pay attention to cross-border issues pointer!

// then the above code 
P = A;
 for (I = 0 ; I < . 5 ; I ++ ) 
    the printf ( " % D " , * (P + I)); // * (P + I) == P [I ] == * (a + I) == a [I]
 // converted address 
the while (P <+ a . 5 ) 
{ 
  the printf ( " % D " , * P); 
   P ++ ; 
} 
// At this point p = a + 5, for normal operation in order p, p needs to redirect a known address! 
= A & P [ 2 ]; // where P [0] = A [2]; 
for (I = 0 ; I < . 3 ; I ++ )
            the printf ( " % D% D " , P [ 2 + I], P [ 2 -i]); // are a [2], a [2 ], a [3], a [1], a [ 4], a [0]

3. Using a one-dimensional array of characters

A one-dimensional array of characters is a string!

 1 #include<stdio.h>
 2 void main()
 3 {
 4     int a[[]={1,2,3,4,5},*p,i;
 5     char c[]="abcde",*cp;
 6     p=&a[2];
 7     cp=&c[2];
 8     for(i=0;i<3;++i)
 9         printf("%d%c%d%c",*(p+i),*(cp+i),*(p-i),*(cp-i));
10     priintf("\n%d%s%c\n",*p,*cp,cp);

Compile error-free, but a run-error (Code tenth row), * cp represent a character you want to use% c, and cp is the first address of the memory strings, so the output from the address pointed to by cp beginning of the string, you need to use % s format, it will be changed

printf("\n%d%c%s\n",*p,*cp,cp);

Please note that the issue of the string end position '\ 0'

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