ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rotate>:
  endbr64
  mov    %rdi,%rax
  movslq %esi,%rsi
  movslq %edx,%rdx
  mov    0x8(%rsp),%rdi
  lea    (%rax,%rsi,8),%rsi
  movaps %xmm0,%xmm3
  movslq %ecx,%rcx
  movslq %r8d,%r8
  movss  (%rsi,%rdx,4),%xmm0
  lea    (%rax,%rcx,8),%rax
  movss  %xmm0,(%rdi)
  movss  (%rax,%r8,4),%xmm4
  movss  %xmm4,(%r9)
  movss  (%rdi),%xmm0
  movaps %xmm0,%xmm2
  mulss  %xmm1,%xmm2
  addss  %xmm4,%xmm2
  mulss  %xmm3,%xmm2
  subss  %xmm2,%xmm0
  movss  %xmm0,(%rsi,%rdx,4)
  movss  (%r9),%xmm2
  movss  (%rdi),%xmm0
  mulss  %xmm2,%xmm1
  subss  %xmm1,%xmm0
  mulss  %xmm3,%xmm0
  addss  %xmm2,%xmm0
  movss  %xmm0,(%rax,%r8,4)
  ret
```

assistant: ```c
void rotate(float a[2][2], int i, int j, int k, int l, float *h, float *g, float s, float tau) {
  (*g) = a[i][j];
  (*h) = a[k][l];
  a[i][j] = (*g) - s * ((*h) + (*g) * tau);
  a[k][l] = (*h) + s * ((*g) - (*h) * tau);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<preadv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <__set_errno@plt>
  mov    $0xffffffff,%eax
  leave
  ret
```

assistant: ```c
ssize_t preadv(int fd, const struct iovec *vector, int count, off_t offset) {
  __set_errno(ENOSYS);
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<preadv>:
  endbr64
  sub    $0x8,%rsp
  mov    D5(%rip),%rax
  mov    (%rax),%edi
  call   <__set_errno@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
ssize_t preadv(int fd, const struct iovec *vector, int count, off_t offset) {
  __set_errno(ENOSYS);
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<preadv>:
  endbr64
  sub    $0x8,%rsp
  mov    D3(%rip),%rax
  mov    (%rax),%edi
  call   <__set_errno@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
ssize_t preadv(int fd, const struct iovec *vector, int count, off_t offset) {
  __set_errno(ENOSYS);
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<preadv>:
  endbr64
  sub    $0x8,%rsp
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <__set_errno@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
ssize_t preadv(int fd, const struct iovec *vector, int count, off_t offset) {
  __set_errno(ENOSYS);
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L2
L10:
  mov    -0x8(%rbp),%eax
  imul   -0x14(%rbp),%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L2:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L10
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
__attribute__((no, cold)) int sum(int i) {
  int j, s;
  s = 0;
  for (j = 0; j < i; j++) {
    s += j * i;
  }
  return s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum>:
  endbr64
  mov    $0x0,%ecx
  mov    $0x0,%eax
  mov    $0x0,%edx
L1:
  cmp    %edi,%edx
  jl     L15
  ret
L15:
  add    %ecx,%eax
  inc    %edx
  add    %edi,%ecx
  jmp    L1
```

assistant: ```c
__attribute__((no, cold)) int sum(int i) {
  int j, s;
  s = 0;
  for (j = 0; j < i; j++) {
    s += j * i;
  }
  return s;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sum>:
  endbr64
  xor    %ecx,%ecx
  xor    %eax,%eax
  xor    %edx,%edx
L7:
  cmp    %edi,%edx
  jge    L10
  add    %ecx,%eax
  inc    %edx
  add    %edi,%ecx
  jmp    L7
L10:
  ret
```

assistant: ```c
__attribute__((no, cold)) int sum(int i) {
  int j, s;
  s = 0;
  for (j = 0; j < i; j++) {
    s += j * i;
  }
  return s;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sum>:
  endbr64
  xor    %ecx,%ecx
  xor    %eax,%eax
  xor    %edx,%edx
L13:
  cmp    %edi,%edx
  jge    L17
  add    %ecx,%eax
  inc    %edx
  add    %edi,%ecx
  jmp    L13
L17:
  ret
```

assistant: ```c
__attribute__((no, cold)) int sum(int i) {
  int j, s;
  s = 0;
  for (j = 0; j < i; j++) {
    s += j * i;
  }
  return s;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<os_pid_get_pid_assoc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <_pqos_cap_get@plt>
  mov    -0x18(%rbp),%rax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <resctrl_alloc_get_grps_num@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L60
  mov    $0x0,%eax
  jmp    L0
L60:
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0x24(%rbp)
  jb     L36
  mov    $0x0,%eax
  jmp    L0
L36:
  mov    $0x0,%eax
  call   <resctrl_lock_shared@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L17
  mov    $0x0,%eax
  jmp    L0
L17:
  mov    -0x30(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <resctrl_alloc_task_read@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <resctrl_lock_release@plt>
  mov    -0x10(%rbp),%rax
L0:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L62
  call   <__stack_chk_fail@plt>
L62:
  leave
  ret
```

