ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<NumericsArrays_merge_test0>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x1,0x8(%rsp)
  movq   $0x2,0x10(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x1,0x28(%rsp)
  movq   $0x2,0x30(%rsp)
  movq   $0x63,0x40(%rsp)
  movq   $0x63,0x48(%rsp)
  movq   $0x63,0x50(%rsp)
  movq   $0x63,0x58(%rsp)
  movq   $0x63,0x60(%rsp)
  movq   $0x63,0x68(%rsp)
  lea    0x40(%rsp),%rbx
  lea    0x20(%rsp),%r13
  mov    %rsp,%r12
  mov    %rbx,%r8
  mov    $0x3,%ecx
  mov    $0x3,%edx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <NA_merge_sorted_arrays@plt>
  mov    $0x6,%esi
  mov    %rbx,%rdi
  call   <NA_display@plt>
  mov    $0x6,%esi
  mov    %rbx,%rdi
  call   <NA_rm_duplicate@plt>
  mov    %eax,%ebp
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <NA_display@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <NA_sort_bubble@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <NA_display@plt>
  mov    %rbx,%r8
  mov    $0x3,%ecx
  mov    $0x3,%edx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <NA_merge_and_sort_sorted_arrays@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <NA_display@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int NumericsArrays_merge_test0() {
  int info = 0;
  int n1 = 3;
  size_t arr1[] = {0, 1, 2};
  int n2 = 3;
  size_t arr2[] = {0, 1, 2};
  int n3 = 6;
  size_t arr3[] = {99, 99, 99, 99, 99, 99};
  NA_merge_sorted_arrays(arr1, arr2, n1, n2, arr3);
  NA_display(arr3, n3);
  int n_rm = NA_rm_duplicate(arr3, n3);
  NA_display(arr3, n_rm);
  NA_sort_bubble(arr3, n_rm);
  NA_display(arr3, n_rm);
  int n = NA_merge_and_sort_sorted_arrays(arr1, arr2, n1, n2, arr3);
  NA_display(arr3, n);
  return info;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<NumericsArrays_merge_test0>:
  endbr64
  push   %r14
  mov    $0x3,%ecx
  mov    $0x3,%edx
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rbp
  lea    0x20(%rsp),%r14
  mov    %rsp,%r13
  mov    %rbp,%r8
  mov    %r14,%rsi
  mov    %r13,%rdi
  movq   $0x0,(%rsp)
  movq   $0x1,0x8(%rsp)
  movq   $0x2,0x10(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x1,0x28(%rsp)
  movq   $0x2,0x30(%rsp)
  movq   $0x63,0x40(%rsp)
  movq   $0x63,0x48(%rsp)
  movq   $0x63,0x50(%rsp)
  movq   $0x63,0x58(%rsp)
  movq   $0x63,0x60(%rsp)
  movq   $0x63,0x68(%rsp)
  call   <NA_merge_sorted_arrays@plt>
  mov    $0x6,%esi
  mov    %rbp,%rdi
  call   <NA_display@plt>
  mov    $0x6,%esi
  mov    %rbp,%rdi
  call   <NA_rm_duplicate@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  mov    %eax,%esi
  call   <NA_display@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <NA_sort_bubble@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <NA_display@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    %rbp,%r8
  mov    $0x3,%ecx
  mov    $0x3,%edx
  call   <NA_merge_and_sort_sorted_arrays@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <NA_display@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int NumericsArrays_merge_test0() {
  int info = 0;
  int n1 = 3;
  size_t arr1[] = {0, 1, 2};
  int n2 = 3;
  size_t arr2[] = {0, 1, 2};
  int n3 = 6;
  size_t arr3[] = {99, 99, 99, 99, 99, 99};
  NA_merge_sorted_arrays(arr1, arr2, n1, n2, arr3);
  NA_display(arr3, n3);
  int n_rm = NA_rm_duplicate(arr3, n3);
  NA_display(arr3, n_rm);
  NA_sort_bubble(arr3, n_rm);
  NA_display(arr3, n_rm);
  int n = NA_merge_and_sort_sorted_arrays(arr1, arr2, n1, n2, arr3);
  NA_display(arr3, n);
  return info;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<NumericsArrays_merge_test0>:
  endbr64
  push   %r14
  mov    $0x3,%ecx
  mov    $0x3,%edx
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x88,%rsp
  movdqa D13(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rbp
  lea    0x20(%rsp),%r14
  mov    %rsp,%r13
  movaps %xmm0,(%rsp)
  mov    %rbp,%r8
  mov    %r14,%rsi
  mov    %r13,%rdi
  movaps %xmm0,0x20(%rsp)
  movdqa D5(%rip),%xmm0
  movq   $0x2,0x10(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movq   $0x2,0x30(%rsp)
  call   <NA_merge_sorted_arrays@plt>
  mov    $0x6,%esi
  mov    %rbp,%rdi
  call   <NA_display@plt>
  mov    $0x6,%esi
  mov    %rbp,%rdi
  call   <NA_rm_duplicate@plt>
  mov    %rbp,%rdi
  mov    %eax,%r12d
  mov    %eax,%esi
  call   <NA_display@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <NA_sort_bubble@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <NA_display@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    %rbp,%r8
  mov    $0x3,%ecx
  mov    $0x3,%edx
  call   <NA_merge_and_sort_sorted_arrays@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <NA_display@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D13
data_type: byte[16]
name: disassemble_data
arguments: label: D5
data_type: byte[16]

