ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<BinarySearch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  mov    %rdi,-0x20(%rbp)
  mov    %esi,-0x24(%rbp)
  mov    -0x20(%rbp),%rax
  movsd  (%rax),%xmm0
  comisd -0x18(%rbp),%xmm0
  jbe    L138
  mov    $0xffffffff,%eax
  jmp    L96
L138:
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movsd  -0x18(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jb     L40
  mov    -0x24(%rbp),%eax
  jmp    L96
L40:
  movl   $0x0,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L28
L119:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  comisd -0x18(%rbp),%xmm0
  jbe    L127
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L28
L127:
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movsd  -0x18(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jbe    L46
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L28
L46:
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L28:
  mov    -0x8(%rbp),%eax
  sub    -0xc(%rbp),%eax
  cmp    $0x1,%eax
  jg     L119
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L120
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movsd  -0x18(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jb     L23
  mov    -0x8(%rbp),%eax
  jmp    L96
L23:
  mov    -0xc(%rbp),%eax
  jmp    L96
L120:
  mov    -0xc(%rbp),%eax
L96:
  pop    %rbp
  ret
```

assistant: ```c
int BinarySearch(double x, double wl[], int n) {
  int low, high;
  int k;
  if (x < wl[0])
    return (-1);
  if (x >= wl[n - 1])
    return (n);
  low = 0;
  high = n - 1;
  while (high - low > 1) {
    k = (low + high) / 2;
    if (x < wl[k]) {
      high = k;
    } else if (x > wl[k + 1]) {
      low = k;
    } else {
      high = k;
      low = k;
    }
  }
  if (low < high) {
    if (wl[high] <= x)
      return (high);
    else
      return (low);
  } else {
    return (low);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<BinarySearch>:
  endbr64
  mov    %rdi,%rcx
  movsd  (%rdi),%xmm1
  comisd %xmm0,%xmm1
  ja     L98
  movslq %esi,%rax
  mov    %esi,%r8d
  comisd -0x8(%rdi,%rax,8),%xmm0
  jae    L32
  sub    $0x1,%esi
  cmp    $0x1,%esi
  jle    L69
  mov    $0x0,%edi
  jmp    L63
L31:
  mov    %eax,%esi
L78:
  mov    %esi,%eax
  sub    %edi,%eax
  cmp    $0x1,%eax
  jle    L82
L63:
  lea    (%rdi,%rsi,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,%r8d
  movslq %eax,%rdx
  lea    0x0(,%rdx,8),%r9
  movsd  (%rcx,%rdx,8),%xmm1
  comisd %xmm0,%xmm1
  ja     L31
  comisd 0x8(%rcx,%r9,1),%xmm0
  jbe    L32
  mov    %eax,%edi
  jmp    L78
L82:
  mov    %esi,%r8d
  jmp    L62
L69:
  mov    %esi,%r8d
  mov    $0x0,%edi
L62:
  cmp    %r8d,%edi
  jge    L46
  movslq %r8d,%rax
  comisd (%rcx,%rax,8),%xmm0
  cmovb  %edi,%r8d
L32:
  mov    %r8d,%eax
  ret
L98:
  mov    $0xffffffff,%r8d
  jmp    L32
L46:
  mov    %edi,%r8d
  jmp    L32
```

assistant: ```c
int BinarySearch(double x, double wl[], int n) {
  int low, high;
  int k;
  if (x < wl[0])
    return (-1);
  if (x >= wl[n - 1])
    return (n);
  low = 0;
  high = n - 1;
  while (high - low > 1) {
    k = (low + high) / 2;
    if (x < wl[k]) {
      high = k;
    } else if (x > wl[k + 1]) {
      low = k;
    } else {
      high = k;
      low = k;
    }
  }
  if (low < high) {
    if (wl[high] <= x)
      return (high);
    else
      return (low);
  } else {
    return (low);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<BinarySearch>:
  endbr64
  movsd  (%rdi),%xmm1
  mov    %rdi,%rcx
  comisd %xmm0,%xmm1
  ja     L64
  movslq %esi,%rax
  mov    %esi,%r9d
  comisd -0x8(%rdi,%rax,8),%xmm0
  jae    L73
  sub    $0x1,%esi
  xor    %edi,%edi
  cmp    $0x1,%esi
  jle    L12
  nopl   (%rax)
L9:
  lea    (%rdi,%rsi,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  movslq %eax,%rdx
  mov    %eax,%r9d
  movsd  (%rcx,%rdx,8),%xmm1
  lea    0x0(,%rdx,8),%r8
  comisd %xmm0,%xmm1
