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
<copy_file>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <rewind@plt>
  jmp    L9
L13:
  mov    D10(%rip),%rax
  mov    (%rax),%esi
  call   <fputc@plt>
L9:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  mov    %eax,%edi
  mov    D16(%rip),%rax
  cmp    %edi,(%rax)
  jne    L13
  pop    %rbx
  ret
```

assistant: ```c
void copy_file(FILE *fout, char *tar) {
  int ch;
  rewind(fout);
  while ((ch = fgetc(fout)) != EOF) {
    fputc(ch, stdout);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<copy_file>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <rewind@plt>
  mov    D12(%rip),%rbp
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L4:
  mov    D16(%rip),%rax
  mov    (%rax),%esi
  call   <fputc@plt>
L3:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  mov    %eax,%edi
  cmp    %eax,0x0(%rbp)
  jne    L4
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void copy_file(FILE *fout, char *tar) {
  int ch;
  rewind(fout);
  while ((ch = fgetc(fout)) != EOF) {
    fputc(ch, stdout);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<copy_file>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <rewind@plt>
  mov    D13(%rip),%rbp
  jmp    L13
  nopl   0x0(%rax,%rax,1)
L17:
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  call   <fputc@plt>
L13:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  mov    %eax,%edi
  cmp    %eax,0x0(%rbp)
  jne    L17
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void copy_file(FILE *fout, char *tar) {
  int ch;
  rewind(fout);
  while ((ch = fgetc(fout)) != EOF) {
    fputc(ch, stdout);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<L2_cN_so3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x24(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x2c(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  je     L8
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%ecx
  mov    -0x24(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <L2_bN_so3@plt>
  movq   %xmm0,%rax
  movq   D19(%rip),%xmm0
  movq   %rax,%xmm3
  xorpd  %xmm0,%xmm3
  movapd %xmm3,%xmm0
  mulsd  -0x10(%rbp),%xmm0
  mulsd  -0x8(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm2
  movsd  D0(%rip),%xmm1
  addsd  %xmm2,%xmm1
  mulsd  -0x18(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  jmp    L15
L8:
  pxor   %xmm0,%xmm0
L15:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D19
type: f32[2]
name: read_data
arguments: label: D0
type: f64

tool: {"D19": [0.0,-0.0]}

tool: D0: 1

assistant: ```c
double L2_cN_so3(int j, int m1, int m2) {
  double dj, dm1, dm2;
  dj = (double)j;
  dm1 = (double)m1;
  dm2 = (double)m2;
  if (j != 0)
    return (-L2_bN_so3(j, m1, m2) * dm1 * dm2 / (dj * (dj + 1.)));
  else
    return 0.;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<L2_cN_so3>:
  endbr64
  pxor   %xmm0,%xmm0
  test   %edi,%edi
  jne    L8
  ret
L8:
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %esi,%ebp
  mov    %edx,%ebx
  pxor   %xmm2,%xmm2
  cvtsi2sd %edi,%xmm2
  movsd  %xmm2,0x8(%rsp)
  call   <L2_bN_so3@plt>
  xorpd  D13(%rip),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebp,%xmm1
  mulsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebx,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  0x8(%rsp),%xmm2
  movapd %xmm2,%xmm1
  addsd  D19(%rip),%xmm1
  mulsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D13
data_type: f32[4]
name: unpack_data
arguments: tag: D19
data_type: f64

tool: {"D13": [0.0,-0.0,0.0,0.0]}

tool: D19: 1

assistant: ```c
double L2_cN_so3(int j, int m1, int m2) {
  double dj, dm1, dm2;
  dj = (double)j;
  dm1 = (double)m1;
  dm2 = (double)m2;
  if (j != 0)
    return (-L2_bN_so3(j, m1, m2) * dm1 * dm2 / (dj * (dj + 1.)));
  else
    return 0.;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<L2_cN_so3>:
  endbr64
  test   %edi,%edi
  jne    L2
  pxor   %xmm0,%xmm0
  ret
  nopl   (%rax)
L2:
  pxor   %xmm2,%xmm2
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  cvtsi2sd %edi,%xmm2
  mov    %edx,%ebx
  sub    $0x18,%rsp
  movsd  %xmm2,0x8(%rsp)
  call   <L2_bN_so3@plt>
  pxor   %xmm1,%xmm1
  xorpd  D17(%rip),%xmm0
  movsd  0x8(%rsp),%xmm2
