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

```asm
<pr70085>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L1
  mov    -0x4(%rbp),%eax
  sub    -0x8(%rbp),%eax
  jmp    L12
L1:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jne    L4
  mov    $0x0,%eax
  jmp    L12
L4:
  mov    $0xffffffff,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int pr70085(int x, int y) {
  if (x > y)
    return x - y;
  if (x == y)
    return 0;
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pr70085>:
  endbr64
  cmp    %esi,%edi
  jg     L5
  setne  %dil
  movzbl %dil,%edi
  mov    %edi,%eax
  neg    %eax
  ret
L5:
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int pr70085(int x, int y) {
  if (x > y)
    return x - y;
  if (x == y)
    return 0;
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<pr70085>:
  endbr64
  cmp    %esi,%edi
  jg     L6
  setne  %dil
  movzbl %dil,%edi
  mov    %edi,%eax
  neg    %eax
  ret
  nopl   (%rax)
L6:
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int pr70085(int x, int y) {
  if (x > y)
    return x - y;
  if (x == y)
    return 0;
  return -1;
}
```
user: What is the c source code of the assembly code below:

```asm
<pr70085>:
  endbr64
  cmp    %esi,%edi
  jg     L9
  setne  %dil
  movzbl %dil,%edi
  mov    %edi,%eax
  neg    %eax
  ret
  nopl   (%rax)
L9:
  mov    %edi,%eax
  sub    %esi,%eax
  ret
```

assistant: ```c
int pr70085(int x, int y) {
  if (x > y)
    return x - y;
  if (x == y)
    return 0;
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<normdistribution>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x28(%rbp)
  mov    %rdi,-0x30(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x1,-0x14(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x14(%rbp),%xmm1
  movsd  D12(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x14(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  mov    %eax,%ecx
  mov    $0x1,%eax
  cltd
  idiv   %ecx
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x18(%rbp),%xmm1
  movsd  -0x28(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movq   D3(%rip),%xmm1
  xorpd  %xmm0,%xmm1
  pxor   %xmm2,%xmm2
  cvtsi2sdl -0x18(%rbp),%xmm2
  movsd  -0x28(%rbp),%xmm0
  subsd  %xmm2,%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x14(%rbp),%eax
  imul   %eax,%eax
  add    %eax,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  mov    D22(%rip),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <pow@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  mulsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D12
type_infered: byte[8]
name: unpack_data
arguments: data_label: D3
type_infered: byte[8]
name: unpack_data
arguments: data_label: D22
type_infered: f64

tool: {"D12": [0x18,0x2d,0x44,0x54,0xfb,0x21,0x19,0x40]}

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

tool: D22: 2.718281828459045

assistant: ```c
double normdistribution(double x, void *param) {
  int u = 1;
  int s = 1;
  double coef = 1 / (sqrt(2 * 3.14159265358979323846264338327950 * s * s));
  double exp = -1 * (x - u) * (x - u) / (2 * s * s);
  return coef * pow(2.718281828459045235360287471352662497757247093699959574966, exp);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<normdistribution>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movq   %xmm0,%rbx
  mov    $0x6,%edi
  call   <sqrt@plt>
  mov    %eax,%ebp
  movq   %rbx,%xmm0
  subsd  D31(%rip),%xmm0
  movapd %xmm0,%xmm1
  xorpd  D25(%rip),%xmm1
  mulsd  %xmm0,%xmm1
  mulsd  D36(%rip),%xmm1
  movsd  D3(%rip),%xmm0
  call   <pow@plt>
  movapd %xmm0,%xmm1
  mov    $0x1,%eax
  cltd
  idiv   %ebp
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  %xmm1,%xmm0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D31
type: f64
name: parse_data
arguments: tag: D25
type: byte[16]
name: parse_data
arguments: tag: D36
type: byte[8]
name: parse_data
arguments: tag: D3
type: f64

tool: D31: 1

tool: {"D25": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D36": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: D3: 2.718281828459045

assistant: ```c
double normdistribution(double x, void *param) {
  int u = 1;
  int s = 1;
  double coef = 1 / (sqrt(2 * 3.14159265358979323846264338327950 * s * s));
  double exp = -1 * (x - u) * (x - u) / (2 * s * s);
  return coef * pow(2.718281828459045235360287471352662497757247093699959574966, exp);
}
```
user: What is the c source code of the assembly code below:

```asm
<normdistribution>:
  endbr64
  push   %rbx
  mov    $0x6,%edi
  sub    $0x10,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <sqrt@plt>
  movsd  0x8(%rsp),%xmm0
  subsd  D15(%rip),%xmm0
  mov    %eax,%ebx
  movapd %xmm0,%xmm1
  xorpd  D29(%rip),%xmm1
  mulsd  %xmm0,%xmm1
  movsd  D20(%rip),%xmm0
  mulsd  D7(%rip),%xmm1
  call   <pow@plt>
  mov    $0x1,%eax
  add    $0x10,%rsp
  cltd
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  idiv   %ebx
  pop    %rbx
  cvtsi2sd %eax,%xmm0
  mulsd  %xmm1,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D15
data_type: f64
name: parse_data
arguments: data_tag: D29
data_type: byte[16]
name: parse_data
arguments: data_tag: D20
data_type: f64
name: parse_data
arguments: data_tag: D7
data_type: byte[8]

tool: D15: 1

tool: {"D29": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D20: 2.718281828459045

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

assistant: ```c
double normdistribution(double x, void *param) {
  int u = 1;
  int s = 1;
  double coef = 1 / (sqrt(2 * 3.14159265358979323846264338327950 * s * s));
  double exp = -1 * (x - u) * (x - u) / (2 * s * s);
  return coef * pow(2.718281828459045235360287471352662497757247093699959574966, exp);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<normdistribution>:
  endbr64
  push   %rbx
  mov    $0x6,%edi
  sub    $0x10,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <sqrt@plt>
  movsd  0x8(%rsp),%xmm0
  subsd  D37(%rip),%xmm0
  mov    %eax,%ebx
  movapd %xmm0,%xmm1
  xorpd  D23(%rip),%xmm1
  mulsd  %xmm0,%xmm1
  movsd  D36(%rip),%xmm0
  mulsd  D24(%rip),%xmm1
  call   <pow@plt>
  mov    $0x1,%eax
  add    $0x10,%rsp
  cltd
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  idiv   %ebx
  pop    %rbx
  cvtsi2sd %eax,%xmm0
  mulsd  %xmm1,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D37
data_type: f64
name: retrieve_data
arguments: data_tag: D23
data_type: byte[16]
name: retrieve_data
arguments: data_tag: D36
data_type: f64
name: retrieve_data
arguments: data_tag: D24
data_type: byte[8]

tool: D37: 1

tool: {"D23": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D36: 2.718281828459045

