好吧，所以扭曲有点不同，我需要帮助实际崩溃这个程序。别担心，这是我教授的任务，虽然我不确定这是否是提出这个问题的地方。他向我们提供了这个代码以及崩溃它的指令，并通过为两个secret[0]函数提供输入来修改secret[1]和scanf。我尝试用不同的输入来崩溃它，但还没有。
（如果这是本网站上不需要的东西，请告诉我，我会删除这个问题。感谢您提前抽出时间！）
/* vul_prog.c */
#include <stdio.h>
#include <stdlib.h>
#define SECRET1 0x44
#define SECRET2 0x55
int main(int argc, char *argv[])
{
char user_input[100];
int *secret;
int int_input;
int a, b, c, d; /* other variables, not used here.*/
/* The secret value is stored on the heap */
secret = (int *) malloc(2*sizeof(int));
/* getting the secret */
secret[0] = SECRET1; secret[1] = SECRET2;
printf("The variable secret's address is 0x%8x (on stack)\n", (unsigned int)&secret);
printf("The variable secret's value is 0x%8x (on heap)\n", (unsigned int)secret);
printf("secret[0]'s address is 0x%8x (on heap)\n", (unsigned int)&secret[0]);
printf("secret[1]'s address is 0x%8x (on heap)\n", (unsigned int)&secret[1]);
printf("Please enter a decimal integer\n");
scanf("%d", &int_input); /* getting an input from user */
printf("Please enter a string\n");
scanf("%s", user_input); /* getting a string from user */
/* Vulnerable place */
printf(user_input);
printf("\n");
/* Verify whether your attack is successful */
printf("The original secrets: 0x%x -- 0x%x\n", SECRET1, SECRET2);
printf("The new secrets: 0x%x -- 0x%x\n", secret[0], secret[1]);
return 0;
}