assistant: ```c
unsigned *os_pid_get_pid_assoc(const unsigned class_id, unsigned *count) {
  unsigned grps;
  int ret;
  unsigned *tasks;
  const struct pqos_cap *cap;
  ;
  _pqos_cap_get(&cap, NULL);
  ret = resctrl_alloc_get_grps_num(cap, &grps);
  if (ret != 0)
    return NULL;
  if (class_id >= grps)
    return NULL;
  ret = resctrl_lock_shared();
  if (ret != 0)
    return NULL;
  tasks = resctrl_alloc_task_read(class_id, count);
  resctrl_lock_release();
  return tasks;
}
```
user: Translate the below assembly code into C programming language:

```asm
<os_pid_get_pid_assoc>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%ebx
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    $0x0,%esi
  call   <_pqos_cap_get@plt>
  lea    0xc(%rsp),%rsi
  mov    0x10(%rsp),%rdi
  call   <resctrl_alloc_get_grps_num@plt>
  test   %eax,%eax
  jne    L8
  mov    $0x0,%ebp
  cmp    %ebx,0xc(%rsp)
  jbe    L17
  call   <resctrl_lock_shared@plt>
  test   %eax,%eax
  jne    L17
  mov    %r12,%rsi
  mov    %ebx,%edi
  call   <resctrl_alloc_task_read@plt>
  mov    %rax,%rbp
  mov    $0x0,%eax
  call   <resctrl_lock_release@plt>
L17:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  mov    %rbp,%rax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  mov    $0x0,%ebp
  jmp    L17
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned *os_pid_get_pid_assoc(const unsigned class_id, unsigned *count) {
  unsigned grps;
  int ret;
  unsigned *tasks;
  const struct pqos_cap *cap;
  ;
  _pqos_cap_get(&cap, NULL);
  ret = resctrl_alloc_get_grps_num(cap, &grps);
  if (ret != 0)
    return NULL;
  if (class_id >= grps)
    return NULL;
  ret = resctrl_lock_shared();
  if (ret != 0)
    return NULL;
  tasks = resctrl_alloc_task_read(class_id, count);
  resctrl_lock_release();
  return tasks;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<os_pid_get_pid_assoc>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  call   <_pqos_cap_get@plt>
  mov    0x10(%rsp),%rdi
  lea    0xc(%rsp),%rsi
  call   <resctrl_alloc_get_grps_num@plt>
  test   %eax,%eax
  jne    L39
  cmp    %ebp,0xc(%rsp)
  jbe    L39
  xor    %eax,%eax
  call   <resctrl_lock_shared@plt>
  test   %eax,%eax
  jne    L39
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <resctrl_alloc_task_read@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <resctrl_lock_release@plt>
  jmp    L24
  cs nopw 0x0(%rax,%rax,1)
L39:
  xor    %r12d,%r12d
L24:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x28,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned *os_pid_get_pid_assoc(const unsigned class_id, unsigned *count) {
  unsigned grps;
  int ret;
  unsigned *tasks;
  const struct pqos_cap *cap;
  ;
  _pqos_cap_get(&cap, NULL);
  ret = resctrl_alloc_get_grps_num(cap, &grps);
  if (ret != 0)
    return NULL;
  if (class_id >= grps)
    return NULL;
  ret = resctrl_lock_shared();
  if (ret != 0)
    return NULL;
  tasks = resctrl_alloc_task_read(class_id, count);
  resctrl_lock_release();
  return tasks;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<os_pid_get_pid_assoc>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  call   <_pqos_cap_get@plt>
  mov    0x10(%rsp),%rdi
  lea    0xc(%rsp),%rsi
  call   <resctrl_alloc_get_grps_num@plt>
  test   %eax,%eax
  jne    L33
  cmp    %ebp,0xc(%rsp)
  jbe    L33
  xor    %eax,%eax
  call   <resctrl_lock_shared@plt>
  test   %eax,%eax
  jne    L33
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <resctrl_alloc_task_read@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <resctrl_lock_release@plt>
  jmp    L30
  cs nopw 0x0(%rax,%rax,1)
L33:
  xor    %r12d,%r12d
L30:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x28,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned *os_pid_get_pid_assoc(const unsigned class_id, unsigned *count) {
  unsigned grps;
  int ret;
  unsigned *tasks;
  const struct pqos_cap *cap;
  ;
  _pqos_cap_get(&cap, NULL);
  ret = resctrl_alloc_get_grps_num(cap, &grps);
  if (ret != 0)
    return NULL;
  if (class_id >= grps)
    return NULL;
  ret = resctrl_lock_shared();
  if (ret != 0)
    return NULL;
  tasks = resctrl_alloc_task_read(class_id, count);
  resctrl_lock_release();
  return tasks;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<imprimeDireccion>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x0,-0x20(%rbp)
  jne    L59
  movl   $0x0,-0xc(%rbp)
  jmp    L15
L68:
  mov    -0xc(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L79
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L79:
  addl   $0x1,-0xc(%rbp)
L15:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L68
  jmp    L49
L59:
  movl   $0x0,-0xc(%rbp)
  jmp    L70
L28:
  movl   $0x0,-0x8(%rbp)
  jmp    L42
L10:
  mov    -0x4(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L84
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L84:
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L42:
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L10
  addl   $0x1,-0xc(%rbp)
L70:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L28
L49:
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D18
type_guessed: string
name: extract_symbolic_data
arguments: label: D1
type_guessed: string

