ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<EncodeFiles>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x470,%rsp
  mov    %rdi,-0x468(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x458(%rbp)
  movl   $0x0,-0x454(%rbp)
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x468(%rbp),%rax
  lea    D18(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x450(%rbp)
  jmp    L13
L7:
  lea    -0x410(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  cltq
  movb   $0x0,-0x410(%rbp,%rax,1)
  lea    -0x440(%rbp),%rdx
  lea    -0x410(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <MD5File@plt>
  mov    %rax,-0x448(%rbp)
  cmpq   $0x0,-0x448(%rbp)
  je     L13
  mov    -0x448(%rbp),%rdx
  lea    -0x410(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L13:
  mov    -0x450(%rbp),%rdx
  lea    -0x410(%rbp),%rax
  mov    $0x400,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L7
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0xffffffff,-0x458(%rbp)
  mov    -0x458(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D38
data_type: string
name: extract_symbolic_data
arguments: data_tag: D18
data_type: string
name: extract_symbolic_data
arguments: data_tag: D5
data_type: string
name: extract_symbolic_data
arguments: data_tag: D46
data_type: string
name: extract_symbolic_data
arguments: data_tag: D33
data_type: string

tool: D38: char* fileList[] = {

tool: D18: r

tool: D5: "%s = %s",


tool: D46: "LASTFILE = 0",

tool: D33: };

assistant: ```c
int EncodeFiles(char *fileList) {
  FILE *filePtr;
  int retval = 0;
  char fileName[1024];
  char buf[33];
  char *p;
  int count = 0;
  printf("char* fileList[] = {\n");
  filePtr = fopen(fileList, "r");
  while (fgets(fileName, 1024, filePtr)) {
    fileName[strlen(fileName) - 1] = 0;
    p = MD5File(fileName, buf);
    if (p)
      printf("\"%s = %s\",\n", fileName, p);
  }
  printf("\"LASTFILE = 0\",\n");
  printf("};\n");
  retval = -1;
  return retval;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<EncodeFiles>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x440,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x438(%rsp)
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <puts@plt>
  lea    D36(%rip),%rsi
  mov    %rbx,%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  lea    D16(%rip),%r12
L30:
  lea    0x30(%rsp),%rbx
L16:
  mov    %rbp,%rdx
  mov    $0x400,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L27
  mov    %rbx,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  cltq
  movb   $0x0,0x30(%rsp,%rax,1)
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  call   <MD5File@plt>
  test   %rax,%rax
  je     L16
  lea    0x30(%rsp),%rsi
  mov    %rax,%rdx
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L30
L27:
  lea    D21(%rip),%rdi
  call   <puts@plt>
  lea    D38(%rip),%rdi
  call   <puts@plt>
  mov    0x438(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  mov    $0xffffffff,%eax
  add    $0x440,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D11
data_type: string
name: unpack_data
arguments: data_tag: D36
data_type: string
name: unpack_data
arguments: data_tag: D16
data_type: string
name: unpack_data
arguments: data_tag: D21
data_type: string
name: unpack_data
arguments: data_tag: D38
data_type: string

tool: D11: char* fileList[] = {

tool: D36: r

tool: D16: "%s = %s",


tool: D21: "LASTFILE = 0",

tool: D38: };

