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

```asm
<QuaternionInverse>:
  endbr64
  mov    %rsi,%rax
  movss  (%rdi),%xmm0
  movss  %xmm0,(%rsi)
  movss  0x4(%rdi),%xmm0
  movss  D3(%rip),%xmm1
  xorps  %xmm1,%xmm0
  movss  %xmm0,0x4(%rsi)
  movss  0x8(%rdi),%xmm0
  xorps  %xmm1,%xmm0
  movss  %xmm0,0x8(%rsi)
  movss  0xc(%rdi),%xmm0
  xorps  %xmm1,%xmm0
  movss  %xmm0,0xc(%rsi)
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
datatype: byte[4]

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

assistant: ```c
float *QuaternionInverse(const float quat[4], float result[4]) {
  result[0] = quat[0];
  result[1] = -quat[1];
  result[2] = -quat[2];
  result[3] = -quat[3];
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<QuaternionInverse>:
  endbr64
  movss  (%rdi),%xmm0
  movss  D4(%rip),%xmm1
  mov    %rsi,%rax
  movss  %xmm0,(%rsi)
  movss  0x4(%rdi),%xmm0
  xorps  %xmm1,%xmm0
  movss  %xmm0,0x4(%rsi)
  movss  0x8(%rdi),%xmm0
  xorps  %xmm1,%xmm0
  movss  %xmm0,0x8(%rsi)
  movss  0xc(%rdi),%xmm0
  xorps  %xmm1,%xmm0
  movss  %xmm0,0xc(%rsi)
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
type_infered: byte[4]

tool: {"D4": [0x00,0x00,0x00,0x80]}

assistant: ```c
float *QuaternionInverse(const float quat[4], float result[4]) {
  result[0] = quat[0];
  result[1] = -quat[1];
  result[2] = -quat[2];
  result[3] = -quat[3];
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<QuaternionInverse>:
  endbr64
  movss  (%rdi),%xmm0
  movss  D8(%rip),%xmm1
  mov    %rsi,%rax
  movss  %xmm0,(%rsi)
  movss  0x4(%rdi),%xmm0
  xorps  %xmm1,%xmm0
  movss  %xmm0,0x4(%rsi)
  movss  0x8(%rdi),%xmm0
  xorps  %xmm1,%xmm0
  movss  %xmm0,0x8(%rsi)
  movss  0xc(%rdi),%xmm0
  xorps  %xmm1,%xmm0
  movss  %xmm0,0xc(%rsi)
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type_infered: byte[4]

tool: {"D8": [0x00,0x00,0x00,0x80]}

assistant: ```c
float *QuaternionInverse(const float quat[4], float result[4]) {
  result[0] = quat[0];
  result[1] = -quat[1];
  result[2] = -quat[2];
  result[3] = -quat[3];
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1a4(%rbp)
  jmp    L27
L51:
  mov    -0x1a4(%rbp),%edx
  lea    -0x1a0(%rbp),%rcx
  mov    -0x1a4(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <foo@plt>
  mov    -0x1a4(%rbp),%edx
  movslq %edx,%rdx
  mov    %eax,-0x1a0(%rbp,%rdx,4)
  addl   $0x1,-0x1a4(%rbp)
L27:
  cmpl   $0x63,-0x1a4(%rbp)
  jle    L51
  movl   $0x0,-0x1a4(%rbp)
  jmp    L8
L19:
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  cmp    %eax,-0x1a4(%rbp)
  je     L8
  nop
  mov    $0x1,%eax
  jmp    L39
L8:
  cmpl   $0x63,-0x1a4(%rbp)
  jle    L19
  mov    $0x0,%eax
L39:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int main() {
  int i, b[100];
  for (i = 0; i < 100; i++) {
    b[i] = foo(i, b, i);
  }
  for (i = 0; i < 100; i) {
    if (b[i] != i) {
    ERROR:
      return 1;
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x1a8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebx
  mov    %rsp,%rbp
L0:
  mov    %ebx,%edi
  mov    %ebx,%edx
  mov    %rbp,%rsi
  call   <foo@plt>
  mov    %eax,0x0(%rbp,%rbx,4)
  add    $0x1,%rbx
  cmp    $0x64,%rbx
  jne    L0
  mov    (%rsp),%eax
L5:
  test   %eax,%eax
  je     L5
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  mov    $0x1,%eax
  add    $0x1a8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i, b[100];
  for (i = 0; i < 100; i++) {
    b[i] = foo(i, b, i);
  }
  for (i = 0; i < 100; i) {
    if (b[i] != i) {
    ERROR:
      return 1;
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x1a8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  nopl   (%rax)
L35:
  mov    %ebx,%edi
  mov    %ebx,%edx
  mov    %rbp,%rsi
  call   <foo@plt>
  mov    %eax,0x0(%rbp,%rbx,4)
  add    $0x1,%rbx
  cmp    $0x64,%rbx
  jne    L35
  mov    (%rsp),%eax
  test   %eax,%eax
