ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x14(%rsp)
  jle    L9
  mov    $0x1,%ebx
  lea    0x10(%rsp),%r14
  lea    0xc(%rsp),%r13
  lea    D4(%rip),%r12
  lea    D23(%rip),%rbp
L1:
  lea    0x8(%rsp),%rsi
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  add    0x8(%rsp),%eax
  mov    0x10(%rsp),%edx
  cmp    %edx,%eax
  jl     L25
  mov    $0x0,%esi
L22:
  add    $0x1,%eax
  sub    %edx,%eax
  add    $0x1,%esi
  cmp    %edx,%eax
  jge    L22
L26:
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x14(%rsp)
  jge    L1
L9:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L25:
  mov    $0x0,%esi
  jmp    L26
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type: string
name: extract_symbolic_data
arguments: label: D4
type: string
name: extract_symbolic_data
arguments: label: D23
type: string

tool: D1: %d

tool: D4: %d %d %d

tool: D23: %d


assistant: ```c
int main() {
  int i, e, f, c, n, s, test;
  scanf("%d", &test);
  for (i = 1; i <= test; i++) {
    scanf("%d %d %d", &e, &f, &c);
    n = e + f;
    s = 0;
    while (n >= c) {
      n = (n - c) + 1;
      s++;
    }
    printf("%d\n", s);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D24(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L3
  mov    $0x1,%ebx
  lea    0x10(%rsp),%r15
  lea    0xc(%rsp),%r14
  lea    0x8(%rsp),%r13
  lea    D0(%rip),%r12
  lea    D10(%rip),%rbp
  nopl   (%rax)
L28:
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r15,%rcx
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%eax
  xor    %esi,%esi
  add    0x8(%rsp),%eax
  cmp    %edx,%eax
  jl     L31
  nopl   (%rax)
L1:
  add    $0x1,%eax
  add    $0x1,%esi
  sub    %edx,%eax
  cmp    %edx,%eax
  jge    L1
L31:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x14(%rsp)
  jge    L28
L3:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D24
data_type: string
name: read_data
arguments: label: D0
data_type: string
name: read_data
arguments: label: D10
data_type: string

tool: D24: %d

tool: D0: %d %d %d

tool: D10: %d


assistant: ```c
int main() {
  int i, e, f, c, n, s, test;
  scanf("%d", &test);
  for (i = 1; i <= test; i++) {
    scanf("%d %d %d", &e, &f, &c);
    n = e + f;
    s = 0;
    while (n >= c) {
      n = (n - c) + 1;
      s++;
    }
    printf("%d\n", s);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D16(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L46
  mov    $0x1,%ebx
  lea    0x10(%rsp),%r15
  lea    0xc(%rsp),%r14
  lea    0x8(%rsp),%r13
  lea    D29(%rip),%r12
  lea    D11(%rip),%rbp
  nopl   (%rax)
L41:
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r15,%rcx
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%eax
  xor    %esi,%esi
  add    0x8(%rsp),%eax
  cmp    %edx,%eax
  jl     L15
  nopl   (%rax)
L40:
  add    $0x1,%eax
  add    $0x1,%esi
  sub    %edx,%eax
  cmp    %edx,%eax
  jge    L40
L15:
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0x14(%rsp)
  jge    L41
L46:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D16
type: string
name: extract_symbolic_data
arguments: data_tag: D29
type: string
name: extract_symbolic_data
arguments: data_tag: D11
type: string

tool: D16: %d

tool: D29: %d %d %d

tool: D11: %d


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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D7(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x68,%al
  setne  %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
data_type: string

tool: D7: hello is better than bye

assistant: ```c
int main(void) {
  char *p;
  p = "hello"
      " is better than bye";
  return (*p == 'h') ? 0 : 1;
}
```
user: What would the C source look like for the assembly code provided here?

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

assistant: ```c
int main(void) {
  char *p;
  p = "hello"
      " is better than bye";
  return (*p == 'h') ? 0 : 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

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

assistant: ```c
int main(void) {
  char *p;
  p = "hello"
      " is better than bye";
  return (*p == 'h') ? 0 : 1;
}
```
user: What would the C source look like for the assembly code provided here?

