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

```asm
<firstMissingPositive>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x9000(%rsp),%r11
L85:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L85
  sub    $0xc70,%rsp
  mov    %rdi,-0x9c68(%rbp)
  mov    %esi,-0x9c6c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x0,-0x9c6c(%rbp)
  jne    L115
  mov    $0x1,%eax
  jmp    L139
L115:
  cmpl   $0x1,-0x9c6c(%rbp)
  jne    L50
  mov    -0x9c68(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  jne    L129
  mov    $0x2,%eax
  jmp    L139
L129:
  mov    $0x1,%eax
  jmp    L139
L50:
  lea    -0x9c50(%rbp),%rax
  mov    $0x9c40,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x9c5c(%rbp)
  jmp    L130
L97:
  mov    -0x9c5c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x9c68(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x9c54(%rbp)
  cmpl   $0x0,-0x9c54(%rbp)
  jle    L43
  mov    -0x9c54(%rbp),%eax
  cltq
  movl   $0x1,-0x9c50(%rbp,%rax,4)
  jmp    L53
L43:
  nop
L53:
  addl   $0x1,-0x9c5c(%rbp)
L130:
  mov    -0x9c5c(%rbp),%eax
  cmp    -0x9c6c(%rbp),%eax
  jl     L97
  movl   $0x1,-0x9c58(%rbp)
  jmp    L5
L149:
  mov    -0x9c58(%rbp),%eax
  cltq
  mov    -0x9c50(%rbp,%rax,4),%eax
  test   %eax,%eax
  jne    L146
  mov    -0x9c58(%rbp),%eax
  jmp    L139
L146:
  addl   $0x1,-0x9c58(%rbp)
L5:
  cmpl   $0x3e7,-0x9c58(%rbp)
  jle    L149
  mov    $0x1,%eax
L139:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

assistant: ```c
int firstMissingPositive(int *nums, int numsSize) {
  if (numsSize == 0)
    return 1;
  if (numsSize == 1)
    return (nums[0] == 1) ? 2 : 1;
  int buf[10000] = {0};
  for (int i = 0; i < numsSize; ++i) {
    int val = nums[i];
    if (val <= 0)
      continue;
    buf[val] = 1;
  }
  for (int i = 1; i < 1000; ++i) {
    if (!buf[i])
      return i;
  }
  return 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
<firstMissingPositive>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L33:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L33
  sub    $0xc58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  mov    $0x1,%edx
  test   %esi,%esi
  je     L54
  mov    %rdi,%rbp
  mov    %esi,%ebx
  cmp    $0x1,%esi
  je     L119
  mov    %rsp,%rdi
  mov    $0x9c40,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  test   %ebx,%ebx
  jle    L39
  mov    %rbp,%rdx
  lea    -0x1(%rbx),%eax
  lea    0x4(%rbp,%rax,4),%rcx
  jmp    L27
L119:
  cmpl   $0x1,(%rdi)
  mov    $0x2,%edx
  cmovne %esi,%edx
L54:
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L91
  mov    %edx,%eax
  add    $0x9c58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L57:
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  je     L39
L27:
  mov    (%rdx),%eax
  test   %eax,%eax
  jle    L57
  cltq
  movl   $0x1,(%rsp,%rax,4)
  jmp    L57
L39:
  mov    $0x1,%eax
  mov    %rsp,%rcx
L75:
  mov    %eax,%edx
  cmpl   $0x0,(%rcx,%rax,4)
  je     L54
  add    $0x1,%rax
  cmp    $0x3e8,%rax
  jne    L75
  mov    $0x1,%edx
  jmp    L54
L91:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int firstMissingPositive(int *nums, int numsSize) {
  if (numsSize == 0)
    return 1;
  if (numsSize == 1)
    return (nums[0] == 1) ? 2 : 1;
  int buf[10000] = {0};
  for (int i = 0; i < numsSize; ++i) {
    int val = nums[i];
    if (val <= 0)
      continue;
    buf[val] = 1;
  }
  for (int i = 1; i < 1000; ++i) {
    if (!buf[i])
      return i;
  }
  return 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
<firstMissingPositive>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L38:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L38
  sub    $0xc58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  je     L126
  mov    %rdi,%rbx
  mov    %esi,%ebp
  cmp    $0x1,%esi
  je     L103
  mov    %rsp,%rcx
  mov    $0x9c40,%edx
  xor    %esi,%esi
  mov    %rcx,%rdi
  call   <memset@plt>
  mov    %rax,%rcx
  test   %ebp,%ebp
  jle    L71
  lea    -0x1(%rbp),%eax
  mov    %rbx,%rdi
  lea    0x4(%rbx,%rax,4),%rdx
  nop
L4:
  movslq (%rdi),%rax
  test   %eax,%eax
  jle    L54
  movl   $0x1,(%rsp,%rax,4)
L54:
  add    $0x4,%rdi
  cmp    %rdx,%rdi
  jne    L4
  mov    0x4(%rsp),%edx
L91:
  mov    $0x1,%eax
  jmp    L69
  nopw   0x0(%rax,%rax,1)
L129:
  mov    (%rcx,%rax,4),%edx
L69:
  mov    %eax,%r8d
  test   %edx,%edx
  je     L66
  add    $0x1,%rax
  cmp    $0x3e8,%rax
  jne    L129
  xchg   %ax,%ax
L126:
  mov    $0x1,%r8d
L66:
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L110
  add    $0x9c58,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
L71:
  xor    %edx,%edx
  jmp    L91
  nopw   0x0(%rax,%rax,1)
L103:
  cmpl   $0x1,(%rdi)
  mov    $0x2,%r8d
  jne    L126
  jmp    L66
L110:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int firstMissingPositive(int *nums, int numsSize) {
  if (numsSize == 0)
    return 1;
  if (numsSize == 1)
    return (nums[0] == 1) ? 2 : 1;
  int buf[10000] = {0};
  for (int i = 0; i < numsSize; ++i) {
    int val = nums[i];
    if (val <= 0)
      continue;
    buf[val] = 1;
  }
  for (int i = 1; i < 1000; ++i) {
    if (!buf[i])
      return i;
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<firstMissingPositive>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L79:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L79
  sub    $0xc58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  je     L117
  mov    %rdi,%rbx
  mov    %esi,%ebp
  cmp    $0x1,%esi
  je     L17
  mov    %rsp,%rcx
  mov    $0x9c40,%edx
  xor    %esi,%esi
  mov    %rcx,%rdi
  call   <memset@plt>
  mov    %rax,%rcx
  test   %ebp,%ebp
  jle    L114
  lea    -0x1(%rbp),%eax
  mov    %rbx,%rdi
  lea    0x4(%rbx,%rax,4),%rdx
  nopl   0x0(%rax,%rax,1)
L73:
  movslq (%rdi),%rax
  test   %eax,%eax
  jle    L86
  movl   $0x1,(%rsp,%rax,4)
L86:
  add    $0x4,%rdi
  cmp    %rdx,%rdi
  jne    L73
L114:
  mov    $0x1,%eax
  jmp    L5
  cs nopw 0x0(%rax,%rax,1)
L103:
  add    $0x1,%rax
  cmp    $0x3e8,%rax
  je     L117
L5:
  mov    (%rcx,%rax,4),%edx
  mov    %eax,%r8d
  test   %edx,%edx
  jne    L103
L119:
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L82
  add    $0x9c58,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L17:
  cmpl   $0x1,(%rdi)
  mov    $0x2,%r8d
  je     L119
  nopl   0x0(%rax,%rax,1)
L117:
  mov    $0x1,%r8d
  jmp    L119
L82:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int firstMissingPositive(int *nums, int numsSize) {
  if (numsSize == 0)
    return 1;
  if (numsSize == 1)
    return (nums[0] == 1) ? 2 : 1;
  int buf[10000] = {0};
  for (int i = 0; i < numsSize; ++i) {
    int val = nums[i];
    if (val <= 0)
      continue;
    buf[val] = 1;
  }
  for (int i = 1; i < 1000; ++i) {
    if (!buf[i])
      return i;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<yahoo_f01_22>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  imul   $0x57f27c5f,%eax,%eax
  add    $0x2433,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <yahoo_M1Common@plt>
  leave
  ret
```

assistant: ```c
int yahoo_f01_22(int Salt, int Parameter) {
  unsigned b = Salt * 0x57f27c5F + 0x2433;
  return yahoo_M1Common(b, Parameter);
}
```
user: What is the c source code of the assembly code below:

```asm
<yahoo_f01_22>:
  endbr64
  sub    $0x8,%rsp
  imul   $0x57f27c5f,%edi,%edi
  add    $0x2433,%edi
  call   <yahoo_M1Common@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int yahoo_f01_22(int Salt, int Parameter) {
  unsigned b = Salt * 0x57f27c5F + 0x2433;
  return yahoo_M1Common(b, Parameter);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<yahoo_f01_22>:
  endbr64
  imul   $0x57f27c5f,%edi,%edi
  add    $0x2433,%edi
  jmp    1050 <yahoo_M1Common@plt>
```

assistant: ```c
int yahoo_f01_22(int Salt, int Parameter) {
  unsigned b = Salt * 0x57f27c5F + 0x2433;
  return yahoo_M1Common(b, Parameter);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<yahoo_f01_22>:
  endbr64
  imul   $0x57f27c5f,%edi,%edi
  add    $0x2433,%edi
  jmp    1050 <yahoo_M1Common@plt>
```

assistant: ```c
int yahoo_f01_22(int Salt, int Parameter) {
  unsigned b = Salt * 0x57f27c5F + 0x2433;
  return yahoo_M1Common(b, Parameter);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L8:
  addl   $0x1,-0x4(%rbp)
L19:
  cmpl   $0x1,-0x4(%rbp)
  jle    L8
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void bar(char *x) {
  int i;
  for (i = 0; i < 2; i++)
    ;
  x[i + i + i + i] = '\0';
}
```
user: What is the c source code of the assembly code below:

```asm
<bar>:
  endbr64
  movb   $0x0,0x8(%rdi)
  ret
```

assistant: ```c
void bar(char *x) {
  int i;
  for (i = 0; i < 2; i++)
    ;
  x[i + i + i + i] = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<bar>:
  endbr64
  movb   $0x0,0x8(%rdi)
  ret
```

assistant: ```c
void bar(char *x) {
  int i;
  for (i = 0; i < 2; i++)
    ;
  x[i + i + i + i] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bar>:
  endbr64
  movb   $0x0,0x8(%rdi)
  ret
```

assistant: ```c
void bar(char *x) {
  int i;
  for (i = 0; i < 2; i++)
    ;
  x[i + i + i + i] = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<str_home_replace_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L12
L18:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x30(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %rax,%rdi
  call   <str_home_replace@plt>
  mov    %rax,(%rbx)
  addl   $0x1,-0x14(%rbp)
L12:
  mov    -0x14(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L18
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void str_home_replace_array(int argc, char **argv) {
  int i;
  for (i = 0; i < argc; i++)
    argv[i] = str_home_replace(argv[i]);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<str_home_replace_array>:
  endbr64
  test   %edi,%edi
  jle    L7
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  lea    -0x1(%rdi),%eax
  lea    0x8(%rsi,%rax,8),%rbp
L19:
  mov    (%rbx),%rdi
  call   <str_home_replace@plt>
  mov    %rax,(%rbx)
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L19
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  ret
```

assistant: ```c
void str_home_replace_array(int argc, char **argv) {
  int i;
  for (i = 0; i < argc; i++)
    argv[i] = str_home_replace(argv[i]);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<str_home_replace_array>:
  endbr64
  test   %edi,%edi
  jle    L13
  push   %rbp
  lea    -0x1(%rdi),%eax
  push   %rbx
  lea    0x8(%rsi,%rax,8),%rbp
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L17:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <str_home_replace@plt>
  mov    %rax,-0x8(%rbx)
  cmp    %rbp,%rbx
  jne    L17
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L13:
  ret
```

assistant: ```c
void str_home_replace_array(int argc, char **argv) {
  int i;
  for (i = 0; i < argc; i++)
    argv[i] = str_home_replace(argv[i]);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<str_home_replace_array>:
  endbr64
  test   %edi,%edi
  jle    L5
  push   %rbp
  lea    -0x1(%rdi),%eax
  push   %rbx
  lea    0x8(%rsi,%rax,8),%rbp
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L4:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <str_home_replace@plt>
  mov    %rax,-0x8(%rbx)
  cmp    %rbp,%rbx
  jne    L4
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L5:
  ret
```

assistant: ```c
void str_home_replace_array(int argc, char **argv) {
  int i;
  for (i = 0; i < argc; i++)
    argv[i] = str_home_replace(argv[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pprint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <len@plt>
  mov    %eax,%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x1c(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L14
  mov    $0x54,%eax
  jmp    L18
L14:
  mov    $0x0,%eax
L18:
  leave
  ret
```

assistant: ```c
int pprint(const char *str, int fd) {
  ssize_t wr;
  wr = 0;
  if ((wr = write(fd, str, len(str))) == -1)
    return (84);
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<pprint>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  call   <len@plt>
  mov    %eax,%edx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <write@plt>
  cmp    $0xffffffff,%eax
  mov    $0x0,%eax
  mov    $0x54,%edx
  cmove  %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int pprint(const char *str, int fd) {
  ssize_t wr;
  wr = 0;
  if ((wr = write(fd, str, len(str))) == -1)
    return (84);
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<pprint>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <len@plt>
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <write@plt>
  mov    $0x54,%edx
  cmp    $0xffffffff,%eax
  mov    $0x0,%eax
  cmove  %edx,%eax
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int pprint(const char *str, int fd) {
  ssize_t wr;
  wr = 0;
  if ((wr = write(fd, str, len(str))) == -1)
    return (84);
  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
<pprint>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <len@plt>
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <write@plt>
  mov    $0x54,%edx
  cmp    $0xffffffff,%eax
  mov    $0x0,%eax
  cmove  %edx,%eax
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int pprint(const char *str, int fd) {
  ssize_t wr;
  wr = 0;
  if ((wr = write(fd, str, len(str))) == -1)
    return (84);
  return (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<hea_cq_timestamp_double>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  movsd  D5(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    $0x20,%eax
  mov    %eax,%ecx
  shll   %cl,-0x14(%rbp)
  mov    -0x28(%rbp),%eax
  or     %eax,-0x14(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x14(%rbp),%xmm0
  divsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_guessed: f64

tool: D5: 4294967296

assistant: ```c
double hea_cq_timestamp_double(__u32 s, __u32 frac) {
  const double div = 4294967296.;
  __u64 v;
  double ret;
  v = s;
  v <<= 32;
  v |= frac;
  ret = (double)v / div;
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<hea_cq_timestamp_double>:
  endbr64
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm0
  mulsd  D0(%rip),%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type_guessed: byte[8]

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3d]}

assistant: ```c
double hea_cq_timestamp_double(__u32 s, __u32 frac) {
  const double div = 4294967296.;
  __u64 v;
  double ret;
  v = s;
  v <<= 32;
  v |= frac;
  ret = (double)v / div;
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<hea_cq_timestamp_double>:
  endbr64
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm0
  mulsd  D3(%rip),%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
type_guessed: byte[8]

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3d]}

assistant: ```c
double hea_cq_timestamp_double(__u32 s, __u32 frac) {
  const double div = 4294967296.;
  __u64 v;
  double ret;
  v = s;
  v <<= 32;
  v |= frac;
  ret = (double)v / div;
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hea_cq_timestamp_double>:
  endbr64
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm0
  mulsd  D0(%rip),%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
type_guessed: byte[8]

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3d]}

assistant: ```c
double hea_cq_timestamp_double(__u32 s, __u32 frac) {
  const double div = 4294967296.;
  __u64 v;
  double ret;
  v = s;
  v <<= 32;
  v |= frac;
  ret = (double)v / div;
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<aysm_tiling>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <tiling@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x24(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L15
  mov    -0x24(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edi
  call   <tiling@plt>
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sub    %edx,%eax
  mov    %eax,%ebx
  mov    -0x24(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <tiling@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  jmp    L5
L15:
  mov    -0x24(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edi
  call   <tiling@plt>
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sub    %edx,%eax
L5:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int aysm_tiling(int n) {
  int all = tiling(n);
  if (n % 2 == 0) {
    return all - tiling(n / 2) - tiling(n / 2 - 1);
  } else {
    return all - tiling(n / 2);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<aysm_tiling>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  call   <tiling@plt>
  mov    %eax,%ebx
  test   $0x1,%bpl
  jne    L5
  mov    %ebp,%edi
  shr    $0x1f,%edi
  add    %edi,%ebp
  sar    %ebp
  mov    %ebp,%edi
  call   <tiling@plt>
  sub    %eax,%ebx
  lea    -0x1(%rbp),%edi
  call   <tiling@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
L12:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  mov    %ebp,%edi
  shr    $0x1f,%edi
  add    %ebp,%edi
  sar    %edi
  call   <tiling@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  jmp    L12
```

assistant: ```c
int aysm_tiling(int n) {
  int all = tiling(n);
  if (n % 2 == 0) {
    return all - tiling(n / 2) - tiling(n / 2 - 1);
  } else {
    return all - tiling(n / 2);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<aysm_tiling>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %r12d,%ebp
  push   %rbx
  shr    $0x1f,%ebp
  add    %r12d,%ebp
  sar    %ebp
  call   <tiling@plt>
  and    $0x1,%r12d
  mov    %ebp,%edi
  mov    %eax,%ebx
  jne    L8
  call   <tiling@plt>
  lea    -0x1(%rbp),%edi
  sub    %eax,%ebx
  call   <tiling@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L8:
  call   <tiling@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  pop    %r12
  ret
```

assistant: ```c
int aysm_tiling(int n) {
  int all = tiling(n);
  if (n % 2 == 0) {
    return all - tiling(n / 2) - tiling(n / 2 - 1);
  } else {
    return all - tiling(n / 2);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<aysm_tiling>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %r12d,%ebp
  push   %rbx
  shr    $0x1f,%ebp
  add    %r12d,%ebp
  sar    %ebp
  call   <tiling@plt>
  and    $0x1,%r12d
  mov    %ebp,%edi
  mov    %eax,%ebx
  jne    L5
  call   <tiling@plt>
  lea    -0x1(%rbp),%edi
  sub    %eax,%ebx
  call   <tiling@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L5:
  call   <tiling@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  sub    %r8d,%eax
  pop    %r12
  ret
```

assistant: ```c
int aysm_tiling(int n) {
  int all = tiling(n);
  if (n % 2 == 0) {
    return all - tiling(n / 2) - tiling(n / 2 - 1);
  } else {
    return all - tiling(n / 2);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<ntruenc_encode_msg>:
  endbr64
  mov    %edx,%r8d
  lea    0x10(,%rsi,8),%r10d
  mov    $0x3,%edx
  cmp    %r8d,%r10d
  jg     L94
  push   %rbp
  push   %rbx
  mov    %rdi,%rax
  mov    %esi,%ebx
  mov    %rcx,%r9
  sub    %r10d,%r8d
  lea    0x7(%r8),%edi
  cmovns %r8d,%edi
  sar    $0x3,%edi
  movzbl %sil,%esi
  mov    $0x0,%ecx
L97:
  mov    %esi,%edx
  sar    %cl,%edx
  add    %edx,%edx
  and    $0x2,%edx
  sub    $0x1,%edx
  mov    %dx,(%r9,%rcx,2)
  add    $0x1,%rcx
  cmp    $0x8,%rcx
  jne    L97
  movzbl %bh,%ebp
  lea    0x10(%r9),%rsi
  mov    $0x0,%ecx
L137:
  mov    %ebp,%edx
  sar    %cl,%edx
  add    %edx,%edx
  and    $0x2,%edx
  sub    $0x1,%edx
  mov    %dx,(%rsi,%rcx,2)
  add    $0x1,%rcx
  cmp    $0x8,%rcx
  jne    L137
  test   %ebx,%ebx
  jle    L65
  mov    %rax,%rsi
  mov    %r9,%rdx
  shl    $0x3,%ebx
  mov    $0x0,%r11d
  jmp    L71
L24:
  add    $0x1,%rsi
  add    $0x8,%r11d
  add    $0x10,%rdx
  cmp    %ebx,%r11d
  je     L65
L71:
  mov    $0x0,%ecx
L130:
  movzbl (%rsi),%eax
  sar    %cl,%eax
  add    %eax,%eax
  and    $0x2,%eax
  sub    $0x1,%eax
  mov    %ax,0x20(%rdx,%rcx,2)
  add    $0x1,%rcx
  cmp    $0x8,%rcx
  jne    L130
  jmp    L24
L65:
  lea    -0x10(%r10),%eax
  cltq
  lea    0x20(%r9,%rax,2),%rsi
  cmp    $0x7,%r8d
  jle    L106
  lea    0x10(%rsi),%rdx
  mov    $0x0,%ecx
  jmp    L135
L0:
  add    $0x1,%ecx
  add    $0x10,%rdx
  cmp    %ecx,%edi
  jle    L106
L135:
  lea    -0x10(%rdx),%rax
L127:
  movw   $0x0,(%rax)
  add    $0x2,%rax
  cmp    %rdx,%rax
  jne    L127
  jmp    L0
L106:
  lea    0x0(,%rdi,8),%eax
  cltq
  lea    (%rsi,%rax,2),%rax
  and    $0x7,%r8d
  lea    D9(%rip),%rcx
  movslq (%rcx,%r8,4),%rdx
  add    %rcx,%rdx
  notrack jmp *%rdx
  movw   $0x0,0xc(%rax)
  movw   $0x0,0xa(%rax)
  movw   $0x0,0x8(%rax)
  movw   $0x0,0x6(%rax)
  movw   $0x0,0x4(%rax)
  movw   $0x0,0x2(%rax)
  movw   $0x0,(%rax)
  mov    $0x0,%edx
L38:
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  ret
  mov    $0x0,%edx
  jmp    L38
L94:
  mov    %edx,%eax
  ret
```

assistant: ```c
int ntruenc_encode_msg(unsigned char *data, int len, int n, short *m) {
  int ret = 0;
  int i, j;
  unsigned char *l = (unsigned char *)&len;
  int zlen;
  if ((len + 2) * 8 > n) {
    ret = 3;
    goto end;
  }
  zlen = (n - ((len + 2) * 8)) / 8;
  for (i = 0; i < 2; i++) {
    for (j = 0; j < 8; j++)
      m[i * 8 + j] = (((l[i] >> j) & 1) << 1) - 1;
  }
  m += 2 * 8;
  for (i = 0; i < len; i++) {
    for (j = 0; j < 8; j++)
      m[i * 8 + j] = (((data[i] >> j) & 1) << 1) - 1;
  }
  m += len * 8;
  for (i = 0; i < zlen; i++) {
    for (j = 0; j < 8; j++)
      m[i * 8 + j] = 0;
  }
  m += zlen * 8;
  switch ((n - ((len + 2) * 8)) & 7) {
  case 7:
    m[6] = 0;
  case 6:
    m[5] = 0;
  case 5:
    m[4] = 0;
  case 4:
    m[3] = 0;
  case 3:
    m[2] = 0;
  case 2:
    m[1] = 0;
  case 1:
    m[0] = 0;
  default:
    break;
  }
end:
  return ret;
}
```