ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<left_count>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edx,0xc(%rsp)
  lea    -0x1(%rdx),%eax
  lea    -0x1(%rsi),%ebx
  test   %esi,%esi
  cmove  %eax,%ebx
  movslq %ebx,%rax
  movzbl (%rdi,%rax,1),%r12d
  lea    -0x1(%rdx),%r15d
  test   %r15d,%r15d
  jle    L63
  mov    %rdi,%r13
  mov    $0x0,%r14d
  mov    $0x1,%ebp
  jmp    L22
L67:
  cltq
  movzbl 0x0(%r13,%rax,1),%r12d
  cmp    $0x77,%r12b
  setne  %r14b
  movzbl %r14b,%r14d
L70:
  movslq %ebx,%rax
  movsbl 0x0(%r13,%rax,1),%esi
  movsbl %r12b,%edi
  call   <is_same_color@plt>
  test   %rax,%rax
  je     L62
  add    $0x1,%ebp
  cmp    %ebp,0xc(%rsp)
  je     L9
L22:
  lea    -0x1(%rbx),%eax
  test   %ebx,%ebx
  cmove  %r15d,%eax
  mov    %eax,%ebx
  test   %r14d,%r14d
  jne    L70
  cmp    $0x77,%r12b
  je     L67
  jmp    L70
L9:
  mov    0xc(%rsp),%ebp
L62:
  mov    %ebp,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L63:
  mov    $0x1,%ebp
  jmp    L62
```

assistant: ```c
int left_count(char necklace[], int point, int len) {
  int i, count = 1;
  int left_point;
  char original_color;
  int is_change = 0;
  if (point == 0)
    point = len - 1;
  else
    point--;
  original_color = necklace[point];
  for (i = 0; i < len - 1; i++) {
    left_point = point == 0 ? len - 1 : point - 1;
    if (!is_change && original_color == 'w' && necklace[left_point] != 'w') {
      original_color = necklace[left_point];
      is_change = 1;
    }
    if (is_same_color(original_color, necklace[left_point]))
      count++;
    else
      break;
    point = left_point;
  }
  return count;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<left_count>:
  endbr64
  push   %r15
  push   %r14
  lea    -0x1(%rsi),%r14d
  push   %r13
  lea    -0x1(%rdx),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %esi,%esi
  cmove  %r13d,%r14d
  mov    %edx,0xc(%rsp)
  movslq %r14d,%rax
  movzbl (%rdi,%rax,1),%r12d
  test   %r13d,%r13d
  jle    L4
  mov    %rdi,%rbp
  xor    %ebx,%ebx
  mov    $0x1,%r15d
  jmp    L10
  xchg   %ax,%ax
L41:
  add    $0x1,%r15d
  cmp    %r15d,0xc(%rsp)
  je     L18
L10:
  lea    -0x1(%r14),%eax
  test   %r14d,%r14d
  cmove  %r13d,%eax
  mov    %eax,%r14d
  cltq
  movsbl 0x0(%rbp,%rax,1),%esi
  test   %ebx,%ebx
  jne    L64
  cmp    $0x77,%r12b
  jne    L64
  xor    %ebx,%ebx
  cmp    $0x77,%sil
  mov    %esi,%r12d
  setne  %bl
L64:
  movsbl %r12b,%edi
  call   <is_same_color@plt>
  test   %rax,%rax
  jne    L41
L18:
  add    $0x18,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  mov    $0x1,%r15d
  jmp    L18
```