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

assistant: ```c
int main(void) {
  char *p;
  p = "hello"
      " is better than bye";
  return (*p == 'h') ? 0 : 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<factoriel>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jle    L12
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <factoriel>
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  imul   %rdx,%rax
  jmp    L1
L12:
  mov    $0x1,%eax
L1:
  leave
  ret
```

assistant: ```c
long long unsigned int factoriel(int n) {
  if (n > 1) {
    return factoriel(n - 1) * n;
  } else {
    return 1;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<factoriel>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jg     L8
  ret
L8:
  push   %rbx
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <factoriel>
  mov    %rax,%rdx
  movslq %ebx,%rax
  imul   %rdx,%rax
  pop    %rbx
  ret
```

assistant: ```c
long long unsigned int factoriel(int n) {
  if (n > 1) {
    return factoriel(n - 1) * n;
  } else {
    return 1;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<factoriel>:
  endbr64
  cmp    $0x1,%edi
  jle    L7
  movslq %edi,%rdx
  lea    -0x2(%rdi),%esi
  mov    $0x1,%r8d
  lea    -0x1(%rdx),%rax
  mov    %rax,%rcx
  sub    %rsi,%rcx
  jmp    L20
  nopl   0x0(%rax)
L21:
  sub    $0x1,%rax
L20:
  imul   %rdx,%r8
  mov    %rax,%rdx
  cmp    %rcx,%rax
  jne    L21
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L7:
  mov    $0x1,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
long long unsigned int factoriel(int n) {
  if (n > 1) {
    return factoriel(n - 1) * n;
  } else {
    return 1;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<factoriel>:
  endbr64
  cmp    $0x1,%edi
  jle    L25
  movslq %edi,%rdx
  lea    -0x2(%rdi),%esi
  mov    $0x1,%r8d
  lea    -0x1(%rdx),%rax
  mov    %rax,%rcx
  sub    %rsi,%rcx
  jmp    L3
  nopl   0x0(%rax)
L19:
  sub    $0x1,%rax
L3:
  imul   %rdx,%r8
  mov    %rax,%rdx
  cmp    %rcx,%rax
  jne    L19
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L25:
  mov    $0x1,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
long long unsigned int factoriel(int n) {
  if (n > 1) {
    return factoriel(n - 1) * n;
  } else {
    return 1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f_print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,-0x8(%rbp)
  mov    %r8,-0x20(%rbp)
  mov    %r9,-0x28(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  js     L2
  mov    -0x20(%rbp),%rax
  mov    0x10(%rbp),%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L2:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int f_print(int mode, unsigned char **sec, float *data, unsigned int ndata, char *inv_out, void **local, const char *arg1) {
  if (mode >= 0) {
    strcpy(inv_out, arg1);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<f_print>:
  endbr64
  test   %edi,%edi
  jns    L1
  mov    $0x0,%eax
  ret
L1:
  sub    $0x8,%rsp
  mov    0x10(%rsp),%rsi
  mov    %r8,%rdi
  call   <strcpy@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int f_print(int mode, unsigned char **sec, float *data, unsigned int ndata, char *inv_out, void **local, const char *arg1) {
  if (mode >= 0) {
    strcpy(inv_out, arg1);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<f_print>:
  endbr64
  test   %edi,%edi
  jns    L9
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  sub    $0x8,%rsp