  ja     L88
  comisd 0x8(%rcx,%r8,1),%xmm0
  ja     L63
L73:
  mov    %r9d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L88:
  mov    %eax,%esi
L0:
  mov    %esi,%eax
  sub    %edi,%eax
  cmp    $0x1,%eax
  jg     L9
  nopl   0x0(%rax,%rax,1)
L12:
  cmp    %edi,%esi
  jle    L83
  movslq %esi,%rax
  comisd (%rcx,%rax,8),%xmm0
  cmovae %esi,%edi
  mov    %edi,%r9d
  mov    %r9d,%eax
  ret
  xchg   %ax,%ax
L63:
  mov    %eax,%edi
  jmp    L0
L64:
  mov    $0xffffffff,%r9d
  jmp    L73
L83:
  mov    %edi,%r9d
  jmp    L73
```

assistant: ```c
int BinarySearch(double x, double wl[], int n) {
  int low, high;
  int k;
  if (x < wl[0])
    return (-1);
  if (x >= wl[n - 1])
    return (n);
  low = 0;
  high = n - 1;
  while (high - low > 1) {
    k = (low + high) / 2;
    if (x < wl[k]) {
      high = k;
    } else if (x > wl[k + 1]) {
      low = k;
    } else {
      high = k;
      low = k;
    }
  }
  if (low < high) {
    if (wl[high] <= x)
      return (high);
    else
      return (low);
  } else {
    return (low);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<BinarySearch>:
  endbr64
  movsd  (%rdi),%xmm1
  mov    %rdi,%rcx
  comisd %xmm0,%xmm1
  ja     L10
  movslq %esi,%rax
  mov    %esi,%r9d
  comisd -0x8(%rdi,%rax,8),%xmm0
  jae    L3
  sub    $0x1,%esi
  xor    %edi,%edi
  cmp    $0x1,%esi
  jle    L13
  nopl   (%rax)
L23:
  lea    (%rdi,%rsi,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  movslq %eax,%rdx
  mov    %eax,%r9d
  movsd  (%rcx,%rdx,8),%xmm1
  lea    0x0(,%rdx,8),%r8
  comisd %xmm0,%xmm1
  ja     L73
  comisd 0x8(%rcx,%r8,1),%xmm0
  ja     L0
L3:
  mov    %r9d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L73:
  mov    %eax,%esi
L31:
  mov    %esi,%eax
  sub    %edi,%eax
  cmp    $0x1,%eax
  jg     L23
  nopl   0x0(%rax,%rax,1)
L13:
  cmp    %edi,%esi
  jle    L60
  movslq %esi,%rax
  comisd (%rcx,%rax,8),%xmm0
  cmovae %esi,%edi
  mov    %edi,%r9d
  mov    %r9d,%eax
  ret
  xchg   %ax,%ax
L0:
  mov    %eax,%edi
  jmp    L31
L10:
  mov    $0xffffffff,%r9d
  jmp    L3
L60:
  mov    %edi,%r9d
  jmp    L3
```

assistant: ```c
int BinarySearch(double x, double wl[], int n) {
  int low, high;
  int k;
  if (x < wl[0])
    return (-1);
  if (x >= wl[n - 1])
    return (n);
  low = 0;
  high = n - 1;
  while (high - low > 1) {
    k = (low + high) / 2;
    if (x < wl[k]) {
      high = k;
    } else if (x > wl[k + 1]) {
      low = k;
    } else {
      high = k;
      low = k;
    }
  }
  if (low < high) {
    if (wl[high] <= x)
      return (high);
    else
      return (low);
  } else {
    return (low);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<must_copy_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L17
  mov    $0x0,%eax
  jmp    L7
L17:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L17
  mov    -0x8(%rbp),%rax
L7:
  leave
  ret
```

assistant: ```c
char *must_copy_string(const char *entry) {
  char *ret;
  if (!entry)
    return NULL;
  do {
    ret = strdup(entry);
  } while (!ret);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<must_copy_string>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L20
L8:
  mov    %rbx,%rdi
  call   <strdup@plt>
  test   %rax,%rax
  je     L8
L9:
  pop    %rbx
  ret
L20:
  mov    %rdi,%rax
  jmp    L9
```

assistant: ```c
char *must_copy_string(const char *entry) {
  char *ret;
  if (!entry)
    return NULL;
  do {
    ret = strdup(entry);
  } while (!ret);
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<must_copy_string>:
  endbr64
  test   %rdi,%rdi
  je     L4
  push   %rbx
  mov    %rdi,%rbx
L6:
  mov    %rbx,%rdi
  call   <strdup@plt>
  test   %rax,%rax
  je     L6
  pop    %rbx
  ret
  nopl   0x0(%rax)
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *must_copy_string(const char *entry) {
  char *ret;
  if (!entry)
    return NULL;
  do {
    ret = strdup(entry);