assistant: ```c
int EncodeFiles(char *fileList) {
  FILE *filePtr;
  int retval = 0;
  char fileName[1024];
  char buf[33];
  char *p;
  int count = 0;
  printf("char* fileList[] = {\n");
  filePtr = fopen(fileList, "r");
  while (fgets(fileName, 1024, filePtr)) {
    fileName[strlen(fileName) - 1] = 0;
    p = MD5File(fileName, buf);
    if (p)
      printf("\"%s = %s\",\n", fileName, p);
  }
  printf("\"LASTFILE = 0\",\n");
  printf("};\n");
  retval = -1;
  return retval;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<EncodeFiles>:
  endbr64
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    D39(%rip),%rdi
  push   %rbx
  sub    $0x448,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x438(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rbx
  mov    %rsp,%r12
  call   <puts@plt>
  mov    %rbp,%rdi
  lea    D24(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  jmp    L8
  cs nopw 0x0(%rax,%rax,1)
L28:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  sub    $0x1,%eax
  cltq
  movb   $0x0,0x30(%rsp,%rax,1)
  call   <MD5File@plt>
  test   %rax,%rax
  jne    L30
L8:
  mov    %rbp,%rdx
  mov    $0x400,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L28
  lea    D10(%rip),%rdi
  call   <puts@plt>
  lea    D29(%rip),%rdi
  call   <puts@plt>
  mov    0x438(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x448,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L30:
  mov    %rax,%rdx
  mov    %rbx,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L8
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D1
type: string
name: read_data
arguments: data_label: D39
type: string
name: read_data
arguments: data_label: D24
type: string
name: read_data
arguments: data_label: D10
type: string
name: read_data
arguments: data_label: D29
type: string

tool: D1: "%s = %s",


tool: D39: char* fileList[] = {

tool: D24: r

tool: D10: "LASTFILE = 0",

tool: D29: };

assistant: ```c
int EncodeFiles(char *fileList) {
  FILE *filePtr;
  int retval = 0;
  char fileName[1024];
  char buf[33];
  char *p;
  int count = 0;
  printf("char* fileList[] = {\n");
  filePtr = fopen(fileList, "r");
  while (fgets(fileName, 1024, filePtr)) {
    fileName[strlen(fileName) - 1] = 0;
    p = MD5File(fileName, buf);
    if (p)
      printf("\"%s = %s\",\n", fileName, p);
  }
  printf("\"LASTFILE = 0\",\n");
  printf("};\n");
  retval = -1;
  return retval;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<EncodeFiles>:
  endbr64
  push   %r13
  lea    D5(%rip),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    D4(%rip),%rdi
  push   %rbx
  sub    $0x448,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x438(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rbx
  mov    %rsp,%r12
  call   <puts@plt>
  mov    %rbp,%rdi
  lea    D40(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  jmp    L5
  cs nopw 0x0(%rax,%rax,1)
L48:
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  sub    $0x1,%eax
  cltq
  movb   $0x0,0x30(%rsp,%rax,1)
  call   <MD5File@plt>
  test   %rax,%rax
  jne    L32
L5:
  mov    %rbp,%rdx
  mov    $0x400,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L48
  lea    D25(%rip),%rdi
  call   <puts@plt>
  lea    D8(%rip),%rdi
  call   <puts@plt>
  mov    0x438(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  add    $0x448,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L32:
  mov    %rax,%rdx
  mov    %rbx,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L5
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
type: string
name: disassemble_data
arguments: tag: D4
type: string
name: disassemble_data
arguments: tag: D40
type: string
name: disassemble_data
arguments: tag: D25
type: string
name: disassemble_data
arguments: tag: D8
type: string

tool: D5: "%s = %s",


tool: D4: char* fileList[] = {

tool: D40: r

tool: D25: "LASTFILE = 0",

tool: D8: };

