ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<CL_Version_f>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <VersionString@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rdi
  mov    $0x0,%eax
  call   <Com_Printf@plt>
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <Com_Printf@plt>
  lea    D23(%rip),%r8
  mov    $0x0,%ecx
  mov    $0x4,%edx
  mov    $0x9,%esi
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <Com_Printf@plt>
  lea    D42(%rip),%rdi
  mov    $0x0,%eax
  call   <Com_Printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D7
type_guessed: string
name: unpack_data
arguments: tag: D8
type_guessed: string
name: unpack_data
arguments: tag: D23
type_guessed: string
name: unpack_data
arguments: tag: D27
type_guessed: string
name: unpack_data
arguments: tag: D42
type_guessed: string

tool: D7: ezQuake %s


tool: D8: Exe: Mar 17 2022 03:18:05


tool: D23: 9.4.0

tool: D27: Compiled with GCC version: %i.%i.%i (%i)


tool: D42: GCC Optimization: Functions currently not inlined into their callers


assistant: ```c
void CL_Version_f(void) {
  Com_Printf("ezQuake %s\n", VersionString());
  Com_Printf("Exe: "
             "Mar 17 2022"
             " "
             "03:18:05"
             "\n");
  Com_Printf("Compiled with GCC version: %i.%i.%i (%i)\n", 9, 4, 0, "9.4.0");
  Com_Printf("GCC Optimization: Functions currently not inlined into their callers\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<CL_Version_f>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <VersionString@plt>
  lea    D39(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <Com_Printf@plt>
  lea    D48(%rip),%rdi
  xor    %eax,%eax
  call   <Com_Printf@plt>
  lea    D8(%rip),%rdi
  xor    %ecx,%ecx
  xor    %eax,%eax
  lea    D45(%rip),%r8
  mov    $0x4,%edx
  mov    $0x9,%esi
  call   <Com_Printf@plt>
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <Com_Printf@plt>
```

tool_calls: name: read_data
arguments: data_tag: D39
datatype: string
name: read_data
arguments: data_tag: D48
datatype: string
name: read_data
arguments: data_tag: D8
datatype: string
name: read_data
arguments: data_tag: D45
datatype: string
name: read_data
arguments: data_tag: D27
datatype: string

tool: D39: ezQuake %s


tool: D48: Exe: Mar 17 2022 03:18:05


tool: D8: Compiled with GCC version: %i.%i.%i (%i)


tool: D45: 9.4.0

tool: D27: GCC Optimization: Functions currently not inlined into their callers


assistant: ```c
void CL_Version_f(void) {
  Com_Printf("ezQuake %s\n", VersionString());
  Com_Printf("Exe: "
             "Mar 17 2022"
             " "
             "03:18:05"
             "\n");
  Com_Printf("Compiled with GCC version: %i.%i.%i (%i)\n", 9, 4, 0, "9.4.0");
  Com_Printf("GCC Optimization: Functions currently not inlined into their callers\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<CL_Version_f>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <VersionString@plt>
  lea    D44(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <Com_Printf@plt>
  lea    D46(%rip),%rdi
  xor    %eax,%eax
  call   <Com_Printf@plt>
  lea    D17(%rip),%rdi
  xor    %ecx,%ecx
  xor    %eax,%eax
  lea    D26(%rip),%r8
  mov    $0x4,%edx
  mov    $0x9,%esi
  call   <Com_Printf@plt>
  lea    D31(%rip),%rdi
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <Com_Printf@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D44
datatype: string
name: disassemble_data
arguments: data_label: D46
datatype: string
name: disassemble_data
arguments: data_label: D17
datatype: string
name: disassemble_data
arguments: data_label: D26
datatype: string
name: disassemble_data
arguments: data_label: D31
datatype: string

tool: D44: ezQuake %s


tool: D46: Exe: Mar 17 2022 03:18:05


tool: D17: Compiled with GCC version: %i.%i.%i (%i)


tool: D26: 9.4.0

tool: D31: GCC Optimization: Functions currently not inlined into their callers


assistant: ```c
void CL_Version_f(void) {
  Com_Printf("ezQuake %s\n", VersionString());
  Com_Printf("Exe: "
             "Mar 17 2022"
             " "
             "03:18:05"
             "\n");
  Com_Printf("Compiled with GCC version: %i.%i.%i (%i)\n", 9, 4, 0, "9.4.0");
  Com_Printf("GCC Optimization: Functions currently not inlined into their callers\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x38,%edi
  call   <myallocinit@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0x14,%edi
  call   <myalloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0xa,%edi
  call   <myalloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <myfree@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <myfree@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0x14,%edi
  call   <myalloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0x18,%edi
  call   <myalloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <myfree@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <myfree@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0x38,%edi
  call   <myalloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  double *p1, *p2;
  myallocinit(56);
  printallocation();
  p1 = (double *)myalloc(20);
  printallocation();
  p2 = (double *)myalloc(10);
  printallocation();
  myfree(p1);
  printallocation();
  myfree(p2);
  printallocation();
  p1 = (double *)myalloc(20);
  printallocation();
  p2 = (double *)myalloc(24);
  printallocation();
  myfree(p1);
  printallocation();
  myfree(p2);