  cvtsi2sd %ebp,%xmm1
  mulsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebx,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  D0(%rip),%xmm1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  addsd  %xmm2,%xmm1
  mulsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: tag: D17
type_guessed: f32[4]
name: unpack_data
arguments: tag: D0
type_guessed: f64

tool: {"D17": [0.0,-0.0,0.0,0.0]}

tool: D0: 1

assistant: ```c
double L2_cN_so3(int j, int m1, int m2) {
  double dj, dm1, dm2;
  dj = (double)j;
  dm1 = (double)m1;
  dm2 = (double)m2;
  if (j != 0)
    return (-L2_bN_so3(j, m1, m2) * dm1 * dm2 / (dj * (dj + 1.)));
  else
    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
<L2_cN_so3>:
  endbr64
  test   %edi,%edi
  jne    L5
  pxor   %xmm0,%xmm0
  ret
  nopl   (%rax)
L5:
  pxor   %xmm2,%xmm2
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  cvtsi2sd %edi,%xmm2
  mov    %edx,%ebx
  sub    $0x18,%rsp
  movsd  %xmm2,0x8(%rsp)
  call   <L2_bN_so3@plt>
  pxor   %xmm1,%xmm1
  xorpd  D0(%rip),%xmm0
  movsd  0x8(%rsp),%xmm2
  cvtsi2sd %ebp,%xmm1
  mulsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebx,%xmm1
  mulsd  %xmm1,%xmm0
  movsd  D3(%rip),%xmm1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  addsd  %xmm2,%xmm1
  mulsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
data_type: f32[4]
name: parse_data
arguments: data_tag: D3
data_type: f64

tool: {"D0": [0.0,-0.0,0.0,0.0]}

tool: D3: 1

assistant: ```c
double L2_cN_so3(int j, int m1, int m2) {
  double dj, dm1, dm2;
  dj = (double)j;
  dm1 = (double)m1;
  dm2 = (double)m2;
  if (j != 0)
    return (-L2_bN_so3(j, m1, m2) * dm1 * dm2 / (dj * (dj + 1.)));
  else
    return 0.;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ui2a>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L8
L46:
  mov    -0x8(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L8:
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  divl   -0x8(%rbp)
  cmp    %eax,-0x18(%rbp)
  jbe    L46
  jmp    L13
L31:
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  divl   -0x8(%rbp)
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  divl   -0x8(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x8(%rbp),%eax
  mov    $0x0,%edx
  divl   -0x18(%rbp)
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L88
  cmpl   $0x0,-0x4(%rbp)
  jg     L88
  cmpl   $0x0,-0x8(%rbp)
  jne    L13
L88:
  cmpl   $0x9,-0x4(%rbp)
  jle    L42
  cmpl   $0x0,-0x1c(%rbp)
  je     L2
  mov    $0x37,%edx
  jmp    L20
L2:
  mov    $0x57,%edx
  jmp    L20
L42:
  mov    $0x30,%edx
L20:
  mov    -0x4(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0xc(%rbp)
L13:
  cmpl   $0x0,-0x8(%rbp)
  jne    L31
  mov    -0x28(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ui2a(unsigned int num, unsigned int base, int uc, char *bf) {
  int n = 0;
  unsigned int d = 1;
  while (num / d >= base)
    d *= base;
  while (d != 0) {
    int dgt = num / d;
    num %= d;
    d /= base;
    if (n || dgt > 0 || d == 0) {
      *bf++ = dgt + (dgt < 10 ? '0' : (uc ? 'A' : 'a') - 10);
      ++n;
    }
  }
  *bf = 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ui2a>:
  endbr64
  push   %rbx
  mov    %edx,%ebx
  mov    %rcx,%r10
  cmp    %esi,%edi
  jb     L82
  mov    $0x1,%ecx
L109:
  imul   %esi,%ecx
  mov    %edi,%eax
  mov    $0x0,%edx
  div    %ecx
  cmp    %esi,%eax
  jae    L109
  test   %ecx,%ecx
  je     L51
L40:
  mov    $0x0,%r9d
  jmp    L93
L82:
  mov    $0x1,%ecx
  jmp    L40
L25:
  lea    0x1(%r10),%r11
  add    %r8d,%edx
  mov    %dl,(%r10)
  add    $0x1,%r9d
  cmp    %ecx,%esi
  ja     L67
  mov    %r11,%r10
L34:
  mov    %eax,%ecx
L93:
  mov    %edi,%eax
  mov    $0x0,%edx
  div    %ecx
  mov    %edx,%edi
  mov    %eax,%r8d
  mov    %ecx,%eax
  mov    $0x0,%edx
  div    %esi
  test   %r9d,%r9d
  setne  %dl
  cmp    %ecx,%esi
  seta   %r11b
  or     %r11b,%dl
  jne    L48
  test   %r8d,%r8d
  jle    L32
L48:
  mov    $0x30,%edx
  cmp    $0x9,%r8d
  jle    L25
  cmp    $0x1,%ebx
  sbb    %edx,%edx
  and    $0x20,%edx
  add    $0x37,%edx
  jmp    L25
L32:
  mov    $0x0,%r9d
  jmp    L34
L51:
  mov    %r10,%r11
L67:
  movb   $0x0,(%r11)
  pop    %rbx
  ret
```