tool: {"D24": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

assistant: ```c
double normdistribution(double x, void *param) {
  int u = 1;
  int s = 1;
  double coef = 1 / (sqrt(2 * 3.14159265358979323846264338327950 * s * s));
  double exp = -1 * (x - u) * (x - u) / (2 * s * s);
  return coef * pow(2.718281828459045235360287471352662497757247093699959574966, exp);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getbits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L2
L38:
  subl   $0x1,-0xc(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  and    %eax,%edx
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  mov    -0x4(%rbp),%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  add    %eax,-0x8(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L29
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x30(%rbp),%rax
  movl   $0x7,(%rax)
  jmp    L2
L29:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
L2:
  cmpl   $0x0,-0xc(%rbp)
  jg     L38
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int getbits(unsigned char *buffer, int bits, int *byte_ptr, int *bit_ptr) {
  int bit, nr = bits, out = 0;
  while (nr > 0) {
    nr--;
    bit = ((int)buffer[*byte_ptr] & (1 << *bit_ptr)) >> *bit_ptr;
    out += bit << nr;
    if (*bit_ptr == 0) {
      (*byte_ptr)++;
      *bit_ptr = 7;
    } else {
      (*bit_ptr)--;
    }
  }
  return (out);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<getbits>:
  endbr64
  test   %esi,%esi
  jle    L36
  push   %r14
  push   %rbp
  push   %rbx
  mov    %rdi,%r11
  mov    %rdx,%r10
  mov    %rcx,%r8
  mov    $0x0,%r9d
  mov    $0x1,%ebx
  mov    $0x7,%ebp
  jmp    L25
L9:
  mov    %eax,(%r8)
  test   %esi,%esi
  je     L40
L25:
  sub    $0x1,%esi
  mov    (%r10),%edi
  mov    (%r8),%edx
  movslq %edi,%rax
  movzbl (%r11,%rax,1),%eax
  mov    %ebx,%r14d
  mov    %edx,%ecx
  shl    %cl,%r14d
  and    %r14d,%eax
  sar    %cl,%eax
  mov    %esi,%ecx
  shl    %cl,%eax
  add    %eax,%r9d
  lea    -0x1(%rdx),%eax
  test   %edx,%edx
  jne    L9
  add    $0x1,%edi
  mov    %edi,(%r10)
  mov    %ebp,%eax
  jmp    L9
L36:
  mov    $0x0,%r9d
  mov    %r9d,%eax
  ret
L40:
  mov    %r9d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r14
  ret
```

assistant: ```c
int getbits(unsigned char *buffer, int bits, int *byte_ptr, int *bit_ptr) {
  int bit, nr = bits, out = 0;
  while (nr > 0) {
    nr--;
    bit = ((int)buffer[*byte_ptr] & (1 << *bit_ptr)) >> *bit_ptr;
    out += bit << nr;
    if (*bit_ptr == 0) {
      (*byte_ptr)++;
      *bit_ptr = 7;
    } else {
      (*bit_ptr)--;
    }
  }
  return (out);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<getbits>:
  endbr64
  test   %esi,%esi
  jle    L9
  push   %r14
  mov    %rdi,%r10
  mov    %rdx,%r11
  mov    %rcx,%r8
  push   %rbx
  xor    %r9d,%r9d
  mov    $0x1,%ebx
  nopl   0x0(%rax)
L19:
  movslq (%r11),%rax
  mov    (%r8),%edx
  mov    %ebx,%r14d
  sub    $0x1,%esi
  mov    %edx,%ecx
  mov    %rax,%rdi
  movzbl (%r10,%rax,1),%eax
  shl    %cl,%r14d
  and    %r14d,%eax
  sar    %cl,%eax
  mov    %esi,%ecx
  shl    %cl,%eax
  add    %eax,%r9d
  lea    -0x1(%rdx),%eax
  test   %edx,%edx
  jne    L16
  add    $0x1,%edi
  mov    $0x7,%eax
  mov    %edi,(%r11)
L16:
  mov    %eax,(%r8)
  test   %esi,%esi
  jne    L19
  mov    %r9d,%eax
  pop    %rbx
  pop    %r14
  ret
  nopl   (%rax)
L9:
  xor    %eax,%eax
  ret
```