  printallocation();
  p1 = (double *)myalloc(56);
  printallocation();
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x38,%edi
  call   <myallocinit@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0x14,%edi
  call   <myalloc@plt>
  mov    %rax,%rbp
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0xa,%edi
  call   <myalloc@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    %rbp,%rdi
  call   <myfree@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    %rbx,%rdi
  call   <myfree@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0x14,%edi
  call   <myalloc@plt>
  mov    %rax,%rbp
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0x18,%edi
  call   <myalloc@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    %rbp,%rdi
  call   <myfree@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    %rbx,%rdi
  call   <myfree@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0x38,%edi
  call   <myalloc@plt>
  mov    $0x0,%eax
  call   <printallocation@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  double *p1, *p2;
  myallocinit(56);
  printallocation();
  p1 = (double *)myalloc(20);
  printallocation();
  p2 = (double *)myalloc(10);
  printallocation();
  myfree(p1);
  printallocation();
  myfree(p2);
  printallocation();
  p1 = (double *)myalloc(20);
  printallocation();
  p2 = (double *)myalloc(24);
  printallocation();
  myfree(p1);
  printallocation();
  myfree(p2);
  printallocation();
  p1 = (double *)myalloc(56);
  printallocation();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x38,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <myallocinit@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    $0x14,%edi
  call   <myalloc@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    $0xa,%edi
  call   <myalloc@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    %r12,%rdi
  call   <myfree@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    %rbp,%rdi
  call   <myfree@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    $0x14,%edi
  call   <myalloc@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    $0x18,%edi
  call   <myalloc@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    %r12,%rdi
  call   <myfree@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    %rbp,%rdi
  call   <myfree@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    $0x38,%edi
  call   <myalloc@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  double *p1, *p2;
  myallocinit(56);
  printallocation();
  p1 = (double *)myalloc(20);
  printallocation();
  p2 = (double *)myalloc(10);
  printallocation();
  myfree(p1);
  printallocation();
  myfree(p2);
  printallocation();
  p1 = (double *)myalloc(20);
  printallocation();
  p2 = (double *)myalloc(24);
  printallocation();
  myfree(p1);
  printallocation();
  myfree(p2);
  printallocation();
  p1 = (double *)myalloc(56);
  printallocation();
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x38,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <myallocinit@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    $0x14,%edi
  call   <myalloc@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    $0xa,%edi
  call   <myalloc@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    %r12,%rdi
  call   <myfree@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    %rbp,%rdi
  call   <myfree@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    $0x14,%edi
  call   <myalloc@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    $0x18,%edi
  call   <myalloc@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    %r12,%rdi
  call   <myfree@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    %rbp,%rdi
  call   <myfree@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  mov    $0x38,%edi
  call   <myalloc@plt>
  xor    %eax,%eax
  call   <printallocation@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  double *p1, *p2;
  myallocinit(56);
  printallocation();
  p1 = (double *)myalloc(20);
  printallocation();
  p2 = (double *)myalloc(10);
  printallocation();
  myfree(p1);
  printallocation();
  myfree(p2);
  printallocation();
  p1 = (double *)myalloc(20);
  printallocation();
  p2 = (double *)myalloc(24);
  printallocation();
  myfree(p1);
  printallocation();
  myfree(p2);
  printallocation();
  p1 = (double *)myalloc(56);
  printallocation();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sayhello>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L14
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L11
L14:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L11:
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D4
type_infered: string
name: unpack_data
arguments: label: D11
type_infered: string

tool: D4: Hello2, World!

tool: D11: Hello2 from %s


assistant: ```c
void sayhello(const char *msg) {
  if (!msg)
    printf("Hello2, World!\n");
  else
    printf("Hello2 from %s\n", msg);
}
```
user: What is the c source code of the assembly code below?

```asm
<sayhello>:
  endbr64
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L3
  mov    %rdi,%rsi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L17:
  add    $0x8,%rsp
  ret
L3:
  lea    D16(%rip),%rdi
  call   <puts@plt>
  jmp    L17
```

tool_calls: name: parse_data
arguments: tag: D15
datatype: string
name: parse_data
arguments: tag: D16
datatype: string

tool: D15: Hello2 from %s


tool: D16: Hello2, World!

