ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<Threshold_Frame>:
  endbr64
  imul   %edx,%esi
  test   %esi,%esi
  jle    L34
  lea    (%rsi,%rsi,2),%esi
  mov    $0x0,%eax
  jmp    L15
L12:
  movslq %eax,%rdx
  movb   $0x0,0x2(%rdi,%rdx,1)
  movb   $0x0,0x1(%rdi,%rdx,1)
  movb   $0x0,(%rdi,%rdx,1)
L28:
  add    $0x1,%r8
  add    $0x3,%eax
  cmp    %esi,%eax
  je     L34
L15:
  movzbl (%r8),%edx
  cmp    %ecx,%edx
  jl     L12
  movslq %eax,%rdx
  movb   $0xff,0x2(%rdi,%rdx,1)
  movb   $0xff,0x1(%rdi,%rdx,1)
  movb   $0xff,(%rdi,%rdx,1)
  jmp    L28
L34:
  ret
```

assistant: ```c
void Threshold_Frame(unsigned char *Frame, int Width, int Height, int Threshold, unsigned char *DensityMap) {
  int I;
  for (I = 0; I < Width * Height; I++) {
    if (DensityMap[I] >= Threshold)
      Frame[3 * I] = Frame[3 * I + 1] = Frame[3 * I + 2] = 255;
    else
      Frame[3 * I] = Frame[3 * I + 1] = Frame[3 * I + 2] = 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Threshold_Frame>:
  endbr64
  mov    %edx,%r9d
  mov    %ecx,%edx
  imul   %r9d,%esi
  test   %esi,%esi
  jle    L18
  lea    -0x1(%rsi),%eax
  lea    (%rax,%rax,2),%rax
  lea    0x3(%rdi,%rax,1),%rcx
  nopl   (%rax)
L17:
  movzbl (%r8),%eax
  cmp    %edx,%eax
  setge  %al
  add    $0x3,%rdi
  add    $0x1,%r8
  neg    %eax
  mov    %al,-0x1(%rdi)
  mov    %al,-0x2(%rdi)
  mov    %al,-0x3(%rdi)
  cmp    %rdi,%rcx
  jne    L17
L18:
  ret
```

assistant: ```c
void Threshold_Frame(unsigned char *Frame, int Width, int Height, int Threshold, unsigned char *DensityMap) {
  int I;
  for (I = 0; I < Width * Height; I++) {
    if (DensityMap[I] >= Threshold)
      Frame[3 * I] = Frame[3 * I + 1] = Frame[3 * I + 2] = 255;
    else
      Frame[3 * I] = Frame[3 * I + 1] = Frame[3 * I + 2] = 0;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<Threshold_Frame>:
  endbr64
  imul   %edx,%esi
  test   %esi,%esi
  jle    L33
  lea    -0x1(%rsi),%eax
  lea    (%rax,%rax,2),%rax
  lea    0x3(%rdi,%rax,1),%rdx
  nopw   0x0(%rax,%rax,1)
L24:
  movzbl (%r8),%eax
  cmp    %eax,%ecx
  jle    L4
  movb   $0x0,0x2(%rdi)
  add    $0x3,%rdi
  add    $0x1,%r8
  movb   $0x0,-0x2(%rdi)
  movb   $0x0,-0x3(%rdi)
  cmp    %rdx,%rdi
  jne    L24
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  movb   $0xff,0x2(%rdi)
  add    $0x3,%rdi
  add    $0x1,%r8
  movb   $0xff,-0x2(%rdi)
  movb   $0xff,-0x3(%rdi)
  cmp    %rdx,%rdi
  jne    L24
L33:
  ret
```

assistant: ```c
void Threshold_Frame(unsigned char *Frame, int Width, int Height, int Threshold, unsigned char *DensityMap) {
  int I;
  for (I = 0; I < Width * Height; I++) {
    if (DensityMap[I] >= Threshold)
      Frame[3 * I] = Frame[3 * I + 1] = Frame[3 * I + 2] = 255;
    else
      Frame[3 * I] = Frame[3 * I + 1] = Frame[3 * I + 2] = 0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<backword>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jns    L38
  mov    -0x8(%rbp),%eax
  neg    %eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <forwword@plt>
  jmp    L4
L38:
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <backchar@plt>
  test   %rax,%rax
  jne    L6
  mov    $0x0,%eax
  jmp    L4
L84:
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <backchar@plt>
  test   %rax,%rax
  jne    L113
  mov    $0x0,%eax
  jmp    L4
L113:
  mov    $0x0,%eax
  call   <inword@plt>
  test   %rax,%rax
  je     L84
  jmp    L112
L117:
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <backchar@plt>
  test   %rax,%rax
  jne    L112
  mov    $0x0,%eax
  jmp    L4
L112:
  mov    $0x0,%eax
  call   <inword@plt>
  test   %rax,%rax
  jne    L117
L6:
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  test   %eax,%eax
  jne    L113
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <forwchar@plt>
L4:
  leave
  ret
```

assistant: ```c
int backword(int f, int n) {
  if (n < 0)
    return (forwword(f, -n));
  if (backchar(0, 1) == 0)
    return (0);
  while (n--) {
    while (inword() == 0) {
      if (backchar(0, 1) == 0)
        return (0);
    }
    while (inword() != 0) {
      if (backchar(0, 1) == 0)
        return (0);
    }
  }