assistant: ```c
int getbits(unsigned char *buffer, int bits, int *byte_ptr, int *bit_ptr) {
  int bit, nr = bits, out = 0;
  while (nr > 0) {
    nr--;
    bit = ((int)buffer[*byte_ptr] & (1 << *bit_ptr)) >> *bit_ptr;
    out += bit << nr;
    if (*bit_ptr == 0) {
      (*byte_ptr)++;
      *bit_ptr = 7;
    } else {
      (*bit_ptr)--;
    }
  }
  return (out);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getbits>:
  endbr64
  test   %esi,%esi
  jle    L40
  push   %r14
  mov    %rdx,%r9
  mov    %rdi,%r10
  mov    %rcx,%r11
  push   %rbx
  mov    (%rcx),%edx
  xor    %r8d,%r8d
  mov    $0x1,%ebx
  xchg   %ax,%ax
L22:
  movslq (%r9),%rax
  mov    %edx,%ecx
  mov    %ebx,%r14d
  sub    $0x1,%esi
  shl    %cl,%r14d
  mov    %rax,%rdi
  movzbl (%r10,%rax,1),%eax
  and    %r14d,%eax
  sar    %cl,%eax
  mov    %esi,%ecx
  shl    %cl,%eax
  add    %eax,%r8d
  test   %edx,%edx
  jne    L46
  add    $0x1,%edi
  mov    %edi,(%r9)
  movl   $0x7,(%r11)
  test   %esi,%esi
  je     L25
  mov    $0x7,%edx
  jmp    L22
  xchg   %ax,%ax
L46:
  sub    $0x1,%edx
  mov    %edx,(%r11)
  test   %esi,%esi
  jne    L22
L25:
  mov    %r8d,%eax
  pop    %rbx
  pop    %r14
  ret
  nopl   0x0(%rax)
L40:
  xor    %eax,%eax
  ret
```

assistant: ```c
int getbits(unsigned char *buffer, int bits, int *byte_ptr, int *bit_ptr) {
  int bit, nr = bits, out = 0;
  while (nr > 0) {
    nr--;
    bit = ((int)buffer[*byte_ptr] & (1 << *bit_ptr)) >> *bit_ptr;
    out += bit << nr;
    if (*bit_ptr == 0) {
      (*byte_ptr)++;
      *bit_ptr = 7;
    } else {
      (*bit_ptr)--;
    }
  }
  return (out);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sump>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L19:
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    (%rax),%eax
  add    %eax,-0x4(%rbp)
L13:
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L19
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int sump(int *start, int *end) {
  int total = 0;
  while (start < end)
    total += *start++;
  return total;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sump>:
  endbr64
  cmp    %rsi,%rdi
  jae    L6
  mov    $0x0,%eax
L10:
  add    $0x4,%rdi
  add    -0x4(%rdi),%eax
  cmp    %rdi,%rsi
  ja     L10
  ret
L6:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int sump(int *start, int *end) {
  int total = 0;
  while (start < end)
    total += *start++;
  return total;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sump>:
  endbr64
  xor    %eax,%eax
  cmp    %rsi,%rdi
  jae    L12
  nopl   0x0(%rax,%rax,1)
L0:
  add    $0x4,%rdi
  add    -0x4(%rdi),%eax
  cmp    %rdi,%rsi
  ja     L0
  ret
  nopl   (%rax)
L12:
  ret
```

assistant: ```c
int sump(int *start, int *end) {
  int total = 0;
  while (start < end)
    total += *start++;
  return total;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sump>:
  endbr64
  mov    %rdi,%rdx
  cmp    %rsi,%rdi
  jae    L9
  lea    -0x1(%rsi),%rcx
  mov    %rdi,%rax
  sub    %rdi,%rcx
  mov    %rcx,%rdi
  shr    $0x2,%rdi
  add    $0x1,%rdi
  cmp    $0xb,%rcx
  jbe    L53
  mov    %rdi,%rcx
  pxor   %xmm0,%xmm0
  shr    $0x2,%rcx
  shl    $0x4,%rcx
  add    %rdx,%rcx
  nopl   (%rax)
L15:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rax,%rcx
  jne    L15
  movdqa %xmm0,%xmm1
  mov    %rdi,%rcx
  psrldq $0x8,%xmm1
  and    $0xfffffffffffffffc,%rcx
  paddd  %xmm1,%xmm0
  lea    (%rdx,%rcx,4),%rdx
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %rcx,%rdi
  je     L11
L24:
  lea    0x4(%rdx),%rcx
  add    (%rdx),%eax
  cmp    %rcx,%rsi
  jbe    L34
  lea    0x8(%rdx),%rcx
  add    0x4(%rdx),%eax
  cmp    %rcx,%rsi
  jbe    L34
  add    0x8(%rdx),%eax
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  xor    %eax,%eax
L34:
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  ret
L53:
  xor    %eax,%eax
  jmp    L24
```

assistant: ```c
int sump(int *start, int *end) {
  int total = 0;
  while (start < end)
    total += *start++;
  return total;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L18:
  addq   $0x1,-0x8(%rbp)
L4:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L18
  subq   $0x1,-0x8(%rbp)
  jmp    L35
L14:
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
  addq   $0x1,-0x18(%rbp)
  subq   $0x1,-0x8(%rbp)
L35:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L14
  nop
  nop
  leave
  ret
```

