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

```asm
<irc_target_host>:
  endbr64
  push   %r12
  mov    $0x21,%esi
  lea    D1(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strchr@plt>
  test   %rax,%rax
  mov    %rax,%rdi
  cmove  %rbp,%rdi
  call   <strlen@plt>
  mov    $0x7f,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  cmp    %rdx,%rax
  mov    %rdx,%rbx
  cmovbe %rax,%rbx
  mov    %rbx,%rdx
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *irc_target_host(const char *target) {
  static char buffer[128];
  const char *split = strstr(target, "!");
  if (!split)
    split = target;
  size_t length = strlen(split);
  if (length > sizeof(buffer) - 1)
    length = sizeof(buffer) - 1;
  memcpy(buffer, target, length);
  buffer[length] = '\0';
  return buffer;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<irc_target_host>:
  endbr64
  push   %r12
  mov    $0x21,%esi
  lea    D8(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strchr@plt>
  test   %rax,%rax
  mov    %rax,%rdi
  cmove  %rbp,%rdi
  call   <strlen@plt>
  mov    $0x7f,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  cmp    %rdx,%rax
  mov    %rdx,%rbx
  cmovbe %rax,%rbx
  mov    %rbx,%rdx
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *irc_target_host(const char *target) {
  static char buffer[128];
  const char *split = strstr(target, "!");
  if (!split)
    split = target;
  size_t length = strlen(split);
  if (length > sizeof(buffer) - 1)
    length = sizeof(buffer) - 1;
  memcpy(buffer, target, length);
  buffer[length] = '\0';
  return buffer;
}
```
user: Translate the below assembly code into C programming language:

```asm
<recipe>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,%eax
  movsd  %xmm0,-0x58(%rbp)
  mov    %fs:0x28,%rdx
  mov    %rdx,-0x38(%rbp)
  xor    %edx,%edx
  movslq %edi,%rcx
  lea    (%rcx,%rcx,4),%rdx
  lea    0xf(%rcx,%rdx,4),%rdx
  mov    %rdx,%rdi
  and    $0xfffffffffffffff0,%rdi
  and    $0xfffffffffffff000,%rdx
  mov    %rsp,%rsi
  sub    %rdx,%rsi
L36:
  cmp    %rsi,%rsp
  je     L63
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L36
L63:
  mov    %rdi,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L68
  orq    $0x0,-0x8(%rsp,%rdx,1)
L68:
  mov    %rsp,%r13
  lea    0xf(,%rcx,8),%rdx
  mov    %rdx,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rdx
  mov    %rsp,%rcx
  sub    %rdx,%rcx
L35:
  cmp    %rcx,%rsp
  je     L128
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L35
L128:
  mov    %rsi,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L27
  orq    $0x0,-0x8(%rsp,%rdx,1)
L27:
  mov    %rsp,%rbx
  test   %eax,%eax
  jle    L73
  mov    %r13,%r14
  mov    %rbx,%r12
  mov    %eax,%eax
  lea    (%rbx,%rax,8),%r15
  movsd  D26(%rip),%xmm2
  movsd  %xmm2,-0x50(%rbp)
  jmp    L65
L100:
  divsd  D22(%rip),%xmm0
  mov    -0x48(%rbp),%rax
  movsd  %xmm0,(%rax)
  add    $0x15,%r13
  add    $0x8,%rbx
  cmp    %r15,%rbx
  je     L41
L65:
  mov    %rbx,-0x48(%rbp)
  lea    -0x40(%rbp),%rdx
  mov    %rbx,%rcx
  mov    %r13,%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  movsd  (%rbx),%xmm0
  ucomisd D22(%rip),%xmm0
  jp     L100
  jne    L100
  movsd  -0x58(%rbp),%xmm1
  mulsd  -0x40(%rbp),%xmm1
  movsd  %xmm1,-0x50(%rbp)
  jmp    L100
L41:
  lea    D11(%rip),%rbx
L12:
  movsd  -0x50(%rbp),%xmm0
  mulsd  (%r12),%xmm0
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  add    $0x8,%r12
  add    $0x15,%r14
  cmp    %r15,%r12
  jne    L12
L73:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L94
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L94:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D26
data_type: byte[8]
name: parse_data
arguments: data_tag: D22
data_type: f64
name: parse_data
arguments: data_tag: D24
data_type: string
name: parse_data
arguments: data_tag: D11
data_type: string

tool: {"D26": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

tool: D22: 100

tool: D24: %s %lf %lf

tool: D11: %s %.1lf


assistant: ```c
void recipe(int ingredients, double scale) {
  char names[ingredients][21];
  double percentage[ingredients];
  double migs = -1;
  for (int i = 0; i < ingredients; i++) {
    double g;
    scanf("%s %lf %lf", names[i], &g, percentage + i);
    if (percentage[i] == 100.0)
      migs = g * scale;
    percentage[i] /= 100.0;
  }
  for (int i = 0; i < ingredients; i++)
    printf("%s %.1lf\n", names[i], migs * percentage[i]);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<recipe>:
  endbr64
  push   %rbp
  movslq %edi,%rcx
  mov    %rcx,%rax
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  movsd  %xmm0,-0x58(%rbp)
  mov    %fs:0x28,%rdx
  mov    %rdx,-0x38(%rbp)
  xor    %edx,%edx
  lea    (%rcx,%rcx,4),%rdx
  mov    %rsp,%rdi
  lea    0xf(%rcx,%rdx,4),%rdx
  mov    %rdx,%rsi
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rdi
  and    $0xfffffffffffffff0,%rsi
  cmp    %rdi,%rsp
  je     L98
L30:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdi,%rsp
  jne    L30
L98:
  and    $0xfff,%esi
  sub    %rsi,%rsp
  test   %rsi,%rsi
  jne    L10
L91:
  lea    0xf(,%rcx,8),%rdx
  mov    %rsp,%rsi
  mov    %rsp,%r12
  mov    %rdx,%rcx
  and    $0xfffffffffffff000,%rdx
  sub    %rdx,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L35
L25:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L25