assistant: ```c
int left_count(char necklace[], int point, int len) {
  int i, count = 1;
  int left_point;
  char original_color;
  int is_change = 0;
  if (point == 0)
    point = len - 1;
  else
    point--;
  original_color = necklace[point];
  for (i = 0; i < len - 1; i++) {
    left_point = point == 0 ? len - 1 : point - 1;
    if (!is_change && original_color == 'w' && necklace[left_point] != 'w') {
      original_color = necklace[left_point];
      is_change = 1;
    }
    if (is_same_color(original_color, necklace[left_point]))
      count++;
    else
      break;
    point = left_point;
  }
  return count;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<left_count>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    -0x1(%rdx),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x1(%rsi),%ebx
  sub    $0x18,%rsp
  test   %esi,%esi
  cmove  %r13d,%ebx
  movslq %ebx,%rax
  movzbl (%rdi,%rax,1),%r8d
  test   %r13d,%r13d
  jle    L88
  movl   $0x0,0xc(%rsp)
  mov    %rdi,%r14
  mov    %edx,%r15d
  mov    $0x1,%r12d
  nopl   0x0(%rax)
L86:
  test   %ebx,%ebx
  lea    -0x1(%rbx),%eax
  movzbl 0xc(%rsp),%ebp
  cmove  %r13d,%eax
  xor    $0x1,%ebp
  cmp    $0x77,%r8b
  movslq %eax,%rcx
  sete   %al
  and    %al,%bpl
  movsbl (%r14,%rcx,1),%esi
  mov    %rcx,%rbx
  mov    %bpl,0xb(%rsp)
  je     L67
  xor    %eax,%eax
  cmp    $0x77,%sil
  mov    %esi,%r8d
  setne  %al
  mov    %eax,0xc(%rsp)
L67:
  movsbl %r8b,%ebp
  mov    %r8b,0xa(%rsp)
  mov    %ebp,%edi
  call   <is_same_color@plt>
  test   %rax,%rax
  je     L48
  add    $0x1,%r12d
  movzbl 0xa(%rsp),%r8d
  cmp    %r12d,%r15d
  je     L95
  cmpb   $0x0,0xb(%rsp)
  jne    L86
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L44:
  lea    0x1(%r12),%r15d
  cmp    %r13d,%r12d
  jge    L95
  mov    %r15d,%r12d
L4:
  test   %ebx,%ebx
  lea    -0x1(%rbx),%eax
  mov    %ebp,%edi
  cmove  %r13d,%eax
  mov    %eax,%ebx
  cltq
  movsbl (%r14,%rax,1),%esi
  call   <is_same_color@plt>
  test   %rax,%rax
  jne    L44
L48:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L95:
  mov    %r15d,%r12d
  jmp    L48
L88:
  mov    $0x1,%r12d
  jmp    L48
```

assistant: ```c
int left_count(char necklace[], int point, int len) {
  int i, count = 1;
  int left_point;
  char original_color;
  int is_change = 0;
  if (point == 0)
    point = len - 1;
  else
    point--;
  original_color = necklace[point];
  for (i = 0; i < len - 1; i++) {
    left_point = point == 0 ? len - 1 : point - 1;
    if (!is_change && original_color == 'w' && necklace[left_point] != 'w') {
      original_color = necklace[left_point];
      is_change = 1;
    }
    if (is_same_color(original_color, necklace[left_point]))
      count++;
    else
      break;
    point = left_point;
  }
  return count;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L12:
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    (%rax),%rax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L19:
  cmpl   $0x3,-0x4(%rbp)
  jle    L12
  nop
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D9
datatype: string

tool: D9: %s


assistant: ```c
void test(char **str) {
  int i;
  for (i = 0; i < 4; i++) {
    printf("%s\n", *str++);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<test>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x20(%rdi),%r12
  lea    D5(%rip),%rbp
L0:
  add    $0x8,%rbx
  mov    -0x8(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_label: D5
type_infered: string

tool: D5: %s


assistant: ```c
void test(char **str) {
  int i;
  for (i = 0; i < 4; i++) {
    printf("%s\n", *str++);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<test>:
  endbr64
  push   %r12
  lea    0x20(%rdi),%r12
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
L4:
  mov    (%rbx),%esi
  add    $0x8,%rbx
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D5
data_type: string

tool: D5: %s


assistant: ```c
void test(char **str) {
  int i;
  for (i = 0; i < 4; i++) {
    printf("%s\n", *str++);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<test>:
  endbr64
  push   %r12
  lea    0x20(%rdi),%r12
  push   %rbp
  lea    D1(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
L7:
  mov    (%rbx),%esi
  add    $0x8,%rbx
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L7
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D1
type: string

tool: D1: %s


assistant: ```c
void test(char **str) {
  int i;
  for (i = 0; i < 4; i++) {
    printf("%s\n", *str++);
  }
}
```
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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L15
L0:
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jne    L33
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  je     L39
L33:
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <calculate@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    -0x10(%rbp),%rcx
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L15:
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L0
  jmp    L8
L39:
  nop
L8:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type: string
name: disassemble_data
arguments: data_tag: D5
type: string

tool: D1: %i things taken %i at a time is %lld exactly.


