ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```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
  mov    %rsp,%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jle    L8
  lea    0x4(%rsp),%r13
  lea    D12(%rip),%r12
  mov    $0x1,%ebp
  lea    D1(%rip),%rbx
L10:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  mov    %ebp,%esi
  shl    %cl,%esi
  sub    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jg     L10
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D12
type: string
name: retrieve_data
arguments: label: D1
type: string

tool: D12: %d

tool: D1: %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  while (n-- > 0) {
    int k;
    scanf("%d", &k);
    printf("%d\n", (1 << k) - 1);
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D2(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jle    L4
  lea    0x4(%rsp),%r13
  lea    D7(%rip),%r12
  mov    $0x1,%ebp
  xchg   %ax,%ax
L3:
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  mov    %ebp,%esi
  mov    %r12,%rdi
  shl    %cl,%esi
  sub    $0x1,%esi
  call   <printf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jg     L3
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D2
data_type: string
name: disassemble_data
arguments: tag: D7
data_type: string

tool: D2: %d

tool: D7: %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  while (n-- > 0) {
    int k;
    scanf("%d", &k);
    printf("%d\n", (1 << k) - 1);
  }
  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   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D11(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jle    L19
  lea    0x4(%rsp),%r13
  lea    D19(%rip),%r12
  mov    $0x1,%ebp
  xchg   %ax,%ax
L12:
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  mov    %ebp,%esi
  mov    %r12,%rdi
  shl    %cl,%esi
  sub    $0x1,%esi
  call   <printf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jg     L12
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D11
type: string
name: extract_symbolic_data
arguments: data_label: D19
type: string