L35:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L79
L71:
  mov    %rsp,%rbx
  test   %eax,%eax
  jle    L76
  mov    %eax,%eax
  movsd  D7(%rip),%xmm2
  mov    %rbx,%r14
  mov    %r12,%r13
  lea    (%rbx,%rax,8),%rax
  lea    -0x40(%rbp),%r15
  mov    %rax,-0x48(%rbp)
  movsd  %xmm2,-0x50(%rbp)
  nop
L38:
  mov    %r14,%rcx
  mov    %r15,%rdx
  lea    D31(%rip),%rdi
  mov    %r13,%rsi
  call   <scanf@plt>
  movsd  (%r14),%xmm0
  ucomisd D35(%rip),%xmm0
  jp     L0
  jne    L0
  movsd  -0x58(%rbp),%xmm1
  mulsd  -0x40(%rbp),%xmm1
  movsd  %xmm1,-0x50(%rbp)
L0:
  divsd  D35(%rip),%xmm0
  add    $0x15,%r13
  movsd  %xmm0,(%r14)
  add    $0x8,%r14
  cmp    -0x48(%rbp),%r14
  jne    L38
  lea    D20(%rip),%r14
  nopw   0x0(%rax,%rax,1)
L70:
  movsd  -0x50(%rbp),%xmm0
  mulsd  (%rbx),%xmm0
  mov    %r12,%rsi
  mov    %r14,%rdi
  add    $0x8,%rbx
  add    $0x15,%r12
  call   <printf@plt>
  cmp    -0x48(%rbp),%rbx
  jne    L70
L76:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L122
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L10:
  orq    $0x0,-0x8(%rsp,%rsi,1)
  jmp    L91
L79:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L71
L122:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D7
type_infered: byte[8]
name: parse_data
arguments: data_label: D31
type_infered: string
name: parse_data
arguments: data_label: D35
type_infered: f64
name: parse_data
arguments: data_label: D20
type_infered: string

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

tool: D31: %s %lf %lf

tool: D35: 100

tool: D20: %s %.1lf


assistant: ```c
void recipe(int ingredients, double scale) {
  char names[ingredients][21];
  double percentage[ingredients];
  double migs = -1;
  for (int i = 0; i < ingredients; i++) {
    double g;
    scanf("%s %lf %lf", names[i], &g, percentage + i);
    if (percentage[i] == 100.0)
      migs = g * scale;
    percentage[i] /= 100.0;
  }
  for (int i = 0; i < ingredients; i++)
    printf("%s %.1lf\n", names[i], migs * percentage[i]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xd,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
type: string
name: extract_symbolic_data
arguments: data_tag: D26
type: string
name: extract_symbolic_data
arguments: data_tag: D11
type: string

tool: D9: x = %d


tool: D26: x_pointer in memory = %d


tool: D11: x_pointer real value = %d


assistant: ```c
int main() {
  int x = 13;
  printf("x = %d\n", x);
  int *x_pointer = &x;
  printf("x_pointer in memory = %d\n", x_pointer);
  printf("x_pointer real value = %d\n", *x_pointer);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0xd,0x4(%rsp)
  mov    $0xd,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  lea    D25(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D7
type: string
name: disassemble_data
arguments: label: D11
type: string
name: disassemble_data
arguments: label: D25
type: string

tool: D7: x = %d


tool: D11: x_pointer in memory = %d


tool: D25: x_pointer real value = %d


assistant: ```c
int main() {
  int x = 13;
  printf("x = %d\n", x);
  int *x_pointer = &x;
  printf("x_pointer in memory = %d\n", x_pointer);
  printf("x_pointer real value = %d\n", *x_pointer);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xd,%esi
  lea    D7(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0xd,0x4(%rsp)
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D7
data_type: string
name: disassemble_data
arguments: label: D0
data_type: string
name: disassemble_data
arguments: label: D18
data_type: string

tool: D7: x = %d


tool: D0: x_pointer in memory = %d


tool: D18: x_pointer real value = %d


assistant: ```c
int main() {
  int x = 13;
  printf("x = %d\n", x);
  int *x_pointer = &x;
  printf("x_pointer in memory = %d\n", x_pointer);
  printf("x_pointer real value = %d\n", *x_pointer);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xd,%esi
  lea    D15(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0xd,0x4(%rsp)
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D15
data_type: string
name: disassemble_data
arguments: data_label: D3
data_type: string
name: disassemble_data
arguments: data_label: D23
data_type: string

tool: D15: x = %d


tool: D3: x_pointer in memory = %d


tool: D23: x_pointer real value = %d


assistant: ```c
int main() {
  int x = 13;
  printf("x = %d\n", x);
  int *x_pointer = &x;
  printf("x_pointer in memory = %d\n", x_pointer);
  printf("x_pointer real value = %d\n", *x_pointer);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  subl   $0x1,-0x1c(%rbp)
  jmp    L8
L1:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x8(%rbp)
  subl   $0x1,-0x1c(%rbp)
L8:
  mov    -0x1c(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jg     L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void reverse(char input[], int size) {
  int tmp;
  int i = 0;
  size--;
  while (size > i) {
    tmp = input[i];
    input[i] = input[size];
    input[size] = tmp;
    ++i;
    --size;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<reverse>:
  endbr64
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L6
  movslq %esi,%rsi
  mov    $0x0,%eax
L10:
  movzbl (%rdi,%rax,1),%edx
  movzbl (%rdi,%rsi,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  mov    %dl,(%rdi,%rsi,1)
  add    $0x1,%rax
  sub    $0x1,%rsi
  cmp    %esi,%eax
  jl     L10
L6:
  ret
```