tool: {"D13": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D5": [0x63,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x63,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
int NumericsArrays_merge_test0() {
  int info = 0;
  int n1 = 3;
  size_t arr1[] = {0, 1, 2};
  int n2 = 3;
  size_t arr2[] = {0, 1, 2};
  int n3 = 6;
  size_t arr3[] = {99, 99, 99, 99, 99, 99};
  NA_merge_sorted_arrays(arr1, arr2, n1, n2, arr3);
  NA_display(arr3, n3);
  int n_rm = NA_rm_duplicate(arr3, n3);
  NA_display(arr3, n_rm);
  NA_sort_bubble(arr3, n_rm);
  NA_display(arr3, n_rm);
  int n = NA_merge_and_sort_sorted_arrays(arr1, arr2, n1, n2, arr3);
  NA_display(arr3, n);
  return info;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getvalue>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x2f,-0x4(%rbp)
  jle    L8
  cmpb   $0x39,-0x4(%rbp)
  jg     L8
  movsbl -0x4(%rbp),%eax
  sub    $0x30,%eax
  jmp    L17
L8:
  cmpb   $0x40,-0x4(%rbp)
  jle    L6
  cmpb   $0x46,-0x4(%rbp)
  jg     L6
  movsbl -0x4(%rbp),%eax
  sub    $0x37,%eax
  jmp    L17
L6:
  mov    $0xffffffff,%eax
L17:
  pop    %rbp
  ret
```

assistant: ```c
int getvalue(char digit) {
  if (digit >= '0' && digit <= '9')
    return digit - '0';
  if (digit >= 'A' && digit <= 'F')
    return digit - 'A' + 10;
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<getvalue>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L0
  lea    -0x41(%rdi),%edx
  movsbl %dil,%eax
  sub    $0x37,%eax
  cmp    $0x6,%dl
  mov    $0xffffffff,%edx
  cmovae %edx,%eax
  ret
L0:
  movsbl %dil,%edi
  lea    -0x30(%rdi),%eax
  ret
```

assistant: ```c
int getvalue(char digit) {
  if (digit >= '0' && digit <= '9')
    return digit - '0';
  if (digit >= 'A' && digit <= 'F')
    return digit - 'A' + 10;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getvalue>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L7
  lea    -0x41(%rdi),%edx
  lea    -0x37(%rdi),%eax
  cmp    $0x6,%dl
  movsbl %al,%eax
  mov    $0xffffffff,%edx
  cmovae %edx,%eax
  ret
L7:
  movsbl %al,%eax
  ret
```

assistant: ```c
int getvalue(char digit) {
  if (digit >= '0' && digit <= '9')
    return digit - '0';
  if (digit >= 'A' && digit <= 'F')
    return digit - 'A' + 10;
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<getvalue>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L1
  lea    -0x41(%rdi),%edx
  lea    -0x37(%rdi),%eax
  cmp    $0x6,%dl
  movsbl %al,%eax
  mov    $0xffffffff,%edx
  cmovae %edx,%eax
  ret
L1:
  movsbl %al,%eax
  ret
```

