ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<options_print_version>:
  endbr64
  sub    $0x8,%rsp
  lea    D12(%rip),%rdi
  call   <printf@plt>
  lea    D28(%rip),%rdi
  call   <printf@plt>
  lea    D16(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: label: D12
datatype: string
name: retrieve_data
arguments: label: D28
datatype: string
name: retrieve_data
arguments: label: D16
datatype: string

tool: D12: Package version 1.01 


tool: D28: Report bugs to rick@cs.washington.edu


tool: D16: Available under the GNU General Public License


assistant: ```c
void options_print_version() {
  printf("Package version 1.01 \n");
  printf("Report bugs to rick@cs.washington.edu\n");
  printf("Available under the GNU General Public License\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sklearn_lgamma>:
  endbr64
  movapd %xmm0,%xmm2
  sub    $0x18,%rsp
  movsd  D32(%rip),%xmm0
  comisd %xmm2,%xmm0
  movapd %xmm2,%xmm0
  ja     L6
  mulsd  %xmm2,%xmm0
  movsd  D77(%rip),%xmm1
  movsd  %xmm2,(%rsp)
  divsd  %xmm0,%xmm1
  movsd  D22(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  addsd  D118(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  subsd  D98(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  addsd  D95(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  subsd  D28(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  addsd  D47(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  subsd  D54(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movapd %xmm2,%xmm0
  movsd  %xmm1,0x8(%rsp)
  call   <log@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  (%rsp),%xmm2
  addsd  D91(%rip),%xmm1
  movapd %xmm0,%xmm3
  movapd %xmm2,%xmm0
  subsd  D106(%rip),%xmm0
  divsd  %xmm2,%xmm1
  mulsd  %xmm3,%xmm0
  subsd  %xmm2,%xmm0
  addsd  D97(%rip),%xmm0
  add    $0x18,%rsp
  addsd  %xmm1,%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  call   <sklearn_gamma@plt>
  mov    %eax,%edi
  call   <fabs@plt>
  add    $0x18,%rsp
  jmp    1080 <log@plt>
```

tool_calls: name: read_data
arguments: data_tag: D32
type_guessed: f64
name: read_data
arguments: data_tag: D77
type_guessed: f64
name: read_data
arguments: data_tag: D22
type_guessed: byte[8]
name: read_data
arguments: data_tag: D118
type_guessed: byte[8]
name: read_data
arguments: data_tag: D98
type_guessed: byte[8]
name: read_data
arguments: data_tag: D95
type_guessed: byte[8]
name: read_data
arguments: data_tag: D28
type_guessed: byte[8]
name: read_data
arguments: data_tag: D47
type_guessed: byte[8]
name: read_data
arguments: data_tag: D54
type_guessed: byte[8]
name: read_data
arguments: data_tag: D91
type_guessed: byte[8]
name: read_data
arguments: data_tag: D106
type_guessed: f64
name: read_data
arguments: data_tag: D97
type_guessed: f64

tool: D32: 12

tool: D77: 1

tool: {"D22": [0x98,0x53,0x0f,0xcb,0x86,0x42,0x9e,0xbf]}

tool: {"D118": [0x1a,0xa4,0x41,0x1a,0xa4,0x41,0x7a,0x3f]}

tool: {"D98": [0x7d,0x3c,0x99,0xd9,0xb0,0x6a,0x5f,0x3f]}

tool: {"D95": [0x23,0xff,0x18,0x2b,0x1e,0x95,0x4b,0x3f]}

tool: {"D28": [0x14,0x38,0x81,0x13,0x38,0x81,0x43,0x3f]}

tool: {"D47": [0x1a,0xa0,0x01,0x1a,0xa0,0x01,0x4a,0x3f]}

tool: {"D54": [0x17,0x6c,0xc1,0x16,0x6c,0xc1,0x66,0x3f]}

tool: {"D91": [0x55,0x55,0x55,0x55,0x55,0x55,0xb5,0x3f]}

tool: D106: 0.5

tool: D97: 0.9189385332046728

assistant: ```c
double sklearn_lgamma(double x) {
  static const double c[8] = {1.0 / 12.0, -1.0 / 360.0, 1.0 / 1260.0, -1.0 / 1680.0, 1.0 / 1188.0, -691.0 / 360360.0, 1.0 / 156.0, -3617.0 / 122400.0};
  double z, sum;
  int i;
  if (x < 12.0)
    return log(fabs(sklearn_gamma(x)));
  z = 1.0 / (x * x);
  sum = c[7];
  for (i = 6; i >= 0; i--) {
    sum *= z;
    sum += c[i];
  }