  } while (!ret);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<must_copy_string>:
  endbr64
  test   %rdi,%rdi
  je     L10
  push   %rbx
  mov    %rdi,%rbx
L1:
  mov    %rbx,%rdi
  call   <strdup@plt>
  test   %rax,%rax
  je     L1
  pop    %rbx
  ret
  nopl   0x0(%rax)
L10:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *must_copy_string(const char *entry) {
  char *ret;
  if (!entry)
    return NULL;
  do {
    ret = strdup(entry);
  } while (!ret);
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<BatchInterpolatedGradX2DTranMirrWarp>:
  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  %xmm2,0x8(%rsp)
  comisd %xmm7,%xmm0
  movsd  %xmm4,0x30(%rsp)
  movsd  %xmm5,0x38(%rsp)
  movsd  %xmm0,0x48(%rsp)
  jbe    L13
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L31:
  movl   $0x0,(%r15,%rax,4)
  pxor   %xmm0,%xmm0
  add    $0x1,%rax
  movsd  0x48(%rsp),%xmm6
  cvtsi2sd %rax,%xmm0
  comisd %xmm0,%xmm6
  ja     L31
  cmpq   $0x0,0x50(%rsp)
  jle    L95
L39:
  mov    0x18(%rsp),%rax
  test   %rax,%rax
  jle    L4
  shl    $0x2,%rax
  mov    %r8,%rbp
  mov    %r9,%r12
  movq   $0x0,0x40(%rsp)
  mov    %rax,0x58(%rsp)
  nopl   0x0(%rax,%rax,1)
L26:
  pxor   %xmm6,%xmm6
  xor    %r13d,%r13d
  cvtsi2ssq 0x40(%rsp),%xmm6
  movss  %xmm6,0x14(%rsp)
  nopl   0x0(%rax)
L50:
  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  0x30(%rsp),%xmm4
  divsd  0x38(%rsp),%xmm4
  cvtss2sd %xmm3,%xmm3
  subsd  0x28(%rsp),%xmm3
  divsd  0x8(%rsp),%xmm3
  call   <InterpolatedGradXTranMirr@plt>
  cmp    %r13,0x18(%rsp)
  movsd  (%rsp),%xmm1
  jne    L50
  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     L26
L4:
  movsd  0x48(%rsp),%xmm7
  pxor   %xmm5,%xmm5
  comisd %xmm5,%xmm7
  jbe    L57
L95:
  xor    %eax,%eax
  xchg   %ax,%ax
L36:
  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     L36
L57:
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  cmpq   $0x0,0x50(%rsp)
  jg     L39
  jmp    L4
```

assistant: ```c
extern void BatchInterpolatedGradX2DTranMirrWarp(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;
      InterpolatedGradXTranMirr(Bout, nx, ny, Bcoeff[idx++], x, y, SplineDegree);
    }
  }
  for (idx = 0; idx < nx * ny; idx++)
    Bout[idx] = Bout[idx] / mx;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<put_hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %edi,%eax
  mov    %al,-0x64(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D7(%rip),%rax
  mov    %rax,-0x58(%rbp)
  movl   $0x0,-0x5c(%rbp)
  jmp    L47
L44:
  movzbl -0x64(%rbp),%eax
  mov    %eax,%edx
  sar    $0x7,%dl
  shr    $0x4,%dl
  add    %edx,%eax
  and    $0xf,%eax
  sub    %edx,%eax
  movsbq %al,%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x5c(%rbp),%eax
  cltq
  mov    %dl,-0x50(%rbp,%rax,1)
  movzbl -0x64(%rbp),%eax
  lea    0xf(%rax),%edx
  test   %al,%al
  cmovs  %edx,%eax
  sar    $0x4,%al
  mov    %al,-0x64(%rbp)
  addl   $0x1,-0x5c(%rbp)
L47:
  cmpb   $0x0,-0x64(%rbp)
  jne    L44
  cmpl   $0x1,-0x5c(%rbp)
  jg     L20
  mov    $0x1,%edx
  lea    D15(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L20
L51:
  lea    -0x50(%rbp),%rdx
  mov    -0x5c(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L20:
  mov    -0x5c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x5c(%rbp)
  test   %eax,%eax
  jne    L51
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D7
datatype: string
name: unpack_data
arguments: tag: D15
datatype: string

tool: D7: 0123456789abcdef

tool: D15: 0

assistant: ```c
void put_hex(char c) {
  int i;
  char str[64];
  char *hex;
  hex = "0123456789abcdef";
  i = 0;
  while (c) {
    str[i] = hex[c % 16];
    c = c / 16;
    ++i;
  }
  if (i < 2)
    write(1, "0", 1);
  while (i--) {
    write(1, &str[i], 1);
  }
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<put_hex>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  test   %dil,%dil