assistant: ```c
int getvalue(char digit) {
  if (digit >= '0' && digit <= '9')
    return digit - '0';
  if (digit >= 'A' && digit <= 'F')
    return digit - 'A' + 10;
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bzero3_char_ptr_null>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D1(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x4,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <clang_analyzer_eval@plt>
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <clang_analyzer_eval@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D1
data_type: string

tool: D1: abcd

assistant: ```c
void bzero3_char_ptr_null() {
  char *str = "abcd";
  clang_analyzer_eval(strlen(str) == 4);
  bzero(str + 2, 2);
  clang_analyzer_eval(strlen(str) == 0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<bzero3_char_ptr_null>:
  endbr64
  push   %rbx
  lea    D6(%rip),%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    $0x4,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <clang_analyzer_eval@plt>
  mov    $0x2,%esi
  lea    0x2(%rbx),%rdi
  call   <bzero@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <clang_analyzer_eval@plt>
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_infered: string

tool: D6: abcd

assistant: ```c
void bzero3_char_ptr_null() {
  char *str = "abcd";
  clang_analyzer_eval(strlen(str) == 4);
  bzero(str + 2, 2);
  clang_analyzer_eval(strlen(str) == 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bzero3_char_ptr_null>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %edi,%edi
  cmp    $0x4,%eax
  sete   %dil
  call   <clang_analyzer_eval@plt>
  lea    0x2(%rbp),%rdi
  mov    $0x2,%esi
  call   <bzero@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %edi,%edi
  pop    %rbp
  test   %eax,%eax
  sete   %dil
  jmp    1090 <clang_analyzer_eval@plt>
```

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

tool: D2: abcd

assistant: ```c
void bzero3_char_ptr_null() {
  char *str = "abcd";
  clang_analyzer_eval(strlen(str) == 4);
  bzero(str + 2, 2);
  clang_analyzer_eval(strlen(str) == 0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<bzero3_char_ptr_null>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %edi,%edi
  cmp    $0x4,%eax
  sete   %dil
  call   <clang_analyzer_eval@plt>
  lea    0x2(%rbp),%rdi
  mov    $0x2,%esi
  call   <bzero@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %edi,%edi
  pop    %rbp
  test   %eax,%eax
  sete   %dil
  jmp    1090 <clang_analyzer_eval@plt>
```

tool_calls: name: parse_data
arguments: data_label: D4
type_guessed: string

tool: D4: abcd

assistant: ```c
void bzero3_char_ptr_null() {
  char *str = "abcd";
  clang_analyzer_eval(strlen(str) == 4);
  bzero(str + 2, 2);
  clang_analyzer_eval(strlen(str) == 0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<seq_gen_rndgamma>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  comisd -0x18(%rbp),%xmm0
  jb     L4
  pxor   %xmm0,%xmm0
  jmp    L47
L4:
  movsd  D10(%rip),%xmm0
  comisd -0x18(%rbp),%xmm0
  jbe    L38
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <rndgamma1@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L1
L38:
  movsd  -0x18(%rbp),%xmm0
  movsd  D10(%rip),%xmm1
  comisd %xmm1,%xmm0
  jbe    L22
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <rndgamma2@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L1
L22:
  mov    $0x0,%eax
  call   <seq_gen_rndu@plt>
  mov    %eax,%edi
  call   <log@plt>
  movq   %xmm0,%rax
  movq   D6(%rip),%xmm0
  movq   %rax,%xmm2
  xorpd  %xmm0,%xmm2
  movapd %xmm2,%xmm0
  movsd  %xmm0,-0x8(%rbp)
L1:
  movsd  -0x8(%rbp),%xmm0
L47:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D10
data_type: f64
name: disassemble_data
arguments: label: D6
data_type: f32[2]