  return (forwchar(0, 1));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<backword>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  test   %esi,%esi
  js     L3
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <backchar@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L44
  test   %ebx,%ebx
  je     L108
L103:
  mov    $0x0,%eax
  call   <inword@plt>
  test   %rax,%rax
  jne    L100
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <backchar@plt>
  test   %rax,%rax
  jne    L103
  mov    $0x0,%eax
L44:
  pop    %rbx
  ret
L3:
  neg    %ebx
  mov    %ebx,%esi
  call   <forwword@plt>
  jmp    L44
L100:
  mov    $0x0,%eax
  call   <inword@plt>
  test   %rax,%rax
  je     L54
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <backchar@plt>
  test   %rax,%rax
  jne    L100
  mov    $0x0,%eax
  jmp    L44
L54:
  sub    $0x1,%ebx
  jne    L103
L108:
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <forwchar@plt>
  jmp    L44
```

assistant: ```c
int backword(int f, int n) {
  if (n < 0)
    return (forwword(f, -n));
  if (backchar(0, 1) == 0)
    return (0);
  while (n--) {
    while (inword() == 0) {
      if (backchar(0, 1) == 0)
        return (0);
    }
    while (inword() != 0) {
      if (backchar(0, 1) == 0)
        return (0);
    }
  }
  return (forwchar(0, 1));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<backword>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  test   %esi,%esi
  js     L83
  xor    %edi,%edi
  mov    $0x1,%esi
  call   <backchar@plt>
  test   %rax,%rax
  je     L40
  test   %ebx,%ebx
  jne    L8
  jmp    L82
  cs nopw 0x0(%rax,%rax,1)
L14:
  xor    %edi,%edi
  mov    $0x1,%esi
  call   <backchar@plt>
  test   %rax,%rax
  je     L40
L8:
  xor    %eax,%eax
  call   <inword@plt>
  test   %rax,%rax
  je     L14
  jmp    L25
  nop
L40:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L39:
  xor    %edi,%edi
  mov    $0x1,%esi
  call   <backchar@plt>
  test   %rax,%rax
  je     L40
L25:
  xor    %eax,%eax
  call   <inword@plt>
  test   %rax,%rax
  jne    L39
  sub    $0x1,%ebx
  jne    L8
L82:
  mov    $0x1,%esi
  xor    %edi,%edi
  pop    %rbx
  jmp    1090 <forwchar@plt>
  nopw   0x0(%rax,%rax,1)
L83:
  neg    %ebx
  mov    %ebx,%esi
  pop    %rbx
  jmp    1080 <forwword@plt>
```

assistant: ```c
int backword(int f, int n) {
  if (n < 0)
    return (forwword(f, -n));
  if (backchar(0, 1) == 0)
    return (0);
  while (n--) {
    while (inword() == 0) {
      if (backchar(0, 1) == 0)
        return (0);
    }
    while (inword() != 0) {
      if (backchar(0, 1) == 0)
        return (0);
    }
  }
  return (forwchar(0, 1));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<backword>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  test   %esi,%esi
  js     L59
  xor    %edi,%edi
  mov    $0x1,%esi
  call   <backchar@plt>
  test   %rax,%rax
  je     L54
  test   %ebx,%ebx
  jne    L74
  jmp    L98
  cs nopw 0x0(%rax,%rax,1)
L96:
  xor    %edi,%edi
  mov    $0x1,%esi
  call   <backchar@plt>
  test   %rax,%rax
  je     L54
L74:
  xor    %eax,%eax
  call   <inword@plt>
  test   %rax,%rax
  je     L96
  jmp    L44
  nop
L54:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L40:
  xor    %edi,%edi
  mov    $0x1,%esi
  call   <backchar@plt>
  test   %rax,%rax
  je     L54
L44:
  xor    %eax,%eax
  call   <inword@plt>
  test   %rax,%rax
  jne    L40
  sub    $0x1,%ebx
  jne    L74
L98:
  mov    $0x1,%esi
  xor    %edi,%edi
  pop    %rbx
  jmp    1090 <forwchar@plt>
  nopw   0x0(%rax,%rax,1)
L59:
  neg    %ebx
  mov    %ebx,%esi
  pop    %rbx
  jmp    1080 <forwword@plt>
```

assistant: ```c
int backword(int f, int n) {
  if (n < 0)
    return (forwword(f, -n));
  if (backchar(0, 1) == 0)
    return (0);
  while (n--) {
    while (inword() == 0) {
      if (backchar(0, 1) == 0)
        return (0);
    }
    while (inword() != 0) {
      if (backchar(0, 1) == 0)
        return (0);
    }
  }
  return (forwchar(0, 1));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<SumWithRecursion>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x1,-0x1c(%rbp)
  jne    L13
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  jmp    L18
