用十六进制覆盖exe的字节

I want to change the logic of an exe by changing the binary. One of the hex values in the exe is 75 which I have to change to 74 (JNE to JE in x86 assembly). I know it's the 1276th byte of the file, but how do I do this?

我想通过更改二进制来更改exe的逻辑。 exe中的一个十六进制值是75，我必须将其更改为74（x86汇编中的JNE到JE）。我知道这是文件的第1276个字节，但我该怎么做？

Here's what I have:

这就是我所拥有的：

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    FILE *fileH = fopen ("foo", "r+");
    fseek(fileH, 1276, SEEK_SET);
    fwrite("74", 1, 1, fileH);
    fclose(fileH);
    return 0;
}

Also for some reason I'm getting 'fileH' undeclared, even though I included stdio and have FILE in all uppercase. I wasn't able to find anyone else with this problem. Running this on ubuntu

也是出于某些原因我得到'fileH'未声明，即使我包含stdio并且全部都是FILE。我无法找到有这个问题的其他人。在ubuntu上运行它

2 个解决方案

#1

You must define a byte value to write to the file. And the 8086 opcode for JE is 74 in hexadecimal, not decimal.

您必须定义一个字节值以写入该文件。 JE的8086操作码是十六进制的74，而不是十进制。

#include <stdio.h>

int main(void)
{
    unsigned char byt = 0x74;
    FILE *fileH = fopen ("foo.txt", "r+");
    if (fileH == NULL)
        return 1;
    if (fseek(fileH, 3, SEEK_SET))
        return 1;
    if (fwrite(&byt, 1, 1, fileH) != 1)
        return 1;
    if (fclose(fileH))
        return 1;
    printf("File updated\n");
    return 0;
}

As a demo using a small text file, content before:

作为使用小文本文件的演示，内容之前：

0123456789

and after:

之后：

012t456789

I have no idea why your compiler refuses fileH except that sometimes a text editor can leave a rogue unseen character where there should not be one. The solution there, is to delete and retype the offending line.

我不知道为什么你的编译器拒绝fileH，除了有时文本编辑器可以留下一个流氓看不见的字符，其中不应该有一个。那里的解决方案是删除并重新键入违规行。

#2

   int main()
    {
        FILE *fileH = fopen ("foo", "r+");
        fseek(fileH, 1276, SEEK_SET);
        putc(0x74,fileH);
    }

and xxd -p foo to dump in hex your file

和xxd -p foo以十六进制形式转储文件

#1