assistant: ```c
int EncodeFiles(char *fileList) {
  FILE *filePtr;
  int retval = 0;
  char fileName[1024];
  char buf[33];
  char *p;
  int count = 0;
  printf("char* fileList[] = {\n");
  filePtr = fopen(fileList, "r");
  while (fgets(fileName, 1024, filePtr)) {
    fileName[strlen(fileName) - 1] = 0;
    p = MD5File(fileName, buf);
    if (p)
      printf("\"%s = %s\",\n", fileName, p);
  }
  printf("\"LASTFILE = 0\",\n");
  printf("};\n");
  retval = -1;
  return retval;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<write_fd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
L13:
  movq   $0x0,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jle    L25
  mov    -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  sub    %rax,-0x18(%rbp)
  jmp    L48
L25:
  mov    D9(%rip),%rax
  mov    (%rax),%rdx
  mov    D7(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L5
L48:
  cmpq   $0x0,-0x18(%rbp)
  jne    L13
  jmp    L6
L5:
  nop
L6:
  mov    -0x38(%rbp),%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  leave
  ret
```

assistant: ```c
int write_fd(int sock, char *buffer, size_t number_of_bytes) {
  size_t bytes_left_to_write;
  size_t total_bytes_written = 0;
  bytes_left_to_write = number_of_bytes;
  do {
    ssize_t bytes_written = 0;
    bytes_written = write(sock, ((uint8_t *)buffer) + total_bytes_written, bytes_left_to_write);
    if (bytes_written > 0) {
      total_bytes_written += bytes_written;
      bytes_left_to_write -= bytes_written;
    } else if ((errno != EINTR))
      break;
  } while (bytes_left_to_write > 0);
  return number_of_bytes - bytes_left_to_write;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<write_fd>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r13d
  mov    %rsi,%r12
  mov    %rdx,%r14
  mov    %rdx,%rbx
  mov    $0x0,%ebp
  jmp    L6
L29:
  add    %rax,%rbp
  sub    %rax,%rbx
L36:
  test   %rbx,%rbx
  je     L8
L6:
  lea    (%r12,%rbp,4),%rsi
  mov    %rbx,%rdx
  mov    %r13d,%edi
  call   <write@plt>
  test   %rax,%rax
  jg     L29
  mov    D6(%rip),%rdx
  mov    D14(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L36
L8:
  mov    %r14d,%eax
  sub    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int write_fd(int sock, char *buffer, size_t number_of_bytes) {
  size_t bytes_left_to_write;
  size_t total_bytes_written = 0;
  bytes_left_to_write = number_of_bytes;
  do {
    ssize_t bytes_written = 0;
    bytes_written = write(sock, ((uint8_t *)buffer) + total_bytes_written, bytes_left_to_write);
    if (bytes_written > 0) {
      total_bytes_written += bytes_written;
      bytes_left_to_write -= bytes_written;
    } else if ((errno != EINTR))
      break;
  } while (bytes_left_to_write > 0);
  return number_of_bytes - bytes_left_to_write;
}
```
user: Translate the below assembly code into C programming language:

```asm
<write_fd>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L29
  nopl   0x0(%rax)
L19:
  add    %rax,%r14
  sub    %rax,%r13
L24:
  test   %r13,%r13
  je     L6
L29:
  lea    (%rbx,%r14,4),%rsi
  mov    %r13,%rdx
  mov    %ebp,%edi
  call   <write@plt>
  test   %rax,%rax
  jg     L19
  mov    D9(%rip),%rax
  mov    D8(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L24
L6:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  sub    %r13d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int write_fd(int sock, char *buffer, size_t number_of_bytes) {
  size_t bytes_left_to_write;
  size_t total_bytes_written = 0;
  bytes_left_to_write = number_of_bytes;
  do {
    ssize_t bytes_written = 0;
    bytes_written = write(sock, ((uint8_t *)buffer) + total_bytes_written, bytes_left_to_write);
    if (bytes_written > 0) {
      total_bytes_written += bytes_written;
      bytes_left_to_write -= bytes_written;
    } else if ((errno != EINTR))
      break;
  } while (bytes_left_to_write > 0);
  return number_of_bytes - bytes_left_to_write;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<write_fd>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L39
  nopl   0x0(%rax)
L7:
  add    %rax,%r14
  sub    %rax,%r13
L5:
  test   %r13,%r13
  je     L3
L39:
  lea    (%rbx,%r14,4),%rsi
  mov    %r13,%rdx
  mov    %ebp,%edi
  call   <write@plt>
  test   %rax,%rax
  jg     L7
  mov    D16(%rip),%rax
  mov    D10(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L5
L3:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  sub    %r13d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int write_fd(int sock, char *buffer, size_t number_of_bytes) {
  size_t bytes_left_to_write;
  size_t total_bytes_written = 0;
  bytes_left_to_write = number_of_bytes;
  do {
    ssize_t bytes_written = 0;
    bytes_written = write(sock, ((uint8_t *)buffer) + total_bytes_written, bytes_left_to_write);
    if (bytes_written > 0) {
      total_bytes_written += bytes_written;
      bytes_left_to_write -= bytes_written;
    } else if ((errno != EINTR))
      break;
  } while (bytes_left_to_write > 0);
  return number_of_bytes - bytes_left_to_write;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<DotProduct>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L4
L16:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L4:
  cmpl   $0x13,-0xc(%rbp)
  jle    L16
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double DotProduct(A, B)
double *A;
double *B;
{
  int i;
  double score;
  score = 0;
  for (i = 0; i < 20; i++) {
    score = score + A[i] * B[i];
  }
  return score;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<DotProduct>:
  endbr64
  mov    $0x0,%eax
  pxor   %xmm1,%xmm1
L7:
  movsd  (%rdi,%rax,1),%xmm0
  mulsd  (%rsi,%rax,1),%xmm0
  addsd  %xmm0,%xmm1
  add    $0x8,%rax
  cmp    $0xa0,%rax
  jne    L7
  movapd %xmm1,%xmm0
  ret
```

