有关strcpy函数的问题

朋乌龟

1. #include <string.h>
   
2. #include <stdio.h>
   
3. void main()
   
4. {
   
5.      int i,j,c;
   
6.      char p[10];
   
7.      char *arr1[3]={"123","321","213"};
   
8.      char **arr=arr1;   
   
9.      for(i=0;i<2;i++)
   
10.      {
    
11.           for(j=0;j<2-i;j++)
    
12.           {
    
13.               if(  (c=strcmp( *(arr+j),*(arr+j+1) )) <0)
    
14.               {
    
15.                   strcpy(p,*(arr+j));   /*这里能复制过去
    
16.                   strcpy(*(arr+j),*(arr+j+1));  /*这里为什么不行
    
17.                   strcpy(*(arr+j+1),p);
    
18. 
    
19.               }
    
20.           }
    
21.      }
    
22.      printf("%s %s %s",*(arr+0),*(arr+1),*(arr+2));
    
23.      
    
24. }

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xypmyp

The problem related to how pointer works in 2D array.

1. /*
   
2.         int i, j, c;
   
3.         char arr[3][4] = {
   
4.                 {"123"},
   
5.                 {"321"},
   
6.                 {"213"}
   
7.         };
   
8. 
   
9.         char* pointerTo2DArray = &arr[0][0];
   
10. */
    
11. 
    
12.         int i, j, c;
    
13.         char p[10];
    
14.         char *arr1[3] = { "123","321","213" };
    
15.         char **arr = arr1;
    
16.        
    
17.         strcpy((char*)((int*)arr +0), (char*)((int*)arr + 1));
    
18. 
    
19.         printf("%s %s %s", (char*)*((int*)arr + 0), (char*)*((int*)arr + 1),(char*)*((int*)arr + 2));
    
20.         getchar();
    
21.         return 0;
    

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As the lever 1 address is pointing to a memory address which 4 Byte long = sizeof(int), than you convert it to (char*) which strcpy() required [optional, it's the start address of your string].

朋乌龟

xypmyp 发表于 2019-10-16 10:59
The problem related to how pointer works in 2D array.

虽然看不大懂，但还是要谢谢你

xypmyp

朋乌龟 发表于 2019-10-16 22:34
虽然看不大懂，但还是要谢谢你

Which part you don't understand?

朋乌龟

xypmyp 发表于 2019-10-17 10:46
Which part you don't understand?

我可以交换首地址来交换位置，不需要strcpy()，
我理解的是：*（arr+0）这是整形指针？需要把它转换成字符型？？

xypmyp

Yes, as arr is a type of (char**), you have to convert it to (char*)

*(arr+0) is not a (int*) pointer, it's a (char**) pointer which you declare at the beginning,
so you have to convert it to (int*) for pointing to an valid address. which is the start address of the string.
then, you have to convert it to (char*), as the function strcpy() needs a (char*) pointer.

Try +1 or -1 in between, you may understand what i'm saying
printf("%s",((char*)*((int*)arr + 0) +1));

朋乌龟

为什么我一定要将其转换为（int *）才能指向有效地址
我这样也可以，对吗？

1. int i, j, c;
   
2.         char p[10];
   
3.         char *arr1[3] = { "123","321","213" };
   
4.         char **arr = arr1;
   
5.         
   
6.         strcpy((arr +0), (arr + 1));
   
7. 
   
8.         printf("%s %s %s", *(arr + 0), *(arr + 1),*(arr + 2));
   
9.         getchar();

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