  return (x - 0.5) * log(x) - x + 0.91893853320467274178032973640562 + sum / x;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x120,%rsp
  mov    %edi,-0x114(%rbp)
  mov    %rsi,-0x120(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D4(%rip),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x2,-0x114(%rbp)
  jne    L4
  mov    -0x120(%rbp),%rax
  mov    0x8(%rax),%rax
  jmp    L55
L4:
  lea    D3(%rip),%rax
L55:
  lea    -0x110(%rbp),%rcx
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <read_config_file@plt>
  test   %eax,%eax
  je     L33
  nop
L28:
  lea    -0x110(%rbp),%rax
  mov    $0xff,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <cli_read_line@plt>
  test   %eax,%eax
  je     L33
  lea    -0x110(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <cli_parse_line@plt>
  test   %eax,%eax
  jne    L28
L33:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D4
data_type: string
name: read_data
arguments: tag: D13
data_type: string
name: read_data
arguments: tag: D3
data_type: string

tool: D4: v1.60 : 2017-02-27

tool: D13: PostMark %s


tool: D3: .pmrc

assistant: ```c
main(argc, argv) int argc;
char *argv[];
{
  char buffer[255 + 1];
  printf("PostMark %s\n", "v1.60 : 2017-02-27");
  if (read_config_file((argc == 2) ? argv[1] : ".pmrc", buffer, 1))
    while (cli_read_line(buffer, 255) && cli_parse_line(buffer))
      ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  lea    D20(%rip),%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D24(%rip),%rdi
  cmp    $0x2,%ebx
  je     L30
L56:
  mov    %rsp,%rsi
  mov    $0x1,%edx
  mov    $0x0,%eax
  call   <read_config_file@plt>
  test   %eax,%eax
  je     L47
  mov    %rsp,%rbx
L49:
  mov    $0xff,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <cli_read_line@plt>
  test   %eax,%eax
  je     L47
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <cli_parse_line@plt>
  test   %eax,%eax
  jne    L49
L47:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  ret
L30:
  mov    0x8(%rbp),%rdi
  jmp    L56
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D20
data_type: string
name: parse_data
arguments: label: D4
data_type: string
name: parse_data
arguments: label: D24
data_type: string

tool: D20: v1.60 : 2017-02-27

tool: D4: PostMark %s


tool: D24: .pmrc

assistant: ```c
main(argc, argv) int argc;
char *argv[];
{
  char buffer[255 + 1];
  printf("PostMark %s\n", "v1.60 : 2017-02-27");
  if (read_config_file((argc == 2) ? argv[1] : ".pmrc", buffer, 1))
    while (cli_read_line(buffer, 255) && cli_parse_line(buffer))
      ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  lea    D14(%rip),%rsi
  push   %rbx
  mov    %edi,%ebx
  lea    D10(%rip),%rdi
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D2(%rip),%rdi
  cmp    $0x2,%ebx
  jne    L24
  mov    0x8(%rbp),%rdi
L24:
  mov    %rsp,%rbx
  xor    %eax,%eax
  mov    $0x1,%edx
  mov    %rbx,%rsi
  call   <read_config_file@plt>
  test   %eax,%eax
  jne    L27
L28:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  add    $0x118,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L7:
  xor    %eax,%eax
  mov    %rbx,%rdi
  call   <cli_parse_line@plt>
  test   %eax,%eax
  je     L28
L27:
  xor    %eax,%eax
  mov    $0xff,%esi
  mov    %rbx,%rdi
  call   <cli_read_line@plt>
  test   %eax,%eax
  jne    L7
  jmp    L28
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D14
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D10
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D2
type_guessed: string

tool: D14: v1.60 : 2017-02-27

tool: D10: PostMark %s


tool: D2: .pmrc

assistant: ```c
main(argc, argv) int argc;
char *argv[];
{
  char buffer[255 + 1];
  printf("PostMark %s\n", "v1.60 : 2017-02-27");
  if (read_config_file((argc == 2) ? argv[1] : ".pmrc", buffer, 1))
