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

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L17
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edx
  mov    D0(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jb     L17
  mov    $0x0,%eax
  call   <link_failure@plt>
L17:
  nop
  leave
  ret
```

assistant: ```c
void foo(int var) {
  if (!(var <= 0 || ((long unsigned)(unsigned)(var - 1) < UINT_MAX)))
    link_failure();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<foo>:
  endbr64
  test   %edi,%edi
  jle    L3
  lea    -0x1(%rdi),%edx
  mov    D6(%rip),%rax
  cmp    (%rax),%rdx
  jae    L1
L3:
  ret
L1:
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <link_failure@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void foo(int var) {
  if (!(var <= 0 || ((long unsigned)(unsigned)(var - 1) < UINT_MAX)))
    link_failure();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<foo>:
  endbr64
  test   %edi,%edi
  jle    L2
  mov    D8(%rip),%rax
  sub    $0x1,%edi
  movslq %edi,%rdi
  cmp    (%rax),%rdi
  jae    L4
L2:
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  jmp    1050 <link_failure@plt>
```

assistant: ```c
void foo(int var) {
  if (!(var <= 0 || ((long unsigned)(unsigned)(var - 1) < UINT_MAX)))
    link_failure();
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<foo>:
  endbr64
  test   %edi,%edi
  jle    L11
  mov    D6(%rip),%rax
  sub    $0x1,%edi
  movslq %edi,%rdi
  cmp    (%rax),%rdi
  jae    L18
L11:
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  xor    %eax,%eax
  jmp    1050 <link_failure@plt>
```

assistant: ```c
void foo(int var) {
  if (!(var <= 0 || ((long unsigned)(unsigned)(var - 1) < UINT_MAX)))
    link_failure();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<integer_length>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x4,-0x4(%rbp)
  jmp    L6
L11:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  shl    $0x3,%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  movzbl %al,%eax
  test   %eax,%eax
  je     L16
  mov    -0x4(%rbp),%eax
  jmp    L8
L16:
  subl   $0x1,-0x4(%rbp)
L6:
  cmpl   $0x1,-0x4(%rbp)
  ja     L11
  mov    $0x1,%eax
L8:
  pop    %rbp
  ret
```

assistant: ```c
int integer_length(int x) {
  unsigned int i;
  for (i = sizeof(x); i >= 2; i--) {
    if ((x >> (8 * (i - 1))) & 0xFF)
      return i;
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<integer_length>:
  endbr64
  mov    %edi,%eax
  shr    $0x18,%eax
  jne    L18
  mov    %edi,%eax
  sar    $0x10,%eax
  test   %al,%al
  jne    L10
  mov    $0x1,%eax
  test   $0xff00,%edi
  jne    L39
L23:
  ret
L39:
  mov    $0x2,%eax
  ret
L18:
  mov    $0x4,%eax
  ret
L10:
  mov    $0x3,%eax
  jmp    L23
```

assistant: ```c
int integer_length(int x) {
  unsigned int i;
  for (i = sizeof(x); i >= 2; i--) {
    if ((x >> (8 * (i - 1))) & 0xFF)
      return i;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<integer_length>:
  endbr64
  mov    %edi,%ecx
  mov    %edi,%eax
  shr    $0x18,%ecx
  jne    L12
  mov    %edi,%edx
  sar    $0x10,%edx
  test   %dl,%dl
  jne    L9
  movzbl %ah,%eax
  cmp    $0x1,%eax
  mov    $0x1,%eax
  sbb    $0xffffffff,%eax
  ret
  nopl   (%rax)
L12:
  mov    $0x4,%eax
  ret
  xchg   %ax,%ax
L9:
  mov    $0x3,%eax
  ret
```