tool: D18: p[%d] = %p 


tool: D1: p[%d][%d] = %p


assistant: ```c
void imprimeDireccion(int *dir, int largo, int ancho) {
  int i;
  int j;
  int aux = 0;
  if (ancho == 0) {
    for (i = 0; i < largo; i++) {
      if (i % 2 == 1)
        printf("p[%d] = %p \n", i, dir + i);
    }
  } else {
    for (i = 0; i < largo; i++) {
      for (j = 0; j < ancho; j++) {
        if (aux % 2 == 1) {
          printf("p[%d][%d] = %p\n", i, j, dir + aux);
        }
        aux++;
      }
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<imprimeDireccion>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,%r15d
  mov    %edx,%r13d
  test   %edx,%edx
  je     L66
  mov    $0x0,%r12d
  mov    $0x0,%r14d
  test   %esi,%esi
  jg     L92
L90:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L66:
  test   %esi,%esi
  jle    L90
  mov    0x8(%rsp),%rbx
  add    $0x4,%rbx
  lea    D12(%rip),%rbp
  jmp    L128
L103:
  add    $0x4,%rbx
L128:
  add    $0x1,%r13d
  cmp    %r13d,%r15d
  je     L90
  mov    %r13d,%edx
  shr    $0x1f,%edx
  lea    0x0(%r13,%rdx,1),%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L103
  mov    %rbx,%rdx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L103
L14:
  mov    %rbp,%rcx
  mov    %ebx,%edx
  mov    %r14d,%esi
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L59:
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,%r13d
  je     L46
L100:
  mov    %r12d,0x4(%rsp)
  lea    (%rbx,%r12,1),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L59
  jmp    L14
L46:
  mov    0x4(%rsp),%r12d
  add    %ebx,%r12d
L50:
  add    $0x1,%r14d
  cmp    %r14d,%r15d
  je     L90
L92:
  test   %r13d,%r13d
  jle    L50
  movslq %r12d,%rax
  mov    0x8(%rsp),%rcx
  lea    (%rcx,%rax,4),%rbp
  mov    $0x0,%ebx
  jmp    L100
```

tool_calls: name: retrieve_data
arguments: data_label: D12
type_guessed: string
name: retrieve_data
arguments: data_label: D0
type_guessed: string

tool: D12: p[%d] = %p 


tool: D0: p[%d][%d] = %p


assistant: ```c
void imprimeDireccion(int *dir, int largo, int ancho) {
  int i;
  int j;
  int aux = 0;
  if (ancho == 0) {
    for (i = 0; i < largo; i++) {
      if (i % 2 == 1)
        printf("p[%d] = %p \n", i, dir + i);
    }
  } else {
    for (i = 0; i < largo; i++) {
      for (j = 0; j < ancho; j++) {
        if (aux % 2 == 1) {
          printf("p[%d][%d] = %p\n", i, j, dir + aux);
        }
        aux++;
      }
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<imprimeDireccion>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,0x4(%rsp)
  test   %edx,%edx
  je     L13
  xor    %r13d,%r13d
  xor    %r12d,%r12d
  lea    D12(%rip),%r14
  test   %esi,%esi
  jle    L34
  nopl   0x0(%rax,%rax,1)
L47:
  test   %r15d,%r15d
  jle    L50
  mov    0x8(%rsp),%rcx
  movslq %r13d,%rax
  xor    %ebx,%ebx
  lea    (%rcx,%rax,4),%rbp
  jmp    L98
  nopl   (%rax)
L107:
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,%r15d
  je     L73
L98:
  lea    (%rbx,%r13,1),%eax
  test   $0x1,%al
  je     L107
  mov    %rbp,%rcx
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %ebx,%r15d
  jne    L98
L73:
  add    %r15d,%r13d
L50:
  add    $0x1,%r12d
  cmp    %r12d,0x4(%rsp)
  jne    L47
L34:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L13:
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L34
  mov    0x8(%rsp),%r12
  lea    D5(%rip),%rbx
  add    $0x4,%r12
  jmp    L68
  nopw   0x0(%rax,%rax,1)
L60:
  add    $0x4,%r12
L68:
  add    $0x1,%r15d
  cmp    %r15d,0x4(%rsp)
  je     L34
  test   $0x1,%r15b
  je     L60
  mov    %r12,%rdx
  mov    %r15d,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L60
```