tool: D10: 1

tool: {"D6": [0.0,-0.0]}

assistant: ```c
double seq_gen_rndgamma(double s) {
  double r = 0.0;
  if (s <= 0.0)
    return 0;
  else if (s < 1.0)
    r = rndgamma1(s);
  else if (s > 1.0)
    r = rndgamma2(s);
  else
    r = -log(seq_gen_rndu());
  return (r);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<seq_gen_rndgamma>:
  endbr64
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  jae    L19
  sub    $0x8,%rsp
  movsd  D5(%rip),%xmm1
  comisd %xmm0,%xmm1
  ja     L0
  comisd D5(%rip),%xmm0
  jbe    L30
  call   <rndgamma2@plt>
L10:
  add    $0x8,%rsp
  ret
L0:
  call   <rndgamma1@plt>
  jmp    L10
L30:
  mov    $0x0,%eax
  call   <seq_gen_rndu@plt>
  mov    %eax,%edi
  call   <log@plt>
  xorpd  D24(%rip),%xmm0
  jmp    L10
L19:
  pxor   %xmm0,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: label: D5
data_type: f64[2]
name: unpack_data
arguments: label: D24
data_type: f32[4]

tool: {"D5": [1.0,0.0]}

tool: {"D24": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double seq_gen_rndgamma(double s) {
  double r = 0.0;
  if (s <= 0.0)
    return 0;
  else if (s < 1.0)
    r = rndgamma1(s);
  else if (s > 1.0)
    r = rndgamma2(s);
  else
    r = -log(seq_gen_rndu());
  return (r);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<seq_gen_rndgamma>:
  endbr64
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  jae    L23
  movsd  D1(%rip),%xmm1
  comisd %xmm0,%xmm1
  ja     L24
  comisd %xmm1,%xmm0
  jbe    L28
  jmp    10b0 <rndgamma2@plt>
  nopw   0x0(%rax,%rax,1)
L28:
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <seq_gen_rndu@plt>
  mov    %eax,%edi
  call   <log@plt>
  xorpd  D11(%rip),%xmm0
  add    $0x8,%rsp
  ret
  nop
L23:
  movapd %xmm1,%xmm0
  ret
  nopl   (%rax)
L24:
  jmp    1090 <rndgamma1@plt>
```

tool_calls: name: parse_data
arguments: data_label: D1
data_type: f64
name: parse_data
arguments: data_label: D11
data_type: f32[4]

tool: D1: 1

tool: {"D11": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double seq_gen_rndgamma(double s) {
  double r = 0.0;
  if (s <= 0.0)
    return 0;
  else if (s < 1.0)
    r = rndgamma1(s);
  else if (s > 1.0)
    r = rndgamma2(s);
  else
    r = -log(seq_gen_rndu());
  return (r);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<seq_gen_rndgamma>:
  endbr64
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  jae    L29
  movsd  D4(%rip),%xmm1
  comisd %xmm0,%xmm1
  ja     L15
  comisd %xmm1,%xmm0
  jbe    L16
  jmp    10b0 <rndgamma2@plt>
  nopw   0x0(%rax,%rax,1)
L16:
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <seq_gen_rndu@plt>
  mov    %eax,%edi
  call   <log@plt>
  xorpd  D2(%rip),%xmm0
  add    $0x8,%rsp
  ret
  nop
L29:
  movapd %xmm1,%xmm0
  ret
  nopl   (%rax)
L15:
  jmp    1090 <rndgamma1@plt>
```

tool_calls: name: read_data
arguments: tag: D4
type: f64
name: read_data
arguments: tag: D2
type: f32[4]