  mov    %r8,%rdi
  mov    0x10(%rsp),%rsi
  call   <strcpy@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int f_print(int mode, unsigned char **sec, float *data, unsigned int ndata, char *inv_out, void **local, const char *arg1) {
  if (mode >= 0) {
    strcpy(inv_out, arg1);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f_print>:
  endbr64
  test   %edi,%edi
  jns    L8
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L8:
  sub    $0x8,%rsp
  mov    %r8,%rdi
  mov    0x10(%rsp),%rsi
  call   <strcpy@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int f_print(int mode, unsigned char **sec, float *data, unsigned int ndata, char *inv_out, void **local, const char *arg1) {
  if (mode >= 0) {
    strcpy(inv_out, arg1);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<binary_search>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    -0x20(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jle    L7
  mov    $0xffffffff,%eax
  jmp    L26
L7:
  mov    -0x20(%rbp),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L50
  mov    -0x4(%rbp),%eax
  jmp    L26
L50:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  jge    L53
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <binary_search>
  jmp    L26
L53:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edi
  mov    -0x24(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %edx,%ecx
  mov    %edi,%edx
  mov    %rax,%rdi
  call   <binary_search>
L26:
  leave
  ret
```

assistant: ```c
int binary_search(int s[], int key, int low, int high) {
  int middle;
  if (low > high)
    return -1;
  middle = (low + high) / 2;
  if (s[middle] == key)
    return middle;
  else if (s[middle] > key)
    return (binary_search(s, key, low, middle - 1));
  else
    return (binary_search(s, key, middle + 1, high));
}
```
user: Translate the below assembly code into C programming language:

```asm
<binary_search>:
  endbr64
  cmp    %ecx,%edx
  jg     L19
  sub    $0x8,%rsp
  mov    %rdi,%r8
  mov    %esi,%edi
  lea    (%rdx,%rcx,1),%eax
  mov    %eax,%esi
  shr    $0x1f,%esi
  add    %eax,%esi
  sar    %esi
  mov    %esi,%eax
  movslq %esi,%r9
  mov    (%r8,%r9,4),%r9d
  cmp    %edi,%r9d
  je     L38
  jle    L33
  lea    -0x1(%rsi),%ecx
  mov    %edi,%esi
  mov    %r8,%rdi
  call   <binary_search>
L38:
  add    $0x8,%rsp
  ret
L33:
  lea    0x1(%rsi),%edx
  mov    %edi,%esi
  mov    %r8,%rdi
  call   <binary_search>
  jmp    L38
L19:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int binary_search(int s[], int key, int low, int high) {
  int middle;
  if (low > high)
    return -1;
  middle = (low + high) / 2;
  if (s[middle] == key)
    return middle;
  else if (s[middle] > key)
    return (binary_search(s, key, low, middle - 1));
  else
    return (binary_search(s, key, middle + 1, high));
}
```
user: What is the c source code of the assembly code below?

```asm
<binary_search>:
  endbr64
  mov    %rdi,%r8
  cmp    %ecx,%edx
  jle    L27
  jmp    L13
  nopl   (%rax)
L2:
  lea    -0x1(%rax),%ecx
  cmp    %edx,%ecx
  jl     L13
L27:
  lea    (%rdx,%rcx,1),%edi
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %edi,%eax
  sar    %eax
  movslq %eax,%rdi
  mov    %eax,%r9d
  cmp    %esi,(%r8,%rdi,4)
  je     L4
  jg     L2
  lea    0x1(%rax),%edx
  cmp    %edx,%ecx
  jge    L27
L13:
  mov    $0xffffffff,%r9d
L4:
  mov    %r9d,%eax
  ret
```