    while (cli_read_line(buffer, 255) && cli_parse_line(buffer))
      ;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  lea    D19(%rip),%rsi
  push   %rbx
  mov    %edi,%ebx
  lea    D9(%rip),%rdi
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D11(%rip),%rdi
  cmp    $0x2,%ebx
  jne    L14
  mov    0x8(%rbp),%rdi
L14:
  mov    %rsp,%rbx
  xor    %eax,%eax
  mov    $0x1,%edx
  mov    %rbx,%rsi
  call   <read_config_file@plt>
  test   %eax,%eax
  jne    L24
L54:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x118,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L30:
  xor    %eax,%eax
  mov    %rbx,%rdi
  call   <cli_parse_line@plt>
  test   %eax,%eax
  je     L54
L24:
  xor    %eax,%eax
  mov    $0xff,%esi
  mov    %rbx,%rdi
  call   <cli_read_line@plt>
  test   %eax,%eax
  jne    L30
  jmp    L54
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D19
type_infered: string
name: unpack_data
arguments: label: D9
type_infered: string
name: unpack_data
arguments: label: D11
type_infered: string

tool: D19: v1.60 : 2017-02-27

tool: D9: PostMark %s


tool: D11: .pmrc

assistant: ```c
main(argc, argv) int argc;
char *argv[];
{
  char buffer[255 + 1];
  printf("PostMark %s\n", "v1.60 : 2017-02-27");
  if (read_config_file((argc == 2) ? argv[1] : ".pmrc", buffer, 1))
    while (cli_read_line(buffer, 255) && cli_parse_line(buffer))
      ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<charep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %edi,%eax
  lea    D19(%rip),%rbx
  mov    %eax,%edx
  and    $0x100,%edx
  test   %edx,%edx
  je     L37
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  movb   $0x58,(%rdx)
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  movb   $0x2d,(%rdx)
  and    $0xfe,%ah
L37:
  mov    %eax,%edx
  and    $0x80,%edx
  test   %edx,%edx
  je     L27
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  movb   $0x4d,(%rdx)
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  movb   $0x2d,(%rdx)
  and    $0x7f,%al
L27:
  cmp    $0x1f,%eax
  jg     L48
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  movb   $0x5e,(%rdx)
  add    $0x40,%eax
L48:
  cmp    $0x7f,%eax
  jne    L49
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  movb   $0x44,(%rax)
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  movb   $0x45,(%rax)
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  movb   $0x4c,(%rax)
  jmp    L16
L49:
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  mov    %al,(%rdx)
L16:
  movb   $0x0,(%rbx)
  lea    D19(%rip),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *charep(c)
register int c;
{
  static char chrbuf[10];
  register char *cp;
  cp = chrbuf;
  if (c & 0400) {
    *cp++ = 'X';
    *cp++ = '-';
    c &= ~0400;
  }
  if (c & 0200) {
    *cp++ = 'M';
    *cp++ = '-';
    c &= ~0200;
  }
  if (c <= 037) {
    *cp++ = '^';
    c += 0100;
  }
  if (c == 0177) {
    *cp++ = 'D';
    *cp++ = 'E';
    *cp++ = 'L';
  } else
    *cp++ = c;
  *cp = 0;
  return (chrbuf);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<charep>:
  endbr64
  lea    D22(%rip),%rax
  test   $0x100,%edi
  je     L29
  movb   $0x58,D22(%rip)
  movb   $0x2d,D31(%rip)
  and    $0xfffffeff,%edi
  lea    0x2(%rax),%rax
L29:
  test   $0x80,%dil
  je     L3
  movb   $0x4d,(%rax)
  movb   $0x2d,0x1(%rax)
  and    $0x7f,%dil
  lea    0x2(%rax),%rax
L3:
  cmp    $0x1f,%edi
  jg     L49
  lea    0x1(%rax),%rcx
  movb   $0x5e,(%rax)
  add    $0x40,%edi
L35:
  lea    0x1(%rcx),%rdx
  mov    %dil,(%rcx)
L27:
  movb   $0x0,(%rdx)
  lea    D22(%rip),%rax
  ret
L49:
  cmp    $0x7f,%edi
  jne    L37
  movb   $0x44,(%rax)
  movb   $0x45,0x1(%rax)
  lea    0x3(%rax),%rdx
  movb   $0x4c,0x2(%rax)
  jmp    L27
L37:
  mov    %rax,%rcx
  jmp    L35
```