assistant: ```c
int integer_length(int x) {
  unsigned int i;
  for (i = sizeof(x); i >= 2; i--) {
    if ((x >> (8 * (i - 1))) & 0xFF)
      return i;
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<integer_length>:
  endbr64
  mov    %edi,%ecx
  mov    %edi,%eax
  shr    $0x18,%ecx
  jne    L19
  mov    %edi,%edx
  sar    $0x10,%edx
  test   %dl,%dl
  jne    L11
  movzbl %ah,%eax
  cmp    $0x1,%eax
  mov    $0x1,%eax
  sbb    $0xffffffff,%eax
  ret
  nopl   (%rax)
L19:
  mov    $0x4,%eax
  ret
  xchg   %ax,%ax
L11:
  mov    $0x3,%eax
  ret
```

assistant: ```c
int integer_length(int x) {
  unsigned int i;
  for (i = sizeof(x); i >= 2; i--) {
    if ((x >> (8 * (i - 1))) & 0xFF)
      return i;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L46
L59:
  lea    -0x20(%rbp),%rcx
  lea    -0x28(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  D3(%rip),%xmm1
  divsd  %xmm1,%xmm0
  mov    -0x28(%rbp),%rax
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  comisd -0x18(%rbp),%xmm0
  jbe    L37
  movsd  -0x18(%rbp),%xmm0
  movq   D20(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
L37:
  movsd  -0x18(%rbp),%xmm0
  cvttsd2si %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdq -0x10(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jbe    L13
  movsd  -0x18(%rbp),%xmm0
  cvttsd2si %xmm0,%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L46
L13:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L46:
  mov    -0x38(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x38(%rbp)
  test   %rax,%rax
  jne    L59
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
type_guessed: string
name: retrieve_data
arguments: label: D26
type_guessed: string
name: retrieve_data
arguments: label: D3
type_guessed: f64
name: retrieve_data
arguments: label: D20
type_guessed: byte[8]
name: retrieve_data
arguments: label: D28
type_guessed: string
name: retrieve_data
arguments: label: D39
type_guessed: string

tool: D4: %lld

tool: D26: %lld %lld %lld

tool: D3: 2

tool: {"D20": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

tool: D28: %.2lf


tool: D39: %lld


assistant: ```c
int main() {
  long long int n, a, b, c, e;
  double d;
  scanf("%lld", &n);
  while (n--) {
    scanf("%lld %lld %lld", &a, &b, &c);
    d = ((a + c) / 2.0) - b;
    if (d < 0)
      d *= -1;
    e = d;
    if (d > e)
      printf("%.2lf\n", d);
    else
      printf("%lld\n", e);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L10
  lea    D16(%rip),%rbx
  jmp    L13
L73:
  xorpd  D21(%rip),%xmm0
  jmp    L67
L21:
  lea    D45(%rip),%rdi
  call   <printf@plt>
L65:
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L10
L13:
  lea    0x20(%rsp),%rcx
  lea    0x18(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x20(%rsp),%rax
  add    0x10(%rsp),%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  mulsd  D6(%rip),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sdq 0x18(%rsp),%xmm1
  subsd  %xmm1,%xmm0
  pxor   %xmm2,%xmm2
  comisd %xmm0,%xmm2
  ja     L73
L67:
  cvttsd2si %xmm0,%rsi
  pxor   %xmm1,%xmm1
  cvtsi2sd %rsi,%xmm1
  comisd %xmm1,%xmm0
  jbe    L21
  lea    D40(%rip),%rdi
  call   <printf@plt>
  jmp    L65
L10:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L72
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L72:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type: string
name: extract_symbolic_data
arguments: data_label: D16
type: string
name: extract_symbolic_data
arguments: data_label: D21
type: byte[16]
name: extract_symbolic_data
arguments: data_label: D45
type: string
name: extract_symbolic_data
arguments: data_label: D6
type: byte[8]
name: extract_symbolic_data
arguments: data_label: D40
type: string

tool: D0: %lld

tool: D16: %lld %lld %lld

tool: {"D21": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D45: %lld


tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: D40: %.2lf


assistant: ```c
int main() {
  long long int n, a, b, c, e;
  double d;
  scanf("%lld", &n);
  while (n--) {
    scanf("%lld %lld %lld", &a, &b, &c);
    d = ((a + c) / 2.0) - b;
    if (d < 0)
      d *= -1;
    e = d;
    if (d > e)
      printf("%.2lf\n", d);
    else
      printf("%lld\n", e);
  }
  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   %r15
  lea    D20(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L25
  lea    0x20(%rsp),%r13
  lea    0x18(%rsp),%r12
  lea    0x10(%rsp),%rbp
  lea    D23(%rip),%rbx
  lea    D45(%rip),%r15
  lea    D18(%rip),%r14
  jmp    L61
L29:
  mov    %r14,%rdi
  call   <printf@plt>
L69:
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L25