L13:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%ebx
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <SumWithRecursion>
  add    %ebx,%eax
L18:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int SumWithRecursion(int *arr, int len) {
  if (len == 1)
    return arr[0];
  return arr[0] + SumWithRecursion(arr + 1, len - 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<SumWithRecursion>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  cmp    $0x1,%esi
  je     L9
  sub    $0x1,%esi
  lea    0x4(%rdi),%rdi
  call   <SumWithRecursion>
  add    (%rbx),%eax
L13:
  pop    %rbx
  ret
L9:
  mov    (%rdi),%eax
  jmp    L13
```

assistant: ```c
int SumWithRecursion(int *arr, int len) {
  if (len == 1)
    return arr[0];
  return arr[0] + SumWithRecursion(arr + 1, len - 1);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<SumWithRecursion>:
  endbr64
  mov    %rdi,%rcx
  xor    %eax,%eax
  cmp    $0x1,%esi
  je     L5
  lea    -0x1(%rsi),%eax
  lea    (%rdi,%rax,4),%rcx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L0:
  mov    (%rdi),%edx
  add    $0x4,%rdi
  add    %edx,%eax
  cmp    %rcx,%rdi
  jne    L0
L5:
  add    (%rcx),%eax
  ret
```

assistant: ```c
int SumWithRecursion(int *arr, int len) {
  if (len == 1)
    return arr[0];
  return arr[0] + SumWithRecursion(arr + 1, len - 1);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<SumWithRecursion>:
  endbr64
  mov    %rdi,%rdx
  xor    %eax,%eax
  cmp    $0x1,%esi
  je     L53
  lea    -0x2(%rsi),%eax
  mov    %esi,%r9d
  lea    -0x1(%rsi),%edi
  cmp    $0x2,%eax
  jbe    L48
  mov    %edi,%ecx
  mov    %rdx,%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdx,%rcx,1),%rcx
  nopl   0x0(%rax)
L33:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rax,%rcx
  jne    L33
  movdqa %xmm0,%xmm1
  mov    %edi,%ecx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%ecx
  paddd  %xmm1,%xmm0
  mov    %ecx,%r8d
  sub    %ecx,%esi
  movdqa %xmm0,%xmm1
  lea    (%rdx,%r8,4),%r8
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %ecx,%edi
  je     L49
  lea    -0x1(%rsi),%edi
L62:
  add    (%r8),%eax
  cmp    $0x1,%edi
  je     L49
  add    0x4(%r8),%eax
  cmp    $0x3,%esi
  je     L49
  add    0x8(%r8),%eax
L49:
  lea    -0x1(%r9),%ecx
  lea    (%rdx,%rcx,4),%rdx
L53:
  add    (%rdx),%eax
  ret
L48:
  mov    %rdx,%r8
  xor    %eax,%eax
  jmp    L62
```

assistant: ```c
int SumWithRecursion(int *arr, int len) {
  if (len == 1)
    return arr[0];
  return arr[0] + SumWithRecursion(arr + 1, len - 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<break_continue_return>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x18(%rbp)
L0:
  addl   $0x1,-0x18(%rbp)
  cmpl   $0x9,-0x18(%rbp)
  jg     L11
  cmpl   $0x1,-0x18(%rbp)
  jle    L17
  mov    -0x18(%rbp),%eax
  cltq
  movb   $0x61,-0x12(%rbp,%rax,1)
  jmp    L0
L17:
  nop
  jmp    L0
L11:
  nop
  movl   $0x0,-0x18(%rbp)
L35:
  cmpl   $0xa,-0x18(%rbp)
  jle    L63
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  jmp    L38
L63:
  mov    -0x18(%rbp),%eax
  cltq
  movb   $0x61,-0x12(%rbp,%rax,1)
  addl   $0x1,-0x18(%rbp)
  jmp    L35
L38:
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
int break_continue_return() {
  int i = 0;
  char a[10];
  while (1) {
    i++;
    if (i >= 10)
      break;
    if (i < 2)
      continue;
    a[i] = 'a';
  }
  i = 0;
  while (1) {
    if (i > 10)
      return 0;
    a[i] = 'a';
    i++;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<break_continue_return>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int break_continue_return() {
  int i = 0;
  char a[10];
  while (1) {
    i++;
    if (i >= 10)
      break;
    if (i < 2)
      continue;
    a[i] = 'a';
  }
  i = 0;
  while (1) {
    if (i > 10)
      return 0;
    a[i] = 'a';
    i++;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<break_continue_return>:
  endbr64
  xor    %eax,%eax
L19:
  add    $0x1,%eax
L12:
  cmp    $0x1,%eax
  je     L19
  add    $0x1,%eax
  cmp    $0xa,%eax
  jne    L12
  xor    %eax,%eax
  ret
```

assistant: ```c
int break_continue_return() {
  int i = 0;
  char a[10];