assistant: ```c
char *charep(c)
register int c;
{
  static char chrbuf[10];
  register char *cp;
  cp = chrbuf;
  if (c & 0400) {
    *cp++ = 'X';
    *cp++ = '-';
    c &= ~0400;
  }
  if (c & 0200) {
    *cp++ = 'M';
    *cp++ = '-';
    c &= ~0200;
  }
  if (c <= 037) {
    *cp++ = '^';
    c += 0100;
  }
  if (c == 0177) {
    *cp++ = 'D';
    *cp++ = 'E';
    *cp++ = 'L';
  } else
    *cp++ = c;
  *cp = 0;
  return (chrbuf);
}
```
user: Translate the below assembly code into C programming language:

```asm
<charep>:
  endbr64
  lea    D15(%rip),%r8
  mov    %r8,%rax
  test   $0x100,%edi
  je     L18
  mov    $0x2d58,%r9d
  and    $0xfffffeff,%edi
  add    $0x2,%rax
  mov    %r9w,D9(%rip)
L18:
  test   $0x80,%dil
  je     L12
  mov    $0x2d4d,%esi
  and    $0x7f,%dil
  add    $0x2,%rax
  mov    %si,-0x2(%rax)
L12:
  cmp    $0x1f,%edi
  jg     L33
  movb   $0x5e,(%rax)
  lea    0x1(%rax),%rcx
  add    $0x40,%edi
L24:
  lea    0x1(%rcx),%rdx
  mov    %dil,(%rcx)
  mov    %r8,%rax
  movb   $0x0,(%rdx)
  ret
  nopw   0x0(%rax,%rax,1)
L33:
  cmp    $0x7f,%edi
  jne    L25
  lea    0x3(%rax),%rdx
  mov    $0x4544,%ecx
  movb   $0x4c,0x2(%rax)
  mov    %cx,(%rax)
  mov    %r8,%rax
  movb   $0x0,(%rdx)
  ret
  nopl   0x0(%rax)
L25:
  mov    %rax,%rcx
  jmp    L24
```

assistant: ```c
char *charep(c)
register int c;
{
  static char chrbuf[10];
  register char *cp;
  cp = chrbuf;
  if (c & 0400) {
    *cp++ = 'X';
    *cp++ = '-';
    c &= ~0400;
  }
  if (c & 0200) {
    *cp++ = 'M';
    *cp++ = '-';
    c &= ~0200;
  }
  if (c <= 037) {
    *cp++ = '^';
    c += 0100;
  }
  if (c == 0177) {
    *cp++ = 'D';
    *cp++ = 'E';
    *cp++ = 'L';
  } else
    *cp++ = c;
  *cp = 0;
  return (chrbuf);
}
```
user: Translate the below assembly code into C programming language:

```asm
<charep>:
  endbr64
  lea    D9(%rip),%r8
  mov    %r8,%rax
  test   $0x100,%edi
  je     L3
  mov    $0x2d58,%esi
  and    $0xfffffeff,%edi
  add    $0x2,%rax
  mov    %si,D6(%rip)
L3:
  test   $0x80,%dil
  je     L37
  mov    $0x2d4d,%ecx
  and    $0x7f,%dil
  add    $0x2,%rax
  mov    %cx,-0x2(%rax)
L37:
  cmp    $0x1f,%edi
  jg     L45
  movb   $0x5e,(%rax)
  lea    0x1(%rax),%rcx
  add    $0x40,%edi
L10:
  lea    0x1(%rcx),%rdx
  mov    %dil,(%rcx)
  mov    %r8,%rax
  movb   $0x0,(%rdx)
  ret
L45:
  cmp    $0x7f,%edi
  jne    L43
  mov    $0x4544,%edx
  movb   $0x4c,0x2(%rax)
  mov    %dx,(%rax)
  lea    0x3(%rax),%rdx
  mov    %r8,%rax
  movb   $0x0,(%rdx)
  ret
  nopl   0x0(%rax)
L43:
  mov    %rax,%rcx
  jmp    L10
```

