C实现nasm代码

Question

我想开始转换一个小nasm项目{c}，以了解更多关于这个小软合成器的信息。

问题是我的asm知识非常非常生疏，我想知道从哪里开始。我想也许有一个反编译器可以帮助我，但是我还没有找到任何能够将这些简单的nasm列表转换为c的开源代码。

另一种方法是做转换asm->ç手动，但我努力理解的最简单的功能之一:(

即：

;distortion_machine 
;--------------------------- 
;float a 
;float b 
;--------------------------- 
;ebp: distort definition 
;edi: stackptr 
;ecx: length 
section distcode code align=1 
distortion_machine: 
    pusha 
    add ecx, ecx 
    .sampleloop: 
     fld dword [edi] 
     fld dword [ebp+0] 
     fpatan 
     fmul dword [ebp+4] 
     fstp dword [edi] 
     scasd 
    loop .sampleloop 
    popa 
    add esi, byte 8 
ret 

破尝试：

void distortion_machine(???) { // pusha; saving all registers 
    int ecx = ecx+ecx; // add ecx, ecx; this doesn't make sense 

    while(???) { // .sampleloop; what's the condition? 
     float a = [edi]; // fld dword [edi]; docs says edi is stackptr, what's the meaning? 
     float b = [ebp+0]; // fld dword [ebp+0]; docs says ebp is distort definition, is that an input parameter? 
     float c = atan(a,b); // fpatan; 
     float d = c*[ebp+4]; // fmul dword [ebp+4]; 
     // scasd; what's doing this instruction? 
    } 

    return ???; 

    // popa; restoring all registers 
    // add esi, byte 8; 
} 

我想上面的nasm列表是一个非常简单的循环扭曲一个简单的音频缓冲区，但我不明白哪些是输入和哪些是输出，我做甚至不理解循环条件：'）

任何有关上述例程的帮助，以及如何在这个小小的教育项目上取得进展，我们将非常感激。

Answer 1

有一些猜测在这里：

;distortion_machine 
;--------------------------- 
;float a << input is 2 arrays of floats, a and b, successive on stack 
;float b 
;--------------------------- 
;ebp: distort definition << 2 floats that control distortion 
;edi: stackptr   << what it says 
;ecx: length    << of each input array (a and b) 
section distcode code align=1 
distortion_machine: 
    pusha  ; << save all registers 
    add ecx, ecx ; << 2 arrays, so double for element count of both 
    .sampleloop: 
     fld dword [edi] ; << Load next float from stack 
     fld dword [ebp+0] ; << Load first float of distortion control 
     fpatan    ; << Distort with partial atan. 
     fmul dword [ebp+4] ; << Scale by multiplying with second distortion float 
     fstp dword [edi] ; << Store back to same location 
     scasd    ; << Funky way to incremement stack pointer 
    loop .sampleloop  ; << decrement ecx and jump if not zero 
    popa     ; << restore registers 
    add esi, byte 8  ; << See call site. si purpose here isn't stated 
ret 

这是一个真正的猜测，但esi可以是单独的参数堆栈指针，并a和b地址已经被推那里。此代码通过对数据堆栈布局进行假设来忽略它们，但仍需要从arg堆栈中删除这些指针。

近似C：

struct distortion_control { 
    float level; 
    float scale; 
}; 

// Input: float vectors a and b stored consecutively in buf. 
void distort(struct distortion_control *c, float *buf, unsigned buf_size) { 
    buf_size *= 2; 
    do { // Note both this and the assembly misbehave if buf_size==0 
    *buf = atan2f(*buf, c->level) * c->scale; 
    ++buf; 
    } while (--buf_size); 
} 

在C重新实现，你可能想更明确和解决零大小的缓冲区错误。这将花费不多：

void distort(struct distortion_control *c, float *a, float *b, unsigned size) { 
    for (unsigned n = size; n; --n, ++a) *a = atan2f(*a, c->level) * c->scale; 
    for (unsigned n = size; n; --n, ++b) *b = atan2f(*b, c->level) * c->scale; 
}