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
<sPrintData>:
  endbr64
  cmp    $0x20,%esi
  jg     L2
  test   %esi,%esi
  jle    L34
  cmp    $0x20,%esi
  ja     L34
  lea    D40(%rip),%rcx
  mov    %esi,%esi
  movslq (%rcx,%rsi,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  nopl   (%rax)
L2:
  cmp    $0x40,%esi
  je     L3
  cmp    $0x80,%esi
  jne    L7
  lea    D5(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L34:
  ret
  nopl   0x0(%rax)
  mov    (%rdx),%edx
  lea    D9(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
  movswl (%rdx),%edx
  lea    D68(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
  pxor   %xmm0,%xmm0
  lea    D10(%rip),%rsi
  mov    $0x1,%eax
  cvtss2sd (%rdx),%xmm0
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
  movsd  (%rdx),%xmm0
  lea    D10(%rip),%rsi
  mov    $0x1,%eax
  jmp    1050 <printf@plt>
  nopl   (%rax)
  mov    (%rdx),%rdx
  lea    D16(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
  mov    (%rdx),%rdx
  lea    D63(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L3:
  movsbl (%rdx),%edx
  lea    D23(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
L7:
  ret
```

tool_calls: name: parse_data
arguments: tag: D5
datatype: string
name: parse_data
arguments: tag: D9
datatype: string
name: parse_data
arguments: tag: D68
datatype: string
name: parse_data
arguments: tag: D10
datatype: string
name: parse_data
arguments: tag: D16
datatype: string
name: parse_data
arguments: tag: D63
datatype: string
name: parse_data
arguments: tag: D23
datatype: string

tool: D5: %s

tool: D9: %d

tool: D68: %hd

tool: D10: %f

tool: D16: %ld

tool: D63: %lld

tool: D23: %c

assistant: ```c
void sPrintData(char *str, int type, void *data) {
  switch (type) {
  case 0x00000001:
    printf(str, "%d", *(int *)data);
    return;
  case 0x00000002:
    printf(str, "%hd", *(short *)data);
    return;
  case 0x00000004:
    printf(str, "%f", *(float *)data);
    return;
  case 0x00000008:
    printf(str, "%f", *(double *)data);
    return;
  case 0x00000010:
    printf(str, "%ld", *(long *)data);
    return;
  case 0x00000020:
    printf(str, "%lld", *(long long *)data);
    return;
  case 0x00000040:
    printf(str, "%c", *(char *)data);
    return;
  case 0x00000080:
    printf(str, "%s", (char *)data);
    return;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sPrintData>:
  endbr64
  cmp    $0x20,%esi
  jg     L45
  test   %esi,%esi
  jle    L0
  cmp    $0x20,%esi
  ja     L0
  lea    D12(%rip),%rcx
  mov    %esi,%esi
  movslq (%rcx,%rsi,4),%rax
  add    %rcx,%rax
  notrack jmp *%rax
  nopl   (%rax)
L45:
  cmp    $0x40,%esi
  je     L25
  cmp    $0x80,%esi
  jne    L3
  lea    D7(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L0:
  ret
  nopl   0x0(%rax)
  mov    (%rdx),%edx
  lea    D17(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
  movswl (%rdx),%edx
  lea    D30(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
  pxor   %xmm0,%xmm0
  lea    D62(%rip),%rsi
  mov    $0x1,%eax
  cvtss2sd (%rdx),%xmm0
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
  movsd  (%rdx),%xmm0
  lea    D62(%rip),%rsi
  mov    $0x1,%eax
  jmp    1050 <printf@plt>
  nopl   (%rax)
  mov    (%rdx),%rdx
  lea    D27(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
  mov    (%rdx),%rdx
  lea    D25(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L25:
  movsbl (%rdx),%edx
  lea    D47(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
L3:
  ret
```

tool_calls: name: read_data
arguments: data_label: D7
type_guessed: string
name: read_data
arguments: data_label: D17
type_guessed: string
name: read_data
arguments: data_label: D30
type_guessed: string
name: read_data
arguments: data_label: D62
type_guessed: string
name: read_data
arguments: data_label: D27
type_guessed: string
name: read_data
arguments: data_label: D25
type_guessed: string
name: read_data
arguments: data_label: D47
type_guessed: string