  jne    L28
L39:
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L28:
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x1a8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i, b[100];
  for (i = 0; i < 100; i++) {
    b[i] = foo(i, b, i);
  }
  for (i = 0; i < 100; i) {
    if (b[i] != i) {
    ERROR:
      return 1;
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x1a8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  nopl   (%rax)
L36:
  mov    %ebx,%edi
  mov    %ebx,%edx
  mov    %rbp,%rsi
  call   <foo@plt>
  mov    %eax,0x0(%rbp,%rbx,4)
  add    $0x1,%rbx
  cmp    $0x64,%rbx
  jne    L36
  mov    (%rsp),%eax
  test   %eax,%eax
  jne    L16
L11:
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L16:
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x1a8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i, b[100];
  for (i = 0; i < 100; i++) {
    b[i] = foo(i, b, i);
  }
  for (i = 0; i < 100; i) {
    if (b[i] != i) {
    ERROR:
      return 1;
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<read_client_request>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,0xc(%rsp)
  mov    %rsi,%rbp
  mov    $0x1,%r14d
  mov    $0x0,%r13d
  mov    $0xfff,%r15d
  jmp    L126
L93:
  cmp    $0xffffffffffffffff,%r12
  je     L0
  add    %r12,%r13
L7:
  test   %r12,%r12
  setne  %dl
  cmp    $0xffe,%r13
  setbe  %al
  test   %al,%dl
  je     L36
  test   %r14d,%r14d
  je     L36
L126:
  mov    %r15,%rdx
  sub    %r13,%rdx
  lea    0x0(%rbp,%r13,1),%rsi
  mov    0xc(%rsp),%edi
  call   <read@plt>
  mov    %rax,%r12
  mov    $0x0,%ebx
L33:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L93
  add    $0x1,%rbx
  cmpb   $0xa,-0x1(%rbp,%rbx,1)
  jne    L33
  mov    $0x0,%r14d
  jmp    L93
L0:
  mov    D19(%rip),%rdx
  mov    D4(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L7
  mov    $0xfffffffe,%eax
  jmp    L114
L36:
  movb   $0x0,0x0(%rbp,%r13,1)
  mov    $0xffffffff,%eax
  cmpb   $0xa,-0x1(%rbp,%r13,1)
  jne    L114
  movzbl -0x2(%rbp,%r13,1),%eax
  cmp    $0xa,%al
  je     L91
  cmp    $0xd,%al
  jne    L110
  cmpb   $0xa,-0x3(%rbp,%r13,1)
  mov    $0xffffffff,%eax
  cmove  %r13d,%eax
L114:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L91:
  mov    %r13d,%eax
  jmp    L114
L110:
  mov    $0xffffffff,%eax
  jmp    L114
```

assistant: ```c
int read_client_request(int clientsd, char *buffer) {
  size_t maxread;
  ssize_t reader, readerc;
  int reading;
  reader = -1;
  readerc = 0;
  maxread = 4096 - 1;
  reading = 1;
  while ((reader != 0) && (readerc < maxread) && reading) {
    int i;
    reader = read(clientsd, buffer + readerc, maxread - readerc);
    for (i = 0; i < strlen(buffer); i++) {
      if (buffer[i] == '\n') {
        reading = 0;
        break;
      }
    }
    if (reader == -1) {
      if (errno != EINTR)
        return -2;
    } else
      readerc += reader;
  }
  buffer[readerc] = '\0';
  if (buffer[readerc - 1] == '\n' && buffer[readerc - 2] == '\n') {
    return readerc;
  } else if (buffer[readerc - 1] == '\n' && buffer[readerc - 2] == '\r' && buffer[readerc - 3] == '\n') {
    return readerc;
  } else {
    return -1;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<read_client_request>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %edi,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  xchg   %ax,%ax
L74:
  mov    $0xfff,%edx
  mov    %r13,%rsi
  mov    %r14d,%edi
  xor    %ebx,%ebx
  sub    %r12,%rdx
  call   <read@plt>
  mov    %rax,%rbp
  jmp    L72
  nopw   0x0(%rax,%rax,1)
L115:
  add    $0x1,%rbx
  cmpb   $0xa,-0x1(%r15,%rbx,1)
  je     L25
L72:
  mov    %r15,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L115
  mov    $0x1,%eax
  cmp    $0xffffffffffffffff,%rbp
  je     L80
L116:
  add    %rbp,%r12
  test   %rbp,%rbp
  setne  %dl
  lea    (%r15,%r12,1),%r13
L18:
  cmp    $0xffe,%r12
