ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<IsResourceInFile>:
  endbr64
  push   %r13
  lea    D1(%rip),%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <fopen@plt>
  test   %rax,%rax
  je     L23
  mov    $0x2f,%esi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strrchr@plt>
  mov    $0x2e,%esi
  test   %rax,%rax
  mov    %rax,%rdi
  cmove  %rbp,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L105
  movzbl 0x1(%rax),%edx
  and    $0xffffffdf,%edx
  cmp    $0x42,%dl
  jne    L115
  movzbl 0x2(%rax),%edx
  and    $0xffffffdf,%edx
  cmp    $0x49,%dl
  jne    L105
  movzbl 0x3(%rax),%edx
  and    $0xffffffdf,%edx
  cmp    $0x4e,%dl
  jne    L105
  cmpb   $0x0,0x4(%rax)
  je     L26
  nopl   0x0(%rax,%rax,1)
L105:
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <IsResourceFork@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <fclose@plt>
  test   %r13d,%r13d
  je     L3
L52:
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1110 <HasResourceFork@plt>
  nopl   (%rax)
L23:
  xor    %r13d,%r13d
  pop    %rbp
  pop    %r12
  mov    %r13d,%eax
  pop    %r13
  ret
  nopl   0x0(%rax)
L115:
  cmp    $0x48,%dl
  jne    L105
  movzbl 0x2(%rax),%edx
  and    $0xffffffdf,%edx
  cmp    $0x51,%dl
  jne    L105
  movzbl 0x3(%rax),%edx
  and    $0xffffffdf,%edx
  cmp    $0x58,%dl
  jne    L105
  cmpb   $0x0,0x4(%rax)
  jne    L105
  mov    %r12,%rdi
  call   <IsResourceInHex@plt>
  test   %rax,%rax
  je     L105
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %r12,%rdi
  call   <IsResourceInBinary@plt>
  test   %rax,%rax
  je     L105
L1:
  mov    %r12,%rdi
  mov    $0x1,%r13d
  call   <fclose@plt>
  jmp    L52
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_guessed: string