assistant: ```c
void sayhello(const char *msg) {
  if (!msg)
    printf("Hello2, World!\n");
  else
    printf("Hello2 from %s\n", msg);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<sayhello>:
  endbr64
  mov    %rdi,%rsi
  test   %rdi,%rdi
  je     L4
  lea    D19(%rip),%rdi
  xor    %eax,%eax
  jmp    1070 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L4:
  lea    D10(%rip),%rdi
  jmp    1060 <puts@plt>
```

tool_calls: name: read_data
arguments: data_label: D19
datatype: string
name: read_data
arguments: data_label: D10
datatype: string

tool: D19: Hello2 from %s


tool: D10: Hello2, World!

assistant: ```c
void sayhello(const char *msg) {
  if (!msg)
    printf("Hello2, World!\n");
  else
    printf("Hello2 from %s\n", msg);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sayhello>:
  endbr64
  mov    %rdi,%rsi
  test   %rdi,%rdi
  je     L3
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  jmp    1070 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L3:
  lea    D15(%rip),%rdi
  jmp    1060 <puts@plt>
```

tool_calls: name: disassemble_data
arguments: label: D4
type: string
name: disassemble_data
arguments: label: D15
type: string

tool: D4: Hello2 from %s


tool: D15: Hello2, World!

assistant: ```c
void sayhello(const char *msg) {
  if (!msg)
    printf("Hello2, World!\n");
  else
    printf("Hello2 from %s\n", msg);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x24(%rbp)
  movl   $0x1,-0x20(%rbp)
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L28
L25:
  mov    -0x28(%rbp),%ebx
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  mov    $0x5,%edi
  call   <power@plt>
  mov    %eax,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  add    %eax,-0x24(%rbp)
  addl   $0x1,-0x20(%rbp)
L28:
  cmpl   $0x0,-0x1c(%rbp)
  jg     L25
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D14
type: string
name: retrieve_data
arguments: label: D12
type: string
name: retrieve_data
arguments: label: D28
type: string

tool: D14: Please enter n to know how many trailing zero there are in n!: 

tool: D12: %d

tool: D28: Trailing zeros - > %d


assistant: ```c
int main() {
  int n;
  printf("Please enter n to know how many trailing zero there are in n!: ");
  scanf("%d", &n);
  int trailingZeroesSum = 0, i = 1, result = n / 5;
  while (result > 0) {
    result = n / power(5, i);
    trailingZeroesSum += result;
    i++;
  }
  printf("Trailing zeros - > %d\n", trailingZeroesSum);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x4,0x4(%rsp)
  jle    L12
  mov    $0x1,%ebp
  mov    $0x0,%r12d
L24:
  mov    0x4(%rsp),%ebx
  mov    %ebp,%esi
  mov    $0x5,%edi
  call   <power@plt>
  mov    %eax,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  add    %eax,%r12d
  add    $0x1,%ebp
  test   %eax,%eax
  jg     L24
L32:
  mov    %r12d,%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    $0x0,%r12d
  jmp    L32
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
datatype: string
name: extract_symbolic_data
arguments: tag: D8
datatype: string
name: extract_symbolic_data
arguments: tag: D10
datatype: string

tool: D7: Please enter n to know how many trailing zero there are in n!: 

tool: D8: %d

tool: D10: Trailing zeros - > %d


assistant: ```c
int main() {
  int n;
  printf("Please enter n to know how many trailing zero there are in n!: ");
  scanf("%d", &n);
  int trailingZeroesSum = 0, i = 1, result = n / 5;
  while (result > 0) {
    result = n / power(5, i);
    trailingZeroesSum += result;
    i++;
  }
  printf("Trailing zeros - > %d\n", trailingZeroesSum);
  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   %r12
  lea    D13(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebx
  cmp    $0x4,%ebx
  jle    L25
  mov    $0x1,%ebp
  xor    %r12d,%r12d
  jmp    L10
  nopl   0x0(%rax)
L24:
  mov    0x4(%rsp),%ebx
L10:
  mov    %ebp,%esi
  mov    $0x5,%edi
  add    $0x1,%ebp
  call   <power@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  add    %eax,%r12d
  test   %eax,%eax
  jg     L24
L35:
  xor    %eax,%eax
  mov    %r12d,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  xor    %r12d,%r12d
  jmp    L35
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D13
type_guessed: string
name: unpack_data
arguments: tag: D22
type_guessed: string
name: unpack_data
arguments: tag: D7
type_guessed: string

