ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<slice>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x24(%rbp),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x1c(%rbp),%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L7
L8:
  mov    -0x20(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L7:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L8
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type: string

tool: D2: start=%d end=%d new_n=%d


assistant: ```c
float *slice(float *buf, int n, int start, int end, int *new_n) {
  int i;
  float *new;
  *new_n = n - (start + ((n - 1) - end));
  printf("start=%d end=%d new_n=%d\n", start, end, *new_n);
  new = malloc(sizeof(float) * *new_n);
  for (i = 0; i < *new_n; i++) {
    new[i] = buf[start + i];
  }
  return new;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<slice>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %edx,%ebp
  mov    %ecx,%edx
  mov    %r8,%rbx
  lea    -0x1(%rsi),%eax
  sub    %ecx,%eax
  add    %ebp,%eax
  sub    %eax,%esi
  mov    %esi,%ecx
  mov    %esi,(%r8)
  mov    %ebp,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  cmpl   $0x0,(%rbx)
  jle    L10
  movslq %ebp,%rbp
  lea    (%r12,%rbp,4),%rcx
  mov    $0x0,%edx
L5:
  movss  (%rcx,%rdx,4),%xmm0
  movss  %xmm0,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %edx,(%rbx)
  jg     L5
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
data_type: string

tool: D6: start=%d end=%d new_n=%d


assistant: ```c
float *slice(float *buf, int n, int start, int end, int *new_n) {
  int i;
  float *new;
  *new_n = n - (start + ((n - 1) - end));
  printf("start=%d end=%d new_n=%d\n", start, end, *new_n);
  new = malloc(sizeof(float) * *new_n);
  for (i = 0; i < *new_n; i++) {
    new[i] = buf[start + i];
  }
  return new;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<slice>:
  endbr64
  lea    -0x1(%rsi),%eax
  push   %r12
  mov    %r8,%r12
  push   %rbp
  sub    %ecx,%eax
  movslq %edx,%rbp
  mov    %ecx,%edx
  add    %ebp,%eax
  push   %rbx
  mov    %rdi,%rbx
  lea    D3(%rip),%rdi
  sub    %eax,%esi
  mov    %esi,(%r8)
  mov    %esi,%ecx
  mov    %ebp,%esi
  call   <printf@plt>
  mov    (%r12),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  movslq (%r12),%rcx
  test   %ecx,%ecx
  jle    L0
  lea    (%rbx,%rbp,4),%rsi
  xor    %edx,%edx
  nopl   (%rax)
L1:
  movss  (%rsi,%rdx,4),%xmm0
  movss  %xmm0,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%rcx
  jne    L1
L0:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D3
datatype: string

tool: D3: start=%d end=%d new_n=%d


assistant: ```c
float *slice(float *buf, int n, int start, int end, int *new_n) {
  int i;
  float *new;
  *new_n = n - (start + ((n - 1) - end));
  printf("start=%d end=%d new_n=%d\n", start, end, *new_n);
  new = malloc(sizeof(float) * *new_n);
  for (i = 0; i < *new_n; i++) {
    new[i] = buf[start + i];
  }
  return new;
}
```
user: What is the c source code of the assembly code below:

```asm
<slice>:
  endbr64
  lea    -0x1(%rsi),%eax
  push   %r12
  mov    %r8,%r12
  push   %rbp
  sub    %ecx,%eax
  mov    %rdi,%rbp
  lea    D4(%rip),%rdi
  push   %rbx
  mov    %edx,%ebx
  mov    %ecx,%edx
  add    %ebx,%eax
  sub    %eax,%esi
  mov    %esi,(%r8)
  mov    %esi,%ecx
  mov    %ebx,%esi
  call   <printf@plt>
  mov    (%r12),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    (%r12),%edx
  mov    %rax,%r9
  test   %edx,%edx
  jle    L71
  movslq %ebx,%rax
  mov    %r9,%rcx
  lea    0x4(,%rax,4),%rax
  lea    0x0(%rbp,%rax,1),%rsi
  sub    %rsi,%rcx
  cmp    $0x8,%rcx
  jbe    L72
  lea    -0x1(%rdx),%ecx
  cmp    $0x2,%ecx
  jbe    L72
  mov    %edx,%ecx
  lea    -0x4(%rbp,%rax,1),%rsi
  xor    %eax,%eax
  shr    $0x2,%ecx
  shl    $0x4,%rcx
  nopl   0x0(%rax)
L2:
  movups (%rsi,%rax,1),%xmm1
  movups %xmm1,(%r9,%rax,1)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L2
  mov    %edx,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%dl
  je     L71
  lea    (%rbx,%rax,1),%ecx
  mov    %eax,%esi
  movslq %ecx,%rcx
  lea    0x0(,%rsi,4),%rdi
  movss  0x0(%rbp,%rcx,4),%xmm0
  lea    0x1(%rax),%ecx
  movss  %xmm0,(%r9,%rsi,4)
  cmp    %ecx,%edx
  jle    L71
  add    %ebx,%ecx
  add    $0x2,%eax
  movslq %ecx,%rcx
  movss  0x0(%rbp,%rcx,4),%xmm0
  movss  %xmm0,0x4(%r9,%rdi,1)
  cmp    %eax,%edx
  jle    L71
  add    %eax,%ebx
  movslq %ebx,%rbx
  movss  0x0(%rbp,%rbx,4),%xmm0
  movss  %xmm0,0x8(%r9,%rdi,1)
L71:
  pop    %rbx
  mov    %r9,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L72:
  lea    -0x4(%rbp,%rax,1),%rcx
  xor    %eax,%eax
  nop
L60:
  movss  (%rcx,%rax,4),%xmm0
  movss  %xmm0,(%r9,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L60
  pop    %rbx
  mov    %r9,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D4
type: string