tool: D1: r

assistant: ```c
int IsResourceInFile(char *filename) {
  FILE *f;
  char *spt, *pt;
  int ret;
  f = fopen(filename, "r");
  if (f == NULL)
    return (0);
  spt = strrchr(filename, '/');
  if (spt == NULL)
    spt = filename;
  pt = strrchr(spt, '.');
  if (pt != NULL && (pt[1] == 'b' || pt[1] == 'B') && (pt[2] == 'i' || pt[2] == 'I') && (pt[3] == 'n' || pt[3] == 'N') && pt[4] == '\0') {
    if (IsResourceInBinary(f)) {
      fclose(f);
      return (1);
    }
  } else if (pt != NULL && (pt[1] == 'h' || pt[1] == 'H') && (pt[2] == 'q' || pt[2] == 'Q') && (pt[3] == 'x' || pt[3] == 'X') && pt[4] == '\0') {
    if (IsResourceInHex(f)) {
      fclose(f);
      return (1);
    }
  }
  ret = IsResourceFork(f, 0);
  fclose(f);
  if (!ret)
    ret = HasResourceFork(filename);
  return (ret);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<IsResourceInFile>:
  endbr64
  push   %r13
  lea    D7(%rip),%rsi
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <fopen@plt>
  test   %rax,%rax
  je     L2
  mov    $0x2f,%esi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strrchr@plt>
  mov    $0x2e,%esi
  test   %rax,%rax
  mov    %rax,%rdi
  cmove  %rbp,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L69
  movzbl 0x1(%rax),%edx
  and    $0xffffffdf,%edx
  cmp    $0x42,%dl
  jne    L111
  movzbl 0x2(%rax),%edx
  and    $0xffffffdf,%edx
  cmp    $0x49,%dl
  jne    L69
  movzbl 0x3(%rax),%edx
  and    $0xffffffdf,%edx
  cmp    $0x4e,%dl
  jne    L69
  cmpb   $0x0,0x4(%rax)
  je     L122
  nopl   0x0(%rax,%rax,1)
L69:
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <IsResourceFork@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <fclose@plt>
  test   %r13d,%r13d
  je     L78
L32:
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L78:
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1110 <HasResourceFork@plt>
  nopl   (%rax)
L2:
  xor    %r13d,%r13d
  pop    %rbp
  pop    %r12
  mov    %r13d,%eax
  pop    %r13
  ret
  nopl   0x0(%rax)
L111:
  cmp    $0x48,%dl
  jne    L69
  movzbl 0x2(%rax),%edx
  and    $0xffffffdf,%edx
  cmp    $0x51,%dl
  jne    L69
  movzbl 0x3(%rax),%edx
  and    $0xffffffdf,%edx
  cmp    $0x58,%dl
  jne    L69
  cmpb   $0x0,0x4(%rax)
  jne    L69
  mov    %r12,%rdi
  call   <IsResourceInHex@plt>
  test   %rax,%rax
  je     L69
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L122:
  mov    %r12,%rdi
  call   <IsResourceInBinary@plt>
  test   %rax,%rax
  je     L69
L39:
  mov    %r12,%rdi
  mov    $0x1,%r13d
  call   <fclose@plt>
  jmp    L32
```

tool_calls: name: retrieve_data
arguments: data_label: D7
type_infered: string

tool: D7: r

assistant: ```c
int IsResourceInFile(char *filename) {
  FILE *f;
  char *spt, *pt;
  int ret;
  f = fopen(filename, "r");
  if (f == NULL)
    return (0);
  spt = strrchr(filename, '/');
  if (spt == NULL)
    spt = filename;
  pt = strrchr(spt, '.');
  if (pt != NULL && (pt[1] == 'b' || pt[1] == 'B') && (pt[2] == 'i' || pt[2] == 'I') && (pt[3] == 'n' || pt[3] == 'N') && pt[4] == '\0') {
    if (IsResourceInBinary(f)) {
      fclose(f);
      return (1);
    }
  } else if (pt != NULL && (pt[1] == 'h' || pt[1] == 'H') && (pt[2] == 'q' || pt[2] == 'Q') && (pt[3] == 'x' || pt[3] == 'X') && pt[4] == '\0') {
    if (IsResourceInHex(f)) {
      fclose(f);
      return (1);
    }
  }
  ret = IsResourceFork(f, 0);
  fclose(f);
  if (!ret)
    ret = HasResourceFork(filename);
  return (ret);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<countDigits1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  jmp    L8
L2:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <count1@plt>
  mov    -0x18(%rbp),%rdx
  mov    %eax,(%rdx)
  addq   $0x4,-0x18(%rbp)
  addq   $0x4,-0x10(%rbp)
  subq   $0x1,-0x8(%rbp)
L8:
  cmpq   $0x0,-0x8(%rbp)
  jne    L2
  nop
  nop
  leave
  ret
```

assistant: ```c
void countDigits1(size_t length, const uint32_t *dataArray, int *countArray) {
  while (length) {
    *countArray = count1(*dataArray);
    countArray++;
    dataArray++;
    length--;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<countDigits1>:
  endbr64
  test   %rdi,%rdi
  je     L8
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    $0x0,%ebx
L13:
  mov    (%r12,%rbx,4),%edi
  call   <count1@plt>
  mov    %eax,0x0(%r13,%rbx,4)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L13
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  ret
```

assistant: ```c
void countDigits1(size_t length, const uint32_t *dataArray, int *countArray) {
  while (length) {
    *countArray = count1(*dataArray);
    countArray++;
    dataArray++;
    length--;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<countDigits1>:
  endbr64
  test   %rdi,%rdi
  je     L11
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  xchg   %ax,%ax
L1:
  mov    (%r12,%rbx,4),%edi
  call   <count1@plt>
  mov    %eax,0x0(%r13,%rbx,4)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L1
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  ret
```