assistant: ```c
void reverse(char input[], int size) {
  int tmp;
  int i = 0;
  size--;
  while (size > i) {
    tmp = input[i];
    input[i] = input[size];
    input[size] = tmp;
    ++i;
    --size;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<reverse>:
  endbr64
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L4
  movslq %esi,%rsi
  xor    %eax,%eax
L13:
  movzbl (%rdi,%rax,1),%edx
  movzbl (%rdi,%rsi,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  mov    %dl,(%rdi,%rsi,1)
  sub    $0x1,%rsi
  cmp    %esi,%eax
  jl     L13
L4:
  ret
```

assistant: ```c
void reverse(char input[], int size) {
  int tmp;
  int i = 0;
  size--;
  while (size > i) {
    tmp = input[i];
    input[i] = input[size];
    input[size] = tmp;
    ++i;
    --size;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L17
  movslq %esi,%rsi
  xor    %eax,%eax
L0:
  movzbl (%rdi,%rax,1),%edx
  movzbl (%rdi,%rsi,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  mov    %dl,(%rdi,%rsi,1)
  sub    $0x1,%rsi
  cmp    %eax,%esi
  jg     L0
L17:
  ret
```

assistant: ```c
void reverse(char input[], int size) {
  int tmp;
  int i = 0;
  size--;
  while (size > i) {
    tmp = input[i];
    input[i] = input[size];
    input[size] = tmp;
    ++i;
    --size;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1a,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L7:
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  and    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L14:
  cmpl   $0x0,-0x8(%rbp)
  jne    L7
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D2: count: %d


assistant: ```c
int main() {
  unsigned int x = 0b11010;
  int count = 0;
  while (x != 0) {
    x = x & (x - 1);
    count++;
  }
  printf("count: %d\n", count);
  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
  sub    $0x8,%rsp
  mov    $0x3,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D2: count: %d