tool: D4: start=%d end=%d new_n=%d


assistant: ```c
float *slice(float *buf, int n, int start, int end, int *new_n) {
  int i;
  float *new;
  *new_n = n - (start + ((n - 1) - end));
  printf("start=%d end=%d new_n=%d\n", start, end, *new_n);
  new = malloc(sizeof(float) * *new_n);
  for (i = 0; i < *new_n; i++) {
    new[i] = buf[start + i];
  }
  return new;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<BatchInterpolatedGradY2DTranMirrWarp>:
  endbr64
  push   %r15
  pxor   %xmm7,%xmm7
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x68,%rsp
  movsd  %xmm0,0x20(%rsp)
  mulsd  %xmm3,%xmm0
  mov    0xa0(%rsp),%r14
  movsd  %xmm1,0x28(%rsp)
  movapd %xmm3,%xmm1
  mov    %rdx,0x18(%rsp)
  mov    %rcx,0x50(%rsp)
  movsd  %xmm2,0x30(%rsp)
  comisd %xmm7,%xmm0
  movsd  %xmm4,0x38(%rsp)
  movsd  %xmm5,0x8(%rsp)
  movsd  %xmm0,0x48(%rsp)
  jbe    L74
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L56:
  movl   $0x0,(%r15,%rax,4)
  pxor   %xmm0,%xmm0
  add    $0x1,%rax
  movsd  0x48(%rsp),%xmm6
  cvtsi2sd %rax,%xmm0
  comisd %xmm0,%xmm6
  ja     L56
  cmpq   $0x0,0x50(%rsp)
  jle    L17
L6:
  mov    0x18(%rsp),%rax
  test   %rax,%rax
  jle    L42
  shl    $0x2,%rax
  mov    %r8,%rbp
  mov    %r9,%r12
  movq   $0x0,0x40(%rsp)
  mov    %rax,0x58(%rsp)
  nopl   0x0(%rax,%rax,1)
L15:
  pxor   %xmm6,%xmm6
  xor    %r13d,%r13d
  cvtsi2ssq 0x40(%rsp),%xmm6
  movss  %xmm6,0x14(%rsp)
  nopl   0x0(%rax)
L29:
  pxor   %xmm3,%xmm3
  movss  0x14(%rsp),%xmm4
  mov    %r14,%rsi
  mov    %r15,%rdi
  cvtsi2ss %r13,%xmm3
  addss  (%r12,%r13,4),%xmm4
  movss  (%rbx,%r13,4),%xmm2
  movsd  %xmm1,(%rsp)
  addss  0x0(%rbp,%r13,4),%xmm3
  movsd  0x20(%rsp),%xmm0
  add    $0x1,%r13
  cvtss2sd %xmm4,%xmm4
  subsd  0x38(%rsp),%xmm4
  divsd  0x8(%rsp),%xmm4
  cvtss2sd %xmm3,%xmm3
  subsd  0x28(%rsp),%xmm3
  divsd  0x30(%rsp),%xmm3
  call   <InterpolatedGradYTranMirr@plt>
  cmp    %r13,0x18(%rsp)
  movsd  (%rsp),%xmm1
  jne    L29
  mov    0x58(%rsp),%rdx
  addq   $0x1,0x40(%rsp)
  mov    0x40(%rsp),%rax
  add    %rdx,%rbp
  add    %rdx,%r12
  add    %rdx,%rbx
  cmp    %rax,0x50(%rsp)
  jg     L15
L42:
  movsd  0x48(%rsp),%xmm7
  pxor   %xmm5,%xmm5
  comisd %xmm5,%xmm7
  jbe    L98
L17:
  xor    %eax,%eax
  xchg   %ax,%ax
L37:
  pxor   %xmm0,%xmm0
  movsd  0x48(%rsp),%xmm5
  cvtss2sd (%r15,%rax,4),%xmm0
  divsd  0x8(%rsp),%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,(%r15,%rax,4)
  add    $0x1,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  comisd %xmm0,%xmm5
  ja     L37
L98:
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L74:
  cmpq   $0x0,0x50(%rsp)
  jg     L6
  jmp    L42
```

assistant: ```c
extern void BatchInterpolatedGradY2DTranMirrWarp(float *Bout, float *Bcoeff, long Width, long Height, double nx, double offx, double mx, float *WarpX, double ny, double offy, double my, float *WarpY, long SplineDegree) {
  double x, y;
  long idx;
  long idcoordx, idcoordy;
  for (idx = 0; idx < nx * ny; idx++)
    Bout[idx] = 0;
  idx = 0;
  for (idcoordy = 0; idcoordy < Height; idcoordy++) {
    for (idcoordx = 0; idcoordx < Width; idcoordx++) {
      x = (idcoordx + WarpX[idx] - offx) / mx;
      y = (idcoordy + WarpY[idx] - offy) / my;
      InterpolatedGradYTranMirr(Bout, nx, ny, Bcoeff[idx++], x, y, SplineDegree);
    }
  }
  for (idx = 0; idx < nx * ny; idx++)
    Bout[idx] = Bout[idx] / my;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_message>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  jmp    L30
L37:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x51,%eax
  je     L10
L30:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,(%rbx)
  mov    (%rbx),%eax
  cmp    $0xa,%eax
  jne    L37
  jmp    L35
L10:
  nop
L35:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void read_message(int *array, int *c) {
  while ((array[*c] = getchar()) != '\n') {
    *c += 1;
    if (*c == 81) {
      break;
    }
  }
  *c -= 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read_message>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
L4:
  movslq (%rbx),%rax
  lea    (%r12,%rax,4),%rbp
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,0x0(%rbp)
  cmp    $0xa,%eax
  je     L18
  mov    (%rbx),%eax
  add    $0x1,%eax
  mov    %eax,(%rbx)
  cmp    $0x51,%eax
  jne    L4
L18:
  subl   $0x1,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void read_message(int *array, int *c) {
  while ((array[*c] = getchar()) != '\n') {
    *c += 1;
    if (*c == 81) {
      break;
    }
  }
  *c -= 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_message>:
  endbr64
  push   %r12
  mov    (%rsi),%eax
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L22:
  add    $0x1,%eax
  mov    %eax,(%rbx)
  cmp    $0x51,%eax
  je     L23
L12:
  cltq
  lea    0x0(%rbp,%rax,4),%r12
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  mov    %eax,(%r12)
  mov    (%rbx),%eax
  jne    L22
  sub    $0x1,%eax
  mov    %eax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L23:
  mov    $0x50,%eax
  mov    %eax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void read_message(int *array, int *c) {
  while ((array[*c] = getchar()) != '\n') {
    *c += 1;
    if (*c == 81) {
      break;
    }
  }
  *c -= 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<read_message>:
  endbr64
  push   %r12
  mov    (%rsi),%eax
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L22
  nopw   0x0(%rax,%rax,1)
L16:
  add    $0x1,%eax
  mov    %eax,(%rbx)
  cmp    $0x51,%eax
  je     L9
L22:
  cltq
  lea    0x0(%rbp,%rax,4),%r12
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  mov    %eax,(%r12)
  mov    (%rbx),%eax
  jne    L16
  sub    $0x1,%eax
  mov    %eax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L9:
  mov    $0x50,%eax
  mov    %eax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void read_message(int *array, int *c) {
  while ((array[*c] = getchar()) != '\n') {
    *c += 1;
    if (*c == 81) {
      break;
    }
  }
  *c -= 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sockutil_addrlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x80,%eax
  je     L12
  cmp    $0x81,%eax
  jne    L29
  mov    $0x4,%eax
  jmp    L36
L12:
  mov    $0x4,%eax
  jmp    L36
L29:
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%eax
L36:
  leave
  ret
```