  ja     L81
  test   %al,%dl
  jne    L74
L81:
  movb   $0x0,0x0(%r13)
  cmpb   $0xa,-0x1(%r15,%r12,1)
  mov    $0xffffffff,%eax
  je     L96
L124:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L25:
  xor    %eax,%eax
  cmp    $0xffffffffffffffff,%rbp
  jne    L116
L80:
  mov    D18(%rip),%rdx
  mov    D11(%rip),%rsi
  mov    (%rdx),%rcx
  cmp    %rcx,(%rsi)
  jne    L60
  mov    $0x1,%edx
  jmp    L18
L96:
  movzbl -0x2(%r15,%r12,1),%edx
  cmp    $0xa,%dl
  je     L2
  cmp    $0xd,%dl
  jne    L124
  cmpb   $0xa,-0x3(%r15,%r12,1)
  cmove  %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L60:
  add    $0x8,%rsp
  mov    $0xfffffffe,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  mov    %r12d,%eax
  jmp    L124
```

assistant: ```c
int read_client_request(int clientsd, char *buffer) {
  size_t maxread;
  ssize_t reader, readerc;
  int reading;
  reader = -1;
  readerc = 0;
  maxread = 4096 - 1;
  reading = 1;
  while ((reader != 0) && (readerc < maxread) && reading) {
    int i;
    reader = read(clientsd, buffer + readerc, maxread - readerc);
    for (i = 0; i < strlen(buffer); i++) {
      if (buffer[i] == '\n') {
        reading = 0;
        break;
      }
    }
    if (reader == -1) {
      if (errno != EINTR)
        return -2;
    } else
      readerc += reader;
  }
  buffer[readerc] = '\0';
  if (buffer[readerc - 1] == '\n' && buffer[readerc - 2] == '\n') {
    return readerc;
  } else if (buffer[readerc - 1] == '\n' && buffer[readerc - 2] == '\r' && buffer[readerc - 3] == '\n') {
    return readerc;
  } else {
    return -1;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_client_request>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %edi,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  xchg   %ax,%ax
L52:
  mov    $0xfff,%edx
  mov    %r13,%rsi
  mov    %r14d,%edi
  xor    %ebx,%ebx
  sub    %r12,%rdx
  call   <read@plt>
  mov    %rax,%rbp
  jmp    L43
  nopw   0x0(%rax,%rax,1)
L23:
  add    $0x1,%rbx
  cmpb   $0xa,-0x1(%r15,%rbx,1)
  je     L28
L43:
  mov    %r15,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L23
  mov    $0x1,%eax
  cmp    $0xffffffffffffffff,%rbp
  je     L96
L104:
  add    %rbp,%r12
  test   %rbp,%rbp
  setne  %dl
  lea    (%r15,%r12,1),%r13
L74:
  cmp    $0xffe,%r12
  ja     L72
  test   %al,%dl
  jne    L52
L72:
  movb   $0x0,0x0(%r13)
  cmpb   $0xa,-0x1(%r15,%r12,1)
  mov    $0xffffffff,%eax
  je     L42
L45:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L28:
  xor    %eax,%eax
  cmp    $0xffffffffffffffff,%rbp
  jne    L104
L96:
  mov    D16(%rip),%rdx
  mov    D3(%rip),%rsi
  mov    (%rdx),%rcx
  cmp    %rcx,(%rsi)
  jne    L85
  mov    $0x1,%edx
  jmp    L74
L42:
  movzbl -0x2(%r15,%r12,1),%edx
  cmp    $0xa,%dl
  je     L91
  cmp    $0xd,%dl
  jne    L45
  cmpb   $0xa,-0x3(%r15,%r12,1)
  cmove  %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L85:
  add    $0x8,%rsp
  mov    $0xfffffffe,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L91:
  mov    %r12d,%eax
  jmp    L45
```

assistant: ```c
int read_client_request(int clientsd, char *buffer) {
  size_t maxread;
  ssize_t reader, readerc;
  int reading;
  reader = -1;
  readerc = 0;
  maxread = 4096 - 1;
  reading = 1;
  while ((reader != 0) && (readerc < maxread) && reading) {
    int i;
    reader = read(clientsd, buffer + readerc, maxread - readerc);
    for (i = 0; i < strlen(buffer); i++) {
      if (buffer[i] == '\n') {
        reading = 0;
        break;
      }
    }
    if (reader == -1) {
      if (errno != EINTR)
        return -2;
    } else
      readerc += reader;
  }
  buffer[readerc] = '\0';
  if (buffer[readerc - 1] == '\n' && buffer[readerc - 2] == '\n') {
    return readerc;
  } else if (buffer[readerc - 1] == '\n' && buffer[readerc - 2] == '\r' && buffer[readerc - 3] == '\n') {
    return readerc;
  } else {