tool_calls: name: unpack_data
arguments: label: D12
type_infered: string
name: unpack_data
arguments: label: D5
type_infered: string

tool: D12: p[%d][%d] = %p


tool: D5: p[%d] = %p 


assistant: ```c
void imprimeDireccion(int *dir, int largo, int ancho) {
  int i;
  int j;
  int aux = 0;
  if (ancho == 0) {
    for (i = 0; i < largo; i++) {
      if (i % 2 == 1)
        printf("p[%d] = %p \n", i, dir + i);
    }
  } else {
    for (i = 0; i < largo; i++) {
      for (j = 0; j < ancho; j++) {
        if (aux % 2 == 1) {
          printf("p[%d][%d] = %p\n", i, j, dir + aux);
        }
        aux++;
      }
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<imprimeDireccion>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,0x4(%rsp)
  test   %edx,%edx
  je     L42
  test   %esi,%esi
  jle    L49
  mov    %edx,%r12d
  xor    %r15d,%r15d
  lea    D12(%rip),%r14
  xor    %r13d,%r13d
  test   %edx,%edx
  jle    L49
  nopl   0x0(%rax,%rax,1)
L87:
  mov    0x8(%rsp),%rcx
  movslq %r15d,%rax
  xor    %ebx,%ebx
  lea    (%rcx,%rax,4),%rbp
  jmp    L47
L65:
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,%r12d
  je     L34
L47:
  lea    (%rbx,%r15,1),%eax
  test   $0x1,%al
  je     L65
  mov    %rbp,%rcx
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %ebx,%r12d
  jne    L47
L34:
  add    %r12d,%r15d
  add    $0x1,%r13d
  cmp    %r13d,0x4(%rsp)
  jne    L87
L49:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L42:
  cmpl   $0x1,0x4(%rsp)
  jle    L49
  mov    %rdi,%r12
  mov    $0x1,%ebx
  lea    D0(%rip),%rbp
  add    $0x4,%r12
  jmp    L93
  nopl   0x0(%rax)
L123:
  add    $0x1,%ebx
  add    $0x4,%r12
  cmp    %ebx,0x4(%rsp)
  je     L49
L93:
  test   $0x1,%bl
  je     L123
  mov    %r12,%rdx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L123
```

tool_calls: name: read_data
arguments: label: D12
datatype: string
name: read_data
arguments: label: D0
datatype: string

tool: D12: p[%d][%d] = %p


tool: D0: p[%d] = %p 


assistant: ```c
void imprimeDireccion(int *dir, int largo, int ancho) {
  int i;
  int j;
  int aux = 0;
  if (ancho == 0) {
    for (i = 0; i < largo; i++) {
      if (i % 2 == 1)
        printf("p[%d] = %p \n", i, dir + i);
    }
  } else {
    for (i = 0; i < largo; i++) {
      for (j = 0; j < ancho; j++) {
        if (aux % 2 == 1) {
          printf("p[%d][%d] = %p\n", i, j, dir + aux);
        }
        aux++;
      }
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  D18(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x5f5e100,-0x14(%rbp)
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L16
L13:
  movsd  D64(%rip),%xmm0
  divsd  -0x8(%rbp),%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x8(%rbp),%xmm1
  movsd  D22(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  D64(%rip),%xmm0
  movapd %xmm0,%xmm1
  divsd  -0x8(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x8(%rbp),%xmm1
  movsd  D22(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    -0x18(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  jne    L16
  mov    $0x2e,%edi
  call   <putchar@plt>
L16:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L13