assistant: ```c
void ui2a(unsigned int num, unsigned int base, int uc, char *bf) {
  int n = 0;
  unsigned int d = 1;
  while (num / d >= base)
    d *= base;
  while (d != 0) {
    int dgt = num / d;
    num %= d;
    d /= base;
    if (n || dgt > 0 || d == 0) {
      *bf++ = dgt + (dgt < 10 ? '0' : (uc ? 'A' : 'a') - 10);
      ++n;
    }
  }
  *bf = 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
<ui2a>:
  endbr64
  push   %rbx
  mov    %rcx,%r10
  mov    %edx,%ebx
  mov    $0x1,%ecx
  cmp    %esi,%edi
  jb     L54
  nopl   0x0(%rax,%rax,1)
L40:
  imul   %esi,%ecx
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  cmp    %esi,%eax
  jae    L40
  test   %ecx,%ecx
  je     L88
L54:
  xor    %r9d,%r9d
  nopl   0x0(%rax)
L69:
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  mov    %edx,%edi
  mov    %eax,%r8d
  xor    %edx,%edx
  mov    %ecx,%eax
  div    %esi
  test   %r9d,%r9d
  jne    L63
  test   %r8d,%r8d
  jg     L63
  cmp    %ecx,%esi
  jbe    L15
  add    $0x30,%r8d
  lea    0x1(%r10),%r11
  mov    %r8b,(%r10)
L89:
  movb   $0x0,(%r11)
  pop    %rbx
  ret
L63:
  mov    $0x30,%edx
  cmp    $0x9,%r8d
  jle    L37
  cmp    $0x1,%ebx
  sbb    %edx,%edx
  and    $0x20,%edx
  add    $0x37,%edx
L37:
  add    %r8d,%edx
  lea    0x1(%r10),%r11
  add    $0x1,%r9d
  mov    %dl,(%r10)
  cmp    %ecx,%esi
  ja     L89
  mov    %r11,%r10
L15:
  mov    %eax,%ecx
  jmp    L69
  nop
L88:
  mov    %r10,%r11
  movb   $0x0,(%r11)
  pop    %rbx
  ret
```

assistant: ```c
void ui2a(unsigned int num, unsigned int base, int uc, char *bf) {
  int n = 0;
  unsigned int d = 1;
  while (num / d >= base)
    d *= base;
  while (d != 0) {
    int dgt = num / d;
    num %= d;
    d /= base;
    if (n || dgt > 0 || d == 0) {
      *bf++ = dgt + (dgt < 10 ? '0' : (uc ? 'A' : 'a') - 10);
      ++n;
    }
  }
  *bf = 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ui2a>:
  endbr64
  mov    %rcx,%r8
  mov    %edx,%r9d
  mov    $0x1,%ecx
  cmp    %esi,%edi
  jb     L21
  nopl   0x0(%rax,%rax,1)
L40:
  imul   %esi,%ecx
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  cmp    %esi,%eax
  jae    L40
  test   %ecx,%ecx
  je     L129
L21:
  push   %rbp
  xor    %r10d,%r10d
  push   %rbx
  test   %r9d,%r9d
  je     L48
  mov    $0x30,%ebp
  mov    $0x37,%ebx
  nopl   0x0(%rax)
L55:
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  mov    %edx,%edi
  mov    %eax,%r9d
  xor    %edx,%edx
  mov    %ecx,%eax
  div    %esi
  test   %r10d,%r10d
  jne    L67
  test   %r9d,%r9d
  jg     L67
  cmp    %ecx,%esi
  jbe    L38
L12:
  add    $0x30,%r9d
  lea    0x1(%r8),%rdx
  mov    %r9b,(%r8)
L29:
  movb   $0x0,(%rdx)
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L48:
  mov    $0x57,%ebp
  mov    $0x30,%ebx
  nopw   0x0(%rax,%rax,1)
L52:
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  mov    %edx,%edi
  mov    %eax,%r9d
  xor    %edx,%edx
  mov    %ecx,%eax
  div    %esi
  test   %r9d,%r9d
  jg     L100
  test   %r10d,%r10d
  jne    L100
  cmp    %ecx,%esi
  ja     L12
  xor    %r10d,%r10d
  mov    %eax,%ecx
  jmp    L52
  cs nopw 0x0(%rax,%rax,1)
L67:
  cmp    $0xa,%r9d
  mov    %ebx,%r11d
  lea    0x1(%r8),%rdx
  cmovl  %ebp,%r11d
  add    $0x1,%r10d
  add    %r11d,%r9d
  mov    %r9b,(%r8)
  cmp    %ecx,%esi
  ja     L29
  mov    %rdx,%r8
L38:
  mov    %eax,%ecx
  jmp    L55
  nopw   0x0(%rax,%rax,1)
L100:
  cmp    $0x9,%r9d
  mov    %ebx,%r11d
  lea    0x1(%r8),%rdx
  cmovg  %ebp,%r11d
  add    $0x1,%r10d
  add    %r11d,%r9d
  mov    %r9b,(%r8)
  cmp    %ecx,%esi
  ja     L29
  mov    %rdx,%r8
  mov    %eax,%ecx
  jmp    L52
  nopl   0x0(%rax,%rax,1)
L129:
  movb   $0x0,(%r8)
  ret
```

