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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x20(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L7
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
L7:
  mov    -0x20(%rbp),%eax
  mov    %eax,%edi
  call   <aloca@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,%edi
  call   <aloca@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,%edi
  call   <aloca@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <preenche@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <preenche@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x1c(%rbp),%edi
  mov    -0x20(%rbp),%ecx
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <multiplica@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
data_type: string

tool: D7: TAM MATRIZ < NUM PROCESSOS nao e valido


assistant: ```c
main(int argc, char *argv[]) {
  int TAMMATRIZ = atoi(argv[1]);
  int NPROCESSOS = atoi(argv[2]);
  if (TAMMATRIZ < NPROCESSOS) {
    printf("TAM MATRIZ < NUM PROCESSOS nao e valido\n");
    exit(0);
  }
  int **m1, **m2, **m3;
  m1 = aloca(TAMMATRIZ);
  m2 = aloca(TAMMATRIZ);
  m3 = aloca(TAMMATRIZ);
  m1 = preenche(m1, TAMMATRIZ);
  m2 = preenche(m2, TAMMATRIZ);
  multiplica(m1, m2, m3, TAMMATRIZ, NPROCESSOS);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbp
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%ebx
  mov    0x10(%rbp),%rdi
  call   <atoi@plt>
  mov    %eax,%ebp
  cmp    %eax,%ebx
  jl     L8
L17:
  mov    %ebx,%edi
  call   <aloca@plt>
  mov    %rax,%r12
  mov    %ebx,%edi
  call   <aloca@plt>
  mov    %rax,%r14
  mov    %ebx,%edi
  call   <aloca@plt>
  mov    %rax,%r13
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <preenche@plt>
  mov    %rax,%r12
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <preenche@plt>
  mov    %rax,%rsi
  mov    %ebp,%r8d
  mov    %ebx,%ecx
  mov    %r13,%rdx
  mov    %r12,%rdi
  call   <multiplica@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L8:
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L17
```

tool_calls: name: parse_data
arguments: data_tag: D2
data_type: string

tool: D2: TAM MATRIZ < NUM PROCESSOS nao e valido


assistant: ```c
main(int argc, char *argv[]) {
  int TAMMATRIZ = atoi(argv[1]);
  int NPROCESSOS = atoi(argv[2]);
  if (TAMMATRIZ < NPROCESSOS) {
    printf("TAM MATRIZ < NUM PROCESSOS nao e valido\n");
    exit(0);
  }
  int **m1, **m2, **m3;
  m1 = aloca(TAMMATRIZ);
  m2 = aloca(TAMMATRIZ);
  m3 = aloca(TAMMATRIZ);
  m1 = preenche(m1, TAMMATRIZ);
  m2 = preenche(m2, TAMMATRIZ);
  multiplica(m1, m2, m3, TAMMATRIZ, NPROCESSOS);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  call   <atoi@plt>
  mov    0x10(%rbx),%rdi
  mov    %eax,%r12d
  call   <atoi@plt>
  mov    %eax,%ebx
  cmp    %eax,%r12d
  jl     L8
L12:
  mov    %r12d,%edi
  call   <aloca@plt>
  mov    %r12d,%edi
  mov    %rax,%rbp
  call   <aloca@plt>
  mov    %r12d,%edi
  mov    %rax,%r14
  call   <aloca@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  mov    %rax,%r13
  call   <preenche@plt>
  mov    %r12d,%esi
  mov    %r14,%rdi
  mov    %rax,%rbp
  call   <preenche@plt>
  mov    %ebx,%r8d
  mov    %r12d,%ecx
  mov    %r13,%rdx
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <multiplica@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L8:
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L12
```

tool_calls: name: unpack_data
arguments: data_tag: D2
datatype: string

tool: D2: TAM MATRIZ < NUM PROCESSOS nao e valido


assistant: ```c
main(int argc, char *argv[]) {
  int TAMMATRIZ = atoi(argv[1]);
  int NPROCESSOS = atoi(argv[2]);
  if (TAMMATRIZ < NPROCESSOS) {
    printf("TAM MATRIZ < NUM PROCESSOS nao e valido\n");
    exit(0);
  }
  int **m1, **m2, **m3;
  m1 = aloca(TAMMATRIZ);
  m2 = aloca(TAMMATRIZ);
  m3 = aloca(TAMMATRIZ);
  m1 = preenche(m1, TAMMATRIZ);
  m2 = preenche(m2, TAMMATRIZ);
  multiplica(m1, m2, m3, TAMMATRIZ, NPROCESSOS);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  call   <atoi@plt>
  mov    0x10(%rbx),%rdi
  mov    %eax,%r12d
  call   <atoi@plt>
  mov    %eax,%ebx
  cmp    %eax,%r12d
  jl     L9
L12:
  mov    %r12d,%edi
  call   <aloca@plt>
  mov    %r12d,%edi
  mov    %rax,%rbp
  call   <aloca@plt>
  mov    %r12d,%edi
  mov    %rax,%r14
  call   <aloca@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  mov    %rax,%r13
  call   <preenche@plt>
  mov    %r12d,%esi
  mov    %r14,%rdi
  mov    %rax,%rbp
  call   <preenche@plt>
  mov    %ebx,%r8d
  mov    %r12d,%ecx
  mov    %r13,%rdx
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <multiplica@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L12
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type_infered: string

