Question

我刚刚学习缓冲区溢出。我试图使用GCC复制它。这是我写的代码。

#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[])
{
    int value = 5;
    char buffer_one[8], buffer_two[8];

    strcpy(buffer_one, "one");
    strcpy(buffer_two, "two");

    printf("[BEFORE] buffer_two is at %p and contains %s\n", buffer_two, buffer_two);
    printf("[BEFORE] buffer_one is at %p and contains %s\n", buffer_one, buffer_one);
    printf("[BEFORE] value is at %p and contains %d\n\n", value, value);

    printf("[STRCPY] copying %d bytes into buffer_two\n\n", strlen(argv[1]));
    strcpy(buffer_two, argv[1]);

    printf("[BEFORE] buffer_two is at %p and contains %s\n", buffer_two, buffer_two);
    printf("[BEFORE] buffer_one is at %p and contains %s\n", buffer_one, buffer_one);
    printf("[BEFORE] value is at %p and contains %d\n\n", value, value);

    return 0;
}

好像应该有用，对吗？ Buffer_two和buffer_one在内存中彼此相邻。

[BEFORE] buffer_two is at 0x7fff56ff2b68 and contains two
[BEFORE] buffer_one is at 0x7fff56ff2b70 and contains one
[BEFORE] value is at 0x5 and contains 5

然而，在此之后不久......

[STRCPY] copying 14 bytes into buffer_two

Abort trap: 6

C怎么认识到这一点？一些黑客如何执行更复杂的缓冲区溢出实际工作？

Answer 1

在您的情况下，您已成功通过尝试在14 {{1}的内存区域中编写char 8来产生缓冲区溢出}}第

一旦写入已分配的内存，行为将不确定。所以，char消息就在那里。

为什么GCC不会发生缓冲区溢出？

3 个答案: