#include <stdio.h>
struct Analysis {
int lnlen;
int arr[2];
char* name;
};
int main()
{
struct Analysis ana_space[2];
char *ptr = (void*) &ana_space;
ana_space[0].lnlen = 0;
ana_space[0].arr[0] = 1;
ana_space[0].arr[1] = 2;
ana_space[0].name = "Peter";
printf("\n%d\n", *ptr); // print 0;
*ptr = 10; // how to use memcpy here;
printf("\n%d\n", *ptr); // print 10;
ptr = ptr + sizeof(int); // advance pointer by int;
printf("\n%d\n", *ptr); // print 1;
ptr = ptr + 2*sizeof(int); // advance pointer by 2 ints;
printf("\n%s\n", *ptr); // print "Peter"; --------------not work
//*ptr = "Jim"; // how to assign new name "Jim" into that memory;
return 0;
}
输出:
0
10
1
(null)
我想使用char *作为指针,通过内存地址获取一些数据,并将值存储到内存中。
对于int和int数组,它可以正常工作,但对于字符串则不行。
如何打印字符串并将新的字符串值存储到内存中?
printf("\n%s\n", *ptr);
这样的问题。 - chux - Reinstate Monica