tool: D11: %d

tool: D19: %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  while (n-- > 0) {
    int k;
    scanf("%d", &k);
    printf("%d\n", (1 << k) - 1);
  }
  return 0;
}
```
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
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L26
L15:
  mov    -0x14(%rbp),%eax
  mov    -0x10(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L26:
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  jne    L15
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D14
type: string
name: read_data
arguments: data_label: D19
type: string
name: read_data
arguments: data_label: D2
type: string

tool: D14: Please enter two non-negative integers.

tool: D19: %i%i

tool: D2: They're greatest common divisor is %i.


assistant: ```c
int main(void) {
  int u, v, temp;
  printf("Please enter two non-negative integers.\n");
  scanf("%i%i", &u, &v);
  while (v != 0) {
    temp = u % v;
    u = v;
    v = temp;
  }
  printf("They're greatest common divisor is %i.\n", u);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D8(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  test   %ecx,%ecx
  je     L11
  mov    (%rsp),%eax
  jmp    L10
L34:
  mov    %edx,%ecx
L10:
  cltd
  idiv   %ecx
  mov    %ecx,%eax
  test   %edx,%edx
  jne    L34
  mov    %ecx,(%rsp)
  movl   $0x0,0x4(%rsp)
L11:
  mov    (%rsp),%esi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D8
type: string
name: parse_data
arguments: label: D22
type: string
name: parse_data
arguments: label: D18
type: string

tool: D8: Please enter two non-negative integers.

tool: D22: %i%i

tool: D18: They're greatest common divisor is %i.


assistant: ```c
int main(void) {
  int u, v, temp;
  printf("Please enter two non-negative integers.\n");
  scanf("%i%i", &u, &v);
  while (v != 0) {
    temp = u % v;
    u = v;
    v = temp;
  }
  printf("They're greatest common divisor is %i.\n", u);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D10(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rdx
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%eax
  test   %esi,%esi
  jne    L17
  jmp    L3
  nopl   (%rax)
L44:
  mov    %edx,%esi
L17:
  cltd
  idiv   %esi
  mov    %esi,%eax
  test   %edx,%edx
  jne    L44
  mov    %esi,(%rsp)
  movl   $0x0,0x4(%rsp)
L36:
  xor    %eax,%eax
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  mov    %eax,%esi
  jmp    L36
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D10
type_guessed: string
name: retrieve_data
arguments: label: D9
type_guessed: string
name: retrieve_data
arguments: label: D3
type_guessed: string

tool: D10: Please enter two non-negative integers.

tool: D9: %i%i

tool: D3: They're greatest common divisor is %i.


assistant: ```c
int main(void) {
  int u, v, temp;
  printf("Please enter two non-negative integers.\n");
  scanf("%i%i", &u, &v);
  while (v != 0) {
    temp = u % v;
    u = v;
    v = temp;
  }
  printf("They're greatest common divisor is %i.\n", u);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D11(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rdx
  lea    D26(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%eax
  test   %esi,%esi
  jne    L36
  jmp    L20
  nopl   (%rax)
L37:
  mov    %edx,%esi
L36:
  cltd
  idiv   %esi
  mov    %esi,%eax
  test   %edx,%edx
  jne    L37
  mov    %esi,(%rsp)
  movl   $0x0,0x4(%rsp)
L7:
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L20:
  mov    %eax,%esi
  jmp    L7
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D11
type_guessed: string
name: retrieve_data
arguments: data_label: D26
type_guessed: string
name: retrieve_data
arguments: data_label: D6
type_guessed: string

tool: D11: Please enter two non-negative integers.

tool: D26: %i%i

tool: D6: They're greatest common divisor is %i.


assistant: ```c
int main(void) {
  int u, v, temp;
  printf("Please enter two non-negative integers.\n");
  scanf("%i%i", &u, &v);
  while (v != 0) {
    temp = u % v;
    u = v;
    v = temp;
  }
  printf("They're greatest common divisor is %i.\n", u);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<benchmark>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L4
L47:
  mov    $0x0,%eax
  call   <rand_double@plt>
  movsd  D10(%rip),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  D16(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <cosm1@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  ucomisd -0x20(%rbp),%xmm0
  jnp    L49
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L33
L49:
  addl   $0x1,-0x24(%rbp)
L4:
  cmpl   $0xf423f,-0x24(%rbp)
  jle    L47
L33:
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  subsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x20(%rbp),%xmm0
  ucomisd -0x20(%rbp),%xmm0
  jnp    L32
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L32:
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D10
type: f64
name: unpack_data
arguments: data_label: D16
type: f64
name: unpack_data
arguments: data_label: D25
type: string

tool: D10: 4

tool: D16: 2

tool: D25: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (4.0 * rand_double()) - 2.0;
    y = cosm1(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<benchmark>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%r12
  mov    $0xf4240,%ebx
L28:
  mov    $0x0,%eax
  call   <rand_double@plt>
  mulsd  D18(%rip),%xmm0
  subsd  D5(%rip),%xmm0
  call   <cosm1@plt>
  ucomisd %xmm0,%xmm0
  jp     L13
  sub    $0x1,%ebx
  jne    L28
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r12,%xmm2
  subsd  %xmm2,%xmm0
  movq   %xmm0,%rbx
  jmp    L18
L13:
  lea    D0(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r12,%xmm1
  subsd  %xmm1,%xmm0
  movq   %xmm0,%rbx
  mov    %rbp,%rdi
  call   <printf@plt>
L18:
  movq   %rbx,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D18
type_guessed: f64
name: read_data
arguments: label: D5
type_guessed: f64
name: read_data
arguments: label: D0
type_guessed: string

tool: D18: 4

tool: D5: 2

tool: D0: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (4.0 * rand_double()) - 2.0;
    y = cosm1(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<benchmark>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  mov    $0xf4240,%ebx
  sub    $0x18,%rsp
  call   <tic@plt>
  movsd  %xmm0,0x8(%rsp)
  jmp    L23
  xchg   %ax,%ax
L24:
  sub    $0x1,%ebx
  je     L14
L23:
  xor    %eax,%eax
  call   <rand_double@plt>
  mulsd  D14(%rip),%xmm0
  subsd  D10(%rip),%xmm0
  call   <cosm1@plt>
  ucomisd %xmm0,%xmm0
  jnp    L24
  lea    D12(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L14:
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: label: D14
type: f64
name: unpack_data
arguments: label: D10
type: f64
name: unpack_data
arguments: label: D12
type: string

tool: D14: 4

tool: D10: 2

tool: D12: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (4.0 * rand_double()) - 2.0;
    y = cosm1(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Translate the below assembly code into C programming language:

```asm
<benchmark>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  mov    $0xf4240,%ebx
  sub    $0x18,%rsp
  call   <tic@plt>
  movsd  %xmm0,0x8(%rsp)
  jmp    L14
  xchg   %ax,%ax
L22:
  sub    $0x1,%ebx
  je     L0
L14:
  xor    %eax,%eax
  call   <rand_double@plt>
  mulsd  D17(%rip),%xmm0
  subsd  D24(%rip),%xmm0
  call   <cosm1@plt>
  ucomisd %xmm0,%xmm0
  jnp    L22
  lea    D0(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  xchg   %ax,%ax
L0:
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D17
datatype: f64
name: parse_data
arguments: data_tag: D24
datatype: f64
name: parse_data
arguments: data_tag: D0
datatype: string

tool: D17: 4

tool: D24: 2

tool: D0: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = (4.0 * rand_double()) - 2.0;
    y = cosm1(x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<perf_reset_clock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <perf_disable_clock@plt>
  mov    $0xf8f00200,%eax
  movl   $0x0,(%rax)
  mov    $0xf8f00204,%eax
  movl   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void perf_reset_clock(void) {
  perf_disable_clock();
  *(volatile unsigned int *)0xF8F00200 = 0;
  *(volatile unsigned int *)0xF8F00204 = 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<perf_reset_clock>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <perf_disable_clock@plt>
  mov    $0xf8f00200,%eax
  movl   $0x0,(%rax)
  lea    0x4(%rax),%rax
  movl   $0x0,(%rax)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void perf_reset_clock(void) {
  perf_disable_clock();
  *(volatile unsigned int *)0xF8F00200 = 0;
  *(volatile unsigned int *)0xF8F00204 = 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<perf_reset_clock>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <perf_disable_clock@plt>
  mov    $0xf8f00200,%eax
  movl   $0x0,(%rax)
  movl   $0x0,0x4(%rax)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void perf_reset_clock(void) {
  perf_disable_clock();
  *(volatile unsigned int *)0xF8F00200 = 0;
  *(volatile unsigned int *)0xF8F00204 = 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<perf_reset_clock>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <perf_disable_clock@plt>
  mov    $0xf8f00200,%eax
  movl   $0x0,(%rax)
  movl   $0x0,0x4(%rax)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void perf_reset_clock(void) {
  perf_disable_clock();
  *(volatile unsigned int *)0xF8F00200 = 0;
  *(volatile unsigned int *)0xF8F00204 = 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<expandString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,%eax
  mov    %al,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L9
L0:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x20(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L0
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void expandString(s, len, c) char *s, c;
int len;
{
  int i;
  i = strlen(s);
  while (i < len) {
    s[i] = c;
    i++;
  }
  s[len] = 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<expandString>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %edx,%ebx
  call   <strlen@plt>
  cmp    %ebp,%eax
  jge    L7
  movslq %eax,%rcx
  lea    (%r12,%rcx,1),%rdx
  lea    0x1(%r12,%rcx,1),%rcx
  lea    -0x1(%rbp),%esi
  sub    %eax,%esi
  lea    (%rcx,%rsi,1),%rax
L6:
  mov    %bl,(%rdx)
  add    $0x1,%rdx
  cmp    %rax,%rdx
  jne    L6
L7:
  movslq %ebp,%rbp
  movb   $0x0,(%r12,%rbp,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void expandString(s, len, c) char *s, c;
int len;
{
  int i;
  i = strlen(s);
  while (i < len) {
    s[i] = c;
    i++;
  }
  s[len] = 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<expandString>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  call   <strlen@plt>
  cmp    %ebx,%eax
  jge    L8
  lea    -0x1(%rbx),%edx
  movsbl %r12b,%esi
  sub    %eax,%edx
  cltq
  add    $0x1,%rdx
  lea    0x0(%rbp,%rax,1),%rdi
  call   <memset@plt>
L8:
  movb   $0x0,0x0(%rbp,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void expandString(s, len, c) char *s, c;