L61:
  xor    %eax,%eax
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x20(%rsp),%rax
  pxor   %xmm0,%xmm0
  add    0x10(%rsp),%rax
  cvtsi2sd %rax,%xmm0
  pxor   %xmm1,%xmm1
  pxor   %xmm2,%xmm2
  mulsd  D29(%rip),%xmm0
  cvtsi2sdq 0x18(%rsp),%xmm1
  subsd  %xmm1,%xmm0
  comisd %xmm0,%xmm2
  jbe    L22
  xorpd  D27(%rip),%xmm0
L22:
  cvttsd2si %xmm0,%rsi
  pxor   %xmm1,%xmm1
  cvtsi2sd %rsi,%xmm1
  comisd %xmm1,%xmm0
  ja     L29
  mov    %r15,%rdi
  call   <printf@plt>
  jmp    L69
  nopl   0x0(%rax,%rax,1)
L25:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D20
datatype: string
name: read_data
arguments: data_tag: D23
datatype: string
name: read_data
arguments: data_tag: D45
datatype: string
name: read_data
arguments: data_tag: D18
datatype: string
name: read_data
arguments: data_tag: D29
datatype: byte[8]
name: read_data
arguments: data_tag: D27
datatype: byte[16]

tool: D20: %lld

tool: D23: %lld %lld %lld

tool: D45: %lld


tool: D18: %.2lf


tool: {"D29": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: {"D27": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
int main() {
  long long int n, a, b, c, e;
  double d;
  scanf("%lld", &n);
  while (n--) {
    scanf("%lld %lld %lld", &a, &b, &c);
    d = ((a + c) / 2.0) - b;
    if (d < 0)
      d *= -1;
    e = d;
    if (d > e)
      printf("%.2lf\n", d);
    else
      printf("%lld\n", e);
  }
  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   %r15
  lea    D14(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L48
  lea    0x20(%rsp),%r13
  lea    0x18(%rsp),%r12
  lea    0x10(%rsp),%rbp
  lea    D53(%rip),%rbx
  lea    D46(%rip),%r15
  lea    D5(%rip),%r14
  xchg   %ax,%ax
L32:
  xor    %eax,%eax
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x20(%rsp),%rax
  pxor   %xmm0,%xmm0
  add    0x10(%rsp),%rax
  cvtsi2sd %rax,%xmm0
  pxor   %xmm1,%xmm1
  pxor   %xmm2,%xmm2
  mulsd  D6(%rip),%xmm0
  cvtsi2sdq 0x18(%rsp),%xmm1
  subsd  %xmm1,%xmm0
  comisd %xmm0,%xmm2
  jbe    L1
  xorpd  D45(%rip),%xmm0
L1:
  cvttsd2si %xmm0,%rsi
  pxor   %xmm1,%xmm1
  mov    %r14,%rdi
  cvtsi2sd %rsi,%xmm1
  comisd %xmm1,%xmm0
  ja     L38
  mov    %r15,%rdi
L38:
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  jne    L32
L48:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D14
data_type: string
name: read_data
arguments: data_label: D53
data_type: string
name: read_data
arguments: data_label: D46
data_type: string
name: read_data
arguments: data_label: D5
data_type: string
name: read_data
arguments: data_label: D6
data_type: byte[8]
name: read_data
arguments: data_label: D45
data_type: byte[16]

tool: D14: %lld

tool: D53: %lld %lld %lld

tool: D46: %lld


tool: D5: %.2lf


tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: {"D45": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
int main() {
  long long int n, a, b, c, e;
  double d;
  scanf("%lld", &n);
  while (n--) {
    scanf("%lld %lld %lld", &a, &b, &c);
    d = ((a + c) / 2.0) - b;
    if (d < 0)
      d *= -1;
    e = d;
    if (d > e)
      printf("%.2lf\n", d);
    else
      printf("%lld\n", e);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<str_trim>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %rax,%r13
  mov    %rbp,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  je     L26
  cmp    $0x1,%eax
  je     L81
  mov    %r13d,%ebx
  cmpb   $0x20,0x0(%rbp)
  jne    L102
  mov    $0x1,%edx
L45:
  mov    %r13d,%ebx
  sub    %edx,%ebx
  mov    %edx,%r12d
  add    $0x1,%rdx
  cmpb   $0x20,-0x1(%rbp,%rdx,1)
  je     L45
L57:
  sub    $0x1,%eax
  cltq
  add    %rbp,%rax
  cmpb   $0x20,(%rax)
  jne    L128
L89:
  sub    $0x1,%ebx
  sub    $0x1,%rax
  cmpb   $0x20,(%rax)
  je     L89
L128:
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %ebx,%ebx
  jle    L86
  mov    %ebx,%ecx
  mov    $0x0,%eax
  movslq %r12d,%r12
  add    %rbp,%r12
L85:
  movzbl (%r12,%rax,1),%edx
  mov    %dl,0x0(%r13,%rax,1)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L85
L86:
  movslq %ebx,%rbx
  movb   $0x0,0x0(%r13,%rbx,1)
  mov    %rbp,%rdi
  call   <free@plt>
L93:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L26:
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  movb   $0x0,(%rax)
  jmp    L93
L81:
  cmpb   $0x20,0x0(%rbp)
  je     L39
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  movzbl 0x0(%rbp),%eax
  mov    %al,0x0(%r13)
  movb   $0x0,0x1(%r13)
L124:
  mov    %rbp,%rdi
  call   <free@plt>
  jmp    L93
L39:
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  movb   $0x0,(%rax)
  jmp    L124
L102:
  mov    $0x0,%r12d
  jmp    L57
```

