ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x4,%esi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x10,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D49
type_guessed: string
name: read_data
arguments: data_label: D34
type_guessed: string
name: read_data
arguments: data_label: D41
type_guessed: string
name: read_data
arguments: data_label: D50
type_guessed: string
name: read_data
arguments: data_label: D35
type_guessed: string
name: read_data
arguments: data_label: D38
type_guessed: string

tool: D49: Type int has a size of %zd bytes.


tool: D34: Type char has a size of %zd bytes.


tool: D41: Type long has a size of %zd bytes.


tool: D50: Type long long has a size of %zd bytes.


tool: D35: Type double has a size of %zd bytes.


tool: D38: Type long double has a size of %zd bytes.


assistant: ```c
int main(void) {
  printf("Type int has a size of %zd bytes.\n", sizeof(int));
  printf("Type char has a size of %zd bytes.\n", sizeof(char));
  printf("Type long has a size of %zd bytes.\n", sizeof(long));
  printf("Type long long has a size of %zd bytes.\n", sizeof(long long));
  printf("Type double has a size of %zd bytes.\n", sizeof(double));
  printf("Type long double has a size of %zd bytes.\n", sizeof(long double));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%esi
  lea    D35(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D54(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D40(%rip),%rdi
  call   <printf@plt>
  mov    $0x10,%esi
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D35
data_type: string
name: retrieve_data
arguments: data_tag: D54
data_type: string
name: retrieve_data
arguments: data_tag: D7
data_type: string
name: retrieve_data
arguments: data_tag: D2
data_type: string
name: retrieve_data
arguments: data_tag: D40
data_type: string
name: retrieve_data
arguments: data_tag: D23
data_type: string

tool: D35: Type int has a size of %zd bytes.


tool: D54: Type char has a size of %zd bytes.


tool: D7: Type long has a size of %zd bytes.


tool: D2: Type long long has a size of %zd bytes.


tool: D40: Type double has a size of %zd bytes.


tool: D23: Type long double has a size of %zd bytes.


assistant: ```c
int main(void) {
  printf("Type int has a size of %zd bytes.\n", sizeof(int));
  printf("Type char has a size of %zd bytes.\n", sizeof(char));
  printf("Type long has a size of %zd bytes.\n", sizeof(long));
  printf("Type long long has a size of %zd bytes.\n", sizeof(long long));
  printf("Type double has a size of %zd bytes.\n", sizeof(double));
  printf("Type long double has a size of %zd bytes.\n", sizeof(long double));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%esi
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D46(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D51(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D57(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    $0x10,%esi
  lea    D31(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D23
type: string
name: disassemble_data
arguments: data_label: D46
type: string
name: disassemble_data
arguments: data_label: D51
type: string
name: disassemble_data
arguments: data_label: D57
type: string
name: disassemble_data
arguments: data_label: D12
type: string
name: disassemble_data
arguments: data_label: D31
type: string

tool: D23: Type int has a size of %zd bytes.


tool: D46: Type char has a size of %zd bytes.


tool: D51: Type long has a size of %zd bytes.


tool: D57: Type long long has a size of %zd bytes.


tool: D12: Type double has a size of %zd bytes.


tool: D31: Type long double has a size of %zd bytes.


assistant: ```c
int main(void) {
  printf("Type int has a size of %zd bytes.\n", sizeof(int));
  printf("Type char has a size of %zd bytes.\n", sizeof(char));
  printf("Type long has a size of %zd bytes.\n", sizeof(long));
  printf("Type long long has a size of %zd bytes.\n", sizeof(long long));
  printf("Type double has a size of %zd bytes.\n", sizeof(double));
  printf("Type long double has a size of %zd bytes.\n", sizeof(long double));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4,%esi
  lea    D35(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D51(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D59(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D22(%rip),%rdi
  call   <printf@plt>
  mov    $0x10,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D35
datatype: string
name: unpack_data
arguments: tag: D51
datatype: string
name: unpack_data
arguments: tag: D30
datatype: string
name: unpack_data
arguments: tag: D59
datatype: string
name: unpack_data
arguments: tag: D22
datatype: string
name: unpack_data
arguments: tag: D4
datatype: string

tool: D35: Type int has a size of %zd bytes.


tool: D51: Type char has a size of %zd bytes.


tool: D30: Type long has a size of %zd bytes.


tool: D59: Type long long has a size of %zd bytes.


tool: D22: Type double has a size of %zd bytes.


tool: D4: Type long double has a size of %zd bytes.


