ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<parseFileName>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,%rbx
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,%rdx
  mov    %rbx,%rax
  sub    %rdx,%rax
  mov    %rax,%rbx
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  lea    0x1(%rax),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rbx,%rdx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *parseFileName(const char *path, char *parsed) {
  memcpy(parsed, strrchr(path, '/') + 1, (size_t)(strrchr(path, 0) - strrchr(path, '/')));
  return parsed;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<parseFileName>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    $0x0,%esi
  call   <strrchr@plt>
  mov    %rax,%rbx
  mov    $0x2f,%esi
  mov    %rbp,%rdi
  call   <strrchr@plt>
  mov    %rax,%r13
  mov    $0x2f,%esi
  mov    %rbp,%rdi
  call   <strrchr@plt>
  sub    %r13,%rbx
  mov    %rbx,%rdx
  lea    0x1(%rax),%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *parseFileName(const char *path, char *parsed) {
  memcpy(parsed, strrchr(path, '/') + 1, (size_t)(strrchr(path, 0) - strrchr(path, '/')));
  return parsed;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<parseFileName>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strrchr@plt>
  mov    %rbp,%rdi
  mov    $0x2f,%esi
  mov    %rax,%rbx
  call   <strrchr@plt>
  mov    %rbp,%rdi
  mov    $0x2f,%esi
  mov    %rax,%r13
  call   <strrchr@plt>
  sub    %r13,%rbx
  mov    %r12,%rdi
  mov    %rbx,%rdx
  lea    0x1(%rax),%rsi
  call   <memcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *parseFileName(const char *path, char *parsed) {
  memcpy(parsed, strrchr(path, '/') + 1, (size_t)(strrchr(path, 0) - strrchr(path, '/')));
  return parsed;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<parseFileName>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strrchr@plt>
  mov    %rbp,%rdi
  mov    $0x2f,%esi
  mov    %rax,%rbx
  call   <strrchr@plt>
  mov    %rbp,%rdi
  mov    $0x2f,%esi
  mov    %rax,%r13
  call   <strrchr@plt>
  sub    %r13,%rbx
  mov    %r12,%rdi
  mov    %rbx,%rdx
  lea    0x1(%rax),%rsi
  call   <memcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *parseFileName(const char *path, char *parsed) {
  memcpy(parsed, strrchr(path, '/') + 1, (size_t)(strrchr(path, 0) - strrchr(path, '/')));
  return parsed;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xbb8,%edi
  call   <l_usleep@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D27
type_infered: string
name: retrieve_data
arguments: data_tag: D3
type_infered: string
name: retrieve_data
arguments: data_tag: D6
type_infered: string

tool: D27: Start Main:

tool: D3: timer remaining %d usec.


tool: D6: End Main:

assistant: ```c
int main(int argc, char *argv[]) {
  int timer = -1;
  printf("Start Main:\n");
  timer = l_usleep(1000 * 3);
  printf("timer remaining %d usec.\n", timer);
  printf("End Main:\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <puts@plt>
  mov    $0xbb8,%edi
  call   <l_usleep@plt>
  mov    %eax,%esi
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D4
data_type: string
name: unpack_data
arguments: tag: D8
data_type: string
name: unpack_data
arguments: tag: D11
data_type: string

tool: D4: Start Main:

tool: D8: timer remaining %d usec.


tool: D11: End Main:

assistant: ```c
int main(int argc, char *argv[]) {
  int timer = -1;
  printf("Start Main:\n");
  timer = l_usleep(1000 * 3);
  printf("timer remaining %d usec.\n", timer);
  printf("End Main:\n");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <puts@plt>
  mov    $0xbb8,%edi
  call   <l_usleep@plt>
  lea    D28(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D22(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type_guessed: string
name: unpack_data
arguments: data_tag: D28
type_guessed: string
name: unpack_data
arguments: data_tag: D22
type_guessed: string

tool: D3: Start Main:

tool: D28: timer remaining %d usec.