assistant: ```c
double DotProduct(A, B)
double *A;
double *B;
{
  int i;
  double score;
  score = 0;
  for (i = 0; i < 20; i++) {
    score = score + A[i] * B[i];
  }
  return score;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<DotProduct>:
  endbr64
  xor    %eax,%eax
  pxor   %xmm1,%xmm1
  nopw   0x0(%rax,%rax,1)
L5:
  movsd  (%rdi,%rax,1),%xmm0
  mulsd  (%rsi,%rax,1),%xmm0
  add    $0x8,%rax
  addsd  %xmm0,%xmm1
  cmp    $0xa0,%rax
  jne    L5
  movapd %xmm1,%xmm0
  ret
```

assistant: ```c
double DotProduct(A, B)
double *A;
double *B;
{
  int i;
  double score;
  score = 0;
  for (i = 0; i < 20; i++) {
    score = score + A[i] * B[i];
  }
  return score;
}
```
user: What is the c source code of the assembly code below?

```asm
<DotProduct>:
  endbr64
  movupd (%rsi),%xmm3
  movupd (%rdi),%xmm0
  pxor   %xmm1,%xmm1
  movupd 0x10(%rsi),%xmm4
  movupd 0x20(%rsi),%xmm5
  mulpd  %xmm3,%xmm0
  movupd 0x30(%rsi),%xmm6
  movupd 0x40(%rsi),%xmm7
  movupd 0x50(%rsi),%xmm3
  addsd  %xmm0,%xmm1
  unpckhpd %xmm0,%xmm0
  addsd  %xmm0,%xmm1
  movupd 0x10(%rdi),%xmm0
  mulpd  %xmm4,%xmm0
  movupd 0x60(%rsi),%xmm4
  addsd  %xmm0,%xmm1
  unpckhpd %xmm0,%xmm0
  movapd %xmm0,%xmm2
  movupd 0x20(%rdi),%xmm0
  mulpd  %xmm5,%xmm0
  addsd  %xmm1,%xmm2
  movupd 0x70(%rsi),%xmm5
  addsd  %xmm0,%xmm2
  unpckhpd %xmm0,%xmm0
  movapd %xmm0,%xmm1
  movupd 0x30(%rdi),%xmm0
  mulpd  %xmm6,%xmm0
  addsd  %xmm2,%xmm1
  movupd 0x80(%rsi),%xmm6
  addsd  %xmm0,%xmm1
  unpckhpd %xmm0,%xmm0
  movapd %xmm0,%xmm2
  movupd 0x40(%rdi),%xmm0
  mulpd  %xmm7,%xmm0
  addsd  %xmm1,%xmm2
  addsd  %xmm0,%xmm2
  unpckhpd %xmm0,%xmm0
  movapd %xmm0,%xmm1
  movupd 0x50(%rdi),%xmm0
  mulpd  %xmm3,%xmm0
  addsd  %xmm2,%xmm1
  addsd  %xmm0,%xmm1
  unpckhpd %xmm0,%xmm0
  movapd %xmm0,%xmm2
  movupd 0x60(%rdi),%xmm0
  mulpd  %xmm4,%xmm0
  addsd  %xmm1,%xmm2
  addsd  %xmm0,%xmm2
  unpckhpd %xmm0,%xmm0
  movapd %xmm0,%xmm1
  movupd 0x70(%rdi),%xmm0
  mulpd  %xmm5,%xmm0
  addsd  %xmm2,%xmm1
  movupd 0x90(%rdi),%xmm2
  addsd  %xmm0,%xmm1
  unpckhpd %xmm0,%xmm0
  addsd  %xmm1,%xmm0
  movupd 0x80(%rdi),%xmm1
  mulpd  %xmm6,%xmm1
  addsd  %xmm1,%xmm0
  unpckhpd %xmm1,%xmm1
  addsd  %xmm0,%xmm1
  movupd 0x90(%rsi),%xmm0
  mulpd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  unpckhpd %xmm0,%xmm0
  addsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double DotProduct(A, B)
double *A;
double *B;
{
  int i;
  double score;
  score = 0;
  for (i = 0; i < 20; i++) {
    score = score + A[i] * B[i];
  }
  return score;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bubble_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L56
L5:
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L43
L28:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L37
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <swap@plt>
  addl   $0x1,-0x8(%rbp)
L37:
  addl   $0x1,-0x4(%rbp)
L43:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jl     L28
  mov    -0x8(%rbp),%eax
  add    %eax,-0x10(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  je     L27
  addl   $0x1,-0xc(%rbp)
L56:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L5
  jmp    L8
L27:
  nop
L8:
  mov    -0x28(%rbp),%rax
  mov    -0x10(%rbp),%edx
  mov    %edx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void bubble_sort(int *arr, int len, int *count) {
  int total_swaps = 0;
  for (int i = 0; i < len; i++) {
    int curr_swaps = 0;
    for (int j = 0; j < len - 1; j++) {
      if (arr[j] > arr[j + 1]) {
        swap(arr, j, j + 1);
        curr_swaps++;
      }
    }
    total_swaps += curr_swaps;
    if (!curr_swaps) {
      break;
    }
  }
  *count = total_swaps;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<bubble_sort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r14d
  mov    %rdx,0x8(%rsp)
  movl   $0x0,(%rsp)
  mov    $0x0,%r15d
  lea    -0x2(%rsi),%eax
  mov    %eax,0x4(%rsp)
  test   %esi,%esi
  jle    L62
L45:
  cmp    $0x1,%r14d
  jle    L62
  mov    0x4(%rsp),%r12d
  mov    $0x0,%ebx
  mov    $0x0,%r13d
  jmp    L81
L22:
  lea    0x1(%rbx),%rax
  cmp    %r12,%rbx
  je     L56
  mov    %rax,%rbx
L81:
  mov    %ebx,%esi
  mov    0x4(%rbp,%rbx,4),%eax
  cmp    %eax,0x0(%rbp,%rbx,4)
  jle    L22
  lea    0x1(%rbx),%edx
  mov    %rbp,%rdi
  call   <swap@plt>
  add    $0x1,%r13d
  jmp    L22
L56:
  add    %r13d,%r15d
  test   %r13d,%r13d
  jne    L5
L62:
  mov    0x8(%rsp),%rax
  mov    %r15d,(%rax)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  addl   $0x1,(%rsp)
  mov    (%rsp),%eax
  cmp    %eax,%r14d
  jne    L45
  jmp    L62
```