assistant: ```c
void reverse(char *str) {
  char *end;
  for (end = str; *end; ++end)
    ;
  --end;
  while (str < end) {
    swap(str, end);
    ++str;
    --end;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<reverse>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  cmpb   $0x0,(%rdi)
  je     L41
  mov    %rdi,%rbx
L33:
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  jne    L33
L21:
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  jae    L0
L24:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <swap@plt>
  add    $0x1,%rbp
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  jb     L24
L0:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L41:
  mov    %rdi,%rbx
  jmp    L21
```

assistant: ```c
void reverse(char *str) {
  char *end;
  for (end = str; *end; ++end)
    ;
  --end;
  while (str < end) {
    swap(str, end);
    ++str;
    --end;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmpb   $0x0,(%rdi)
  je     L32
  mov    %rdi,%rax
  nopl   (%rax)
L2:
  mov    %rax,%rbx
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L2
L13:
  cmp    %rbp,%rbx
  jbe    L28
  nopl   0x0(%rax)
L21:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  sub    $0x1,%rbx
  add    $0x1,%rbp
  call   <swap@plt>
  cmp    %rbx,%rbp
  jb     L21
L28:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nop
L32:
  lea    -0x1(%rdi),%rbx
  jmp    L13
```

assistant: ```c
void reverse(char *str) {
  char *end;
  for (end = str; *end; ++end)
    ;
  --end;
  while (str < end) {
    swap(str, end);
    ++str;
    --end;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmpb   $0x0,(%rdi)
  je     L15
  mov    %rdi,%rax
  nopl   (%rax)
L43:
  mov    %rax,%rbx
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L43
L3:
  cmp    %rbp,%rbx
  jbe    L37
  nopl   0x0(%rax)
L4:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  sub    $0x1,%rbx
  add    $0x1,%rbp
  call   <swap@plt>
  cmp    %rbx,%rbp
  jb     L4
L37:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nop
L15:
  lea    -0x1(%rdi),%rbx
  jmp    L3
```

assistant: ```c
void reverse(char *str) {
  char *end;
  for (end = str; *end; ++end)
    ;
  --end;
  while (str < end) {
    swap(str, end);
    ++str;
    --end;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<time_units>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movsd  (%rax),%xmm0
  comisd D39(%rip),%xmm0
  jbe    L49
  lea    D19(%rip),%rax
  jmp    L41
L49:
  mov    -0x8(%rbp),%rax
  movsd  (%rax),%xmm0
  comisd D97(%rip),%xmm0
  jb     L47
  mov    -0x8(%rbp),%rax
  movsd  (%rax),%xmm1
  movsd  D99(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x8(%rbp),%rax
  movsd  %xmm0,(%rax)
  lea    D15(%rip),%rax
  jmp    L41
L47:
  mov    -0x8(%rbp),%rax
  movsd  (%rax),%xmm0
  comisd D5(%rip),%xmm0
  jb     L28
  mov    -0x8(%rbp),%rax
  movsd  (%rax),%xmm1
  movsd  D64(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x8(%rbp),%rax
  movsd  %xmm0,(%rax)
  lea    D11(%rip),%rax
  jmp    L41
L28:
  mov    -0x8(%rbp),%rax
  movsd  (%rax),%xmm1
  movsd  D60(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x8(%rbp),%rax
  movsd  %xmm0,(%rax)