tool: D22: End Main:

assistant: ```c
int main(int argc, char *argv[]) {
  int timer = -1;
  printf("Start Main:\n");
  timer = l_usleep(1000 * 3);
  printf("timer remaining %d usec.\n", timer);
  printf("End Main:\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    $0xbb8,%edi
  call   <l_usleep@plt>
  lea    D10(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D3(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D19
data_type: string
name: extract_symbolic_data
arguments: label: D10
data_type: string
name: extract_symbolic_data
arguments: label: D3
data_type: string

tool: D19: Start Main:

tool: D10: timer remaining %d usec.


tool: D3: End Main:

assistant: ```c
int main(int argc, char *argv[]) {
  int timer = -1;
  printf("Start Main:\n");
  timer = l_usleep(1000 * 3);
  printf("timer remaining %d usec.\n", timer);
  printf("End Main:\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<m3f_mul_av3fo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L2
L17:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rax,%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <m3f_mul_v3fo@plt>
  addq   $0x3,-0x8(%rbp)
L2:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L17
  nop
  nop
  leave
  ret
```

assistant: ```c
void m3f_mul_av3fo(float *m, const float *input, size_t count, float *result) {
  size_t i;
  for (i = 0; i < count; i += 3) {
    m3f_mul_v3fo(m, input + i, result + i);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<m3f_mul_av3fo>:
  endbr64
  test   %rdx,%rdx
  je     L0
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rdx,%r13
  mov    %rsi,%rbp
  mov    %rcx,%rbx
  mov    $0x0,%r14d
L1:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <m3f_mul_v3fo@plt>
  add    $0x3,%r14
  add    $0xc,%rbp
  add    $0xc,%rbx
  cmp    %r14,%r13
  ja     L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  ret
```

assistant: ```c
void m3f_mul_av3fo(float *m, const float *input, size_t count, float *result) {
  size_t i;
  for (i = 0; i < count; i += 3) {
    m3f_mul_v3fo(m, input + i, result + i);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<m3f_mul_av3fo>:
  endbr64
  test   %rdx,%rdx
  je     L9
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rcx,%rbx
L6:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x3,%r14
  call   <m3f_mul_v3fo@plt>
  add    $0xc,%rbp
  add    $0xc,%rbx
  cmp    %r14,%r13
  ja     L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  ret
```

assistant: ```c
void m3f_mul_av3fo(float *m, const float *input, size_t count, float *result) {
  size_t i;
  for (i = 0; i < count; i += 3) {
    m3f_mul_v3fo(m, input + i, result + i);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<m3f_mul_av3fo>:
  endbr64
  test   %rdx,%rdx
  je     L4
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rcx,%rbx
L18:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x3,%r14
  call   <m3f_mul_v3fo@plt>
  add    $0xc,%rbp
  add    $0xc,%rbx
  cmp    %r14,%r13
  ja     L18
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  ret
```

