ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <BRRunTests@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
main(int argc, const char *argv[]) {
  int r = BRRunTests();
  return (r) ? 0 : 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <BRRunTests@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main(int argc, const char *argv[]) {
  int r = BRRunTests();
  return (r) ? 0 : 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <BRRunTests@plt>
  test   %eax,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
main(int argc, const char *argv[]) {
  int r = BRRunTests();
  return (r) ? 0 : 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <BRRunTests@plt>
  test   %eax,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
main(int argc, const char *argv[]) {
  int r = BRRunTests();
  return (r) ? 0 : 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<audio_out_resample_stereo>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r8
  mov    %esi,%r10d
  mov    %rdx,%r11
  mov    %ecx,%ebx
  lea    -0x2(%rsi),%eax
  shl    $0x10,%eax
  lea    -0x2(%rcx),%ecx
  cltd
  idiv   %ecx
  cmp    $0x1,%ebx
  jle    L17
  mov    %r11,%rdi
  lea    -0x2(%rbx),%edx
  lea    0x8(%r11,%rdx,8),%r12
  mov    $0x0,%r9d
  mov    $0x10000,%ebp
L3:
  movzwl %r9w,%edx
  mov    %r9d,%ecx
  sar    $0x10,%ecx
  add    %ecx,%ecx
  movslq %ecx,%rcx
  mov    %ebp,%r13d
  sub    %edx,%r13d
  mov    %edx,%esi
  imul   0x8(%r8,%rcx,4),%esi
  mov    %r13d,%r14d
  imul   (%r8,%rcx,4),%r14d
  add    %r14d,%esi
  sar    $0x10,%esi
  mov    %esi,(%rdi)
  imul   0xc(%r8,%rcx,4),%edx
  imul   0x4(%r8,%rcx,4),%r13d
  add    %r13d,%edx
  sar    $0x10,%edx
  mov    %edx,0x4(%rdi)
  add    %eax,%r9d
  add    $0x8,%rdi
  cmp    %r12,%rdi
  jne    L3
L17:
  add    %r10d,%r10d
  movslq %r10d,%r10
  shl    $0x2,%r10
  add    %ebx,%ebx
  movslq %ebx,%rbx
  shl    $0x2,%rbx
  mov    -0x8(%r8,%r10,1),%eax
  mov    %eax,-0x8(%r11,%rbx,1)
  mov    -0x4(%r8,%r10,1),%eax
  mov    %eax,-0x4(%r11,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void audio_out_resample_stereo(int16_t *input_samples, uint32_t in_samples, int16_t *output_samples, uint32_t out_samples) {
  int osample;
  uint32_t isample = 0;
  uint32_t istep = ((in_samples - 2) << 16) / (out_samples - 2);
  for (osample = 0; osample < out_samples - 1; osample++) {
    int s1;
    int s2;
    int16_t os;
    uint32_t t = isample & 0xffff;
    s1 = input_samples[(isample >> 16) * 2];
    s2 = input_samples[(isample >> 16) * 2 + 2];
    os = (s1 * (0x10000 - t) + s2 * t) >> 16;
    output_samples[osample * 2] = os;
    s1 = input_samples[(isample >> 16) * 2 + 1];
    s2 = input_samples[(isample >> 16) * 2 + 3];
    os = (s1 * (0x10000 - t) + s2 * t) >> 16;
    output_samples[(osample * 2) + 1] = os;
    isample += istep;
  }
  output_samples[out_samples * 2 - 2] = input_samples[in_samples * 2 - 2];
  output_samples[out_samples * 2 - 1] = input_samples[in_samples * 2 - 1];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<audio_out_resample_stereo>:
  endbr64
  lea    -0x2(%rsi),%eax
  push   %r14
  mov    %rdx,%r11
  mov    %rdi,%r8
  shl    $0x10,%eax
  push   %r13
  mov    %esi,%r10d
  push   %r12
  cltd
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x2,%ecx
  idiv   %ecx
  cmp    $0x1,%ebx
  jle    L7
  mov    %r11,%rdi
  lea    0x8(%r11,%rcx,8),%r12
  xor    %r9d,%r9d
  mov    $0x10000,%ebp
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %r9d,%ecx
  movzwl %r9w,%edx
  mov    %ebp,%r13d
  add    $0x8,%rdi
  sar    $0x10,%ecx
  sub    %edx,%r13d
  add    %eax,%r9d
  add    %ecx,%ecx
  movslq %ecx,%rcx
  mov    0x8(%r8,%rcx,4),%esi
  mov    (%r8,%rcx,4),%r14d
  imul   %edx,%esi