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D18
datatype: f64
name: parse_data
arguments: data_tag: D11
datatype: string
name: parse_data
arguments: data_tag: D60
datatype: string
name: parse_data
arguments: data_tag: D64
datatype: f32[2]
name: parse_data
arguments: data_tag: D22
datatype: f32[2]
name: parse_data
arguments: data_tag: D7
datatype: string

tool: D18: 1

tool: D11: Series of Leibniz to find PI - programming by Lyes Belhadj

tool: D60: 

Search

tool: {"D64": [0.0,2.25]}

tool: {"D22": [0.0,2.0]}

tool: D7: 

Pi = %.64g


assistant: ```c
int main() {
  double Pi = 0.0, denominateur = 1.0;
  int limite = 0;
  const int iteration = 100000000;
  printf("Series of Leibniz to find PI - programming by Lyes Belhadj");
  printf("\n\nSearch");
  while (limite != iteration) {
    Pi += 4 / denominateur;
    denominateur += 2;
    Pi -= 4 / denominateur;
    denominateur += 2;
    limite++;
    if (limite % (iteration / 10) == 0) {
      printf(".");
    }
  }
  printf("\n\nPi = %.64g\n", Pi);
  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
  push   %rbx
  sub    $0x10,%rsp
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D41(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%ebx
  movsd  D4(%rip),%xmm5
  movsd  %xmm5,0x8(%rsp)
  movsd  D53(%rip),%xmm6
  movsd  %xmm6,(%rsp)
  jmp    L15
L21:
  cmp    $0x5f5e100,%ebx
  je     L8
L15:
  movsd  D45(%rip),%xmm0
  movsd  0x8(%rsp),%xmm3
  divsd  %xmm3,%xmm0
  addsd  (%rsp),%xmm0
  movsd  D76(%rip),%xmm1
  addsd  %xmm3,%xmm1
  movsd  D45(%rip),%xmm2
  divsd  %xmm1,%xmm2
  subsd  %xmm2,%xmm0
  movsd  %xmm0,(%rsp)
  addsd  D76(%rip),%xmm1
  movsd  %xmm1,0x8(%rsp)
  add    $0x1,%ebx
  movslq %ebx,%rax
  imul   $0x6b5fca6b,%rax,%rax
  sar    $0x36,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0x989680,%eax,%eax
  cmp    %eax,%ebx
  jne    L21
  mov    $0x2e,%edi
  call   <putchar@plt>
  jmp    L21
L8:
  movsd  (%rsp),%xmm0
  lea    D73(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D18
datatype: string
name: extract_symbolic_data
arguments: data_label: D41
datatype: string
name: extract_symbolic_data
arguments: data_label: D4
datatype: f32[2]
name: extract_symbolic_data
arguments: data_label: D53
datatype: byte[8]
name: extract_symbolic_data
arguments: data_label: D45
datatype: f32[2]
name: extract_symbolic_data
arguments: data_label: D76
datatype: f32[2]
name: extract_symbolic_data
arguments: data_label: D73
datatype: string

tool: D18: Series of Leibniz to find PI - programming by Lyes Belhadj

tool: D41: 

Search

tool: {"D4": [0.0,2.3125]}

tool: {"D53": [0x56,0x55,0x55,0x55,0x55,0x55,0x05,0x40]}

tool: {"D45": [0.0,2.25]}

tool: {"D76": [0.0,2.0]}

tool: D73: 

Pi = %.64g


assistant: ```c
int main() {
  double Pi = 0.0, denominateur = 1.0;
  int limite = 0;
  const int iteration = 100000000;
  printf("Series of Leibniz to find PI - programming by Lyes Belhadj");
  printf("\n\nSearch");
  while (limite != iteration) {
    Pi += 4 / denominateur;
    denominateur += 2;
    Pi -= 4 / denominateur;
    denominateur += 2;
    limite++;
    if (limite % (iteration / 10) == 0) {
      printf(".");
    }
  }
  printf("\n\nPi = %.64g\n", Pi);
  return 0;
}
```