assistant: ```c
void countDigits1(size_t length, const uint32_t *dataArray, int *countArray) {
  while (length) {
    *countArray = count1(*dataArray);
    countArray++;
    dataArray++;
    length--;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<countDigits1>:
  endbr64
  test   %rdi,%rdi
  je     L18
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  xchg   %ax,%ax
L4:
  mov    (%r12,%rbx,4),%edi
  call   <count1@plt>
  mov    %eax,0x0(%r13,%rbx,4)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L4
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  ret
```

assistant: ```c
void countDigits1(size_t length, const uint32_t *dataArray, int *countArray) {
  while (length) {
    *countArray = count1(*dataArray);
    countArray++;
    dataArray++;
    length--;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<lc_lin_fir>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  cvtsd2ss %xmm0,%xmm0
  mov    (%rdi),%edx
  mov    %edx,%ecx
  shr    $0x1f,%ecx
  lea    (%rdx,%rcx,1),%eax
  and    $0x1,%eax
  sub    %ecx,%eax
  cmp    $0x1,%eax
  je     L56
  add    $0x1,%edx
  mov    %edx,(%rdi)
L56:
  mov    0x0(%r13),%eax
  lea    0x1(%rax),%edx
  mov    %edx,%r12d
  shr    $0x1f,%r12d
  add    %edx,%r12d
  sar    %r12d
  movaps %xmm0,%xmm1
  addss  %xmm0,%xmm1
  movss  %xmm1,0x0(%rbp)
  cvtss2sd %xmm0,%xmm0
  mulsd  D58(%rip),%xmm0
  movsd  %xmm0,0x8(%rsp)
  cmp    $0x2,%eax
  jle    L9
  mov    $0x1,%ebx
L7:
  pxor   %xmm2,%xmm2
  cvtsi2sd %ebx,%xmm2
  movsd  %xmm2,(%rsp)
  movapd %xmm2,%xmm0
  mulsd  0x8(%rsp),%xmm0
  call   <sin@plt>
  movsd  D64(%rip),%xmm1
  mulsd  (%rsp),%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,0x0(%rbp,%rbx,4)
  add    $0x1,%rbx
  cmp    %ebx,%r12d
  jg     L7
  pxor   %xmm0,%xmm0
  cvtsi2sdl 0x0(%r13),%xmm0
  movsd  D58(%rip),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%rsp)
L35:
  movslq %r12d,%rax
  lea    0x0(%rbp,%rax,4),%rbp
  mov    $0x0,%ebx
L30:
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  addsd  D17(%rip),%xmm0
  mulsd  (%rsp),%xmm0
  call   <cos@plt>
  mulsd  D17(%rip),%xmm0
  movsd  D17(%rip),%xmm1
  subsd  %xmm0,%xmm1
  cvtsd2ss %xmm1,%xmm1
  mulss  -0x4(%rbp),%xmm1
  movss  %xmm1,-0x4(%rbp)
  add    $0x1,%ebx
  sub    $0x4,%rbp
  cmp    %ebx,%r12d
  jg     L30
L6:
  mov    $0x1,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  test   %eax,%eax
  jle    L6
  pxor   %xmm0,%xmm0
  cvtsi2sdl 0x0(%r13),%xmm0
  movsd  D58(%rip),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%rsp)
  jmp    L35
```

tool_calls: name: read_data
arguments: data_tag: D58
datatype: byte[8]
name: read_data
arguments: data_tag: D64
datatype: byte[8]
name: read_data
arguments: data_tag: D17
datatype: f64