tool: D7: %s

tool: D17: %d

tool: D30: %hd

tool: D62: %f

tool: D27: %ld

tool: D25: %lld

tool: D47: %c

assistant: ```c
void sPrintData(char *str, int type, void *data) {
  switch (type) {
  case 0x00000001:
    printf(str, "%d", *(int *)data);
    return;
  case 0x00000002:
    printf(str, "%hd", *(short *)data);
    return;
  case 0x00000004:
    printf(str, "%f", *(float *)data);
    return;
  case 0x00000008:
    printf(str, "%f", *(double *)data);
    return;
  case 0x00000010:
    printf(str, "%ld", *(long *)data);
    return;
  case 0x00000020:
    printf(str, "%lld", *(long long *)data);
    return;
  case 0x00000040:
    printf(str, "%c", *(char *)data);
    return;
  case 0x00000080:
    printf(str, "%s", (char *)data);
    return;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ancil_send_fds_with_buffer_to>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x90,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %rsi,-0x80(%rbp)
  mov    %edx,-0x78(%rbp)
  mov    %rcx,-0x88(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9d,-0x94(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movb   $0x21,-0x6d(%rbp)
  lea    -0x6d(%rbp),%rax
  mov    %rax,-0x60(%rbp)
  movl   $0x1,-0x58(%rbp)
  mov    -0x90(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x94(%rbp),%eax
  mov    %eax,-0x30(%rbp)
  lea    -0x60(%rbp),%rax
  mov    %rax,-0x38(%rbp)
  movl   $0x1,-0x50(%rbp)
  movq   $0x0,-0x40(%rbp)
  mov    -0x88(%rbp),%rax
  mov    %rax,-0x48(%rbp)
  mov    -0x78(%rbp),%eax
  add    $0x3,%rax
  shl    $0x2,%eax
  mov    %eax,-0x4c(%rbp)
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <CMSG_FIRSTHDR@plt>
  mov    %rax,-0x68(%rbp)
  mov    -0x4c(%rbp),%edx
  mov    -0x68(%rbp),%rax
  mov    %edx,(%rax)
  mov    D10(%rip),%rax
  mov    (%rax),%edx
  mov    -0x68(%rbp),%rax
  mov    %edx,0x8(%rax)
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    -0x68(%rbp),%rax
  mov    %edx,0x4(%rax)
  movl   $0x0,-0x6c(%rbp)
  jmp    L38
L8:
  mov    -0x6c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x80(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x6c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%r12
  mov    -0x68(%rbp),%rax
  mov    %rax,%rdi
  call   <CMSG_DATA@plt>
  add    %r12,%rax
  mov    %rax,%rdx
  mov    (%rbx),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x6c(%rbp)
L38:
  mov    -0x6c(%rbp),%eax
  cmp    %eax,-0x78(%rbp)
  ja     L8
  lea    -0x50(%rbp),%rcx
  mov    -0x74(%rbp),%eax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <sendmsg@plt>
  test   %rax,%rax
  js     L23
  mov    $0x0,%eax
  jmp    L22
L23:
  mov    $0xffffffff,%eax
L22:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  add    $0x90,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
int ancil_send_fds_with_buffer_to(int sock, const int *fds, unsigned n_fds, void *buffer, struct sockaddr *to, socklen_t tolen) {