assistant: ```c
int binary_search(int s[], int key, int low, int high) {
  int middle;
  if (low > high)
    return -1;
  middle = (low + high) / 2;
  if (s[middle] == key)
    return middle;
  else if (s[middle] > key)
    return (binary_search(s, key, low, middle - 1));
  else
    return (binary_search(s, key, middle + 1, high));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<binary_search>:
  endbr64
  nopl   0x0(%rax)
L5:
  cmp    %ecx,%edx
  jg     L46
L12:
  lea    (%rdx,%rcx,1),%r8d
  mov    %r8d,%eax
  shr    $0x1f,%eax
  add    %r8d,%eax
  sar    %eax
  movslq %eax,%r8
  cmp    %esi,(%rdi,%r8,4)
  je     L34
  jle    L25
  lea    -0x1(%rax),%ecx
  cmp    %edx,%ecx
  jge    L12
L46:
  mov    $0xffffffff,%eax
L34:
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  lea    0x1(%rax),%edx
  jmp    L5
```

assistant: ```c
int binary_search(int s[], int key, int low, int high) {
  int middle;
  if (low > high)
    return -1;
  middle = (low + high) / 2;
  if (s[middle] == key)
    return middle;
  else if (s[middle] > key)
    return (binary_search(s, key, low, middle - 1));
  else
    return (binary_search(s, key, middle + 1, high));
}
```
user: What is the c source code of the assembly code below:

```asm
<polylog_get_zone>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x38(%rbp)
  movsd  %xmm1,-0x40(%rbp)
  movsd  -0x38(%rbp),%xmm0
  movsd  D3(%rip),%xmm2
  movapd %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  movsd  -0x38(%rbp),%xmm0
  movsd  D3(%rip),%xmm2
  subsd  %xmm2,%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x40(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  addsd  %xmm0,%xmm1
  movsd  D3(%rip),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  -0x38(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  -0x40(%rbp),%xmm0
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  subsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x38(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  movsd  -0x40(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x38(%rbp),%xmm0
  movsd  D3(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x20(%rbp),%xmm1
  movsd  -0x40(%rbp),%xmm0
  mulsd  -0x18(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x38(%rbp),%xmm0
  movsd  D3(%rip),%xmm1
  subsd  %xmm1,%xmm0
  mulsd  -0x18(%rbp),%xmm0
  movsd  -0x40(%rbp),%xmm1
  mulsd  -0x20(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x10(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  -0x8(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  addsd  %xmm1,%xmm0
  mulsd  -0x28(%rbp),%xmm0
  movsd  -0x28(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  -0x28(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
type_guessed: f64

tool: D3: 1

assistant: ```c
double polylog_get_zone(double zre, double zim) {
  double den = 1.0 / ((zre - 1.0) * (zre - 1.0) + zim * zim);
  double sre = zre * zre - zim * zim;
  double sim = 2.0 * zre * zim;
  double fre = sre * (zre - 1.0) + zim * sim;
  double fim = sim * (zre - 1.0) - zim * sre;
  den = (fre * fre + fim * fim) * den * den;
  return den;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<polylog_get_zone>:
  endbr64
  movsd  D8(%rip),%xmm4
  movapd %xmm0,%xmm3
  subsd  %xmm4,%xmm3
  movapd %xmm1,%xmm6
  mulsd  %xmm1,%xmm6
  movapd %xmm3,%xmm2
  mulsd  %xmm3,%xmm2
  addsd  %xmm6,%xmm2
  divsd  %xmm2,%xmm4
  movapd %xmm0,%xmm5
  mulsd  %xmm0,%xmm5
  subsd  %xmm6,%xmm5
  addsd  %xmm0,%xmm0
  mulsd  %xmm1,%xmm0
  movapd %xmm3,%xmm2
  mulsd  %xmm5,%xmm2
  movapd %xmm1,%xmm6
  mulsd  %xmm0,%xmm6
  addsd  %xmm6,%xmm2
  mulsd  %xmm0,%xmm3
  mulsd  %xmm5,%xmm1
  subsd  %xmm1,%xmm3
  mulsd  %xmm2,%xmm2
  mulsd  %xmm3,%xmm3
  addsd  %xmm3,%xmm2
  mulsd  %xmm4,%xmm2
  movapd %xmm2,%xmm0
  mulsd  %xmm4,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
datatype: f64

tool: D8: 1

assistant: ```c
double polylog_get_zone(double zre, double zim) {
  double den = 1.0 / ((zre - 1.0) * (zre - 1.0) + zim * zim);
  double sre = zre * zre - zim * zim;
  double sim = 2.0 * zre * zim;