  imul   %r13d,%r14d
  add    %r14d,%esi
  sar    $0x10,%esi
  mov    %esi,-0x8(%rdi)
  imul   0xc(%r8,%rcx,4),%edx
  imul   0x4(%r8,%rcx,4),%r13d
  add    %r13d,%edx
  sar    $0x10,%edx
  mov    %edx,-0x4(%rdi)
  cmp    %rdi,%r12
  jne    L5
L7:
  add    %r10d,%r10d
  add    %ebx,%ebx
  movslq %r10d,%r10
  movslq %ebx,%rbx
  shl    $0x2,%r10
  shl    $0x2,%rbx
  mov    -0x8(%r8,%r10,1),%eax
  mov    %eax,-0x8(%r11,%rbx,1)
  mov    -0x4(%r8,%r10,1),%eax
  mov    %eax,-0x4(%r11,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void audio_out_resample_stereo(int16_t *input_samples, uint32_t in_samples, int16_t *output_samples, uint32_t out_samples) {
  int osample;
  uint32_t isample = 0;
  uint32_t istep = ((in_samples - 2) << 16) / (out_samples - 2);
  for (osample = 0; osample < out_samples - 1; osample++) {
    int s1;
    int s2;
    int16_t os;
    uint32_t t = isample & 0xffff;
    s1 = input_samples[(isample >> 16) * 2];
    s2 = input_samples[(isample >> 16) * 2 + 2];
    os = (s1 * (0x10000 - t) + s2 * t) >> 16;
    output_samples[osample * 2] = os;
    s1 = input_samples[(isample >> 16) * 2 + 1];
    s2 = input_samples[(isample >> 16) * 2 + 3];
    os = (s1 * (0x10000 - t) + s2 * t) >> 16;
    output_samples[(osample * 2) + 1] = os;
    isample += istep;
  }
  output_samples[out_samples * 2 - 2] = input_samples[in_samples * 2 - 2];
  output_samples[out_samples * 2 - 1] = input_samples[in_samples * 2 - 1];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<audio_out_resample_stereo>:
  endbr64
  lea    -0x2(%rsi),%eax
  push   %r14
  mov    %rdx,%r11
  mov    %rdi,%r8
  shl    $0x10,%eax
  push   %r13
  mov    %esi,%r10d
  push   %r12
  cltd
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x2,%ecx
  idiv   %ecx
  cmp    $0x1,%ebx
  jle    L3
  mov    %r11,%rdi
  lea    0x8(%r11,%rcx,8),%r12
  xor    %r9d,%r9d
  mov    $0x10000,%ebp
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %r9d,%ecx
  movzwl %r9w,%edx
  mov    %ebp,%r13d
  add    $0x8,%rdi
  sar    $0x10,%ecx
  sub    %edx,%r13d
  add    %eax,%r9d
  add    %ecx,%ecx
  movslq %ecx,%rcx
  mov    0x8(%r8,%rcx,4),%esi
  mov    (%r8,%rcx,4),%r14d
  imul   %edx,%esi
  imul   %r13d,%r14d
  add    %r14d,%esi
  sar    $0x10,%esi
  mov    %esi,-0x8(%rdi)
  imul   0xc(%r8,%rcx,4),%edx
  imul   0x4(%r8,%rcx,4),%r13d
  add    %r13d,%edx
  sar    $0x10,%edx
  mov    %edx,-0x4(%rdi)
  cmp    %r12,%rdi
  jne    L17
L3:
  add    %r10d,%r10d
  add    %ebx,%ebx
  movslq %r10d,%r10
  movslq %ebx,%rbx
  shl    $0x2,%r10
  shl    $0x2,%rbx
  mov    -0x8(%r8,%r10,1),%eax
  mov    %eax,-0x8(%r11,%rbx,1)
  mov    -0x4(%r8,%r10,1),%eax
  mov    %eax,-0x4(%r11,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void audio_out_resample_stereo(int16_t *input_samples, uint32_t in_samples, int16_t *output_samples, uint32_t out_samples) {
  int osample;
  uint32_t isample = 0;
  uint32_t istep = ((in_samples - 2) << 16) / (out_samples - 2);
  for (osample = 0; osample < out_samples - 1; osample++) {
    int s1;
    int s2;
    int16_t os;
    uint32_t t = isample & 0xffff;
    s1 = input_samples[(isample >> 16) * 2];
    s2 = input_samples[(isample >> 16) * 2 + 2];
    os = (s1 * (0x10000 - t) + s2 * t) >> 16;
    output_samples[osample * 2] = os;
    s1 = input_samples[(isample >> 16) * 2 + 1];
    s2 = input_samples[(isample >> 16) * 2 + 3];
    os = (s1 * (0x10000 - t) + s2 * t) >> 16;
    output_samples[(osample * 2) + 1] = os;
    isample += istep;
  }
  output_samples[out_samples * 2 - 2] = input_samples[in_samples * 2 - 2];
  output_samples[out_samples * 2 - 1] = input_samples[in_samples * 2 - 1];
}
```
user: What is the c source code of the assembly code below:

```asm
<showUsage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
datatype: string
name: disassemble_data
arguments: data_tag: D18
datatype: string

tool: D5: Usage:   

tool: D18: %s 'input file' 'output file'


assistant: ```c
void showUsage(const char *progname) {
  printf("Usage:   ");
  printf("%s 'input file' 'output file'\n", progname);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<showUsage>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: tag: D18
data_type: string
name: parse_data
arguments: tag: D0
data_type: string

tool: D18: Usage:   

tool: D0: %s 'input file' 'output file'


assistant: ```c
void showUsage(const char *progname) {
  printf("Usage:   ");
  printf("%s 'input file' 'output file'\n", progname);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<showUsage>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  pop    %rbp
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: data_tag: D15
type: string
name: read_data
arguments: data_tag: D18
type: string

tool: D15: Usage:   

tool: D18: %s 'input file' 'output file'


assistant: ```c
void showUsage(const char *progname) {
  printf("Usage:   ");
  printf("%s 'input file' 'output file'\n", progname);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<showUsage>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  pop    %rbp
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: tag: D9
type_infered: string
name: read_data
arguments: tag: D0
type_infered: string

tool: D9: Usage:   

tool: D0: %s 'input file' 'output file'


assistant: ```c
void showUsage(const char *progname) {
  printf("Usage:   ");
  printf("%s 'input file' 'output file'\n", progname);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<LogB10>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movsd  %xmm0,-0x8(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <gettext@plt>
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <log10@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
datatype: string
name: disassemble_data
arguments: tag: D10
datatype: string

tool: D0: Enter a number: 

tool: D10: %lf

assistant: ```c
double LogB10(double a) {
  printf(gettext("Enter a number: "));
  scanf("%lf", &a);
  return (log10(a));
}
```
user: Translate the below assembly code into C programming language:

```asm
<LogB10>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm0,0x8(%rsp)
  lea    D5(%rip),%rdi
  call   <gettext@plt>
  mov    %rax,%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  movsd  0x8(%rsp),%xmm0
  call   <log10@plt>
  add    $0x18,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
datatype: string
name: unpack_data
arguments: data_label: D4
datatype: string

tool: D5: Enter a number: 

tool: D4: %lf

assistant: ```c
double LogB10(double a) {
  printf(gettext("Enter a number: "));
  scanf("%lf", &a);
  return (log10(a));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<LogB10>:
  endbr64
  sub    $0x18,%rsp
  lea    D12(%rip),%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <gettext@plt>
  mov    %rax,%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  movsd  0x8(%rsp),%xmm0
  call   <log10@plt>
  add    $0x18,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D12
datatype: string
name: retrieve_data
arguments: data_tag: D0
datatype: string

tool: D12: Enter a number: 

tool: D0: %lf

assistant: ```c
double LogB10(double a) {
  printf(gettext("Enter a number: "));
  scanf("%lf", &a);
  return (log10(a));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<LogB10>:
  endbr64
  sub    $0x18,%rsp
  lea    D8(%rip),%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <gettext@plt>
  mov    %rax,%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  movsd  0x8(%rsp),%xmm0
  call   <log10@plt>
  add    $0x18,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D8
datatype: string
name: disassemble_data
arguments: data_tag: D16
datatype: string

tool: D8: Enter a number: 

tool: D16: %lf

assistant: ```c
double LogB10(double a) {
  printf(gettext("Enter a number: "));
  scanf("%lf", &a);
  return (log10(a));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ft_putstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L6:
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  nop
  leave
  ret
```

assistant: ```c
void ft_putstr(char const *s) {
  int i;
  i = 0;
  while (s[i] != '\0') {
    i++;
  }
  write(1, s, i);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_putstr>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  cmpb   $0x0,(%rdi)
  je     L8
  mov    $0x1,%eax
L26:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rsi,%rax,1)
  jne    L26
L21:
  mov    $0x1,%edi
  call   <write@plt>
  add    $0x8,%rsp
  ret
L8:
  mov    $0x0,%edx
  jmp    L21
```