assistant: ```c
socklen_t sockutil_addrlen(struct sockaddr *a) {
  switch (a->sa_family) {
  case AF_INET:
    return sizeof(struct sockaddr_in);
  case AF_INET6:
    return sizeof(struct sockaddr_in6);
  default:
    assert(0);
    return 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<sockutil_addrlen>:
  endbr64
  mov    (%rdi),%eax
  add    $0xffffff80,%eax
  mov    $0x4,%edx
  cmp    $0x1,%eax
  jbe    L2
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%edx
  mov    %edx,%eax
  add    $0x8,%rsp
  ret
L2:
  mov    %edx,%eax
  ret
```

assistant: ```c
socklen_t sockutil_addrlen(struct sockaddr *a) {
  switch (a->sa_family) {
  case AF_INET:
    return sizeof(struct sockaddr_in);
  case AF_INET6:
    return sizeof(struct sockaddr_in6);
  default:
    assert(0);
    return 0;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<sockutil_addrlen>:
  endbr64
  mov    (%rdi),%eax
  add    $0xffffff80,%eax
  cmp    $0x1,%eax
  jbe    L7
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L7:
  mov    $0x4,%eax
  ret
```

assistant: ```c
socklen_t sockutil_addrlen(struct sockaddr *a) {
  switch (a->sa_family) {
  case AF_INET:
    return sizeof(struct sockaddr_in);
  case AF_INET6:
    return sizeof(struct sockaddr_in6);
  default:
    assert(0);
    return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<sockutil_addrlen>:
  endbr64
  mov    (%rdi),%eax
  add    $0xffffff80,%eax
  cmp    $0x1,%eax
  jbe    L5
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L5:
  mov    $0x4,%eax
  ret
```

assistant: ```c
socklen_t sockutil_addrlen(struct sockaddr *a) {
  switch (a->sa_family) {
  case AF_INET:
    return sizeof(struct sockaddr_in);
  case AF_INET6:
    return sizeof(struct sockaddr_in6);
  default:
    assert(0);
    return 0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x800000(%rsp),%r11
L0:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L0
  sub    $0x40,%rsp
  mov    %edi,-0x800034(%rbp)
  mov    %rsi,-0x800040(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x3,-0x800034(%rbp)
  je     L59
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L6
L59:
  mov    -0x800040(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  lea    D9(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x800028(%rbp)
  mov    -0x800040(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  lea    D0(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x800020(%rbp)
  cmpq   $0x0,-0x800028(%rbp)
  je     L31
  cmpq   $0x0,-0x800020(%rbp)
  jne    L68
L31:
  mov    $0x1,%eax
  jmp    L6
L35:
  mov    -0x800020(%rbp),%rcx
  mov    -0x800018(%rbp),%rdx
  lea    -0x800010(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fwrite@plt>
L68:
  mov    -0x800028(%rbp),%rdx
  lea    -0x800010(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x800000,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fread@plt>
  mov    %rax,-0x800018(%rbp)
  cmpq   $0x0,-0x800018(%rbp)
  jne    L35
  mov    -0x800028(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L6:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L75
  call   <__stack_chk_fail@plt>
L75:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D19
datatype: string
name: read_data
arguments: label: D9
datatype: string
name: read_data
arguments: label: D0
datatype: string

tool: D19: Usage: cp source destination


tool: D9: r

tool: D0: w+

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printf("Usage: cp source destination\n");
    return 1;
  }
  FILE *fp = fopen(argv[1], "r");
  FILE *fcp = fopen(argv[2], "w+");
  if (fp == NULL || fcp == NULL)
    return 1;
  char *buffer[(1024 * 1024)];
  size_t bytes;
  while ((bytes = fread(buffer, 1, sizeof(buffer), fp)) > 0)
    fwrite(buffer, 1, bytes, fcp);
  fclose(fp);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x800000(%rsp),%r11
L26:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L26
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x800008(%rsp)
  xor    %eax,%eax
  cmp    $0x3,%edi
  je     L66
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
L22:
  mov    0x800008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L86
  add    $0x800010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L66:
  mov    %rsi,%rbx
  mov    0x8(%rsi),%rdi
  lea    D29(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  mov    0x10(%rbx),%rdi
  lea    D28(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%r12
  test   %rbp,%rbp
  je     L44
  test   %rax,%rax
  je     L44