assistant: ```c
void ui2a(unsigned int num, unsigned int base, int uc, char *bf) {
  int n = 0;
  unsigned int d = 1;
  while (num / d >= base)
    d *= base;
  while (d != 0) {
    int dgt = num / d;
    num %= d;
    d /= base;
    if (n || dgt > 0 || d == 0) {
      *bf++ = dgt + (dgt < 10 ? '0' : (uc ? 'A' : 'a') - 10);
      ++n;
    }
  }
  *bf = 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<full_pwrite>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L44
L36:
  mov    -0x18(%rbp),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <pwrite@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L6
  mov    D4(%rip),%rax
  mov    (%rax),%rdx
  mov    D9(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L65
  jmp    L44
L65:
  cmpq   $0x0,-0x10(%rbp)
  jne    L5
  mov    $0xffffffffffffffff,%rax
  jmp    L39
L6:
  cmpq   $0x0,-0x8(%rbp)
  je     L42
  mov    -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  sub    %rax,-0x28(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  shl    $0x2,%rax
  add    %rax,-0x20(%rbp)
L44:
  cmpq   $0x0,-0x28(%rbp)
  jne    L36
  jmp    L29
L5:
  nop
  jmp    L29
L42:
  nop
L29:
  mov    -0x10(%rbp),%rax
L39:
  leave
  ret
```

assistant: ```c
ssize_t full_pwrite(int fd, const void *buf, size_t count, off_t offset) {
  do {
    ssize_t total = 0;
    while (count > 0) {
      ssize_t chunk = (pwrite(fd, buf, count, offset));
      if (chunk < 0) {
        if (errno == EINTR) {
          continue;
        } else if (total == 0) {
          return -1;
        } else {
          break;
        }
      } else if (chunk == 0) {
        break;
      } else {
        total += chunk;
        count -= chunk;
        offset += chunk;
        buf = ((uint8_t *)buf) + chunk;
      }
    }
    return total;
  } while (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<full_pwrite>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdx,%rdx
  je     L17
  mov    %edi,%r14d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %ecx,%r12d
  mov    $0x0,%r13d
  jmp    L99
L78:
  mov    D6(%rip),%rdx
  mov    D16(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L90
  test   %r13,%r13
  je     L45
L65:
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L45:
  mov    $0xffffffffffffffff,%r13
  jmp    L65
L90:
  test   %rbx,%rbx
  je     L65
L99:
  mov    %r12d,%ecx
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r14d,%edi
  call   <pwrite@plt>
  test   %rax,%rax
  js     L78
  je     L65
  add    %rax,%r13
  sub    %rax,%rbx
  add    %eax,%r12d
  lea    0x0(%rbp,%rax,4),%rbp
  jmp    L90
L17:
  mov    $0x0,%r13d
  jmp    L65
```