assistant: ```c
char *str_trim(char *string) {
  int final_len = strlen(string);
  int i = 0;
  const int len = strlen(string);
  if (len == 0) {
    char *output = malloc(sizeof(char));
    output[0] = '\0';
    return output;
  }
  if (len == 1) {
    char *output;
    if (string[0] != ' ') {
      output = malloc(sizeof(char) * 2);
      output[0] = string[0];
      output[1] = '\0';
    } else {
      output = malloc(sizeof(char));
      output[0] = '\0';
    }
    free(string);
    return output;
  }
  while (true) {
    if (string[i] != ' ') {
      break;
    } else {
      final_len -= 1;
      i += 1;
    }
  }
  int k = len - 1;
  while (true) {
    if (string[k] != ' ') {
      break;
    } else {
      final_len -= 1;
      k -= 1;
    }
  }
  char *output = malloc(sizeof(char) * (final_len + 1));
  for (int l = 0; l < final_len; l += 1) {
    output[l] = string[l + i];
  }
  output[final_len] = '\0';
  free(string);
  return output;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<str_trim>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %rax,%r13
  call   <strlen@plt>
  test   %eax,%eax
  je     L119
  movzbl 0x0(%rbp),%edx
  cmp    $0x1,%eax
  je     L92
  mov    %r13d,%ebx
  cmp    $0x20,%dl
  jne    L82
  mov    $0x1,%edx
  nopw   0x0(%rax,%rax,1)
L113:
  mov    %r13d,%ebx
  movslq %edx,%r12
  sub    %edx,%ebx
  add    $0x1,%rdx
  cmpb   $0x20,-0x1(%rbp,%rdx,1)
  je     L113
L123:
  sub    $0x1,%eax
  cltq
  add    %rbp,%rax
  cmpb   $0x20,(%rax)
  jne    L3
  nopl   0x0(%rax)
L114:
  sub    $0x1,%rax
  mov    %ebx,%edi
  sub    $0x1,%ebx
  cmpb   $0x20,(%rax)
  je     L114
L35:
  call   <malloc@plt>
  add    %rbp,%r12
  xor    %edx,%edx
  mov    %rax,%r13
  movslq %ebx,%rax
  test   %ebx,%ebx
  jle    L30
  nopw   0x0(%rax,%rax,1)
L32:
  movzbl (%r12,%rdx,1),%ecx
  mov    %cl,0x0(%r13,%rdx,1)
  add    $0x1,%rdx
  cmp    %rdx,%rax
  jne    L32
L30:
  movb   $0x0,0x0(%r13,%rax,1)
L57:
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L119:
  mov    $0x1,%edi
  call   <malloc@plt>
  movb   $0x0,(%rax)
  mov    %rax,%r13
  add    $0x8,%rsp
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L92:
  cmp    $0x20,%dl
  je     L85
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  movzbl 0x0(%rbp),%eax
  movb   $0x0,0x1(%r13)
  mov    %al,0x0(%r13)
  jmp    L57
  nopl   0x0(%rax)
L82:
  xor    %r12d,%r12d
  jmp    L123
L3:
  lea    0x1(%rbx),%edi
  jmp    L35
L85:
  mov    $0x1,%edi
  call   <malloc@plt>
  movb   $0x0,(%rax)
  mov    %rax,%r13
  jmp    L57
```

assistant: ```c
char *str_trim(char *string) {
  int final_len = strlen(string);
  int i = 0;
  const int len = strlen(string);
  if (len == 0) {
    char *output = malloc(sizeof(char));
    output[0] = '\0';
    return output;
  }
  if (len == 1) {
    char *output;
    if (string[0] != ' ') {
      output = malloc(sizeof(char) * 2);
      output[0] = string[0];
      output[1] = '\0';
    } else {
      output = malloc(sizeof(char));
      output[0] = '\0';
    }
    free(string);
    return output;
  }
  while (true) {
    if (string[i] != ' ') {
      break;
    } else {
      final_len -= 1;
      i += 1;
    }
  }
  int k = len - 1;
  while (true) {
    if (string[k] != ' ') {
      break;
    } else {
      final_len -= 1;
      k -= 1;
    }
  }
  char *output = malloc(sizeof(char) * (final_len + 1));
  for (int l = 0; l < final_len; l += 1) {
    output[l] = string[l + i];
  }
  output[final_len] = '\0';
  free(string);
  return output;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ln_grid_search>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %rsi,%r14
  mov    %edx,%r15d
  movsd  %xmm0,0x8(%rsp)
  movq   %xmm1,%r13
  movsd  %xmm2,0x10(%rsp)
  mov    %rcx,%rbx
  mov    %r8d,%r12d
  mov    %r9,0x20(%rsp)
  movsd  %xmm3,0x28(%rsp)
  pxor   %xmm0,%xmm0
  comisd %xmm2,%xmm0
  ja     L30
L46:
  lea    0x0(,%r12,8),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  movq   %r13,%xmm1
  subsd  0x8(%rsp),%xmm1
  lea    -0x1(%r12),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  divsd  %xmm0,%xmm1
  test   %r12d,%r12d
  jle    L19
  mov    %r12d,%edx
  mov    $0x0,%eax
L44:
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  %xmm1,%xmm0
  addsd  0x8(%rsp),%xmm0
  addsd  0x10(%rsp),%xmm0
  movsd  %xmm0,0x0(%rbp,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L44
  lea    D6(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L60
  mov    %rbp,%rbx
  mov    %r12d,%eax
  lea    0x0(%rbp,%rax,8),%r13
L18:
  movsd  (%rbx),%xmm0
  mov    $0xa,%edi
  call   <pow@plt>
  movsd  %xmm0,(%rbx)
  add    $0x8,%rbx
  cmp    %r13,%rbx
  jne    L18
L60:
  movsd  0x28(%rsp),%xmm0
  mov    0x20(%rsp),%r9
  mov    %r12d,%r8d
  mov    %rbp,%rcx
  mov    %r15d,%edx
  mov    %r14,%rsi
  mov    0x18(%rsp),%rdi
  call   <ln_grid_optimal@plt>
  cltq
  mov    0x0(%rbp,%rax,8),%rbx
  mov    %rbp,%rdi
  call   <free@plt>
  movq   %rbx,%xmm0
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L30:
  call   <ln_argmin_asymptotic@plt>
  movsd  %xmm0,0x10(%rsp)
  lea    D6(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L46
  movsd  0x10(%rsp),%xmm0
  call   <log10@plt>
  movsd  %xmm0,0x10(%rsp)
  jmp    L46
L19:
  lea    D6(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  jmp    L60
```