assistant: ```c
int main() {
  unsigned int x = 0b11010;
  int count = 0;
  while (x != 0) {
    x = x & (x - 1);
    count++;
  }
  printf("count: %d\n", count);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D3
type_infered: string

tool: D3: count: %d


assistant: ```c
int main() {
  unsigned int x = 0b11010;
  int count = 0;
  while (x != 0) {
    x = x & (x - 1);
    count++;
  }
  printf("count: %d\n", count);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D8
data_type: string

tool: D8: count: %d


assistant: ```c
int main() {
  unsigned int x = 0b11010;
  int count = 0;
  while (x != 0) {
    x = x & (x - 1);
    count++;
  }
  printf("count: %d\n", count);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L66
L42:
  cmpb   $0x20,-0x9(%rbp)
  je     L59
  cmpb   $0xa,-0x9(%rbp)
  je     L59
  cmpb   $0x9,-0x9(%rbp)
  jne    L29
L59:
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L76
L14:
  mov    $0x2d,%edi
  call   <putchar@plt>
  addl   $0x1,-0x8(%rbp)
L76:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L14
  movl   $0x0,-0x4(%rbp)
  jmp    L66
L29:
  addl   $0x1,-0x4(%rbp)
L66:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x9(%rbp)
  mov    D0(%rip),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x9(%rbp)
  jne    L42
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
main() {
  char c;
  int i, str_len;
  str_len = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ' || c == '\n' || c == '\t') {
      putchar('\n');
      for (i = 0; i < str_len; i++) {
        putchar('-');
      }
      str_len = 0;
    } else {
      str_len++;
    }
  }
  putchar('\n');
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebp
  jmp    L38
L56:
  mov    $0xa,%edi
  call   <putchar@plt>
  test   %ebp,%ebp
  jle    L78
  mov    $0x0,%ebx
L55:
  mov    $0x2d,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L55
  mov    $0x0,%ebp
L38:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D4(%rip),%rdx
  cmp    %al,(%rdx)
  je     L24
  lea    -0x9(%rax),%edx
  cmp    $0x1,%dl
  jbe    L56
  cmp    $0x20,%al
  je     L56
  add    $0x1,%ebp
  jmp    L38
L78:
  mov    $0x0,%ebp
  jmp    L38
L24:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
main() {
  char c;
  int i, str_len;
  str_len = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ' || c == '\n' || c == '\t') {
      putchar('\n');
      for (i = 0; i < str_len; i++) {
        putchar('-');
      }
      str_len = 0;
    } else {
      str_len++;
    }
  }
  putchar('\n');
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  mov    D3(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
L28:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %al,(%r12)
  je     L54
L19:
  lea    -0x9(%rax),%edx
  cmp    $0x1,%dl
  jbe    L43
  cmp    $0x20,%al
  je     L43
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <getchar@plt>
  cmp    %al,(%r12)
  jne    L19
L54:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L43:
  mov    $0xa,%edi
  xor    %ebx,%ebx
  call   <putchar@plt>
  test   %ebp,%ebp
  je     L66
L35:
  mov    $0x2d,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebx,%ebp
  jne    L35
L66:
  xor    %ebp,%ebp
  jmp    L28
```

assistant: ```c
main() {
  char c;
  int i, str_len;
  str_len = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ' || c == '\n' || c == '\t') {
      putchar('\n');
      for (i = 0; i < str_len; i++) {
        putchar('-');
      }
      str_len = 0;
    } else {
      str_len++;
    }
  }
  putchar('\n');
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  mov    D5(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
L9:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %al,(%r12)
  je     L41
L32:
  lea    -0x9(%rax),%edx
  cmp    $0x1,%dl
  jbe    L34
  cmp    $0x20,%al
  je     L34
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <getchar@plt>
  cmp    %al,(%r12)
  jne    L32
L41:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L34:
  mov    $0xa,%edi
  xor    %ebx,%ebx
  call   <putchar@plt>
  test   %ebp,%ebp
  je     L29
L13:
  mov    $0x2d,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  cmp    %ebx,%ebp
  jne    L13
L29:
  xor    %ebp,%ebp
  jmp    L9
```

assistant: ```c
main() {
  char c;
  int i, str_len;
  str_len = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ' || c == '\n' || c == '\t') {
      putchar('\n');
      for (i = 0; i < str_len; i++) {
        putchar('-');
      }
      str_len = 0;
    } else {
      str_len++;
    }
  }
  putchar('\n');
}
```