  je     L12
  mov    %edi,%edx
  mov    $0x1,%esi
  lea    -0x1(%rsp),%r8
  lea    D6(%rip),%rdi
L0:
  mov    %edx,%ecx
  sar    $0x7,%cl
  shr    $0x4,%cl
  lea    (%rdx,%rcx,1),%eax
  and    $0xf,%eax
  sub    %ecx,%eax
  movsbq %al,%rax
  movzbl (%rdi,%rax,1),%eax
  mov    %al,(%r8,%rsi,1)
  lea    0xf(%rdx),%eax
  test   %dl,%dl
  cmovns %edx,%eax
  sar    $0x4,%al
  mov    %eax,%edx
  mov    %esi,%ebp
  add    $0x1,%rsi
  test   %al,%al
  jne    L0
  cmp    $0x1,%ebp
  jle    L4
L60:
  lea    -0x1(%rbp),%eax
  test   %ebp,%ebp
  je     L21
  cltq
  lea    (%rsp,%rax,1),%rbx
  movslq %ebp,%rax
  lea    -0x2(%rsp,%rax,1),%rax
  lea    -0x1(%rbp),%edx
  sub    %rdx,%rax
  mov    %rax,%rbp
L46:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  sub    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L46
L21:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L12:
  mov    $0x0,%ebp
L4:
  mov    $0x1,%edx
  lea    D12(%rip),%rsi
  mov    $0x1,%edi
  call   <write@plt>
  jmp    L60
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
type_infered: string
name: extract_symbolic_data
arguments: label: D12
type_infered: string

tool: D6: 0123456789abcdef

tool: D12: 0

assistant: ```c
void put_hex(char c) {
  int i;
  char str[64];
  char *hex;
  hex = "0123456789abcdef";
  i = 0;
  while (c) {
    str[i] = hex[c % 16];
    c = c / 16;
    ++i;
  }
  if (i < 2)
    write(1, "0", 1);
  while (i--) {
    write(1, &str[i], 1);
  }
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<put_hex>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  test   %dil,%dil
  je     L16
  mov    %rsp,%rbp
  mov    %edi,%edx
  lea    D28(%rip),%r8
  xor    %edi,%edi
  mov    %rbp,%rsi
  nopl   0x0(%rax)
L40:
  mov    %edx,%ecx
  mov    %edi,%ebx
  lea    0x1(%rdi),%edi
  sar    $0x7,%cl
  shr    $0x4,%cl
  lea    (%rdx,%rcx,1),%eax
  and    $0xf,%eax
  sub    %ecx,%eax
  test   %dl,%dl
  movsbq %al,%rax
  movzbl (%r8,%rax,1),%eax
  mov    %al,(%rsi)
  lea    0xf(%rdx),%eax
  cmovns %edx,%eax
  add    $0x1,%rsi
  sar    $0x4,%al
  mov    %eax,%edx
  jne    L40
  cmp    $0x1,%edi
  je     L50
L61:
  movslq %ebx,%rax
  lea    0x0(%rbp,%rax,1),%rbx
  lea    -0x1(%rbp,%rax,1),%rbp
  sub    %rax,%rbp
L11:
  mov    %rbx,%rsi
  mov    $0x1,%edx
  mov    $0x1,%edi
  sub    $0x1,%rbx
  call   <write@plt>
  cmp    %rbp,%rbx
  jne    L11
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L69
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L50:
  mov    $0x1,%edx
  lea    D3(%rip),%rsi
  call   <write@plt>
  jmp    L61
  nopl   0x0(%rax,%rax,1)
L16:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L69
  add    $0x58,%rsp
  mov    $0x1,%edx
  mov    $0x1,%edi
  pop    %rbx
  lea    D3(%rip),%rsi
  pop    %rbp
  jmp    1060 <write@plt>
L69:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D28
type: string
name: unpack_data
arguments: data_label: D3
type: string

tool: D28: 0123456789abcdef

tool: D3: 0

assistant: ```c
void put_hex(char c) {
  int i;
  char str[64];
  char *hex;
  hex = "0123456789abcdef";
  i = 0;
  while (c) {
    str[i] = hex[c % 16];
    c = c / 16;
    ++i;
  }
  if (i < 2)
    write(1, "0", 1);
  while (i--) {
    write(1, &str[i], 1);
  }
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<put_hex>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  test   %dil,%dil
  je     L6
  mov    %rsp,%rbp
  mov    %edi,%edx
  lea    D19(%rip),%r8
  xor    %edi,%edi
  mov    %rbp,%rsi
  nopl   0x0(%rax)
L11:
  mov    %edx,%ecx
  mov    %edi,%ebx
  lea    0x1(%rdi),%edi
  sar    $0x7,%cl
  shr    $0x4,%cl
  lea    (%rdx,%rcx,1),%eax
  and    $0xf,%eax
  sub    %ecx,%eax
  test   %dl,%dl
  movsbq %al,%rax
  movzbl (%r8,%rax,1),%eax
  mov    %al,(%rsi)
  lea    0xf(%rdx),%eax
  cmovns %edx,%eax
  add    $0x1,%rsi
  sar    $0x4,%al
  mov    %eax,%edx
  jne    L11
  cmp    $0x1,%edi
  je     L21