  while (1) {
    i++;
    if (i >= 10)
      break;
    if (i < 2)
      continue;
    a[i] = 'a';
  }
  i = 0;
  while (1) {
    if (i > 10)
      return 0;
    a[i] = 'a';
    i++;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<break_continue_return>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int break_continue_return() {
  int i = 0;
  char a[10];
  while (1) {
    i++;
    if (i >= 10)
      break;
    if (i < 2)
      continue;
    a[i] = 'a';
  }
  i = 0;
  while (1) {
    if (i > 10)
      return 0;
    a[i] = 'a';
    i++;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<curses_end>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <clrtoeol@plt>
  mov    $0x0,%eax
  call   <refresh@plt>
  mov    $0x0,%eax
  call   <endwin@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void curses_end() {
  clrtoeol();
  refresh();
  endwin();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<curses_end>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <clrtoeol@plt>
  mov    $0x0,%eax
  call   <refresh@plt>
  mov    $0x0,%eax
  call   <endwin@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void curses_end() {
  clrtoeol();
  refresh();
  endwin();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<curses_end>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clrtoeol@plt>
  xor    %eax,%eax
  call   <refresh@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <endwin@plt>
```

assistant: ```c
void curses_end() {
  clrtoeol();
  refresh();
  endwin();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<curses_end>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clrtoeol@plt>
  xor    %eax,%eax
  call   <refresh@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <endwin@plt>
```

assistant: ```c
void curses_end() {
  clrtoeol();
  refresh();
  endwin();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<f_6_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x20(%rbp)
  movl   $0x2,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0x4,-0x14(%rbp)
  movl   $0x5,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <doStuff_pointerToConstInt@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void f_6_1(void) {
  int ta[5] = {1, 2, 3, 4, 5};
  doStuff_pointerToConstInt(ta);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<f_6_1>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x4,0xc(%rsp)
  movl   $0x5,0x10(%rsp)
  mov    %rsp,%rdi
  call   <doStuff_pointerToConstInt@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void f_6_1(void) {
  int ta[5] = {1, 2, 3, 4, 5};
  doStuff_pointerToConstInt(ta);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<f_6_1>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x200000001,%rax
  mov    %rsp,%rdi
  movl   $0x5,0x10(%rsp)
  mov    %rax,(%rsp)
  movabs $0x400000003,%rax
  mov    %rax,0x8(%rsp)
  call   <doStuff_pointerToConstInt@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void f_6_1(void) {
  int ta[5] = {1, 2, 3, 4, 5};
  doStuff_pointerToConstInt(ta);
}
```
user: Translate the below assembly code into C programming language:

```asm
<f_6_1>:
  endbr64
  sub    $0x28,%rsp
  movdqa D8(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movl   $0x5,0x10(%rsp)
  movaps %xmm0,(%rsp)
  call   <doStuff_pointerToConstInt@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D8
type: byte[16]

tool: {"D8": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
void f_6_1(void) {
  int ta[5] = {1, 2, 3, 4, 5};
  doStuff_pointerToConstInt(ta);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<dd_mul_d_dd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  movsd  %xmm0,-0x28(%rbp)
  mov    %rdi,-0x30(%rbp)
  mov    %rsi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x30(%rbp),%rax
  movsd  (%rax),%xmm0
  lea    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rdi
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <dd_two_prod@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x28(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  lea    0x8(%rax),%rdx
  movsd  -0x18(%rbp),%xmm0
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rdi