assistant: ```c
ssize_t full_pwrite(int fd, const void *buf, size_t count, off_t offset) {
  do {
    ssize_t total = 0;
    while (count > 0) {
      ssize_t chunk = (pwrite(fd, buf, count, offset));
      if (chunk < 0) {
        if (errno == EINTR) {
          continue;
        } else if (total == 0) {
          return -1;
        } else {
          break;
        }
      } else if (chunk == 0) {
        break;
      } else {
        total += chunk;
        count -= chunk;
        offset += chunk;
        buf = ((uint8_t *)buf) + chunk;
      }
    }
    return total;
  } while (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<full_pwrite>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdx,%rdx
  je     L19
  mov    %edi,%r14d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %ecx,%r12d
  xor    %r13d,%r13d
L46:
  mov    %r12d,%ecx
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r14d,%edi
  call   <pwrite@plt>
  test   %rax,%rax
  js     L61
  je     L74
  add    %rax,%r13
  add    %eax,%r12d
  lea    0x0(%rbp,%rax,4),%rbp
  sub    %rax,%rbx
  jne    L46
L74:
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L61:
  mov    D10(%rip),%rax
  mov    D6(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L46
  test   %r13,%r13
  jne    L74
  mov    $0xffffffffffffffff,%r13
  jmp    L74
  nopl   0x0(%rax)
L19:
  xor    %r13d,%r13d
  jmp    L74
```

assistant: ```c
ssize_t full_pwrite(int fd, const void *buf, size_t count, off_t offset) {
  do {
    ssize_t total = 0;
    while (count > 0) {
      ssize_t chunk = (pwrite(fd, buf, count, offset));
      if (chunk < 0) {
        if (errno == EINTR) {
          continue;
        } else if (total == 0) {
          return -1;
        } else {
          break;
        }
      } else if (chunk == 0) {
        break;
      } else {
        total += chunk;
        count -= chunk;
        offset += chunk;
        buf = ((uint8_t *)buf) + chunk;
      }
    }
    return total;
  } while (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<full_pwrite>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdx,%rdx
  je     L6
  mov    %edi,%r14d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %ecx,%r12d
  xor    %r13d,%r13d
L16:
  mov    %r12d,%ecx
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r14d,%edi
  call   <pwrite@plt>
  test   %rax,%rax
  js     L31
  je     L17
  add    %rax,%r13
  add    %eax,%r12d
  lea    0x0(%rbp,%rax,4),%rbp
  sub    %rax,%rbx
  jne    L16
L17:
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L31:
  mov    D8(%rip),%rax
  mov    D18(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L16
  test   %r13,%r13
  jne    L17
  mov    $0xffffffffffffffff,%r13
  jmp    L17
  nopl   0x0(%rax)
L6:
  xor    %r13d,%r13d
  jmp    L17
```