L26:
  movslq %ebx,%rax
  lea    0x0(%rbp,%rax,1),%rbx
  lea    -0x1(%rbp,%rax,1),%rbp
  sub    %rax,%rbp
L0:
  mov    %rbx,%rsi
  mov    $0x1,%edx
  mov    $0x1,%edi
  sub    $0x1,%rbx
  call   <write@plt>
  cmp    %rbx,%rbp
  jne    L0
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L21:
  mov    $0x1,%edx
  lea    D4(%rip),%rsi
  call   <write@plt>
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L6:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  add    $0x58,%rsp
  mov    $0x1,%edx
  mov    $0x1,%edi
  pop    %rbx
  lea    D4(%rip),%rsi
  pop    %rbp
  jmp    1060 <write@plt>
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D19
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D4
type_infered: string

tool: D19: 0123456789abcdef

tool: D4: 0

assistant: ```c
void put_hex(char c) {
  int i;
  char str[64];
  char *hex;
  hex = "0123456789abcdef";
  i = 0;
  while (c) {
    str[i] = hex[c % 16];
    c = c / 16;
    ++i;
  }
  if (i < 2)
    write(1, "0", 1);
  while (i--) {
    write(1, &str[i], 1);
  }
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<store_lines>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rsi
  lea    -0x1c(%rbp),%rcx
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %rsi,%r8
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <buffered_line_reader@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
char *store_lines(FILE *fp, int k) {
  char *result;
  char *acc;
  int len = 0;
  acc = (char *)calloc(1, sizeof(char));
  acc[0] = '\0';
  result = buffered_line_reader(fp, 0, k, &len, acc);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<store_lines>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,%r8
  movb   $0x0,(%rax)
  lea    0x4(%rsp),%rcx
  mov    %ebp,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <buffered_line_reader@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *store_lines(FILE *fp, int k) {
  char *result;
  char *acc;
  int len = 0;
  acc = (char *)calloc(1, sizeof(char));
  acc[0] = '\0';
  result = buffered_line_reader(fp, 0, k, &len, acc);
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<store_lines>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  mov    $0x1,%esi
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x1,%edi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <calloc@plt>
  mov    %r12d,%edx
  xor    %esi,%esi
  lea    0x4(%rsp),%rcx
  movb   $0x0,(%rax)
  mov    %rax,%r8
  mov    %rbp,%rdi
  call   <buffered_line_reader@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *store_lines(FILE *fp, int k) {
  char *result;
  char *acc;
  int len = 0;
  acc = (char *)calloc(1, sizeof(char));
  acc[0] = '\0';
  result = buffered_line_reader(fp, 0, k, &len, acc);
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<store_lines>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  mov    $0x1,%esi
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x1,%edi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <calloc@plt>
  mov    %r12d,%edx
  xor    %esi,%esi
  lea    0x4(%rsp),%rcx
  movb   $0x0,(%rax)
  mov    %rax,%r8
  mov    %rbp,%rdi
  call   <buffered_line_reader@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *store_lines(FILE *fp, int k) {
  char *result;
  char *acc;
  int len = 0;
  acc = (char *)calloc(1, sizeof(char));
  acc[0] = '\0';
  result = buffered_line_reader(fp, 0, k, &len, acc);
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cdops_read_sectors_dma>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x11,%edi
  mov    $0x0,%eax
  call   <cdops_exec_cmd@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int cdops_read_sectors_dma(char *buf, int sec, int num) {
  struct {
    int sec, num;
    void *buffer;
    int dunno;
  } param;
  param.sec = sec;
  param.num = num;
  param.buffer = buf;
  param.dunno = 0;
  return cdops_exec_cmd(17, &param);
}
```