  struct msghdr msghdr;
  char nothing = '!';
  struct iovec nothing_ptr;
  struct cmsghdr *cmsg;
  int i;
  nothing_ptr.iov_base = &nothing;
  nothing_ptr.iov_len = 1;
  msghdr.msg_name = to;
  msghdr.msg_namelen = tolen;
  msghdr.msg_iov = &nothing_ptr;
  msghdr.msg_iovlen = 1;
  msghdr.msg_flags = 0;
  msghdr.msg_control = buffer;
  msghdr.msg_controllen = sizeof(struct cmsghdr) + sizeof(int) * n_fds;
  cmsg = CMSG_FIRSTHDR(&msghdr);
  cmsg->cmsg_len = msghdr.msg_controllen;
  cmsg->cmsg_level = SOL_SOCKET;
  cmsg->cmsg_type = SCM_RIGHTS;
  for (i = 0; i < n_fds; i++)
    ((int *)CMSG_DATA(cmsg))[i] = fds[i];
  return (sendmsg(sock, &msghdr, 0) >= 0 ? 0 : -1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ancil_send_fds_with_buffer_to>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %edi,%r14d
  mov    %rsi,%r13
  mov    %edx,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movb   $0x21,0xf(%rsp)
  lea    0xf(%rsp),%rax
  mov    %rax,0x10(%rsp)
  movl   $0x1,0x18(%rsp)
  mov    %r8,0x48(%rsp)
  mov    %r9d,0x40(%rsp)
  lea    0x10(%rsp),%rax
  mov    %rax,0x38(%rsp)
  movl   $0x1,0x20(%rsp)
  movq   $0x0,0x30(%rsp)
  mov    %rcx,0x28(%rsp)
  lea    0xc(,%rdx,4),%eax
  mov    %eax,0x24(%rsp)
  lea    0x20(%rsp),%rdi
  call   <CMSG_FIRSTHDR@plt>
  mov    %rax,%r12
  mov    0x24(%rsp),%eax
  mov    %eax,(%r12)
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x8(%r12)
  mov    D10(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x4(%r12)
  test   %ebp,%ebp
  je     L3
  mov    %ebp,%ebp
  shl    $0x2,%rbp
  mov    $0x0,%ebx
L29:
  mov    %r12,%rdi
  call   <CMSG_DATA@plt>
  mov    0x0(%r13,%rbx,1),%edx
  mov    %edx,(%rax,%rbx,1)
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L29
L3:
  lea    0x20(%rsp),%rsi
  mov    $0x0,%edx
  mov    %r14d,%edi
  call   <sendmsg@plt>
  sar    $0x3f,%rax
  mov    0x58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ancil_send_fds_with_buffer_to(int sock, const int *fds, unsigned n_fds, void *buffer, struct sockaddr *to, socklen_t tolen) {
  struct msghdr msghdr;
  char nothing = '!';
  struct iovec nothing_ptr;
  struct cmsghdr *cmsg;
  int i;
  nothing_ptr.iov_base = &nothing;
  nothing_ptr.iov_len = 1;
  msghdr.msg_name = to;
  msghdr.msg_namelen = tolen;
  msghdr.msg_iov = &nothing_ptr;
  msghdr.msg_iovlen = 1;
  msghdr.msg_flags = 0;
  msghdr.msg_control = buffer;
  msghdr.msg_controllen = sizeof(struct cmsghdr) + sizeof(int) * n_fds;
  cmsg = CMSG_FIRSTHDR(&msghdr);
  cmsg->cmsg_len = msghdr.msg_controllen;
  cmsg->cmsg_level = SOL_SOCKET;
  cmsg->cmsg_type = SCM_RIGHTS;
  for (i = 0; i < n_fds; i++)
    ((int *)CMSG_DATA(cmsg))[i] = fds[i];
  return (sendmsg(sock, &msghdr, 0) >= 0 ? 0 : -1);
}
```
user: What is the c source code of the assembly code below:

```asm
<ancil_send_fds_with_buffer_to>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0xf(%rsp),%rax
  lea    0x20(%rsp),%r14
  movb   $0x21,0xf(%rsp)
  mov    %rax,0x10(%rsp)
  lea    0x10(%rsp),%rax
  mov    %r14,%rdi
  mov    %rax,0x38(%rsp)
  lea    0xc(,%rbp,4),%eax
  movl   $0x1,0x18(%rsp)
  mov    %r8,0x48(%rsp)
  mov    %r9d,0x40(%rsp)
  movl   $0x1,0x20(%rsp)
  movq   $0x0,0x30(%rsp)
  mov    %rcx,0x28(%rsp)
  mov    %eax,0x24(%rsp)
  call   <CMSG_FIRSTHDR@plt>
  mov    %rax,%r15
  mov    0x24(%rsp),%eax
  mov    %eax,(%r15)
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x8(%r15)
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x4(%r15)
  test   %ebp,%ebp
  je     L21
  shl    $0x2,%rbp
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L18:
  mov    %r15,%rdi
  call   <CMSG_DATA@plt>
  mov    (%r12,%rbx,1),%edx
  mov    %edx,(%rax,%rbx,1)
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L18
L21:
  xor    %edx,%edx
  mov    %r14,%rsi
  mov    %r13d,%edi
  call   <sendmsg@plt>
  sar    $0x3f,%rax
  mov    0x58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ancil_send_fds_with_buffer_to(int sock, const int *fds, unsigned n_fds, void *buffer, struct sockaddr *to, socklen_t tolen) {
  struct msghdr msghdr;
  char nothing = '!';
  struct iovec nothing_ptr;
  struct cmsghdr *cmsg;
  int i;
  nothing_ptr.iov_base = &nothing;
  nothing_ptr.iov_len = 1;
  msghdr.msg_name = to;
  msghdr.msg_namelen = tolen;
  msghdr.msg_iov = &nothing_ptr;
  msghdr.msg_iovlen = 1;
  msghdr.msg_flags = 0;
  msghdr.msg_control = buffer;
  msghdr.msg_controllen = sizeof(struct cmsghdr) + sizeof(int) * n_fds;
  cmsg = CMSG_FIRSTHDR(&msghdr);
  cmsg->cmsg_len = msghdr.msg_controllen;
  cmsg->cmsg_level = SOL_SOCKET;
  cmsg->cmsg_type = SCM_RIGHTS;
  for (i = 0; i < n_fds; i++)
    ((int *)CMSG_DATA(cmsg))[i] = fds[i];
  return (sendmsg(sock, &msghdr, 0) >= 0 ? 0 : -1);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ancil_send_fds_with_buffer_to>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0xf(%rsp),%rax
  lea    0x20(%rsp),%r14
  movb   $0x21,0xf(%rsp)
  mov    %rax,0x10(%rsp)
  lea    0x10(%rsp),%rax
  mov    %r14,%rdi
  mov    %rax,0x38(%rsp)
  lea    0xc(,%rbp,4),%eax
  movl   $0x1,0x18(%rsp)
  mov    %r8,0x48(%rsp)
  mov    %r9d,0x40(%rsp)
  movl   $0x1,0x20(%rsp)
  movq   $0x0,0x30(%rsp)
  mov    %rcx,0x28(%rsp)
  mov    %eax,0x24(%rsp)
  call   <CMSG_FIRSTHDR@plt>
  movd   0x24(%rsp),%xmm0
  mov    %rax,%r15
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x8(%r15)
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  movd   %eax,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,(%r15)
  test   %ebp,%ebp
  je     L18
  shl    $0x2,%rbp
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L23:
  mov    %r15,%rdi
  call   <CMSG_DATA@plt>
  mov    (%r12,%rbx,1),%edx
  mov    %edx,(%rax,%rbx,1)
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L23
L18:
  xor    %edx,%edx
  mov    %r14,%rsi
  mov    %r13d,%edi
  call   <sendmsg@plt>
  sar    $0x3f,%rax
  mov    0x58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ancil_send_fds_with_buffer_to(int sock, const int *fds, unsigned n_fds, void *buffer, struct sockaddr *to, socklen_t tolen) {
  struct msghdr msghdr;
  char nothing = '!';
  struct iovec nothing_ptr;
  struct cmsghdr *cmsg;
  int i;
  nothing_ptr.iov_base = &nothing;
  nothing_ptr.iov_len = 1;
  msghdr.msg_name = to;
  msghdr.msg_namelen = tolen;
  msghdr.msg_iov = &nothing_ptr;
  msghdr.msg_iovlen = 1;
  msghdr.msg_flags = 0;
  msghdr.msg_control = buffer;
  msghdr.msg_controllen = sizeof(struct cmsghdr) + sizeof(int) * n_fds;
  cmsg = CMSG_FIRSTHDR(&msghdr);
  cmsg->cmsg_len = msghdr.msg_controllen;
  cmsg->cmsg_level = SOL_SOCKET;
  cmsg->cmsg_type = SCM_RIGHTS;
  for (i = 0; i < n_fds; i++)
    ((int *)CMSG_DATA(cmsg))[i] = fds[i];