  mov    %rsp,%rbx
  jmp    L69
L16:
  mov    %r12,%rcx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <fwrite@plt>
L69:
  mov    %rbp,%rcx
  mov    $0x800000,%edx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  jne    L16
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
  jmp    L22
L44:
  mov    $0x1,%eax
  jmp    L22
L86:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D11
data_type: string
name: retrieve_data
arguments: data_tag: D29
data_type: string
name: retrieve_data
arguments: data_tag: D28
data_type: string

tool: D11: Usage: cp source destination


tool: D29: r

tool: D28: w+

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printf("Usage: cp source destination\n");
    return 1;
  }
  FILE *fp = fopen(argv[1], "r");
  FILE *fcp = fopen(argv[2], "w+");
  if (fp == NULL || fcp == NULL)
    return 1;
  char *buffer[(1024 * 1024)];
  size_t bytes;
  while ((bytes = fread(buffer, 1, sizeof(buffer), fp)) > 0)
    fwrite(buffer, 1, bytes, fcp);
  fclose(fp);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x800000(%rsp),%r11
L28:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L28
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x800008(%rsp)
  xor    %eax,%eax
  cmp    $0x3,%edi
  je     L81
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
L54:
  mov    0x800008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L31
  add    $0x800010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L81:
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  lea    D21(%rip),%rsi
  call   <fopen@plt>
  mov    0x10(%rbx),%rdi
  lea    D25(%rip),%rsi
  mov    %rax,%rbp
  call   <fopen@plt>
  mov    %rax,%r12
  test   %rbp,%rbp
  je     L69
  test   %rax,%rax
  je     L69
  mov    %rsp,%rbx
  jmp    L16
  nopl   0x0(%rax)
L36:
  mov    %r12,%rcx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <fwrite@plt>
L16:
  mov    $0x800000,%edx
  mov    %rbp,%rcx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  jne    L36
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
  jmp    L54
L69:
  mov    $0x1,%eax
  jmp    L54
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D27
type: string
name: extract_symbolic_data
arguments: label: D21
type: string
name: extract_symbolic_data
arguments: label: D25
type: string

tool: D27: Usage: cp source destination


tool: D21: r

tool: D25: w+

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printf("Usage: cp source destination\n");
    return 1;
  }
  FILE *fp = fopen(argv[1], "r");
  FILE *fcp = fopen(argv[2], "w+");
  if (fp == NULL || fcp == NULL)
    return 1;
  char *buffer[(1024 * 1024)];
  size_t bytes;
  while ((bytes = fread(buffer, 1, sizeof(buffer), fp)) > 0)
    fwrite(buffer, 1, bytes, fcp);
  fclose(fp);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x800000(%rsp),%r11
L11:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L11
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x800008(%rsp)
  xor    %eax,%eax
  cmp    $0x3,%edi
  je     L31
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
L77:
  mov    0x800008(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x800010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L31:
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  lea    D2(%rip),%rsi
  call   <fopen@plt>
  mov    0x10(%rbx),%rdi
  lea    D15(%rip),%rsi
  mov    %rax,%rbp
  call   <fopen@plt>
  mov    %rax,%r12
  test   %rbp,%rbp
  je     L19
  test   %rax,%rax
  je     L19
  mov    %rsp,%rbx
  jmp    L66
  nopl   0x0(%rax)
L25:
  mov    %r12,%rcx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <fwrite@plt>
L66:
  mov    $0x800000,%edx
  mov    %rbp,%rcx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  jne    L25
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
  jmp    L77
L19:
  mov    $0x1,%eax
  jmp    L77
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D10
data_type: string
name: parse_data
arguments: label: D2
data_type: string
name: parse_data
arguments: label: D15
data_type: string

tool: D10: Usage: cp source destination


tool: D2: r

tool: D15: w+

assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 3) {
    printf("Usage: cp source destination\n");
    return 1;
  }
  FILE *fp = fopen(argv[1], "r");
  FILE *fcp = fopen(argv[2], "w+");
  if (fp == NULL || fcp == NULL)
    return 1;
  char *buffer[(1024 * 1024)];
  size_t bytes;
  while ((bytes = fread(buffer, 1, sizeof(buffer), fp)) > 0)
    fwrite(buffer, 1, bytes, fcp);
  fclose(fp);
  return 0;
}
```