tool: D13: Please enter n to know how many trailing zero there are in n!: 

tool: D22: %d

tool: D7: Trailing zeros - > %d


assistant: ```c
int main() {
  int n;
  printf("Please enter n to know how many trailing zero there are in n!: ");
  scanf("%d", &n);
  int trailingZeroesSum = 0, i = 1, result = n / 5;
  while (result > 0) {
    result = n / power(5, i);
    trailingZeroesSum += result;
    i++;
  }
  printf("Trailing zeros - > %d\n", trailingZeroesSum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  lea    D3(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebx
  cmp    $0x4,%ebx
  jle    L9
  mov    $0x1,%ebp
  xor    %r12d,%r12d
  jmp    L16
  nopl   0x0(%rax)
L15:
  mov    0x4(%rsp),%ebx
L16:
  mov    %ebp,%esi
  mov    $0x5,%edi
  add    $0x1,%ebp
  call   <power@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  add    %eax,%r12d
  test   %eax,%eax
  jg     L15
L47:
  xor    %eax,%eax
  mov    %r12d,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  xor    %r12d,%r12d
  jmp    L47
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
datatype: string
name: extract_symbolic_data
arguments: data_tag: D12
datatype: string
name: extract_symbolic_data
arguments: data_tag: D9
datatype: string

tool: D3: Please enter n to know how many trailing zero there are in n!: 

tool: D12: %d

tool: D9: Trailing zeros - > %d


assistant: ```c
int main() {
  int n;
  printf("Please enter n to know how many trailing zero there are in n!: ");
  scanf("%d", &n);
  int trailingZeroesSum = 0, i = 1, result = n / 5;
  while (result > 0) {
    result = n / power(5, i);
    trailingZeroesSum += result;
    i++;
  }
  printf("Trailing zeros - > %d\n", trailingZeroesSum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<set_tab_to_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L16
L39:
  movl   $0x0,-0x4(%rbp)
  jmp    L35
L28:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  addl   $0x1,-0x4(%rbp)
L35:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L28
  addl   $0x1,-0x8(%rbp)
L16:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L39
  mov    -0x20(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int **set_tab_to_0(int n, int p, int **tab) {
  int i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      tab[i][j] = 0;
    }
  }
  return tab;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<set_tab_to_0>:
  endbr64
  test   %edi,%edi
  jle    L21
  mov    %rdx,%r8
  lea    -0x1(%rdi),%eax
  lea    0x8(%rdx,%rax,8),%r9
  lea    -0x1(%rsi),%eax
  lea    0x4(,%rax,4),%rdi
  jmp    L34
L46:
  mov    (%r8),%rcx
  movl   $0x0,(%rcx,%rax,1)
  add    $0x4,%rax
  cmp    %rdi,%rax
  jne    L46
L2:
  add    $0x8,%r8
  cmp    %r9,%r8
  je     L21
L34:
  mov    $0x0,%eax
  test   %esi,%esi
  jg     L46
  jmp    L2
L21:
  mov    %rdx,%rax
  ret
```

assistant: ```c
int **set_tab_to_0(int n, int p, int **tab) {
  int i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      tab[i][j] = 0;
    }
  }
  return tab;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<set_tab_to_0>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L25
  lea    -0x1(%rdi),%eax
  mov    %esi,%ebp
  mov    %rdx,%rbx
  lea    0x8(%rdx,%rax,8),%r12
  lea    -0x1(%rsi),%eax
  lea    0x4(,%rax,4),%r13
  nopl   0x0(%rax,%rax,1)
L28:
  test   %ebp,%ebp
  jle    L10
  mov    (%rbx),%rdi
  mov    %r13,%rdx
  xor    %esi,%esi
  call   <memset@plt>
L10:
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L28
L25:
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int **set_tab_to_0(int n, int p, int **tab) {
  int i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      tab[i][j] = 0;
    }
  }
  return tab;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<set_tab_to_0>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edi,%edi
  jle    L8
  test   %esi,%esi
  jle    L8
  lea    -0x1(%rdi),%eax
  movslq %esi,%rsi
  mov    %rdx,%rbx
  lea    0x8(%rdx,%rax,8),%r12
  lea    0x0(,%rsi,4),%rbp
  nop
L24:
  mov    (%rbx),%rdi
  mov    %rbp,%rdx
  xor    %esi,%esi
  add    $0x8,%rbx
  call   <memset@plt>
  cmp    %rbx,%r12
  jne    L24
L8:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int **set_tab_to_0(int n, int p, int **tab) {
  int i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      tab[i][j] = 0;
    }
  }
  return tab;
}
```