assistant: ```c
void ft_putstr(char const *s) {
  int i;
  i = 0;
  while (s[i] != '\0') {
    i++;
  }
  write(1, s, i);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ft_putstr>:
  endbr64
  cmpb   $0x0,(%rdi)
  mov    %rdi,%rsi
  je     L1
  mov    $0x1,%eax
  nopl   0x0(%rax)
L6:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rsi,%rax,1)
  jne    L6
  mov    $0x1,%edi
  jmp    1050 <write@plt>
  nop
L1:
  xor    %edx,%edx
  mov    $0x1,%edi
  jmp    1050 <write@plt>
```

assistant: ```c
void ft_putstr(char const *s) {
  int i;
  i = 0;
  while (s[i] != '\0') {
    i++;
  }
  write(1, s, i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_putstr>:
  endbr64
  cmpb   $0x0,(%rdi)
  mov    %rdi,%rsi
  je     L13
  mov    $0x1,%eax
  nopl   0x0(%rax)
L11:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rsi,%rax,1)
  jne    L11
  mov    $0x1,%edi
  jmp    1050 <write@plt>
  nop
L13:
  xor    %edx,%edx
  mov    $0x1,%edi
  jmp    1050 <write@plt>
```

assistant: ```c
void ft_putstr(char const *s) {
  int i;
  i = 0;
  while (s[i] != '\0') {
    i++;
  }
  write(1, s, i);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<bibl_notexify>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D16(%rip),%rax
  mov    %rax,-0x30(%rbp)
  lea    D31(%rip),%rax
  mov    %rax,-0x28(%rbp)
  lea    D48(%rip),%rax
  mov    %rax,-0x20(%rbp)
  lea    D25(%rip),%rax
  mov    %rax,-0x18(%rbp)
  lea    D22(%rip),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x5,-0x34(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L11
L13:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcasecmp@plt>
  test   %eax,%eax
  jne    L2
  mov    $0x1,%eax
  jmp    L1
L2:
  addl   $0x1,-0x38(%rbp)
L11:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L13
  mov    $0x0,%eax
L1:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D16
type: string
name: disassemble_data
arguments: tag: D31
type: string
name: disassemble_data
arguments: tag: D48
type: string
name: disassemble_data
arguments: tag: D25
type: string
name: disassemble_data
arguments: tag: D22
type: string

tool: D16: DOI

tool: D31: URL

tool: D48: REFNUM

tool: D25: FILEATTACH

tool: D22: FILE

assistant: ```c
int bibl_notexify(char *tag) {
  char *protected[] = {"DOI", "URL", "REFNUM", "FILEATTACH", "FILE"};
  int i, nprotected = sizeof(protected) / sizeof(protected[0]);
  for (i = 0; i < nprotected; ++i)
    if (!strcasecmp(tag, protected[i]))
      return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bibl_notexify>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D7(%rip),%rax
  mov    %rax,(%rsp)
  lea    D22(%rip),%rax
  mov    %rax,0x8(%rsp)
  lea    D43(%rip),%rax
  mov    %rax,0x10(%rsp)
  lea    D42(%rip),%rax
  mov    %rax,0x18(%rsp)
  lea    D36(%rip),%rax
  mov    %rax,0x20(%rsp)
  mov    %rsp,%rbx
  lea    0x28(%rsp),%r12
L13:
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  call   <strcasecmp@plt>
  test   %eax,%eax
  je     L23
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L13
  mov    $0x0,%eax
L20:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  mov    $0x1,%eax
  jmp    L20
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D7
datatype: string
name: parse_data
arguments: tag: D22
datatype: string
name: parse_data
arguments: tag: D43
datatype: string
name: parse_data
arguments: tag: D42
datatype: string
name: parse_data
arguments: tag: D36
datatype: string

tool: D7: DOI

tool: D22: URL

tool: D43: REFNUM

tool: D42: FILEATTACH

tool: D36: FILE

assistant: ```c
int bibl_notexify(char *tag) {
  char *protected[] = {"DOI", "URL", "REFNUM", "FILEATTACH", "FILE"};
  int i, nprotected = sizeof(protected) / sizeof(protected[0]);
  for (i = 0; i < nprotected; ++i)
    if (!strcasecmp(tag, protected[i]))
      return 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bibl_notexify>:
  endbr64
  push   %r12
  lea    D14(%rip),%rsi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  lea    D40(%rip),%rax
  mov    %rsp,%rbx
  lea    0x28(%rsp),%r12
  mov    %rax,0x8(%rsp)
  lea    D38(%rip),%rax
  mov    %rax,0x10(%rsp)
  lea    D9(%rip),%rax
  mov    %rax,0x18(%rsp)
  lea    D33(%rip),%rax
  mov    %rax,0x20(%rsp)
  call   <strcasecmp@plt>
  test   %eax,%eax
  je     L44
L11:
  add    $0x8,%rbx
  cmp    %r12,%rbx
  je     L9
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  call   <strcasecmp@plt>
  test   %eax,%eax
  jne    L11
L44:
  mov    $0x1,%eax
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L9:
  xor    %eax,%eax
L17:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L47
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D14
type: string
name: read_data
arguments: tag: D40
type: string
name: read_data
arguments: tag: D38
type: string
name: read_data
arguments: tag: D9
type: string
name: read_data
arguments: tag: D33
type: string