tool: D5: %i %i

assistant: ```c
int main() {
  int n, m;
  while (scanf("%i %i", &n, &m)) {
    if (n == 0 && m == 0) {
      break;
    }
    long long int c = calculate(n, m);
    printf("%i things taken %i at a time is %lld exactly.\n", n, m, c);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  lea    D14(%rip),%rbx
  lea    D17(%rip),%r13
  jmp    L1
L39:
  call   <calculate@plt>
  mov    %rax,%rcx
  mov    0x4(%rsp),%edx
  mov    (%rsp),%esi
  mov    %r13,%rdi
  call   <printf@plt>
L1:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L23
  mov    (%rsp),%edi
  mov    0x4(%rsp),%esi
  mov    %edi,%eax
  or     %esi,%eax
  jne    L39
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D14
datatype: string
name: parse_data
arguments: label: D17
datatype: string

tool: D14: %i %i

tool: D17: %i things taken %i at a time is %lld exactly.


assistant: ```c
int main() {
  int n, m;
  while (scanf("%i %i", &n, &m)) {
    if (n == 0 && m == 0) {
      break;
    }
    long long int c = calculate(n, m);
    printf("%i things taken %i at a time is %lld exactly.\n", n, m, c);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D1(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  jmp    L30
  cs nopw 0x0(%rax,%rax,1)
L13:
  mov    (%rsp),%edi
  mov    0x4(%rsp),%esi
  mov    %edi,%eax
  or     %esi,%eax
  je     L28
  call   <calculate@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%esi
  mov    %r13,%rdi
  mov    %rax,%rcx
  call   <printf@plt>
L30:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L13
L28:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type_infered: string
name: extract_symbolic_data
arguments: data_label: D1
type_infered: string

tool: D0: %i things taken %i at a time is %lld exactly.


tool: D1: %i %i

assistant: ```c
int main() {
  int n, m;
  while (scanf("%i %i", &n, &m)) {
    if (n == 0 && m == 0) {
      break;
    }
    long long int c = calculate(n, m);
    printf("%i things taken %i at a time is %lld exactly.\n", n, m, c);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D16(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D11(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  jmp    L13
  cs nopw 0x0(%rax,%rax,1)
L31:
  mov    (%rsp),%edi
  mov    0x4(%rsp),%esi
  mov    %edi,%eax
  or     %esi,%eax
  je     L6
  call   <calculate@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%esi
  mov    %r13,%rdi
  mov    %rax,%rcx
  call   <printf@plt>
L13:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L31
L6:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D16
type_guessed: string
name: parse_data
arguments: tag: D11
type_guessed: string

tool: D16: %i things taken %i at a time is %lld exactly.


tool: D11: %i %i

assistant: ```c
int main() {
  int n, m;
  while (scanf("%i %i", &n, &m)) {
    if (n == 0 && m == 0) {
      break;
    }
    long long int c = calculate(n, m);
    printf("%i things taken %i at a time is %lld exactly.\n", n, m, c);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wrefresh>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <wnoutrefresh@plt>
  mov    $0x0,%eax
  call   <doupdate@plt>
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  leave
  ret
```

assistant: ```c
int wrefresh(WINDOW *win) {
  wnoutrefresh(win);
  doupdate();
  return OK;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<wrefresh>:
  endbr64
  sub    $0x8,%rsp
  call   <wnoutrefresh@plt>
  mov    $0x0,%eax
  call   <doupdate@plt>
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int wrefresh(WINDOW *win) {
  wnoutrefresh(win);
  doupdate();
  return OK;
}
```
user: Translate the below assembly code into C programming language:

```asm
<wrefresh>:
  endbr64
  sub    $0x8,%rsp
  call   <wnoutrefresh@plt>
  xor    %eax,%eax
  call   <doupdate@plt>
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int wrefresh(WINDOW *win) {
  wnoutrefresh(win);
  doupdate();
  return OK;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<wrefresh>:
  endbr64
  sub    $0x8,%rsp
  call   <wnoutrefresh@plt>
  xor    %eax,%eax
  call   <doupdate@plt>
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int wrefresh(WINDOW *win) {
  wnoutrefresh(win);
  doupdate();
  return OK;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<who_win>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jl     L18
  mov    $0x41,%eax
  jmp    L27
L18:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x8(%rbp),%rax
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L6
  mov    $0x42,%eax
  jmp    L27
L6:
  mov    $0x41,%eax
L27:
  pop    %rbp
  ret
```