assistant: ```c
char *charep(c)
register int c;
{
  static char chrbuf[10];
  register char *cp;
  cp = chrbuf;
  if (c & 0400) {
    *cp++ = 'X';
    *cp++ = '-';
    c &= ~0400;
  }
  if (c & 0200) {
    *cp++ = 'M';
    *cp++ = '-';
    c &= ~0200;
  }
  if (c <= 037) {
    *cp++ = '^';
    c += 0100;
  }
  if (c == 0177) {
    *cp++ = 'D';
    *cp++ = 'E';
    *cp++ = 'L';
  } else
    *cp++ = c;
  *cp = 0;
  return (chrbuf);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L16
L9:
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  add    %eax,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L16:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L9
  mov    -0x1c(%rbp),%ecx
  mov    -0x10(%rbp),%eax
  cltd
  idiv   %ecx
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm1,%xmm1
  cvtss2sd -0xc(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D16
type_infered: string
name: unpack_data
arguments: data_label: D13
type_infered: string
name: unpack_data
arguments: data_label: D3
type_infered: string
name: unpack_data
arguments: data_label: D36
type_infered: string

tool: D16: Inserire quanti numeri si vuole

tool: D13: %d

tool: D3: Inserire numero

tool: D36: La media e':%f

assistant: ```c
int main() {
  int N;
  int I;
  int Num;
  int Somma;
  float Media;
  printf("Inserire quanti numeri si vuole\n");
  scanf("%d", &N);
  Somma = 0;
  I = 0;
  while (I < N) {
    printf("Inserire numero\n");
    scanf("%d", &Num);
    Somma = Somma + Num;
    I++;
  }
  Media = Somma / N;
  printf("La media e':%f", Media);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D43(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%ecx
  test   %ecx,%ecx
  jle    L34
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  lea    D0(%rip),%r14
  lea    0x4(%rsp),%r13
  lea    D14(%rip),%r12
L18:
  mov    %r14,%rdi
  call   <puts@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    0x4(%rsp),%ebp
  add    $0x1,%ebx
  mov    (%rsp),%ecx
  cmp    %ebx,%ecx
  jg     L18
L20:
  mov    %ebp,%eax
  cltd
  idiv   %ecx
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D9(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L34:
  mov    $0x0,%ebp
  jmp    L20
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D43
data_type: string
name: retrieve_data
arguments: label: D14
data_type: string
name: retrieve_data
arguments: label: D0
data_type: string
name: retrieve_data
arguments: label: D9
data_type: string

