如何确定的函数的长度？ [英] How to determine the length of a function?

问题描述

考虑以下code，是以函数f（），复制函数本身以其整体向一个缓冲器，修改其code和运行功能改变。在实践中，原始函数，返回号码22被克隆和修饰的返回号码42

Consider the following code that takes the function f(), copies the function itself in its entirety to a buffer, modifies its code and runs the altered function. In practice, the original function that returns number 22 is cloned and modified to return number 42.

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

#define ENOUGH 1000
#define MAGICNUMBER 22
#define OTHERMAGICNUMBER 42

int f(void)
{
    return MAGICNUMBER;
}

int main(void)
{
    int i,k;
    char buffer[ENOUGH];
    /* Pointer to original function f */
    int (*srcfptr)(void) = f;
    /* Pointer to hold the manipulated function */
    int (*dstfptr)(void) = (void*)buffer;
    char* byte;
    memcpy(dstfptr, srcfptr, ENOUGH);
    /* Replace magic number inside the function with another */
    for (i=0; i < ENOUGH; i++) {
        byte = ((char*)dstfptr)+i;
        if (*byte == MAGICNUMBER) {
            *byte = OTHERMAGICNUMBER;
        }
    }

    k = dstfptr();
    /* Prints the other magic number */
    printf("Hello %d!\n", k);
    return 0;
}

在code现在依靠只是猜测，该功能将适合在1000字节的缓冲区。它还通过复制过多的缓冲器，由于函数f（违反规则）将大于1000字节最可能短了很多

The code now relies on just guessing that the function will fit in the 1000 byte buffer. It also violates rules by copying too much to the buffer, since the function f() will be most likely a lot shorter than 1000 bytes.

这给我们带来了的问题的：有没有搞清楚用C任一给定函数的大小的方法？一些方法包括寻找到中间连接器输出，并根据函数中的指令猜测，不过这只是不太够。有没有什么办法，以确保？

This brings us to the question: Is there a method to figure out the size of any given function in C? Some methods include looking into intermediate linker output, and guessing based on the instructions in the function, but that's just not quite enough. Is there any way to be sure?

请注意：它编译和作品在我的系统上，但并不完全遵循标准，因为函数指针和空*间的转换不完全允许的：

Please note: It compiles and works on my system but doesn't quite adhere to standards because conversions between function pointers and void* aren't exactly allowed:

$ gcc -Wall -ansi -pedantic fptr.c -o fptr
fptr.c: In function 'main':
fptr.c:21: warning: ISO C forbids initialization between function pointer and 'void *'
fptr.c:23: warning: ISO C forbids passing argument 1 of 'memcpy' between function pointer and 'void *'
/usr/include/string.h:44: note: expected 'void * __restrict__' but argument is of type 'int (*)(void)'
fptr.c:23: warning: ISO C forbids passing argument 2 of 'memcpy' between function pointer and 'void *'
/usr/include/string.h:44: note: expected 'const void * __restrict__' but argument is of type 'int (*)(void)'
fptr.c:26: warning: ISO C forbids conversion of function pointer to object pointer type
$ ./fptr
Hello 42!
$

请注意：一些从可写存储器执行系统是不可能的，这个code会崩溃。它已经过测试与x86_64体系运行在Linux上的gcc 4.4.4。

Please note: on some systems executing from writable memory is not possible and this code will crash. It has been tested with gcc 4.4.4 on Linux running on x86_64 architecture.

推荐答案

您不能在C.做到这一点，即使你知道的长度，功能方面的地址，因为函数调用和访问某些类型的数据将使用程序计数器相对寻址。因此，位于不同地址的函数的副本不会做同样的事情原来。当然，还有很多其他的问题了。

You cannot do this in C. Even if you knew the length, the address of a function matters, because function calls and accesses to certain types of data will use program-counter-relative addressing. Thus, a copy of the function located at a different address will not do the same thing as the original. Of course there are many other issues too.

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