  return (sendmsg(sock, &msghdr, 0) >= 0 ? 0 : -1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  char a[10];
  int i;
  for (i = 0; i < 10; ++i)
    a[i] = 0;
#pragma acc data copy(a)
  {
#pragma acc parallel present(a)
    {
      int j;
      for (j = 0; j < 10; ++j)
        a[j] = j;
    }
  }
  for (i = 0; i < 10; ++i) {
    if (a[i] != i)
      abort();
  }
  for (i = 0; i < 10; ++i)
    a[i] = 0;
#pragma acc data copy(a)
  {
#pragma acc kernels present(a)
    {
      int j;
      for (j = 0; j < 10; ++j)
        a[j] = j;
    }
  }
  for (i = 0; i < 10; ++i) {
    if (a[i] != i)
      abort();
  }
  for (i = 0; i < 10; ++i)
    a[i] = 0;
#pragma acc data copy(a)
  {
#pragma acc data present(a)
    {
#pragma acc parallel present(a)
      {
        int j;
        for (j = 0; j < 10; ++j)
          a[j] = j;
      }
    }
  }
  for (i = 0; i < 10; ++i) {
    if (a[i] != i)
      abort();
  }
#pragma acc data copy(a)
  {
#pragma acc data present(a)
    {
#pragma acc kernels present(a)
      {
        int j;
        for (j = 0; j < 10; ++j)
          a[j] = j;
      }
    }
  }
  for (i = 0; i < 10; ++i) {
    if (a[i] != i)
      abort();
  }
  for (i = 0; i < 10; ++i)
    a[i] = 0;
#pragma acc enter data copyin(a)
#pragma acc data present(a)
  {
#pragma acc parallel present(a)
    {
      int j;
      for (j = 0; j < 10; ++j)
        a[j] = j;
    }
  }
#pragma acc exit data copyout(a)
  for (i = 0; i < 10; ++i) {
    if (a[i] != i)
      abort();
  }
#pragma acc enter data copyin(a)
#pragma acc data present(a)
  {
#pragma acc kernels present(a)
    {
      int j;
      for (j = 0; j < 10; ++j)
        a[j] = j;
    }
  }
#pragma acc exit data copyout(a)
  for (i = 0; i < 10; ++i) {
    if (a[i] != i)
      abort();
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<parse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <length@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L26
L10:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L18
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L18
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  lea    -0x30(%rax),%ebx
  mov    -0x20(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <exponentiation@plt>
  movsbl %al,%eax
  imul   %ebx,%eax
  add    %eax,-0x1c(%rbp)
L18:
  addl   $0x1,-0x18(%rbp)
L26:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L10
  mov    -0x1c(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int parse(char *number) {
  int size = length(number);
  int exp = size - 1;
  int total = 0;
  int i;
  for (i = 0; i < size; i++) {
    if (number[i] >= '0' && number[i] <= '9') {
      total += (number[i] - 48) * exponentiation(10, exp--);
    }
  }
  return total;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<parse>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <length@plt>
  test   %eax,%eax
  jle    L31
  lea    -0x1(%rax),%esi
  mov    %rbx,%rbp
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%r12
  mov    $0x0,%r13d
  jmp    L26
L33:
  add    $0x1,%rbp
  cmp    %r12,%rbp
  je     L45
L26:
  movzbl 0x0(%rbp),%ebx
  lea    -0x30(%rbx),%eax
  cmp    $0x9,%al
  ja     L33
  lea    -0x1(%rsi),%r14d
  mov    $0xa,%edi
  call   <exponentiation@plt>
  movsbl %bl,%ebx
  sub    $0x30,%ebx
  movsbl %al,%eax
  imul   %eax,%ebx
  add    %ebx,%r13d
  mov    %r14d,%esi
  jmp    L33
L31:
  mov    $0x0,%r13d
L45:
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int parse(char *number) {
  int size = length(number);
  int exp = size - 1;
  int total = 0;
  int i;
  for (i = 0; i < size; i++) {
    if (number[i] >= '0' && number[i] <= '9') {
      total += (number[i] - 48) * exponentiation(10, exp--);
    }
  }
  return total;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<parse>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <length@plt>
  test   %eax,%eax
  jle    L29
  lea    -0x1(%rax),%esi
  xor    %r14d,%r14d
  mov    %esi,%eax
  lea    0x1(%rbp,%rax,1),%r13
  jmp    L46
