c - 将索引访问转换为指针访问

Question

假设我有一个函数 int access_p_i(uintptr_t i) 可以访问p[i] 其中 char *p; 是一个全局变量。

char *p;
int access_p_i(uintptr_t i)
{
    return printf("p[%lx]=%c\n", i, p[i]);
}

预期用途如下:

char buf[] = "abcde";
p = buf; /*set p once*/
access_p_i(3); //prints p[3]=d
access_p_i(2); //prints p[2]=c

假设我想劫持它(不修改函数)以获取绝对指针:

memset(&p,0,sizeof(char*)); /*set p once*/
access_p_i((uintptr_t)"d");
access_p_i((uintptr_t)"c");

这会是可移植且合法的 C 语言吗？

Answer 1

不，我认为这不太好。来自 ANSI/ISO 9899-1990，关于向指针添加或从指针中减去整数:

If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined. Unless both the pointer operand and the result point to elements of the same array object, or the pointer operand points one past the last element of an array object and the result points to an element of the same array object, the behavior is undelined if the result is used as an operand of the unary * operator.