assistant: ```c
char who_win(long long n, long long k) {
  if (k >= n)
    return 'A';
  else
    return n % (k + 1) == 0 ? 'B' : 'A';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<who_win>:
  endbr64
  mov    %rdi,%rax
  mov    $0x41,%edx
  cmp    %rdi,%rsi
  jge    L8
  add    $0x1,%rsi
  cqto
  idiv   %rsi
  cmp    $0x1,%rdx
  mov    $0x41,%edx
  adc    $0x0,%dl
L8:
  mov    %edx,%eax
  ret
```

assistant: ```c
char who_win(long long n, long long k) {
  if (k >= n)
    return 'A';
  else
    return n % (k + 1) == 0 ? 'B' : 'A';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<who_win>:
  endbr64
  mov    %rdi,%rax
  mov    $0x41,%r8d
  cmp    %rdi,%rsi
  jge    L7
  add    $0x1,%rsi
  cqto
  idiv   %rsi
  cmp    $0x1,%rdx
  adc    $0x0,%r8b
L7:
  mov    %r8d,%eax
  ret
```

assistant: ```c
char who_win(long long n, long long k) {
  if (k >= n)
    return 'A';
  else
    return n % (k + 1) == 0 ? 'B' : 'A';
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<who_win>:
  endbr64
  mov    %rdi,%rax
  mov    $0x41,%r8d
  cmp    %rdi,%rsi
  jge    L0
  add    $0x1,%rsi
  cqto
  idiv   %rsi
  cmp    $0x1,%rdx
  adc    $0x0,%r8b
L0:
  mov    %r8d,%eax
  ret
```

assistant: ```c
char who_win(long long n, long long k) {
  if (k >= n)
    return 'A';
  else
    return n % (k + 1) == 0 ? 'B' : 'A';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<EnterBackground>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <RestoreKeyRepeat@plt>
  mov    $0x0,%eax
  call   <DisconnectKeyCodes3@plt>
  mov    $0x0,%eax
  call   <ForceShowCursor@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void EnterBackground(void) {
  RestoreKeyRepeat();
  DisconnectKeyCodes3();
  ForceShowCursor();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<EnterBackground>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <RestoreKeyRepeat@plt>
  mov    $0x0,%eax
  call   <DisconnectKeyCodes3@plt>
  mov    $0x0,%eax
  call   <ForceShowCursor@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void EnterBackground(void) {
  RestoreKeyRepeat();
  DisconnectKeyCodes3();
  ForceShowCursor();
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<EnterBackground>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <RestoreKeyRepeat@plt>
  xor    %eax,%eax
  call   <DisconnectKeyCodes3@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <ForceShowCursor@plt>
```