tool: D4: 1

tool: {"D2": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double seq_gen_rndgamma(double s) {
  double r = 0.0;
  if (s <= 0.0)
    return 0;
  else if (s < 1.0)
    r = rndgamma1(s);
  else if (s > 1.0)
    r = rndgamma2(s);
  else
    r = -log(seq_gen_rndu());
  return (r);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L1
L12:
  mov    -0x4(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  subl   $0x1,-0x14(%rbp)
L1:
  cmpl   $0x1,-0x14(%rbp)
  jg     L12
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int fat(int n) {
  int fatorial;
  for (fatorial = 1; n > 1; n--) {
    fatorial = fatorial * n;
  }
  return fatorial;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fat>:
  endbr64
  cmp    $0x1,%edi
  jle    L13
  mov    $0x1,%eax
L0:
  imul   %edi,%eax
  sub    $0x1,%edi
  cmp    $0x1,%edi
  jne    L0
  ret
L13:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int fat(int n) {
  int fatorial;
  for (fatorial = 1; n > 1; n--) {
    fatorial = fatorial * n;
  }
  return fatorial;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fat>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jle    L0
  xchg   %ax,%ax
L4:
  imul   %edi,%eax
  sub    $0x1,%edi
  cmp    $0x1,%edi
  jne    L4
  ret
  nopl   0x0(%rax)
L0:
  ret
```

assistant: ```c
int fat(int n) {
  int fatorial;
  for (fatorial = 1; n > 1; n--) {
    fatorial = fatorial * n;
  }
  return fatorial;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fat>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jle    L12
  xchg   %ax,%ax
L11:
  imul   %edi,%eax
  sub    $0x1,%edi
  cmp    $0x1,%edi
  jne    L11
  ret
  nopl   0x0(%rax)
L12:
  ret
```

assistant: ```c
int fat(int n) {
  int fatorial;
  for (fatorial = 1; n > 1; n--) {
    fatorial = fatorial * n;
  }
  return fatorial;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L49
  negl   -0x4(%rbp)
L49:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    %ecx,%eax
  imul   %edx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L49
  cmpl   $0x0,-0x4(%rbp)
  jns    L43
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
  jmp    L43
L11:
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
L43:
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  test   %eax,%eax
  jne    L11
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <reverse@plt>
  nop
  leave
  ret
```

assistant: ```c
void itoa(int n, char s[], int width) {
  int i = 0, sign = 1;
  if (n < 0)
    sign *= -1;
  do {
    s[i++] = sign * (n % 10) + '0';
  } while ((n /= 10) != 0);
  if (sign < 0)
    s[i++] = '-';
  while (width--)
    s[i++] = ' ';
  s[i] = '\0';
  reverse(s);
}
```
user: Translate the below assembly code into C programming language:

```asm
<itoa>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%eax
  mov    %rsi,%r9
  mov    %edx,%r10d
  sar    $0x1f,%edi
  or     $0x1,%edi
  mov    $0x0,%ecx
L42:
  mov    %ecx,%r8d
  add    $0x1,%ecx
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%r11d
  sar    $0x1f,%r11d
  sub    %r11d,%edx
  lea    (%rdx,%rdx,4),%r11d
  add    %r11d,%r11d
  sub    %r11d,%eax
  imul   %edi,%eax
  add    $0x30,%eax
  mov    %al,(%rsi)
  mov    %edx,%eax
  add    $0x1,%rsi
  test   %edx,%edx
  jne    L42
  test   %edi,%edi
  js     L7
L35:
  lea    -0x1(%r10),%edi
  test   %r10d,%r10d
  je     L16
  movslq %ecx,%rsi
  lea    (%r9,%rsi,1),%rax
  lea    -0x1(%r10),%edx
  lea    0x1(%r9,%rdx,1),%rdx
  add    %rsi,%rdx
L8:
  movb   $0x20,(%rax)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L8
  lea    0x1(%rcx,%rdi,1),%ecx
L16:
  movslq %ecx,%rcx
  movb   $0x0,(%r9,%rcx,1)
  mov    %r9,%rdi
  call   <reverse@plt>
  add    $0x8,%rsp
  ret
L7:
  movslq %ecx,%rcx
  movb   $0x2d,(%r9,%rcx,1)
  lea    0x2(%r8),%ecx
  jmp    L35
```

assistant: ```c
void itoa(int n, char s[], int width) {
  int i = 0, sign = 1;
  if (n < 0)
    sign *= -1;
  do {
    s[i++] = sign * (n % 10) + '0';
  } while ((n /= 10) != 0);
  if (sign < 0)
    s[i++] = '-';
  while (width--)
    s[i++] = ' ';
  s[i] = '\0';
  reverse(s);
}
```
user: Translate the below assembly code into C programming language:

```asm
<itoa>:
  endbr64
  mov    %edi,%eax
  sar    $0x1f,%edi
  push   %r12
  mov    %edx,%r9d
  mov    %eax,%r10d
  push   %rbp
  or     $0x1,%edi
  mov    %rsi,%rbp
  sar    $0x1f,%r10d
  push   %rbx
  xor    %ebx,%ebx
  or     $0x1,%r10d
  nopl   0x0(%rax,%rax,1)
L0:
  movslq %eax,%rdx
  mov    %eax,%ecx
  mov    %ebx,%r8d
  add    $0x1,%rsi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  add    $0x1,%ebx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%r11d
  add    %r11d,%r11d
  sub    %r11d,%eax
  imul   %edi,%eax
  add    $0x30,%eax
  mov    %al,-0x1(%rsi)
  mov    %edx,%eax
  test   %edx,%edx
  jne    L0
  cmp    $0xffffffff,%r10d
  jne    L18
  movslq %ebx,%rbx
  movb   $0x2d,0x0(%rbp,%rbx,1)
  lea    0x2(%r8),%ebx
L18:
  lea    -0x1(%r9),%r12d
  test   %r9d,%r9d
  je     L27
  movslq %ebx,%rdi
  mov    %r12d,%edx
  lea    0x1(%rbx,%r12,1),%ebx
  mov    $0x20,%esi
  add    $0x1,%rdx
  add    %rbp,%rdi
  call   <memset@plt>
L27:
  movslq %ebx,%rbx
  mov    %rbp,%rdi
  movb   $0x0,0x0(%rbp,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1060 <reverse@plt>
```

assistant: ```c
void itoa(int n, char s[], int width) {
  int i = 0, sign = 1;
  if (n < 0)
    sign *= -1;
  do {
    s[i++] = sign * (n % 10) + '0';
  } while ((n /= 10) != 0);
  if (sign < 0)
    s[i++] = '-';
  while (width--)
    s[i++] = ' ';
  s[i] = '\0';
  reverse(s);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<itoa>:
  endbr64
  mov    %edi,%eax
  sar    $0x1f,%edi
  push   %r12
  mov    %edx,%r9d
  mov    %eax,%r10d
  push   %rbp
  or     $0x1,%edi
  mov    %rsi,%rbp
  sar    $0x1f,%r10d
  push   %rbx
  xor    %ebx,%ebx
  or     $0x1,%r10d
  nopl   0x0(%rax,%rax,1)
L27:
  movslq %eax,%rdx
  mov    %eax,%ecx
  mov    %ebx,%r8d
  add    $0x1,%rsi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  add    $0x1,%ebx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%r11d
  add    %r11d,%r11d
  sub    %r11d,%eax
  imul   %edi,%eax
  add    $0x30,%eax
  mov    %al,-0x1(%rsi)
  mov    %edx,%eax
  test   %edx,%edx
  jne    L27
  cmp    $0xffffffff,%r10d
  jne    L18
  movslq %ebx,%rbx
  movb   $0x2d,0x0(%rbp,%rbx,1)
  lea    0x2(%r8),%ebx
L18:
  test   %r9d,%r9d
  je     L16
  lea    -0x1(%r9),%edx
  movslq %ebx,%rdi
  mov    $0x20,%esi
  mov    %rdx,%r12
  add    %rbp,%rdi
  add    $0x1,%rdx
  call   <memset@plt>
  lea    0x1(%rbx,%r12,1),%ebx
L16:
  movslq %ebx,%rbx
  mov    %rbp,%rdi
  movb   $0x0,0x0(%rbp,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1060 <reverse@plt>
```

assistant: ```c
void itoa(int n, char s[], int width) {
  int i = 0, sign = 1;
  if (n < 0)
    sign *= -1;
  do {
    s[i++] = sign * (n % 10) + '0';
  } while ((n /= 10) != 0);
  if (sign < 0)
    s[i++] = '-';
  while (width--)
    s[i++] = ' ';
  s[i] = '\0';
  reverse(s);
}
```