The C language uses pointers and arrays to implement reverse string and reverse string order.

The C language uses pointers and arrays to implement reverse string and reverse string order.

1. The Array Implementation (without using string functions) program is as follows:

1 void ReverseByArray (char * s)
 2 {
 3 int len = 0;
 4 while (s [len]! = '\ 0')
 5 len ++;
 6 int t;
 7 for (int i = 0; i <len / 2; i ++)
 8     {
 9 t = s [i];
10 s [i] = s [len-i-1];
11 s [len-i-1] = t;
12}
13}
For operations related to arrays, you must use the string length len (because the length is relatively simple, you don't need a string function)

On line 3-5, first check the '\ 0' character to determine the length of the string and record it with the len variable

I tried the following code. The measured string length is 1 larger than the original

int len = 0;
while (s [len ++]! = '\ 0')
; // Empty statement
Don't use the above method

Then the specific implementation of the reverse order,

There are several common methods, many online are malloc out len + 1 size space, and then put it in reverse order

But I think that since it is an array implementation, I use a temporary variable t to directly exchange the method before and after the exchange (the loop is just the loop len / 2)

Another problem that is prone to occur is when switching s [len-i-1], why do we need to decrease by 1?

 

2. Use the pointer method to reverse the string

 1 void ReverseByPointer (char * s)
 2 {
 3 char buf [100];
 4 char * b = buf;
 5 char * p = s;
 6 while (* p! = '\ 0')
 7 * (b ++) = * (p ++);
 8 while (* s! = '\ 0')
 9 * (s ++) = * (-b);
10}
Use pointers to highlight the characteristics of pointers, unlike arrays

At first glance, the program seems particularly simple.

But this program uses a buffer,

If this buffer size is not enough ↓ as follows


When the buffer is enough, the program will run normally.

The two newly created pointer variables b and p are not superfluous.

buf is the name of the array. Although the value is the first address of the array, this first address cannot be added or subtracted, so a new b pointer is created to point to buf.

s is of type char *, it can be added or subtracted, but how to add it to the end (after finding \ 0)?

Then I chose to define a p pointer variable and assign it to the pointer s, that is, s and p are the same

And p moves back to find '\ 0', its task is completed, there is no need to return to the starting position

Finally, fill the content of buf-(inverted) and fill in the string pointed to by s.

Finally, there is the problem of \ 0. Because the string is only in reverse order, the length does not change, so \ 0 is still the original s of \ 0, and you do n’t need to add \ 0

 

3. Complete program and operation result

 1 #include <stdio.h>
 2 
 3 void ReverseByArray (char * s)
 4 {
 5 int len = 0;
 6 while (s [len]! = '\ 0')
 7 len ++;
 8 int t;
 9 for (int i = 0; i <len / 2; i ++)
10 {
11 t = s [i];
12 s [i] = s [len-i-1];
13 s [len-i-1] = t;
14}
15}
16
17 void ReverseByPointer (char * s)
18 {
19 char buf [100];
20 char * b = buf;
21 char * p = s;
22 while (* p! = '\ 0')
23 * (b ++) = * (p ++);
24 while (* s! = '\ 0')
25 * (s ++) = * (-b);
26}
27
28 int main ()
29 {
30 char s [20] = "I Love you 233";
31 printf ("Source string:% s \ n", s);
32
33 ReverseByArray (s);
34 printf ("Reverse order of array mode:% s \ n", s);
35
36 ReverseByPointer (s);
37 printf ("Reverse order of pointer mode:% s \ n", s);
38}
39
40 / * Running result in VC compiler
41 Source string: I Love you 233
42 Reverse order in array mode: 332 uoy evoL I
43 Reverse pointer mode: I Love you 233
44 Press any key to continue ...
45 * /
 Original Original (Blog Park)