tool_calls: name: unpack_data
arguments: tag: D6
type: string

tool: D6: log

assistant: ```c
double ln_grid_search(const double *data, const double *weight, const int n, const double xmin, const double xmax, double offset, const char *scale, const int k, const char *risk_function, const double threshold) {
  if (offset < 0) {
    offset = ln_argmin_asymptotic(data, weight, n);
    if (strcmp(scale, "log") == 0) {
      offset = log10(offset);
    }
  }
  double *bandwidths = malloc(k * sizeof(double));
  double delta = (xmax - xmin) / (k - 1);
  for (int i = 0; i < k; i++) {
    bandwidths[i] = i * delta + xmin + offset;
  }
  if (strcmp(scale, "log") == 0) {
    for (int i = 0; i < k; i++) {
      bandwidths[i] = pow(10, bandwidths[i]);
    }
  }
  int idx = ln_grid_optimal(data, weight, n, bandwidths, k, risk_function, threshold);
  double result = bandwidths[idx];
  free(bandwidths);
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x20(%rbp)
  movl   $0x1,-0x18(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
void foo(void) {
  unsigned int a[2];
  unsigned int b[2];
  unsigned int c[2];
#pragma acc kernels pcopyout(a, b, c)
  {
    a[0] = 0;
    b[0] = 1;
    c[0] = a[0];
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<foo>:
  endbr64
  ret
```

assistant: ```c
void foo(void) {
  unsigned int a[2];
  unsigned int b[2];
  unsigned int c[2];
#pragma acc kernels pcopyout(a, b, c)
  {
    a[0] = 0;
    b[0] = 1;
    c[0] = a[0];
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  ret
```

assistant: ```c
void foo(void) {
  unsigned int a[2];
  unsigned int b[2];
  unsigned int c[2];
#pragma acc kernels pcopyout(a, b, c)
  {
    a[0] = 0;
    b[0] = 1;
    c[0] = a[0];
  }
}
```