assistant: ```c
int main(void) {
  printf("Type int has a size of %zd bytes.\n", sizeof(int));
  printf("Type char has a size of %zd bytes.\n", sizeof(char));
  printf("Type long has a size of %zd bytes.\n", sizeof(long));
  printf("Type long long has a size of %zd bytes.\n", sizeof(long long));
  printf("Type double has a size of %zd bytes.\n", sizeof(double));
  printf("Type long double has a size of %zd bytes.\n", sizeof(long double));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strtoll>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L13
L71:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x2f,-0xc(%rbp)
  jle    L107
  cmpl   $0x39,-0xc(%rbp)
  jg     L107
  subl   $0x30,-0xc(%rbp)
  jmp    L48
L107:
  cmpl   $0x40,-0xc(%rbp)
  jle    L96
  cmpl   $0x5a,-0xc(%rbp)
  jg     L96
  subl   $0x37,-0xc(%rbp)
  jmp    L48
L96:
  cmpl   $0x60,-0xc(%rbp)
  jle    L93
  cmpl   $0x7a,-0xc(%rbp)
  jg     L93
  subl   $0x57,-0xc(%rbp)
L48:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jge    L47
  mov    -0x24(%rbp),%eax
  cltq
  imul   -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  addq   $0x1,-0x18(%rbp)
L13:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L71
  jmp    L93
L47:
  nop
L93:
  mov    -0x20(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long long int strtoll(const char *nptr, char **endptr, int base) {
  long long val = 0;
  for (; *nptr; nptr++) {
    int v = *nptr;
    if ('0' <= v && v <= '9') {
      v -= '0';
    } else if ('A' <= v && v <= 'Z') {
      v -= 'A' - 10;
    } else if ('a' <= v && v <= 'z') {
      v -= 'a' - 10;
    } else {
      break;
    }
    if (v >= base) {
      break;
    }
    val = val * base + v;
  }
  *endptr = (char *)nptr;
  return val;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strtoll>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L85
  mov    $0x0,%eax
  movslq %edx,%r9
  jmp    L34
L45:
  lea    -0x41(%rcx),%r8d
  cmp    $0x19,%r8d
  ja     L82
  sub    $0x37,%ecx
L46:
  cmp    %edx,%ecx
  jge    L24
  imul   %r9,%rax
  movslq %ecx,%rcx
  add    %rcx,%rax
  add    $0x1,%rdi
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L24
L34:
  movsbl %cl,%ecx
  lea    -0x30(%rcx),%r8d
  cmp    $0x9,%r8d
  ja     L45
  mov    %r8d,%ecx
  jmp    L46
L82:
  lea    -0x61(%rcx),%r8d
  cmp    $0x19,%r8d
  ja     L24
  sub    $0x57,%ecx
  jmp    L46
L85:
  mov    $0x0,%eax
L24:
  mov    %rdi,(%rsi)
  ret
```

assistant: ```c
long long int strtoll(const char *nptr, char **endptr, int base) {
  long long val = 0;
  for (; *nptr; nptr++) {
    int v = *nptr;
    if ('0' <= v && v <= '9') {
      v -= '0';
    } else if ('A' <= v && v <= 'Z') {
      v -= 'A' - 10;
    } else if ('a' <= v && v <= 'z') {
      v -= 'a' - 10;
    } else {
      break;
    }
    if (v >= base) {
      break;
    }
    val = val * base + v;
  }
  *endptr = (char *)nptr;
  return val;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strtoll>:
  endbr64
  mov    %rdi,%r8
  movsbl (%rdi),%edi
  xor    %eax,%eax
  test   %dil,%dil
  je     L37
  movslq %edx,%r9
  jmp    L58
  cs nopw 0x0(%rax,%rax,1)
L51:
  lea    -0x37(%rdi),%ecx
L47:
  cmp    %edx,%ecx
  jge    L37
L55:
  imul   %r9,%rax
  movsbl 0x1(%r8),%edi
  movslq %ecx,%rcx
  add    $0x1,%r8
  add    %rcx,%rax
  test   %dil,%dil
  je     L37
L58:
  lea    -0x30(%rdi),%ecx
  cmp    $0x9,%ecx
  jbe    L47
  lea    -0x41(%rdi),%ecx
  cmp    $0x19,%ecx
  jbe    L51
  lea    -0x61(%rdi),%ecx
  cmp    $0x19,%ecx
  ja     L37
  lea    -0x57(%rdi),%ecx
  cmp    %edx,%ecx
  jl     L55
L37:
  mov    %r8,(%rsi)
  ret
```