assistant: ```c
void m3f_mul_av3fo(float *m, const float *input, size_t count, float *result) {
  size_t i;
  for (i = 0; i < count; i += 3) {
    m3f_mul_v3fo(m, input + i, result + i);
  }
}
```
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   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x40(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D40
type: string
name: extract_symbolic_data
arguments: label: D20
type: string
name: extract_symbolic_data
arguments: label: D19
type: string
name: extract_symbolic_data
arguments: label: D30
type: string

tool: D40: What's your first name?

tool: D20: %s

tool: D19: and your last name?

tool: D30: %s,%s


assistant: ```c
int main() {
  char first[20];
  char last[20];
  printf("What's your first name?\n");
  scanf("%s", first);
  printf("and your last name?\n");
  scanf("%s", last);
  printf("%s,%s\n", last, first);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    D34(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  lea    D32(%rip),%r12
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D6(%rip),%rdi
  call   <puts@plt>
  lea    0x20(%rsp),%rbx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  lea    D25(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D34
type: string
name: extract_symbolic_data
arguments: label: D32
type: string
name: extract_symbolic_data
arguments: label: D6
type: string
name: extract_symbolic_data
arguments: label: D25
type: string

tool: D34: What's your first name?

tool: D32: %s

tool: D6: and your last name?

tool: D25: %s,%s


assistant: ```c
int main() {
  char first[20];
  char last[20];
  printf("What's your first name?\n");
  scanf("%s", first);
  printf("and your last name?\n");
  scanf("%s", last);
  printf("%s,%s\n", last, first);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  lea    D12(%rip),%rdi
  lea    D32(%rip),%r13
  push   %r12
  push   %rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x20(%rsp),%rbp
  call   <puts@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    D36(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D12
datatype: string
name: read_data
arguments: data_tag: D32
datatype: string
name: read_data
arguments: data_tag: D36
datatype: string
name: read_data
arguments: data_tag: D15
datatype: string

tool: D12: What's your first name?

tool: D32: %s

tool: D36: and your last name?

tool: D15: %s,%s


assistant: ```c
int main() {
  char first[20];
  char last[20];
  printf("What's your first name?\n");
  scanf("%s", first);
  printf("and your last name?\n");
  scanf("%s", last);
  printf("%s,%s\n", last, first);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D15(%rip),%rdi
  lea    D36(%rip),%r13
  push   %r12
  push   %rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x20(%rsp),%rbp
  call   <puts@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  lea    D8(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D15
type_infered: string
name: retrieve_data
arguments: tag: D36
type_infered: string
name: retrieve_data
arguments: tag: D8
type_infered: string
name: retrieve_data
arguments: tag: D28
type_infered: string

tool: D15: What's your first name?

tool: D36: %s

tool: D8: and your last name?

tool: D28: %s,%s


assistant: ```c
int main() {
  char first[20];
  char last[20];
  printf("What's your first name?\n");
  scanf("%s", first);
  printf("and your last name?\n");
  scanf("%s", last);
  printf("%s,%s\n", last, first);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<GenPrintInstr1Operand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %ecx,-0x10(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <GenPrintInstr@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <GenPrintOperand@plt>
  mov    $0x0,%eax
  call   <GenPrintNewLine@plt>
  nop
  leave
  ret
```

assistant: ```c
void GenPrintInstr1Operand(int instr, int instrval, int operand, int operandval) {
  GenPrintInstr(instr, instrval);
  GenPrintOperand(operand, operandval);
  GenPrintNewLine();
}
```
user: What is the c source code of the assembly code below?

```asm
<GenPrintInstr1Operand>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edx,%ebx
  mov    %ecx,%ebp
  call   <GenPrintInstr@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <GenPrintOperand@plt>
  mov    $0x0,%eax
  call   <GenPrintNewLine@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void GenPrintInstr1Operand(int instr, int instrval, int operand, int operandval) {
  GenPrintInstr(instr, instrval);
  GenPrintOperand(operand, operandval);
  GenPrintNewLine();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<GenPrintInstr1Operand>:
  endbr64
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %edx,%ebp
  sub    $0x8,%rsp
  call   <GenPrintInstr@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <GenPrintOperand@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    1080 <GenPrintNewLine@plt>
```

assistant: ```c
void GenPrintInstr1Operand(int instr, int instrval, int operand, int operandval) {
  GenPrintInstr(instr, instrval);
  GenPrintOperand(operand, operandval);
  GenPrintNewLine();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<GenPrintInstr1Operand>:
  endbr64
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %edx,%ebp
  sub    $0x8,%rsp
  call   <GenPrintInstr@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <GenPrintOperand@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    1080 <GenPrintNewLine@plt>
```

assistant: ```c
void GenPrintInstr1Operand(int instr, int instrval, int operand, int operandval) {
  GenPrintInstr(instr, instrval);
  GenPrintOperand(operand, operandval);
  GenPrintNewLine();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<iv2string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x9,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%esi
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x8(%rbp),%rax
  mov    %esi,%r8d
  lea    D9(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
type_guessed: string

tool: D9: %02X %02X %02X

assistant: ```c
char *iv2string(unsigned char *iv) {
  char *string = (char *)malloc(9);
  sprintf(string, "%02X %02X %02X", iv[0], iv[1], iv[2]);
  return string;
}
```
user: What is the c source code of the assembly code below:

```asm
<iv2string>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    $0x9,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  movzbl 0x1(%rbx),%ecx
  movzbl (%rbx),%edx
  movzbl 0x2(%rbx),%r8d
  lea    D4(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D4
datatype: string

tool: D4: %02X %02X %02X

assistant: ```c
char *iv2string(unsigned char *iv) {
  char *string = (char *)malloc(9);
  sprintf(string, "%02X %02X %02X", iv[0], iv[1], iv[2]);
  return string;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<iv2string>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x9,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  movzbl 0x1(%rbx),%ecx
  movzbl (%rbx),%edx
  lea    D4(%rip),%rsi
  movzbl 0x2(%rbx),%r8d
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: label: D4
type_infered: string

tool: D4: %02X %02X %02X

assistant: ```c
char *iv2string(unsigned char *iv) {
  char *string = (char *)malloc(9);
  sprintf(string, "%02X %02X %02X", iv[0], iv[1], iv[2]);
  return string;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<iv2string>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x9,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  movzbl 0x1(%rbx),%ecx
  movzbl (%rbx),%edx
  lea    D5(%rip),%rsi
  movzbl 0x2(%rbx),%r8d
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

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

tool: D5: %02X %02X %02X

assistant: ```c
char *iv2string(unsigned char *iv) {
  char *string = (char *)malloc(9);
  sprintf(string, "%02X %02X %02X", iv[0], iv[1], iv[2]);
  return string;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<my_sin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  pxor   %xmm1,%xmm1
  cvtss2sd -0x14(%rbp),%xmm1
  movsd  D30(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  pxor   %xmm3,%xmm3
  cvtsd2ss %xmm0,%xmm3
  movd   %xmm3,%eax
  movd   %eax,%xmm0
  call   <my_rint@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x8(%rbp)
  movss  -0x8(%rbp),%xmm0
  cvttss2si %xmm0,%eax
  mov    %eax,-0x4(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0x14(%rbp),%xmm0
  pxor   %xmm2,%xmm2
  cvtss2sd -0x8(%rbp),%xmm2
  movsd  D16(%rip),%xmm1
  mulsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0xc(%rbp),%xmm0
  pxor   %xmm2,%xmm2
  cvtss2sd -0x8(%rbp),%xmm2
  movsd  D37(%rip),%xmm1
  mulsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L13
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  call   <cos_core@plt>
  movd   %xmm0,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L16
L13:
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  call   <sin_core@plt>
  movd   %xmm0,%eax
  mov    %eax,-0xc(%rbp)
L16:
  mov    -0x4(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L31
  movss  -0xc(%rbp),%xmm0
  movss  D11(%rip),%xmm1
  xorps  %xmm1,%xmm0
  jmp    L35
L31:
  movss  -0xc(%rbp),%xmm0
L35:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D30
data_type: byte[8]
name: extract_symbolic_data
arguments: tag: D16
data_type: byte[8]
name: extract_symbolic_data
arguments: tag: D37
data_type: byte[8]
name: extract_symbolic_data
arguments: tag: D11
data_type: byte[4]

tool: {"D30": [0x83,0xc8,0xc9,0x6d,0x30,0x5f,0xe4,0x3f]}

tool: {"D16": [0x00,0x00,0x44,0x54,0xfb,0x21,0xf9,0x3f]}

tool: {"D37": [0x29,0x66,0x4c,0x4c,0x23,0x8c,0x86,0x3d]}

tool: {"D11": [0x00,0x00,0x00,0x80]}

assistant: ```c
float my_sin(float x) {
  float q, t;
  int quadrant;
  q = my_rint(x * 6.3661977236758138e-1);
  quadrant = (int)q;
  t = x - q * 1.5707963267923333e+00;
  t = t - q * 2.5633441515945189e-12;
  if (quadrant & 1) {
    t = cos_core(t);
  } else {
    t = sin_core(t);
  }
  return (quadrant & 2) ? -t : t;
}
```