tool: {"D58": [0x00,0x00,0x00,0x60,0xfb,0x21,0x19,0x40]}

tool: {"D64": [0x00,0x00,0x00,0x60,0xfb,0x21,0x09,0x40]}

tool: D17: 0.5

assistant: ```c
int lc_lin_fir(fc, nf, coef)
register float fc;
float *coef;
int *nf;
{
  register int i, n;
  register double twopi, fn, c;
  if (((*nf % 2) != 1))
    *nf = *nf + 1;
  n = (*nf + 1) / 2;
  twopi = (3.1415926536f) * 2.0;
  coef[0] = (float)(2.0 * fc);
  c = (3.1415926536f);
  fn = twopi * fc;
  for (i = 1; i < n; i++)
    coef[i] = (float)(sin(i * fn) / (c * i));
  fn = twopi / (double)(*nf);
  for (i = 0; i < n; i++)
    coef[n - i - 1] *= (float)((.5 - (.5 * cos(fn * ((double)i + 0.5)))));
  return (1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<lc_lin_fir>:
  endbr64
  push   %r12
  cvtsd2ss %xmm0,%xmm0
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    (%rdi),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0x1,%edx
  sub    %ecx,%edx
  cmp    $0x1,%edx
  je     L26
  add    $0x1,%eax
  mov    %eax,(%rdi)
L26:
  movaps %xmm0,%xmm1
  lea    0x1(%rax),%edx
  addss  %xmm0,%xmm1
  cvtss2sd %xmm0,%xmm0
  mov    %edx,%ebx
  shr    $0x1f,%ebx
  add    %edx,%ebx
  movss  %xmm1,(%r12)
  movsd  D28(%rip),%xmm1
  sar    %ebx
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x10(%rsp)
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x18(%rsp)
  cmp    $0x2,%eax
  jle    L7
  mov    $0x1,%ebp
  nopl   0x0(%rax,%rax,1)
L9:
  pxor   %xmm1,%xmm1
  movsd  0x10(%rsp),%xmm0
  cvtsi2sd %ebp,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm1,0x8(%rsp)
  call   <sin@plt>
  movsd  0x8(%rsp),%xmm1
  mulsd  D35(%rip),%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,(%r12,%rbp,4)
  add    $0x1,%rbp
  cmp    %ebp,%ebx
  jg     L9
L42:
  movslq %ebx,%rax
  movsd  D3(%rip),%xmm2
  xor    %ebp,%ebp
  lea    (%r12,%rax,4),%r12
  nopw   0x0(%rax,%rax,1)
L29:
  pxor   %xmm0,%xmm0
  sub    $0x4,%r12
  cvtsi2sd %ebp,%xmm0
  add    $0x1,%ebp
  addsd  %xmm2,%xmm0
  mulsd  0x18(%rsp),%xmm0
  call   <cos@plt>
  mov    D3(%rip),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm2
  movq   %rax,%xmm0
  mulsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  mulss  (%r12),%xmm0
  movss  %xmm0,(%r12)
  cmp    %ebp,%ebx
  jg     L29
  add    $0x20,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L7:
  test   %eax,%eax
  jg     L42
  add    $0x20,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D28
type: byte[8]
name: retrieve_data
arguments: data_tag: D35
type: byte[8]
name: retrieve_data
arguments: data_tag: D3
type: f64