assistant: ```c
long long int strtoll(const char *nptr, char **endptr, int base) {
  long long val = 0;
  for (; *nptr; nptr++) {
    int v = *nptr;
    if ('0' <= v && v <= '9') {
      v -= '0';
    } else if ('A' <= v && v <= 'Z') {
      v -= 'A' - 10;
    } else if ('a' <= v && v <= 'z') {
      v -= 'a' - 10;
    } else {
      break;
    }
    if (v >= base) {
      break;
    }
    val = val * base + v;
  }
  *endptr = (char *)nptr;
  return val;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strtoll>:
  endbr64
  mov    %rdi,%r8
  movsbl (%rdi),%edi
  xor    %eax,%eax
  test   %dil,%dil
  je     L75
  movslq %edx,%r9
  jmp    L53
  cs nopw 0x0(%rax,%rax,1)
L17:
  lea    -0x37(%rdi),%ecx
L68:
  cmp    %edx,%ecx
  jge    L75
L15:
  imul   %r9,%rax
  movsbl 0x1(%r8),%edi
  movslq %ecx,%rcx
  add    $0x1,%r8
  add    %rcx,%rax
  test   %dil,%dil
  je     L75
L53:
  lea    -0x30(%rdi),%ecx
  cmp    $0x9,%ecx
  jbe    L68
  lea    -0x41(%rdi),%ecx
  cmp    $0x19,%ecx
  jbe    L17
  lea    -0x61(%rdi),%ecx
  cmp    $0x19,%ecx
  ja     L75
  lea    -0x57(%rdi),%ecx
  cmp    %edx,%ecx
  jl     L15
L75:
  mov    %r8,(%rsi)
  ret
```

assistant: ```c
long long int strtoll(const char *nptr, char **endptr, int base) {
  long long val = 0;
  for (; *nptr; nptr++) {
    int v = *nptr;
    if ('0' <= v && v <= '9') {
      v -= '0';
    } else if ('A' <= v && v <= 'Z') {
      v -= 'A' - 10;
    } else if ('a' <= v && v <= 'z') {
      v -= 'a' - 10;
    } else {
      break;
    }
    if (v >= base) {
      break;
    }
    val = val * base + v;
  }
  *endptr = (char *)nptr;
  return val;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L11
  movl   $0x0,-0x8(%rbp)
L11:
  cmpl   $0x0,-0x8(%rbp)
  jne    L16
  movl   $0x0,-0x8(%rbp)
  jmp    L17
L16:
  nop
L17:
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int foo(int x, int y) {
  if (x)
    y = 0;
  if (y)
    goto lab;
  y = 0;
lab:
  return y;
}
```
user: What is the c source code of the assembly code below?

```asm
<foo>:
  endbr64
  test   %esi,%esi
  je     L7
  test   %edi,%edi
  je     L16
L7:
  mov    $0x0,%eax
  ret
L16:
  mov    %esi,%eax
  ret
```

assistant: ```c
int foo(int x, int y) {
  if (x)
    y = 0;
  if (y)
    goto lab;
  y = 0;
lab:
  return y;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<foo>:
  endbr64
  mov    %esi,%eax
  test   %esi,%esi
  je     L14
  test   %edi,%edi
  jne    L14
  ret
  nop
L14:
  xor    %eax,%eax
  ret
```

assistant: ```c
int foo(int x, int y) {
  if (x)
    y = 0;
  if (y)
    goto lab;
  y = 0;
lab:
  return y;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<foo>:
  endbr64
  mov    %esi,%eax
  test   %esi,%esi
  je     L7
  test   %edi,%edi
  jne    L7
  ret
  nop
L7:
  xor    %eax,%eax
  ret
```

assistant: ```c
int foo(int x, int y) {
  if (x)
    y = 0;
  if (y)
    goto lab;
  y = 0;
lab:
  return y;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <seek_definition@plt>
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L50
  mov    -0x8(%rbp),%eax
  jmp    L38
L50:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  je     L17
  mov    $0x1,%eax
  jmp    L38
L17:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  je     L27
  mov    $0x2,%eax
  jmp    L38
L27:
  nop
L51:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  test   %eax,%eax
  jne    L51
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <fread@plt>
  mov    $0x0,%eax
L38:
  leave
  ret
```

assistant: ```c
int read_int(FILE *f, char *config_path, int32_t *result) {
  int success;
  uint8_t temp;
  success = seek_definition(f, config_path);
  if (success != 0)
    return success;
  temp = fgetc(f);
  if (temp != 1)
    return 1;
  temp = fgetc(f);
  if (temp != 2)
    return 2;
  while (fgetc(f) != 0)
    ;
  fread(result, 4, 1, f);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<read_int>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rdx,%r12
  call   <seek_definition@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  je     L32
L26:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L32:
  mov    %rbp,%rdi
  call   <fgetc@plt>
  mov    $0x1,%ebx
  cmp    $0x1,%eax
  jne    L26
  mov    %rbp,%rdi
  call   <fgetc@plt>
  mov    $0x2,%ebx
  cmp    $0x2,%eax
  jne    L26
L23:
  mov    %rbp,%rdi
  call   <fgetc@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L23
  mov    %rbp,%rcx
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %r12,%rdi
  call   <fread@plt>
  jmp    L26
```