  nopw   0x0(%rax,%rax,1)
L1:
  add    $0x1,%rbp
  cmp    %r13,%rbp
  je     L16
L46:
  movzbl 0x0(%rbp),%eax
  lea    -0x30(%rax),%ebx
  cmp    $0x9,%bl
  ja     L1
  mov    $0xa,%edi
  lea    -0x1(%rsi),%r12d
  movsbl %bl,%ebx
  add    $0x1,%rbp
  call   <exponentiation@plt>
  mov    %r12d,%esi
  movsbl %al,%eax
  imul   %eax,%ebx
  add    %ebx,%r14d
  cmp    %r13,%rbp
  jne    L46
L16:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L29:
  xor    %r14d,%r14d
  pop    %rbx
  pop    %rbp
  mov    %r14d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int parse(char *number) {
  int size = length(number);
  int exp = size - 1;
  int total = 0;
  int i;
  for (i = 0; i < size; i++) {
    if (number[i] >= '0' && number[i] <= '9') {
      total += (number[i] - 48) * exponentiation(10, exp--);
    }
  }
  return total;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parse>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <length@plt>
  test   %eax,%eax
  jle    L25
  lea    -0x1(%rax),%esi
  xor    %r14d,%r14d
  mov    %esi,%eax
  lea    0x1(%rbp,%rax,1),%r13
  jmp    L20
  nopw   0x0(%rax,%rax,1)
L32:
  add    $0x1,%rbp
  cmp    %r13,%rbp
  je     L30
L20:
  movzbl 0x0(%rbp),%eax
  lea    -0x30(%rax),%ebx
  cmp    $0x9,%bl
  ja     L32
  mov    $0xa,%edi
  lea    -0x1(%rsi),%r12d
  movsbl %bl,%ebx
  add    $0x1,%rbp
  call   <exponentiation@plt>
  mov    %r12d,%esi
  movsbl %al,%eax
  imul   %eax,%ebx
  add    %ebx,%r14d
  cmp    %r13,%rbp
  jne    L20
L30:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L25:
  xor    %r14d,%r14d
  pop    %rbx
  pop    %rbp
  mov    %r14d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int parse(char *number) {
  int size = length(number);
  int exp = size - 1;
  int total = 0;
  int i;
  for (i = 0; i < size; i++) {
    if (number[i] >= '0' && number[i] <= '9') {
      total += (number[i] - 48) * exponentiation(10, exp--);
    }
  }
  return total;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<array_sum_2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L11
L6:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    $0x8,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L11:
  cmpl   $0xf9,-0xc(%rbp)
  jle    L6
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double array_sum_2(double ***a, int t) {
  int i;
  double sum = 0;
  for (i = 0; i < 250; i++) {
    sum = sum + a[t][1][i];
  }
  return (sum);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<array_sum_2>:
  endbr64
  movslq %esi,%rax
  mov    (%rdi,%rax,8),%rax
  mov    0x8(%rax),%rax
  lea    0x7d0(%rax),%rdx
  pxor   %xmm0,%xmm0
L5:
  addsd  (%rax),%xmm0
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L5
  ret
```

assistant: ```c
double array_sum_2(double ***a, int t) {
  int i;
  double sum = 0;
  for (i = 0; i < 250; i++) {
    sum = sum + a[t][1][i];
  }
  return (sum);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<array_sum_2>:
  endbr64
  movslq %esi,%rax
  pxor   %xmm0,%xmm0
  mov    (%rdi,%rax,8),%rax
  mov    0x8(%rax),%rax
  lea    0x7d0(%rax),%rdx
  nopw   0x0(%rax,%rax,1)
L1:
  addsd  (%rax),%xmm0
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L1
  ret
```

assistant: ```c
double array_sum_2(double ***a, int t) {
  int i;
  double sum = 0;
  for (i = 0; i < 250; i++) {
    sum = sum + a[t][1][i];
  }
  return (sum);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<array_sum_2>:
  endbr64
  movslq %esi,%rax
  pxor   %xmm0,%xmm0
  mov    (%rdi,%rax,8),%rax
  mov    0x8(%rax),%rax
  lea    0x7d0(%rax),%rdx
  nopw   0x0(%rax,%rax,1)
L1:
  movsd  (%rax),%xmm1
  add    $0x10,%rax
  addsd  %xmm1,%xmm0
  movsd  -0x8(%rax),%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rax,%rdx
  jne    L1
  ret
```

assistant: ```c
double array_sum_2(double ***a, int t) {
  int i;
  double sum = 0;
  for (i = 0; i < 250; i++) {
    sum = sum + a[t][1][i];
  }
  return (sum);
}
```