  double fre = sre * (zre - 1.0) + zim * sim;
  double fim = sim * (zre - 1.0) - zim * sre;
  den = (fre * fre + fim * fim) * den * den;
  return den;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<polylog_get_zone>:
  endbr64
  movsd  D5(%rip),%xmm4
  movapd %xmm0,%xmm3
  movapd %xmm1,%xmm6
  movapd %xmm0,%xmm5
  mulsd  %xmm1,%xmm6
  subsd  %xmm4,%xmm3
  mulsd  %xmm0,%xmm5
  addsd  %xmm0,%xmm0
  mulsd  %xmm1,%xmm0
  movapd %xmm3,%xmm2
  mulsd  %xmm3,%xmm2
  subsd  %xmm6,%xmm5
  addsd  %xmm6,%xmm2
  movapd %xmm1,%xmm6
  mulsd  %xmm0,%xmm6
  mulsd  %xmm5,%xmm1
  divsd  %xmm2,%xmm4
  movapd %xmm3,%xmm2
  mulsd  %xmm5,%xmm2
  mulsd  %xmm0,%xmm3
  addsd  %xmm6,%xmm2
  subsd  %xmm1,%xmm3
  mulsd  %xmm2,%xmm2
  mulsd  %xmm3,%xmm3
  addsd  %xmm3,%xmm2
  mulsd  %xmm4,%xmm2
  movapd %xmm2,%xmm0
  mulsd  %xmm4,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_label: D5
datatype: f64

tool: D5: 1

assistant: ```c
double polylog_get_zone(double zre, double zim) {
  double den = 1.0 / ((zre - 1.0) * (zre - 1.0) + zim * zim);
  double sre = zre * zre - zim * zim;
  double sim = 2.0 * zre * zim;
  double fre = sre * (zre - 1.0) + zim * sim;
  double fim = sim * (zre - 1.0) - zim * sre;
  den = (fre * fre + fim * fim) * den * den;
  return den;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<polylog_get_zone>:
  endbr64
  movsd  D0(%rip),%xmm4
  movapd %xmm0,%xmm3
  movapd %xmm1,%xmm6
  movapd %xmm0,%xmm5
  mulsd  %xmm1,%xmm6
  subsd  %xmm4,%xmm3
  mulsd  %xmm0,%xmm5
  addsd  %xmm0,%xmm0
  mulsd  %xmm1,%xmm0
  movapd %xmm3,%xmm2
  mulsd  %xmm3,%xmm2
  subsd  %xmm6,%xmm5
  addsd  %xmm6,%xmm2
  movapd %xmm1,%xmm6
  mulsd  %xmm0,%xmm6
  mulsd  %xmm5,%xmm1
  divsd  %xmm2,%xmm4
  movapd %xmm3,%xmm2
  mulsd  %xmm5,%xmm2
  mulsd  %xmm0,%xmm3
  addsd  %xmm6,%xmm2
  subsd  %xmm1,%xmm3
  mulsd  %xmm2,%xmm2
  mulsd  %xmm3,%xmm3
  addsd  %xmm3,%xmm2
  mulsd  %xmm4,%xmm2
  movapd %xmm2,%xmm0
  mulsd  %xmm4,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
data_type: f64

tool: D0: 1

assistant: ```c
double polylog_get_zone(double zre, double zim) {
  double den = 1.0 / ((zre - 1.0) * (zre - 1.0) + zim * zim);
  double sre = zre * zre - zim * zim;
  double sim = 2.0 * zre * zim;
  double fre = sre * (zre - 1.0) + zim * sim;
  double fim = sim * (zre - 1.0) - zim * sre;
  den = (fre * fre + fim * fim) * den * den;
  return den;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_truncatable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <num_digits@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <is_prime@plt>
  test   %eax,%eax
  jne    L91
  mov    $0x0,%eax
  jmp    L61
L91:
  movl   $0x1,-0x10(%rbp)
  jmp    L24
L0:
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    -0x24(%rbp),%eax
  mov    $0x0,%edx
  div    %ecx
  mov    %eax,%edi
  call   <is_prime@plt>
  test   %eax,%eax
  jne    L9
  mov    $0x0,%eax
  jmp    L61
L9:
  addl   $0x1,-0x10(%rbp)
L24:
  mov    -0x10(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jb     L0
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L67
L27:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  div    %ecx
  mov    %edx,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <is_prime@plt>
  test   %eax,%eax
  jne    L50
  mov    $0x0,%eax
  jmp    L61
L50:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  subl   $0x1,-0xc(%rbp)
L67:
  cmpl   $0x0,-0xc(%rbp)
  jne    L27
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
L61:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D2
datatype: string

tool: D2: Prime number: %u


assistant: ```c
bool is_truncatable(unsigned num) {
  unsigned tmp = num;
  unsigned rnum = 0;
  unsigned l = num_digits(num);
  if (is_prime(num) == false)
    return false;
  for (unsigned i = 1; i < l; i++) {
    if (is_prime(num / pow(10, i)) == false)
      return false;
  }
  for (unsigned i = l - 1; i >= 1; i--) {
    rnum = tmp % (unsigned)pow(10, i);
    if (is_prime(rnum) == false)
      return false;
    tmp = rnum;
  }
  printf("Prime number: %u\n", num);
  return true;
}
```