使用puts（）在指针指向的位置覆盖新值

Question

我希望得到一些建议。我非常感谢你。

char *p = "abc";
printf("Value at the location the pointer points to = %s\n", p);
printf("The location where the pointer points to = %d\n", p);
printf("The location of the pointer = %d\n", &p);
char s[] = "Hello";
printf("Vi tri cua s = %d\n", &s);
gets(s);  //we must enter less than 5 characters, because we have s[5]
puts(s);
printf("The new value at the location the pointer points to:\n");
gets(p);  //This function cause problem, while the function gets(s) does not
puts(p);

Answer 1

"abc"是文字，并且受写保护（只读）。因此，您无法阅读它。

char s[] = "Hello";创建一个数组并将文字字符串复制到其中。结果数组不是文字，因此可以修改。