assistant: ```c
void EnterBackground(void) {
  RestoreKeyRepeat();
  DisconnectKeyCodes3();
  ForceShowCursor();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<EnterBackground>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <RestoreKeyRepeat@plt>
  xor    %eax,%eax
  call   <DisconnectKeyCodes3@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <ForceShowCursor@plt>
```

assistant: ```c
void EnterBackground(void) {
  RestoreKeyRepeat();
  DisconnectKeyCodes3();
  ForceShowCursor();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<PutSignedInt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %al,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jns    L76
  mov    -0x24(%rbp),%eax
  neg    %eax
  mov    %eax,-0x4(%rbp)
  jmp    L37
L76:
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L37:
  subl   $0x1,-0x20(%rbp)
  cmpl   $0x9,-0x4(%rbp)
  jbe    L35
  cmpl   $0x0,-0x24(%rbp)
  jns    L60
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  neg    %eax
  mov    %eax,%ecx
  movsbl -0x1c(%rbp),%esi
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <PutSignedInt>
  mov    %eax,-0x8(%rbp)
  jmp    L69
L60:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  mov    %eax,%ecx
  movsbl -0x1c(%rbp),%esi
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <PutSignedInt>
  mov    %eax,-0x8(%rbp)
L69:
  mov    -0x8(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
  jmp    L30
L35:
  cmpl   $0x0,-0x24(%rbp)
  jns    L68
  subl   $0x1,-0x20(%rbp)
  jmp    L68
L4:
  movsbl -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <PutChar@plt>
  addq   $0x1,-0x18(%rbp)
  addl   $0x1,-0x8(%rbp)
  subl   $0x1,-0x20(%rbp)
L68:
  cmpl   $0x0,-0x20(%rbp)
  jg     L4
  cmpl   $0x0,-0x24(%rbp)
  jns    L30
  mov    -0x18(%rbp),%rax
  mov    $0x2d,%esi
  mov    %rax,%rdi
  call   <PutChar@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  addq   $0x1,-0x18(%rbp)
L30:
  mov    -0x4(%rbp),%ecx
  mov    %ecx,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  mov    %eax,%edx
  shr    $0x3,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  add    $0x30,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <PutChar@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
signed int PutSignedInt(char *pStr, char fill, signed int width, signed int value) {
  signed int num = 0;
  unsigned int absolute;
  if (value < 0) {
    absolute = -value;
  } else {
    absolute = value;
  }
  width--;
  if ((absolute / 10) > 0) {
    if (value < 0) {
      num = PutSignedInt(pStr, fill, width, -(absolute / 10));
    } else {
      num = PutSignedInt(pStr, fill, width, absolute / 10);
    }
    pStr += num;
  } else {
    if (value < 0) {
      width--;
    }
    while (width > 0) {
      PutChar(pStr, fill);
      pStr++;
      num++;
      width--;
    }
    if (value < 0) {
      num += PutChar(pStr, '-');
      pStr++;
    }
  }
  num += PutChar(pStr, (absolute % 10) + '0');
  return num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<PutSignedInt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %ecx,%r13d
  mov    %esi,%r15d
  test   %ecx,%ecx
  js     L74
  mov    %ecx,%r14d
  lea    -0x1(%rdx),%ebp
  cmp    $0x9,%ecx
  ja     L83
  mov    $0x0,%ebx
  test   %ebp,%ebp
  jle    L46
L89:
  mov    %r12,%rbx
  movsbl %r15b,%r15d
L32:
  mov    %r15d,%esi
  mov    %rbx,%rdi
  call   <PutChar@plt>
  add    $0x1,%rbx
  lea    0x0(%rbp,%r12,1),%eax
  sub    %ebx,%eax
  test   %eax,%eax
  jg     L32
  lea    -0x1(%rbp),%eax
  add    $0x1,%rax
  test   %ebp,%ebp
  mov    $0x1,%edx
  cmovle %rdx,%rax
  add    %rax,%r12
  test   %ebp,%ebp
  mov    $0x1,%ebx
  cmovg  %ebp,%ebx
  test   %r13d,%r13d
  jns    L46
L77:
  mov    $0x2d,%esi
  mov    %r12,%rdi
  call   <PutChar@plt>
  add    %eax,%ebx
  add    $0x1,%r12
  jmp    L46
L74:
  mov    %ecx,%r14d
  neg    %r14d
  lea    -0x1(%rdx),%eax
  cmp    $0x9,%r14d
  jbe    L57
  mov    %r14d,%ecx
  mov    $0xcccccccd,%edx
  imul   %rdx,%rcx
  shr    $0x23,%rcx
  neg    %ecx
  movsbl %sil,%esi
  mov    %eax,%edx
  call   <PutSignedInt>
  mov    %eax,%ebx
  jmp    L45
L83:
  mov    %ecx,%ecx
  mov    $0xcccccccd,%eax
  imul   %rax,%rcx
  shr    $0x23,%rcx
  movsbl %sil,%esi
  mov    %ebp,%edx
  call   <PutSignedInt>
  mov    %eax,%ebx
L45:
  movslq %ebx,%rax
  add    %rax,%r12
L46:
  mov    %r14d,%eax
  mov    $0xcccccccd,%edx
  imul   %rdx,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%r14d
  lea    0x30(%r14),%esi
  mov    %r12,%rdi
  call   <PutChar@plt>
  add    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L57:
  lea    -0x2(%rdx),%ebp
  mov    $0x0,%ebx
  test   %ebp,%ebp
  jg     L89
  jmp    L77
```

assistant: ```c
signed int PutSignedInt(char *pStr, char fill, signed int width, signed int value) {
  signed int num = 0;
  unsigned int absolute;
  if (value < 0) {
    absolute = -value;
  } else {
    absolute = value;
  }
  width--;
  if ((absolute / 10) > 0) {
    if (value < 0) {
      num = PutSignedInt(pStr, fill, width, -(absolute / 10));
    } else {
      num = PutSignedInt(pStr, fill, width, absolute / 10);
    }
    pStr += num;
  } else {
    if (value < 0) {
      width--;
    }
    while (width > 0) {
      PutChar(pStr, fill);
      pStr++;
      num++;
      width--;
    }
    if (value < 0) {
      num += PutChar(pStr, '-');
      pStr++;
    }
  }
  num += PutChar(pStr, (absolute % 10) + '0');
  return num;
}
```