assistant: ```c
ssize_t full_pwrite(int fd, const void *buf, size_t count, off_t offset) {
  do {
    ssize_t total = 0;
    while (count > 0) {
      ssize_t chunk = (pwrite(fd, buf, count, offset));
      if (chunk < 0) {
        if (errno == EINTR) {
          continue;
        } else if (total == 0) {
          return -1;
        } else {
          break;
        }
      } else if (chunk == 0) {
        break;
      } else {
        total += chunk;
        count -= chunk;
        offset += chunk;
        buf = ((uint8_t *)buf) + chunk;
      }
    }
    return total;
  } while (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L45
L29:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jne    L54
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jge    L49
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L49:
  movl   $0x0,-0x10(%rbp)
  jmp    L14
L54:
  addl   $0x1,-0x10(%rbp)
L14:
  addl   $0x1,-0x14(%rbp)
L45:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L29
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jge    L50
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L50:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

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

tool: D10: %d

assistant: ```c
int main() {
  int size, i, inp, count = 0, max_count = 0;
  scanf("%d", &size);
  for (i = 0; i < size; i++) {
    scanf("%d", &inp);
    if (inp == 0) {
      if (max_count < count)
        max_count = count;
      count = 0;
    } else
      count++;
  }
  if (max_count < count)
    max_count = count;
  printf("%d", max_count);
  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
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,(%rsp)
  jle    L34
  mov    $0x0,%r12d
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  lea    D5(%rip),%r13
  jmp    L28
L42:
  add    $0x1,%ebx
L39:
  add    $0x1,%ebp
  cmp    %ebp,(%rsp)
  jle    L20
L28:
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L42
  cmp    %ebx,%r12d
  cmovl  %ebx,%r12d
  mov    %eax,%ebx
  jmp    L39
L34:
  mov    $0x0,%r12d
  mov    $0x0,%ebx
L20:
  cmp    %r12d,%ebx
  mov    %r12d,%esi
  cmovge %ebx,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D5
type_guessed: string

tool: D5: %d

assistant: ```c
int main() {
  int size, i, inp, count = 0, max_count = 0;
  scanf("%d", &size);
  for (i = 0; i < size; i++) {
    scanf("%d", &inp);
    if (inp == 0) {
      if (max_count < count)
        max_count = count;
      count = 0;
    } else
      count++;
  }
  if (max_count < count)
    max_count = count;
  printf("%d", max_count);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D8(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%edx
  test   %edx,%edx
  jle    L45
  xor    %ebp,%ebp
  lea    0x4(%rsp),%r14
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L21:
  cmp    %ebx,%r12d
  cmovl  %ebx,%r12d
  xor    %ebx,%ebx
  add    $0x1,%ebp
  cmp    %ebp,(%rsp)
  jle    L45
L27:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L21
  add    $0x1,%ebx
  add    $0x1,%ebp
  cmp    %ebp,(%rsp)
  jg     L27
L45:
  cmp    %r12d,%ebx
  mov    %r12d,%esi
  mov    %r13,%rdi
  cmovge %ebx,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D8
data_type: string

tool: D8: %d

assistant: ```c
int main() {
  int size, i, inp, count = 0, max_count = 0;
  scanf("%d", &size);
  for (i = 0; i < size; i++) {
    scanf("%d", &inp);
    if (inp == 0) {
      if (max_count < count)
        max_count = count;
      count = 0;
    } else
      count++;
  }
  if (max_count < count)
    max_count = count;
  printf("%d", max_count);
  return 0;
}
```