int len;
{
  int i;
  i = strlen(s);
  while (i < len) {
    s[i] = c;
    i++;
  }
  s[len] = 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<expandString>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  call   <strlen@plt>
  cmp    %ebx,%eax
  jge    L4
  lea    -0x1(%rbx),%edx
  movsbl %r12b,%esi
  sub    %eax,%edx
  cltq
  add    $0x1,%rdx
  lea    0x0(%rbp,%rax,1),%rdi
  call   <memset@plt>
L4:
  movb   $0x0,0x0(%rbp,%rbx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void expandString(s, len, c) char *s, c;
int len;
{
  int i;
  i = strlen(s);
  while (i < len) {
    s[i] = c;
    i++;
  }
  s[len] = 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<convert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movl   $0x0,(%rax)
  jmp    L9
L4:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  and    $0x1,%eax
  sub    %ecx,%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
L17:
  cmpl   $0x0,-0xc(%rbp)
  jg     L4
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
L9:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%esi
  mov    -0xc(%rbp),%eax
  cltd
  idiv   %esi
  cmp    $0x1,%eax
  jg     L17
  nop
  nop
  leave
  ret
```

assistant: ```c
void convert(int bin[], int x, int *len) {
  for (*len = 0; x / pow(2, *len) > 1; ++*len)
    while (x > 0) {
      bin[*len] = x % 2;
      x /= 2;
      ++*len;
    }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<convert>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebx
  mov    %rdx,%rbp
  mov    $0x0,%esi
  jmp    L18
L15:
  movslq 0x0(%rbp),%rcx
  mov    %ebx,%edx
  shr    $0x1f,%edx
  lea    (%rbx,%rdx,1),%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,(%r12,%rcx,4)
  mov    %ebx,%edx
  mov    %ebx,%eax
  shr    $0x1f,%eax
  add    %ebx,%eax
  sar    %eax
  mov    %eax,%ebx
  addl   $0x1,0x0(%rbp)
  cmp    $0x1,%edx
  jg     L15
L28:
  mov    0x0(%rbp),%eax
  lea    0x1(%rax),%esi
L18:
  mov    %esi,0x0(%rbp)
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  cmp    $0x1,%eax
  jle    L42
  test   %ebx,%ebx
  jg     L15
  jmp    L28
L42:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void convert(int bin[], int x, int *len) {
  for (*len = 0; x / pow(2, *len) > 1; ++*len)
    while (x > 0) {
      bin[*len] = x % 2;
      x /= 2;
      ++*len;
    }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<convert>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %esi,%ebx
  xor    %esi,%esi
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %esi,0x0(%rbp)
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  cmp    $0x1,%eax
  jle    L12
  mov    0x0(%rbp),%eax
  test   %ebx,%ebx
  jle    L37
  nopw   0x0(%rax,%rax,1)
L24:
  mov    %ebx,%edx
  cltq
  and    $0x1,%edx
  mov    %edx,(%r12,%rax,4)
  mov    0x0(%rbp),%eax
  add    $0x1,%eax
  sar    %ebx
  mov    %eax,0x0(%rbp)
  jne    L24
L37:
  lea    0x1(%rax),%esi
  jmp    L14
  nopl   (%rax)
L12:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void convert(int bin[], int x, int *len) {
  for (*len = 0; x / pow(2, *len) > 1; ++*len)
    while (x > 0) {
      bin[*len] = x % 2;
      x /= 2;
      ++*len;
    }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<convert>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %esi,%ebx
  xor    %esi,%esi
  nopw   0x0(%rax,%rax,1)
L16:
  mov    %esi,0x0(%rbp)
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  cmp    $0x1,%eax
  jle    L34
  mov    0x0(%rbp),%eax
  test   %ebx,%ebx
  jle    L10
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %ebx,%edx
  cltq
  and    $0x1,%edx
  mov    %edx,(%r12,%rax,4)
  mov    0x0(%rbp),%eax
  add    $0x1,%eax
  sar    %ebx
  mov    %eax,0x0(%rbp)
  jne    L8
L10:
  lea    0x1(%rax),%esi
  jmp    L16
  nopl   (%rax)
L34:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void convert(int bin[], int x, int *len) {
  for (*len = 0; x / pow(2, *len) > 1; ++*len)
    while (x > 0) {
      bin[*len] = x % 2;
      x /= 2;
      ++*len;
    }
}
```