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <dd_quick_two_sum@plt>
  movq   %xmm0,%rax
  mov    -0x38(%rbp),%rdx
  mov    %rax,(%rdx)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void dd_mul_d_dd(double a, const double *b, double *c) {
  double p1, p2;
  p1 = dd_two_prod(a, b[0], &p2);
  p2 += (b[1] * a);
  c[0] = dd_quick_two_sum(p1, p2, &c[1]);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<dd_mul_d_dd>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movq   %xmm0,%rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movsd  (%r12),%xmm1
  call   <dd_two_prod@plt>
  movq   %rbx,%xmm1
  mulsd  0x8(%r12),%xmm1
  addsd  (%rsp),%xmm1
  movsd  %xmm1,(%rsp)
  lea    0x8(%rbp),%rdi
  call   <dd_quick_two_sum@plt>
  movsd  %xmm0,0x0(%rbp)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void dd_mul_d_dd(double a, const double *b, double *c) {
  double p1, p2;
  p1 = dd_two_prod(a, b[0], &p2);
  p2 += (b[1] * a);
  c[0] = dd_quick_two_sum(p1, p2, &c[1]);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<dd_mul_d_dd>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  movsd  0x0(%rbp),%xmm1
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <dd_two_prod@plt>
  movsd  0x8(%rsp),%xmm2
  mulsd  0x8(%rbp),%xmm2
  lea    0x8(%rbx),%rdi
  movsd  0x10(%rsp),%xmm1
  addsd  %xmm2,%xmm1
  movsd  %xmm1,0x10(%rsp)
  call   <dd_quick_two_sum@plt>
  movsd  %xmm0,(%rbx)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void dd_mul_d_dd(double a, const double *b, double *c) {
  double p1, p2;
  p1 = dd_two_prod(a, b[0], &p2);
  p2 += (b[1] * a);
  c[0] = dd_quick_two_sum(p1, p2, &c[1]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<dd_mul_d_dd>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  movsd  0x0(%rbp),%xmm1
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <dd_two_prod@plt>
  movsd  0x8(%rsp),%xmm2
  mulsd  0x8(%rbp),%xmm2
  lea    0x8(%rbx),%rdi
  movsd  0x10(%rsp),%xmm1
  addsd  %xmm2,%xmm1
  movsd  %xmm1,0x10(%rsp)
  call   <dd_quick_two_sum@plt>
  movsd  %xmm0,(%rbx)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void dd_mul_d_dd(double a, const double *b, double *c) {
  double p1, p2;
  p1 = dd_two_prod(a, b[0], &p2);
  p2 += (b[1] * a);
  c[0] = dd_quick_two_sum(p1, p2, &c[1]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<binsearch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  jmp    L68
L24:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jge    L16
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L64
L16:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L64:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
L68:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jg     L0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jne    L24
L0:
  mov    -0x14(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jne    L5
  mov    -0x4(%rbp),%eax
  jmp    L6
L5:
  mov    $0xffffffff,%eax
L6:
  pop    %rbp
  ret
```

assistant: ```c
int binsearch(int x, int v[], int n) {
  int low, high, mid;
  low = 0;
  high = n - 1;
  mid = (low + high) / 2;
  while (low <= high && x != v[mid]) {
    if (x < v[mid])
      high = mid - 1;
    else
      low = mid + 1;
    mid = (low + high) / 2;
  }
  if (x == mid)
    return mid;
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<binsearch>:
  endbr64
  mov    %edi,%r8d
  sub    $0x1,%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  mov    $0x0,%edi
  test   %edx,%edx
  jns    L9
L17:
  cmp    %eax,%r8d
  mov    $0xffffffff,%edx
  cmovne %edx,%eax
  ret
L39:
  lea    0x1(%rax),%edi
L30:
  lea    (%rdi,%rdx,1),%ecx
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  cmp    %edx,%edi
  jg     L17
L9:
  movslq %eax,%rcx
  mov    (%rsi,%rcx,4),%ecx
  cmp    %r8d,%ecx
  je     L17
  jle    L39
  lea    -0x1(%rax),%edx
  jmp    L30
```

assistant: ```c
int binsearch(int x, int v[], int n) {
  int low, high, mid;
  low = 0;
  high = n - 1;
  mid = (low + high) / 2;
  while (low <= high && x != v[mid]) {
    if (x < v[mid])
      high = mid - 1;
    else
      low = mid + 1;
    mid = (low + high) / 2;
  }
  if (x == mid)
    return mid;
  return -1;
}
```