assistant: ```c
int read_int(FILE *f, char *config_path, int32_t *result) {
  int success;
  uint8_t temp;
  success = seek_definition(f, config_path);
  if (success != 0)
    return success;
  temp = fgetc(f);
  if (temp != 1)
    return 1;
  temp = fgetc(f);
  if (temp != 2)
    return 2;
  while (fgetc(f) != 0)
    ;
  fread(result, 4, 1, f);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<read_int>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  call   <seek_definition@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  je     L7
L4:
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L7:
  mov    %r13,%rdi
  mov    $0x1,%r12d
  call   <fgetc@plt>
  cmp    $0x1,%eax
  jne    L4
  mov    %r13,%rdi
  mov    $0x2,%r12d
  call   <fgetc@plt>
  cmp    $0x2,%eax
  jne    L4
  xchg   %ax,%ax
L17:
  mov    %r13,%rdi
  call   <fgetc@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L17
  mov    %r13,%rcx
  mov    %rbp,%rdi
  mov    $0x1,%edx
  mov    $0x4,%esi
  call   <fread@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int read_int(FILE *f, char *config_path, int32_t *result) {
  int success;
  uint8_t temp;
  success = seek_definition(f, config_path);
  if (success != 0)
    return success;
  temp = fgetc(f);
  if (temp != 1)
    return 1;
  temp = fgetc(f);
  if (temp != 2)
    return 2;
  while (fgetc(f) != 0)
    ;
  fread(result, 4, 1, f);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<read_int>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  call   <seek_definition@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  je     L8
L2:
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L8:
  mov    %r13,%rdi
  mov    $0x1,%r12d
  call   <fgetc@plt>
  cmp    $0x1,%eax
  jne    L2
  mov    %r13,%rdi
  mov    $0x2,%r12d
  call   <fgetc@plt>
  cmp    $0x2,%eax
  jne    L2
  xchg   %ax,%ax
L19:
  mov    %r13,%rdi
  call   <fgetc@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L19
  mov    %r13,%rcx
  mov    %rbp,%rdi
  mov    $0x1,%edx
  mov    $0x4,%esi
  call   <fread@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int read_int(FILE *f, char *config_path, int32_t *result) {
  int success;
  uint8_t temp;
  success = seek_definition(f, config_path);
  if (success != 0)
    return success;
  temp = fgetc(f);
  if (temp != 1)
    return 1;
  temp = fgetc(f);
  if (temp != 2)
    return 2;
  while (fgetc(f) != 0)
    ;
  fread(result, 4, 1, f);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test5>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <recip@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  ucomisd D37(%rip),%xmm0
  setnp  %al
  mov    $0x0,%edx
  movsd  -0x8(%rbp),%xmm0
  ucomisd D37(%rip),%xmm0
  cmovne %edx,%eax
  movzbl %al,%eax
  mov    %eax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <do_nothing@plt>
  movsd  -0x8(%rbp),%xmm0
  ucomisd D37(%rip),%xmm0
  setnp  %al
  mov    $0x0,%edx
  movsd  -0x8(%rbp),%xmm0
  ucomisd D37(%rip),%xmm0
  cmovne %edx,%eax
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  je     L6
  mov    $0x3d,%edx
  jmp    L2
L6:
  mov    $0x21,%edx
L2:
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  movq   %rax,%xmm0
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  cmpl   $0x0,-0xc(%rbp)
  je     L13
  mov    $0x3d,%edx
  jmp    L30
L13:
  mov    $0x21,%edx
L30:
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  movq   %rax,%xmm0
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D37
type_infered: f64
name: read_data
arguments: data_label: D31
type_infered: string
name: read_data
arguments: data_label: D36
type_infered: string