tool: D9: TAM MATRIZ < NUM PROCESSOS nao e valido


assistant: ```c
main(int argc, char *argv[]) {
  int TAMMATRIZ = atoi(argv[1]);
  int NPROCESSOS = atoi(argv[2]);
  if (TAMMATRIZ < NPROCESSOS) {
    printf("TAM MATRIZ < NUM PROCESSOS nao e valido\n");
    exit(0);
  }
  int **m1, **m2, **m3;
  m1 = aloca(TAMMATRIZ);
  m2 = aloca(TAMMATRIZ);
  m3 = aloca(TAMMATRIZ);
  m1 = preenche(m1, TAMMATRIZ);
  m2 = preenche(m2, TAMMATRIZ);
  multiplica(m1, m2, m3, TAMMATRIZ, NPROCESSOS);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<uid_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <getpwuid@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L1
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  jmp    L2
L1:
  mov    $0x0,%eax
L2:
  leave
  ret
```

assistant: ```c
char *uid_name(uid_t uid) {
  struct passwd *pw = getpwuid(uid);
  return pw ? pw->pw_name : NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<uid_name>:
  endbr64
  sub    $0x8,%rsp
  call   <getpwuid@plt>
  test   %rax,%rax
  je     L0
  mov    (%rax),%rax
L0:
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *uid_name(uid_t uid) {
  struct passwd *pw = getpwuid(uid);
  return pw ? pw->pw_name : NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<uid_name>:
  endbr64
  sub    $0x8,%rsp
  call   <getpwuid@plt>
  test   %rax,%rax
  je     L8
  mov    (%rax),%rax
L8:
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *uid_name(uid_t uid) {
  struct passwd *pw = getpwuid(uid);
  return pw ? pw->pw_name : NULL;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<uid_name>:
  endbr64
  sub    $0x8,%rsp
  call   <getpwuid@plt>
  test   %rax,%rax
  je     L8
  mov    (%rax),%rax
L8:
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *uid_name(uid_t uid) {
  struct passwd *pw = getpwuid(uid);
  return pw ? pw->pw_name : NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<starts_with>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
L16:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L4
  mov    $0x1,%eax
  jmp    L18
L4:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L48
  mov    $0x0,%eax
  jmp    L18
L48:
  addl   $0x1,-0x4(%rbp)
  jmp    L16
L18:
  pop    %rbp
  ret
```

assistant: ```c
bool starts_with(char *s, char *substr) {
  for (int i = 0;; i++) {
    if (substr[i] == '\0')
      return true;
    else if (s[i] != substr[i])
      return false;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<starts_with>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L16
  mov    $0x0,%eax
L22:
  cmp    %dl,(%rdi,%rax,1)
  jne    L21
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L22
  mov    $0x1,%eax
  ret
L16:
  mov    $0x1,%eax
  ret
L21:
  mov    $0x0,%eax
  ret
```

assistant: ```c
bool starts_with(char *s, char *substr) {
  for (int i = 0;; i++) {
    if (substr[i] == '\0')
      return true;
    else if (s[i] != substr[i])
      return false;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<starts_with>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L10
  xor    %eax,%eax
  jmp    L4
  nop
L24:
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  je     L10
L4:
  cmp    %dl,(%rdi,%rax,1)
  je     L24
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L10:
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool starts_with(char *s, char *substr) {
  for (int i = 0;; i++) {
    if (substr[i] == '\0')
      return true;
    else if (s[i] != substr[i])
      return false;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<starts_with>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L24
  xor    %eax,%eax
  jmp    L29
  nop
L22:
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  je     L24
L29:
  cmp    %dl,(%rdi,%rax,1)
  je     L22
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L24:
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool starts_with(char *s, char *substr) {
  for (int i = 0;; i++) {
    if (substr[i] == '\0')
      return true;
    else if (s[i] != substr[i])
      return false;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<rightrot>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x4(%rbp)
  mov    %al,-0x8(%rbp)
  movzbl -0x4(%rbp),%edx
  movzbl -0x8(%rbp),%ecx
  mov    $0x8,%eax
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,%esi
  movzbl -0x4(%rbp),%edx
  movzbl -0x8(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  or     %esi,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char rightrot(unsigned char x, unsigned char n) {
  enum size { BYTE = (sizeof(unsigned char) * 8) };
  return ((x << (BYTE - n)) | (x >> n));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rightrot>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%ecx
  ror    %cl,%al
  ret
```

