有关strcpy函数的问题
#include <string.h>#include <stdio.h>
void main()
{
int i,j,c;
char p;
char *arr1={"123","321","213"};
char **arr=arr1;
for(i=0;i<2;i++)
{
for(j=0;j<2-i;j++)
{
if((c=strcmp( *(arr+j),*(arr+j+1) )) <0)
{
strcpy(p,*(arr+j)); /*这里能复制过去
strcpy(*(arr+j),*(arr+j+1));/*这里为什么不行
strcpy(*(arr+j+1),p);
}
}
}
printf("%s %s %s",*(arr+0),*(arr+1),*(arr+2));
} 本帖最后由 xypmyp 于 2019-10-16 11:08 编辑
The problem related to how pointer works in 2D array.
/*
int i, j, c;
char arr = {
{"123"},
{"321"},
{"213"}
};
char* pointerTo2DArray = &arr;
*/
int i, j, c;
char p;
char *arr1 = { "123","321","213" };
char **arr = arr1;
strcpy((char*)((int*)arr +0), (char*)((int*)arr + 1));
printf("%s %s %s", (char*)*((int*)arr + 0), (char*)*((int*)arr + 1),(char*)*((int*)arr + 2));
getchar();
return 0;
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 .
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 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? 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？？ 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)); 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？int i, j, c;
char p;
char *arr1 = { "123","321","213" };
char **arr = arr1;
strcpy((arr +0), (arr + 1));
printf("%s %s %s", *(arr + 0), *(arr + 1),*(arr + 2));
getchar();
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