tool: D37: 0.1

tool: D31: test5 t1: r1 %f %c= 0.1


tool: D36: test5 t2: r1 %f %c= 0.1


assistant: ```c
void test5(int denom) {
  double r1;
  int t1, t2;
  r1 = recip(denom);
  t1 = r1 == 0.1;
  do_nothing();
  t2 = r1 == 0.1;
  printf("test5 t1: r1 %f %c= 0.1\n", r1, t1 ? '=' : '!');
  printf("test5 t2: r1 %f %c= 0.1\n", r1, t2 ? '=' : '!');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test5>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <recip@plt>
  movq   %xmm0,%rbx
  mov    $0x0,%eax
  call   <do_nothing@plt>
  movq   %rbx,%xmm1
  ucomisd D23(%rip),%xmm1
  jp     L16
  mov    $0x3d,%ebp
  jne    L16
L10:
  mov    %ebp,%esi
  movq   %rbx,%xmm0
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%esi
  movq   %rbx,%xmm0
  lea    D24(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  mov    $0x21,%ebp
  jmp    L10
```

tool_calls: name: disassemble_data
arguments: label: D23
datatype: f64
name: disassemble_data
arguments: label: D27
datatype: string
name: disassemble_data
arguments: label: D24
datatype: string

tool: D23: 0.1

tool: D27: test5 t1: r1 %f %c= 0.1


tool: D24: test5 t2: r1 %f %c= 0.1


assistant: ```c
void test5(int denom) {
  double r1;
  int t1, t2;
  r1 = recip(denom);
  t1 = r1 == 0.1;
  do_nothing();
  t2 = r1 == 0.1;
  printf("test5 t1: r1 %f %c= 0.1\n", r1, t1 ? '=' : '!');
  printf("test5 t2: r1 %f %c= 0.1\n", r1, t2 ? '=' : '!');
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test5>:
  endbr64
  push   %rbp
  sub    $0x10,%rsp
  call   <recip@plt>
  xor    %eax,%eax
  movsd  %xmm0,0x8(%rsp)
  call   <do_nothing@plt>
  movsd  0x8(%rsp),%xmm0
  ucomisd D10(%rip),%xmm0
  jp     L3
  mov    $0x3d,%ebp
  jne    L3
L22:
  mov    %ebp,%esi
  lea    D15(%rip),%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x10,%rsp
  mov    %ebp,%esi
  lea    D0(%rip),%rdi
  pop    %rbp
  jmp    1070 <printf@plt>
  nop
L3:
  mov    $0x21,%ebp
  jmp    L22
```

tool_calls: name: parse_data
arguments: label: D10
type_infered: f64
name: parse_data
arguments: label: D15
type_infered: string
name: parse_data
arguments: label: D0
type_infered: string

tool: D10: 0.1

tool: D15: test5 t1: r1 %f %c= 0.1


tool: D0: test5 t2: r1 %f %c= 0.1


assistant: ```c
void test5(int denom) {
  double r1;
  int t1, t2;
  r1 = recip(denom);
  t1 = r1 == 0.1;
  do_nothing();
  t2 = r1 == 0.1;
  printf("test5 t1: r1 %f %c= 0.1\n", r1, t1 ? '=' : '!');
  printf("test5 t2: r1 %f %c= 0.1\n", r1, t2 ? '=' : '!');
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<test5>:
  endbr64
  push   %rbp
  sub    $0x10,%rsp
  call   <recip@plt>
  xor    %eax,%eax
  movsd  %xmm0,0x8(%rsp)
  call   <do_nothing@plt>
  movsd  0x8(%rsp),%xmm0
  ucomisd D4(%rip),%xmm0
  jp     L16
  mov    $0x3d,%ebp
  jne    L16
L21:
  mov    %ebp,%esi
  lea    D20(%rip),%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x10,%rsp
  mov    %ebp,%esi
  lea    D18(%rip),%rdi
  pop    %rbp
  jmp    1070 <printf@plt>
  nop
L16:
  mov    $0x21,%ebp
  jmp    L21
```

tool_calls: name: read_data
arguments: data_tag: D4
type: f64
name: read_data
arguments: data_tag: D20
type: string
name: read_data
arguments: data_tag: D18
type: string

tool: D4: 0.1

tool: D20: test5 t1: r1 %f %c= 0.1


tool: D18: test5 t2: r1 %f %c= 0.1


assistant: ```c
void test5(int denom) {
  double r1;
  int t1, t2;
  r1 = recip(denom);
  t1 = r1 == 0.1;
  do_nothing();
  t2 = r1 == 0.1;
  printf("test5 t1: r1 %f %c= 0.1\n", r1, t1 ? '=' : '!');
  printf("test5 t2: r1 %f %c= 0.1\n", r1, t2 ? '=' : '!');
}
```