assistant: ```c
unsigned char rightrot(unsigned char x, unsigned char n) {
  enum size { BYTE = (sizeof(unsigned char) * 8) };
  return ((x << (BYTE - n)) | (x >> n));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<rightrot>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%ecx
  ror    %cl,%al
  ret
```

assistant: ```c
unsigned char rightrot(unsigned char x, unsigned char n) {
  enum size { BYTE = (sizeof(unsigned char) * 8) };
  return ((x << (BYTE - n)) | (x >> n));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rightrot>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%ecx
  ror    %cl,%al
  ret
```

assistant: ```c
unsigned char rightrot(unsigned char x, unsigned char n) {
  enum size { BYTE = (sizeof(unsigned char) * 8) };
  return ((x << (BYTE - n)) | (x >> n));
}
```
user: Translate the below assembly code into C programming language:

```asm
<ill_handler>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type: string

tool: D0: Illegal instruction hit.  Try again


assistant: ```c
void ill_handler(int sig) {
  printf("Illegal instruction hit.  Try again\n");
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ill_handler>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
type_infered: string

tool: D8: Illegal instruction hit.  Try again


assistant: ```c
void ill_handler(int sig) {
  printf("Illegal instruction hit.  Try again\n");
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ill_handler>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: read_data
arguments: label: D6
type: string

tool: D6: Illegal instruction hit.  Try again


assistant: ```c
void ill_handler(int sig) {
  printf("Illegal instruction hit.  Try again\n");
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ill_handler>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type_infered: string

tool: D2: Illegal instruction hit.  Try again


assistant: ```c
void ill_handler(int sig) {
  printf("Illegal instruction hit.  Try again\n");
  exit(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L15
L19:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0x989680,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L0
L27:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D21(%rip),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L0:
  cmpl   $0x2,-0x14(%rbp)
  jle    L27
  addl   $0x1,-0x18(%rbp)
L15:
  cmpl   $0x6,-0x18(%rbp)
  jle    L19
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D21
type_infered: string
name: parse_data
arguments: tag: D0
type_infered: string
name: parse_data
arguments: tag: D12
type_infered: string

tool: D21: dsqrt

tool: D0: # c::%s:len=%d


tool: D12: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dsqrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D18(%rip),%r15
  jmp    L39
L6:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L2
L39:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0x989680,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L23:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L23
  jmp    L6
L2:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: label: D18
type_infered: string
name: read_data
arguments: label: D9
type_infered: string
name: read_data
arguments: label: D14
type_infered: string

tool: D18: dsqrt

tool: D9: # c::%s:len=%d


tool: D14: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dsqrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D2(%rip),%r15
  push   %r14
  lea    D11(%rip),%r14
  push   %r13
  lea    D25(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L14:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0x989680,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L11:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L11
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L14
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
type_infered: string
name: unpack_data
arguments: data_label: D11
type_infered: string
name: unpack_data
arguments: data_label: D25
type_infered: string

tool: D2: dsqrt

tool: D11: # c::%s:len=%d


tool: D25: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dsqrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
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   %r15
  xor    %edi,%edi
  lea    D14(%rip),%r15
  push   %r14
  lea    D27(%rip),%r14
  push   %r13
  lea    D24(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L19:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0x989680,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L15:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L15
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L19
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D14
datatype: string
name: extract_symbolic_data
arguments: label: D27
datatype: string
name: extract_symbolic_data
arguments: label: D24
datatype: string

tool: D14: dsqrt

tool: D27: # c::%s:len=%d


tool: D24: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dsqrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_a_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x1(%rbp)
  cmpb   $0xa,-0x1(%rbp)
  je     L18
  nop
L6:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L6
L18:
  movsbl -0x1(%rbp),%eax
  leave
  ret
```

assistant: ```c
get_a_char() {
  char ch;
  ch = getchar();
  if (ch != '\n')
    while (getchar() != '\n')
      ;
  return (ch);
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_a_char>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    $0xa,%al
  je     L7
L16:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L16
L7:
  movsbl %bl,%eax
  pop    %rbx
  ret
```

assistant: ```c
get_a_char() {
  char ch;
  ch = getchar();
  if (ch != '\n')
    while (getchar() != '\n')
      ;
  return (ch);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_a_char>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    $0xa,%al
  je     L16
  nopw   0x0(%rax,%rax,1)
L10:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L10
L16:
  movsbl %bl,%eax
  pop    %rbx
  ret
```

assistant: ```c
get_a_char() {
  char ch;
  ch = getchar();
  if (ch != '\n')
    while (getchar() != '\n')
      ;
  return (ch);
}
```