  lea    D31(%rip),%rax
L41:
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D39
type_guessed: byte[16]
name: read_data
arguments: tag: D19
type_guessed: string
name: read_data
arguments: tag: D97
type_guessed: byte[16]
name: read_data
arguments: tag: D99
type_guessed: f64
name: read_data
arguments: tag: D15
type_guessed: string
name: read_data
arguments: tag: D5
type_guessed: byte[16]
name: read_data
arguments: tag: D64
type_guessed: f64
name: read_data
arguments: tag: D11
type_guessed: string
name: read_data
arguments: tag: D60
type_guessed: f64
name: read_data
arguments: tag: D31
type_guessed: string

tool: {"D39": [0x9a,0x99,0x99,0x99,0x99,0x99,0xb9,0x3f,0xfc,0xa9,0xf1,0xd2,0x4d,0x62,0x50,0x3f]}

tool: D19: s

tool: {"D97": [0xfc,0xa9,0xf1,0xd2,0x4d,0x62,0x50,0x3f,0x00,0x00,0x00,0x00,0x00,0x40,0x8f,0x40]}

tool: D99: 1000

tool: D15: ms

tool: {"D5": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e,0x00,0x00,0x00,0x00,0x80,0x84,0x2e,0x41]}

tool: D64: 1000000

tool: D11: us

tool: D60: 1000000000

tool: D31: ns

assistant: ```c
const char *time_units(double *secs) {
  if (*secs > 1e-1) {
    return "s";
  } else if (*secs >= 1e-3) {
    *secs *= 1e3;
    return "ms";
  } else if (*secs >= 1e-6) {
    *secs *= 1e6;
    return "us";
  } else {
    *secs *= 1e9;
    return "ns";
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<time_units>:
  endbr64
  movsd  (%rdi),%xmm0
  lea    D2(%rip),%rax
  comisd D70(%rip),%xmm0
  ja     L16
  comisd D10(%rip),%xmm0
  jae    L9
  comisd D41(%rip),%xmm0
  jb     L0
  mulsd  D80(%rip),%xmm0
  movsd  %xmm0,(%rdi)
  lea    D61(%rip),%rax
  ret
L9:
  mulsd  D71(%rip),%xmm0
  movsd  %xmm0,(%rdi)
  lea    D65(%rip),%rax
  ret
L0:
  mulsd  D75(%rip),%xmm0
  movsd  %xmm0,(%rdi)
  lea    D3(%rip),%rax
L16:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
datatype: string
name: extract_symbolic_data
arguments: label: D70
datatype: byte[16]
name: extract_symbolic_data
arguments: label: D10
datatype: byte[16]
name: extract_symbolic_data
arguments: label: D41
datatype: byte[16]
name: extract_symbolic_data
arguments: label: D80
datatype: f64
name: extract_symbolic_data
arguments: label: D61
datatype: string
name: extract_symbolic_data
arguments: label: D71
datatype: f64
name: extract_symbolic_data
arguments: label: D65
datatype: string
name: extract_symbolic_data
arguments: label: D75
datatype: f64
name: extract_symbolic_data
arguments: label: D3
datatype: string

tool: D2: s

tool: {"D70": [0x9a,0x99,0x99,0x99,0x99,0x99,0xb9,0x3f,0xfc,0xa9,0xf1,0xd2,0x4d,0x62,0x50,0x3f]}

tool: {"D10": [0xfc,0xa9,0xf1,0xd2,0x4d,0x62,0x50,0x3f,0x00,0x00,0x00,0x00,0x00,0x40,0x8f,0x40]}

tool: {"D41": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e,0x00,0x00,0x00,0x00,0x80,0x84,0x2e,0x41]}

tool: D80: 1000000

tool: D61: us

tool: D71: 1000

tool: D65: ms

tool: D75: 1000000000

tool: D3: ns

assistant: ```c
const char *time_units(double *secs) {
  if (*secs > 1e-1) {
    return "s";
  } else if (*secs >= 1e-3) {
    *secs *= 1e3;
    return "ms";
  } else if (*secs >= 1e-6) {
    *secs *= 1e6;
    return "us";
  } else {
    *secs *= 1e9;
    return "ns";
  }
}
```