    return -1;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rr_dt_ustime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x18(%rbp),%rax
  imul   $0xf4240,%rax,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  add    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
long long rr_dt_ustime(void) {
  struct timeval tv;
  long long ust;
  gettimeofday(&tv, NULL);
  ust = ((long long)tv.tv_sec) * 1000000;
  ust += tv.tv_usec;
  return ust;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rr_dt_ustime>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long rr_dt_ustime(void) {
  struct timeval tv;
  long long ust;
  gettimeofday(&tv, NULL);
  ust = ((long long)tv.tv_sec) * 1000000;
  ust += tv.tv_usec;
  return ust;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rr_dt_ustime>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long rr_dt_ustime(void) {
  struct timeval tv;
  long long ust;
  gettimeofday(&tv, NULL);
  ust = ((long long)tv.tv_sec) * 1000000;
  ust += tv.tv_usec;
  return ust;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rr_dt_ustime>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long rr_dt_ustime(void) {
  struct timeval tv;
  long long ust;
  gettimeofday(&tv, NULL);
  ust = ((long long)tv.tv_sec) * 1000000;
  ust += tv.tv_usec;
  return ust;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getTimestamp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x18(%rbp),%rax
  imul   $0x3e8,%eax,%ecx
  mov    -0x20(%rbp),%rax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%esi
  mov    %edx,%eax
  sub    %esi,%eax
  add    %ecx,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
uint64_t getTimestamp() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (uint64_t)tv.tv_sec * 1000 + (uint64_t)tv.tv_usec / 1000;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getTimestamp>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  imul   $0x3e8,0x8(%rsp),%eax
  mov    (%rsp),%rcx
  movslq %ecx,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x26,%rdx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t getTimestamp() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (uint64_t)tv.tv_sec * 1000 + (uint64_t)tv.tv_usec / 1000;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getTimestamp>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%rcx
  imul   $0x3e8,0x8(%rsp),%eax
  movslq %ecx,%rdx
  sar    $0x1f,%ecx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x26,%rdx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t getTimestamp() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (uint64_t)tv.tv_sec * 1000 + (uint64_t)tv.tv_usec / 1000;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getTimestamp>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%rcx
  imul   $0x3e8,0x8(%rsp),%eax
  movslq %ecx,%rdx
  sar    $0x1f,%ecx
  imul   $0x10624dd3,%rdx,%rdx
  sar    $0x26,%rdx
  sub    %ecx,%edx
  add    %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t getTimestamp() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (uint64_t)tv.tv_sec * 1000 + (uint64_t)tv.tv_usec / 1000;
}
```
user: What is the c source code of the assembly code below:

```asm
<sqlite3FreeX>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L7
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <sqlite3GenericFree@plt>
L7:
  nop
  leave
  ret
```

assistant: ```c
void sqlite3FreeX(void *p) {
  if (p) {
    ;
    sqlite3GenericFree(p);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sqlite3FreeX>:
  endbr64
  test   %rdi,%rdi
  je     L6
  sub    $0x8,%rsp
  call   <sqlite3GenericFree@plt>
  add    $0x8,%rsp
  ret
L6:
  ret
```

assistant: ```c
void sqlite3FreeX(void *p) {
  if (p) {
    ;
    sqlite3GenericFree(p);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sqlite3FreeX>:
  endbr64
  test   %rdi,%rdi
  je     L7
  jmp    1050 <sqlite3GenericFree@plt>
  xchg   %ax,%ax
L7:
  ret
```

assistant: ```c
void sqlite3FreeX(void *p) {
  if (p) {
    ;
    sqlite3GenericFree(p);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sqlite3FreeX>:
  endbr64
  test   %rdi,%rdi
  je     L7
  jmp    1050 <sqlite3GenericFree@plt>
  xchg   %ax,%ax
L7:
  ret
```

assistant: ```c
void sqlite3FreeX(void *p) {
  if (p) {
    ;
    sqlite3GenericFree(p);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<str_hashcode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    $0x0,%ebx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <assert@plt>
  cmpl   $0x0,-0x1c(%rbp)
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <assert@plt>
  jmp    L3
L19:
  mov    %ebx,%ecx
  shl    $0x4,%ecx
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  lea    (%rcx,%rax,1),%ebx
  mov    %ebx,%r12d
  and    $0xf0000000,%r12d
  test   %r12d,%r12d
  je     L3
  mov    %r12d,%eax
  shr    $0x18,%eax
  xor    %eax,%ebx
  xor    %r12d,%ebx
L3:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L19
  mov    -0x1c(%rbp),%ecx
  mov    %ebx,%eax
  mov    $0x0,%edx
  div    %ecx
  mov    %edx,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
int str_hashcode(const char *str, const int range) {
  register unsigned int h = 0;
  register unsigned int g;
  assert(str);
  assert(*str);
  assert(range > 0);
  while (*str) {
    h = (h << 4) + (*str++);
    if (g = h & 0xf0000000) {
      h = h ^ (g >> 24);
      h = h ^ g;
    }
  }
  return h % range;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<str_hashcode>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    $0x0,%eax
  call   <assert@plt>
  movsbl (%rbx),%edi
  mov    $0x0,%eax
  call   <assert@plt>
  test   %ebp,%ebp
  setg   %dil
  movzbl %dil,%edi
  mov    $0x0,%eax
  call   <assert@plt>
  movzbl (%rbx),%edx
  test   %dl,%dl
  je     L33
  mov    $0x0,%eax
  jmp    L34
L38:
  movzbl (%rbx),%edx
  test   %dl,%dl
  je     L18
L34:
  shl    $0x4,%eax
  add    $0x1,%rbx
  movsbl %dl,%edx
  add    %edx,%eax
  mov    %eax,%edx
  and    $0xf0000000,%edx
  je     L38
  and    $0xfffffff,%eax
  shr    $0x18,%edx
  xor    %edx,%eax
  jmp    L38
L33:
  mov    $0x0,%eax
L18:
  mov    $0x0,%edx
  div    %ebp
  mov    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int str_hashcode(const char *str, const int range) {
  register unsigned int h = 0;
  register unsigned int g;
  assert(str);
  assert(*str);
  assert(range > 0);
  while (*str) {
    h = (h << 4) + (*str++);
    if (g = h & 0xf0000000) {
      h = h ^ (g >> 24);
      h = h ^ g;
    }
  }
  return h % range;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<str_hashcode>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <assert@plt>
  movsbl 0x0(%rbp),%edi
  xor    %eax,%eax
  call   <assert@plt>
  xor    %edi,%edi
  test   %ebx,%ebx
  setg   %dil
  xor    %eax,%eax
  call   <assert@plt>
  movsbl 0x0(%rbp),%edx
  test   %dl,%dl
  je     L2
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L17:
  shl    $0x4,%eax
  add    $0x1,%rbp
  add    %edx,%eax
  mov    %eax,%edx
  and    $0xf0000000,%edx
  je     L24
  and    $0xfffffff,%eax
  shr    $0x18,%edx
  xor    %edx,%eax
L24:
  movsbl 0x0(%rbp),%edx
  test   %dl,%dl
  jne    L17
  xor    %edx,%edx
  add    $0x8,%rsp
  div    %ebx
  pop    %rbx
  pop    %rbp
  mov    %edx,%r8d
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L2:
  add    $0x8,%rsp
  xor    %r8d,%r8d
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int str_hashcode(const char *str, const int range) {
  register unsigned int h = 0;
  register unsigned int g;
  assert(str);
  assert(*str);
  assert(range > 0);
  while (*str) {
    h = (h << 4) + (*str++);
    if (g = h & 0xf0000000) {
      h = h ^ (g >> 24);
      h = h ^ g;
    }
  }
  return h % range;
}
```