tool: D43: Inserire quanti numeri si vuole

tool: D14: %d

tool: D0: Inserire numero

tool: D9: La media e':%f

assistant: ```c
int main() {
  int N;
  int I;
  int Num;
  int Somma;
  float Media;
  printf("Inserire quanti numeri si vuole\n");
  scanf("%d", &N);
  Somma = 0;
  I = 0;
  while (I < N) {
    printf("Inserire numero\n");
    scanf("%d", &Num);
    Somma = Somma + Num;
    I++;
  }
  Media = Somma / N;
  printf("La media e':%f", Media);
}
```
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
  lea    D0(%rip),%rdi
  push   %r13
  push   %r12
  lea    D19(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    (%rsp),%ecx
  test   %ecx,%ecx
  jle    L11
  xor    %ebx,%ebx
  lea    D13(%rip),%r14
  lea    0x4(%rsp),%r13
  nopl   (%rax)
L9:
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <puts@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    (%rsp),%ecx
  add    0x4(%rsp),%ebp
  cmp    %ebx,%ecx
  jg     L9
L11:
  mov    %ebp,%eax
  pxor   %xmm0,%xmm0
  lea    D14(%rip),%rdi
  cltd
  idiv   %ecx
  cvtsi2ss %eax,%xmm0
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
datatype: string
name: extract_symbolic_data
arguments: data_tag: D19
datatype: string
name: extract_symbolic_data
arguments: data_tag: D13
datatype: string
name: extract_symbolic_data
arguments: data_tag: D14
datatype: string

tool: D0: Inserire quanti numeri si vuole

tool: D19: %d

tool: D13: Inserire numero

tool: D14: La media e':%f

assistant: ```c
int main() {
  int N;
  int I;
  int Num;
  int Somma;
  float Media;
  printf("Inserire quanti numeri si vuole\n");
  scanf("%d", &N);
  Somma = 0;
  I = 0;
  while (I < N) {
    printf("Inserire numero\n");
    scanf("%d", &Num);
    Somma = Somma + Num;
    I++;
  }
  Media = Somma / N;
  printf("La media e':%f", Media);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r14
  lea    D36(%rip),%rdi
  push   %r13
  push   %r12
  lea    D34(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    (%rsp),%ecx
  test   %ecx,%ecx
  jle    L17
  xor    %ebx,%ebx
  lea    D12(%rip),%r14
  lea    0x4(%rsp),%r13
  nopl   (%rax)
L5:
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <puts@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    (%rsp),%ecx
  add    0x4(%rsp),%ebp
  cmp    %ebx,%ecx
  jg     L5
L17:
  mov    %ebp,%eax
  pxor   %xmm0,%xmm0
  lea    D8(%rip),%rdi
  cltd
  idiv   %ecx
  cvtsi2ss %eax,%xmm0
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D36
type: string
name: disassemble_data
arguments: label: D34
type: string
name: disassemble_data
arguments: label: D12
type: string
name: disassemble_data
arguments: label: D8
type: string

tool: D36: Inserire quanti numeri si vuole

tool: D34: %d

tool: D12: Inserire numero

tool: D8: La media e':%f

assistant: ```c
int main() {
  int N;
  int I;
  int Num;
  int Somma;
  float Media;
  printf("Inserire quanti numeri si vuole\n");
  scanf("%d", &N);
  Somma = 0;
  I = 0;
  while (I < N) {
    printf("Inserire numero\n");
    scanf("%d", &Num);
    Somma = Somma + Num;
    I++;
  }
  Media = Somma / N;
  printf("La media e':%f", Media);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xa,-0x4(%rbp)
  movb   $0x41,-0x6(%rbp)
  movzbl -0x6(%rbp),%eax
  mov    %al,-0x5(%rbp)
  movsbl -0x5(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  call   <getche@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
main() {
  char c;
  int i;
  i = 10;
  char j;
  c = 'A';
  j = c;
  putchar(j);
  getche();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x41,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  call   <getche@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  char c;
  int i;
  i = 10;
  char j;
  c = 'A';
  j = c;
  putchar(j);
  getche();
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x41,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getche@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  char c;
  int i;
  i = 10;
  char j;
  c = 'A';
  j = c;
  putchar(j);
  getche();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x41,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getche@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  char c;
  int i;
  i = 10;
  char j;
  c = 'A';
  j = c;
  putchar(j);
  getche();
}
```