tool: D14: DOI

tool: D40: URL

tool: D38: REFNUM

tool: D9: FILEATTACH

tool: D33: FILE

assistant: ```c
int bibl_notexify(char *tag) {
  char *protected[] = {"DOI", "URL", "REFNUM", "FILEATTACH", "FILE"};
  int i, nprotected = sizeof(protected) / sizeof(protected[0]);
  for (i = 0; i < nprotected; ++i)
    if (!strcasecmp(tag, protected[i]))
      return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bibl_notexify>:
  endbr64
  push   %rbp
  lea    D36(%rip),%rsi
  mov    %rdi,%rbp
  call   <strcasecmp@plt>
  test   %eax,%eax
  je     L2
  lea    D11(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcasecmp@plt>
  test   %eax,%eax
  je     L2
  lea    D19(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcasecmp@plt>
  test   %eax,%eax
  je     L2
  lea    D17(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcasecmp@plt>
  test   %eax,%eax
  je     L2
  mov    %rbp,%rdi
  lea    D10(%rip),%rsi
  call   <strcasecmp@plt>
  pop    %rbp
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  mov    $0x1,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D36
datatype: string
name: extract_symbolic_data
arguments: data_tag: D11
datatype: string
name: extract_symbolic_data
arguments: data_tag: D19
datatype: string
name: extract_symbolic_data
arguments: data_tag: D17
datatype: string
name: extract_symbolic_data
arguments: data_tag: D10
datatype: string

tool: D36: DOI

tool: D11: URL

tool: D19: REFNUM

tool: D17: FILEATTACH

tool: D10: FILE

assistant: ```c
int bibl_notexify(char *tag) {
  char *protected[] = {"DOI", "URL", "REFNUM", "FILEATTACH", "FILE"};
  int i, nprotected = sizeof(protected) / sizeof(protected[0]);
  for (i = 0; i < nprotected; ++i)
    if (!strcasecmp(tag, protected[i]))
      return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<test4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x3,-0x4(%rbp)
  jmp    L24
L64:
  cmpl   $0x2,-0x4(%rbp)
  jg     L48
  jmp    L18
L42:
  addl   $0x1,-0x4(%rbp)
L18:
  cmpl   $0x0,-0x4(%rbp)
  jg     L42
  jmp    L24
L3:
  addl   $0x1,-0x4(%rbp)
L48:
  cmpl   $0x9,-0x4(%rbp)
  jle    L3
L24:
  cmpl   $0x0,-0x4(%rbp)
  jg     L64
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int test4() {
  int y = 3;
  while (y > 0) {
    if (y < 3) {
      while (y > 0)
        y++;
    } else {
      while (y < 10)
        y++;
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test4>:
  endbr64
  mov    $0x3,%eax
L41:
  cmp    $0x9,%eax
  jg     L46
L42:
  add    $0x1,%eax
  cmp    $0xa,%eax
  jne    L42
  jmp    L41
L46:
  cmp    $0x2,%eax
  jg     L41
  test   %eax,%eax
  jle    L57
L47:
  add    $0x1,%eax
  test   %eax,%eax
  jg     L47
L57:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int test4() {
  int y = 3;
  while (y > 0) {
    if (y < 3) {
      while (y > 0)
        y++;
    } else {
      while (y < 10)
        y++;
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test4>:
  endbr64
  nopl   0x0(%rax)
L1:
  jmp    L1
```

assistant: ```c
int test4() {
  int y = 3;
  while (y > 0) {
    if (y < 3) {
      while (y > 0)
        y++;
    } else {
      while (y < 10)
        y++;
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<test4>:
  endbr64
  nopl   0x0(%rax)
L7:
  jmp    L7
```

assistant: ```c
int test4() {
  int y = 3;
  while (y > 0) {
    if (y < 3) {
      while (y > 0)
        y++;
    } else {
      while (y < 10)
        y++;
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L8:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L8
```

assistant: ```c
int main(void) {
  while (1) {
    sleep(1);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
L0:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L0
```

assistant: ```c
int main(void) {
  while (1) {
    sleep(1);
  }
  return 0;
}
```