tool: {"D28": [0x00,0x00,0x00,0x60,0xfb,0x21,0x19,0x40]}

tool: {"D35": [0x00,0x00,0x00,0x60,0xfb,0x21,0x09,0x40]}

tool: D3: 0.5

assistant: ```c
int lc_lin_fir(fc, nf, coef)
register float fc;
float *coef;
int *nf;
{
  register int i, n;
  register double twopi, fn, c;
  if (((*nf % 2) != 1))
    *nf = *nf + 1;
  n = (*nf + 1) / 2;
  twopi = (3.1415926536f) * 2.0;
  coef[0] = (float)(2.0 * fc);
  c = (3.1415926536f);
  fn = twopi * fc;
  for (i = 1; i < n; i++)
    coef[i] = (float)(sin(i * fn) / (c * i));
  fn = twopi / (double)(*nf);
  for (i = 0; i < n; i++)
    coef[n - i - 1] *= (float)((.5 - (.5 * cos(fn * ((double)i + 0.5)))));
  return (1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<lc_lin_fir>:
  endbr64
  push   %r12
  cvtsd2ss %xmm0,%xmm0
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    (%rdi),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  lea    (%rax,%rcx,1),%edx
  and    $0x1,%edx
  sub    %ecx,%edx
  cmp    $0x1,%edx
  je     L40
  add    $0x1,%eax
  mov    %eax,(%rdi)
L40:
  movaps %xmm0,%xmm1
  lea    0x1(%rax),%edx
  addss  %xmm0,%xmm1
  cvtss2sd %xmm0,%xmm0
  mov    %edx,%ebx
  shr    $0x1f,%ebx
  add    %edx,%ebx
  movss  %xmm1,(%r12)
  movsd  D14(%rip),%xmm1
  sar    %ebx
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,0x10(%rsp)
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x18(%rsp)
  cmp    $0x2,%eax
  jle    L10
  mov    $0x1,%ebp
  nopl   0x0(%rax,%rax,1)
L22:
  pxor   %xmm1,%xmm1
  movsd  0x10(%rsp),%xmm0
  cvtsi2sd %ebp,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm1,0x8(%rsp)
  call   <sin@plt>
  movsd  0x8(%rsp),%xmm1
  mulsd  D22(%rip),%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,(%r12,%rbp,4)
  add    $0x1,%rbp
  cmp    %ebp,%ebx
  jg     L22
L24:
  movslq %ebx,%rax
  movsd  D28(%rip),%xmm2
  xor    %ebp,%ebp
  lea    (%r12,%rax,4),%r12
  nopw   0x0(%rax,%rax,1)
L11:
  pxor   %xmm0,%xmm0
  sub    $0x4,%r12
  cvtsi2sd %ebp,%xmm0
  add    $0x1,%ebp
  addsd  %xmm2,%xmm0
  mulsd  0x18(%rsp),%xmm0
  call   <cos@plt>
  mov    D28(%rip),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm2
  movq   %rax,%xmm0
  mulsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  mulss  (%r12),%xmm0
  movss  %xmm0,(%r12)
  cmp    %ebp,%ebx
  jg     L11
  add    $0x20,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L10:
  test   %eax,%eax
  jg     L24
  add    $0x20,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D14
datatype: byte[8]
name: disassemble_data
arguments: tag: D22
datatype: byte[8]
name: disassemble_data
arguments: tag: D28
datatype: f64

tool: {"D14": [0x00,0x00,0x00,0x60,0xfb,0x21,0x19,0x40]}

tool: {"D22": [0x00,0x00,0x00,0x60,0xfb,0x21,0x09,0x40]}

tool: D28: 0.5

assistant: ```c
int lc_lin_fir(fc, nf, coef)
register float fc;
float *coef;
int *nf;
{
  register int i, n;
  register double twopi, fn, c;
  if (((*nf % 2) != 1))
    *nf = *nf + 1;
  n = (*nf + 1) / 2;
  twopi = (3.1415926536f) * 2.0;
  coef[0] = (float)(2.0 * fc);
  c = (3.1415926536f);
  fn = twopi * fc;
  for (i = 1; i < n; i++)
    coef[i] = (float)(sin(i * fn) / (c * i));
  fn = twopi / (double)(*nf);
  for (i = 0; i < n; i++)
    coef[n - i - 1] *= (float)((.5 - (.5 * cos(fn * ((double)i + 0.5)))));
  return (1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <gcd@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%esi
  mov    $0x2,%edi
  call   <gcd@plt>
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x2,%esi
  mov    $0x0,%edi
  call   <gcd@plt>
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x2,%esi
  mov    $0x2,%edi
  call   <gcd@plt>
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x17,%esi
  mov    $0x1,%edi
  call   <gcd@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x1,%esi
  mov    $0x17,%edi
  call   <gcd@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x5,%esi
  mov    $0x7,%edi
  call   <gcd@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x2ef,%esi
  mov    $0x21,%edi
  call   <gcd@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xa,%esi
  mov    $0x46,%edi
  call   <gcd@plt>
  cmp    $0xa,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x46,%esi
  mov    $0xa,%edi
  call   <gcd@plt>
  cmp    $0xa,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x5a,%esi
  mov    $0x46,%edi
  call   <gcd@plt>
  cmp    $0xa,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x46,%esi
  mov    $0x5a,%edi
  call   <gcd@plt>
  cmp    $0xa,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  assert(0 == gcd(0, 0));
  assert(2 == gcd(2, 0));
  assert(2 == gcd(0, 2));
  assert(2 == gcd(2, 2));
  assert(1 == gcd(1, 23));
  assert(1 == gcd(23, 1));
  assert(1 == gcd(7, 5));
  assert(1 == gcd(33, 751));
  assert(10 == gcd(70, 10));
  assert(10 == gcd(10, 70));
  assert(10 == gcd(70, 90));
  assert(10 == gcd(90, 70));
  return (0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <gcd@plt>
  test   %eax,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%esi
  mov    $0x2,%edi
  call   <gcd@plt>
  cmp    $0x2,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x2,%esi
  mov    $0x0,%edi
  call   <gcd@plt>
  cmp    $0x2,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x2,%esi
  mov    $0x2,%edi
  call   <gcd@plt>
  cmp    $0x2,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x17,%esi
  mov    $0x1,%edi
  call   <gcd@plt>
  cmp    $0x1,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x1,%esi
  mov    $0x17,%edi
  call   <gcd@plt>
  cmp    $0x1,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x5,%esi
  mov    $0x7,%edi
  call   <gcd@plt>
  cmp    $0x1,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x2ef,%esi
  mov    $0x21,%edi
  call   <gcd@plt>
  cmp    $0x1,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0xa,%esi
  mov    $0x46,%edi
  call   <gcd@plt>
  cmp    $0xa,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x46,%esi
  mov    $0xa,%edi
  call   <gcd@plt>
  cmp    $0xa,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x5a,%esi
  mov    $0x46,%edi
  call   <gcd@plt>
  cmp    $0xa,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x46,%esi
  mov    $0x5a,%edi
  call   <gcd@plt>
  cmp    $0xa,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  assert(0 == gcd(0, 0));
  assert(2 == gcd(2, 0));
  assert(2 == gcd(0, 2));
  assert(2 == gcd(2, 2));
  assert(1 == gcd(1, 23));
  assert(1 == gcd(23, 1));
  assert(1 == gcd(7, 5));
  assert(1 == gcd(33, 751));
  assert(10 == gcd(70, 10));
  assert(10 == gcd(10, 70));
  assert(10 == gcd(70, 90));
  assert(10 == gcd(90, 70));
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  xor    %edi,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  xor    %esi,%esi
  mov    $0x2,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0x2,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x2,%esi
  xor    %edi,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0x2,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x2,%esi
  mov    $0x2,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0x2,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x17,%esi
  mov    $0x1,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0x1,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x1,%esi
  mov    $0x17,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0x1,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x5,%esi
  mov    $0x7,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0x1,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x2ef,%esi
  mov    $0x21,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0x1,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0xa,%esi
  mov    $0x46,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0xa,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x46,%esi
  mov    $0xa,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0xa,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x5a,%esi
  mov    $0x46,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0xa,%eax
  sete   %dil
  call   <assert@plt>
  mov    $0x46,%esi
  mov    $0x5a,%edi
  call   <gcd@plt>
  xor    %edi,%edi
  cmp    $0xa,%eax
  sete   %dil
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  assert(0 == gcd(0, 0));
  assert(2 == gcd(2, 0));
  assert(2 == gcd(0, 2));
  assert(2 == gcd(2, 2));
  assert(1 == gcd(1, 23));
  assert(1 == gcd(23, 1));
  assert(1 == gcd(7, 5));
  assert(1 == gcd(33, 751));
  assert(10 == gcd(70, 10));
  assert(10 == gcd(10, 70));
  assert(10 == gcd(70, 90));
  assert(10 == gcd(90, 70));
  return (0);
}
```