assistant: ```c
void bubble_sort(int *arr, int len, int *count) {
  int total_swaps = 0;
  for (int i = 0; i < len; i++) {
    int curr_swaps = 0;
    for (int j = 0; j < len - 1; j++) {
      if (arr[j] > arr[j + 1]) {
        swap(arr, j, j + 1);
        curr_swaps++;
      }
    }
    total_swaps += curr_swaps;
    if (!curr_swaps) {
      break;
    }
  }
  *count = total_swaps;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bubble_sort>:
  endbr64
  push   %r15
  lea    -0x2(%rsi),%eax
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdx,0x8(%rsp)
  movl   $0x0,(%rsp)
  mov    %eax,0x4(%rsp)
  test   %esi,%esi
  jle    L46
  nopl   0x0(%rax)
L6:
  cmp    $0x1,%r12d
  je     L46
  mov    0x4(%rsp),%r13d
  xor    %ebx,%ebx
  xor    %r14d,%r14d
  jmp    L21
  nopw   0x0(%rax,%rax,1)
L28:
  lea    0x1(%rbx),%rax
  cmp    %r13,%rbx
  je     L1
L19:
  mov    %rax,%rbx
L21:
  mov    0x4(%r15,%rbx,4),%eax
  mov    %ebx,%esi
  lea    0x1(%rbx),%edx
  cmp    %eax,(%r15,%rbx,4)
  jle    L28
  mov    %r15,%rdi
  add    $0x1,%r14d
  call   <swap@plt>
  lea    0x1(%rbx),%rax
  cmp    %r13,%rbx
  jne    L19
L1:
  add    %r14d,(%rsp)
  test   %r14d,%r14d
  je     L46
  add    $0x1,%ebp
  cmp    %ebp,%r12d
  jne    L6
L46:
  mov    0x8(%rsp),%rax
  mov    (%rsp),%ecx
  mov    %ecx,(%rax)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void bubble_sort(int *arr, int len, int *count) {
  int total_swaps = 0;
  for (int i = 0; i < len; i++) {
    int curr_swaps = 0;
    for (int j = 0; j < len - 1; j++) {
      if (arr[j] > arr[j + 1]) {
        swap(arr, j, j + 1);
        curr_swaps++;
      }
    }
    total_swaps += curr_swaps;
    if (!curr_swaps) {
      break;
    }
  }
  *count = total_swaps;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bubble_sort>:
  endbr64
  push   %r15
  lea    -0x2(%rsi),%eax
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdx,0x8(%rsp)
  movl   $0x0,(%rsp)
  mov    %eax,0x4(%rsp)
  test   %esi,%esi
  jle    L43
  nopl   0x0(%rax)
L11:
  cmp    $0x1,%r12d
  je     L43
  mov    0x4(%rsp),%r13d
  xor    %ebx,%ebx
  xor    %r14d,%r14d
  jmp    L66
  nopw   0x0(%rax,%rax,1)
L68:
  lea    0x1(%rbx),%rax
  cmp    %rbx,%r13
  je     L77
L15:
  mov    %rax,%rbx
L66:
  mov    0x4(%r15,%rbx,4),%eax
  mov    %ebx,%esi
  lea    0x1(%rbx),%edx
  cmp    %eax,(%r15,%rbx,4)
  jle    L68
  mov    %r15,%rdi
  add    $0x1,%r14d
  call   <swap@plt>
  lea    0x1(%rbx),%rax
  cmp    %rbx,%r13
  jne    L15
L77:
  add    %r14d,(%rsp)
  test   %r14d,%r14d
  je     L43
  add    $0x1,%ebp
  cmp    %ebp,%r12d
  jne    L11
L43:
  mov    0x8(%rsp),%rax
  mov    (%rsp),%ecx
  mov    %ecx,(%rax)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void bubble_sort(int *arr, int len, int *count) {
  int total_swaps = 0;
  for (int i = 0; i < len; i++) {
    int curr_swaps = 0;
    for (int j = 0; j < len - 1; j++) {
      if (arr[j] > arr[j + 1]) {
        swap(arr, j, j + 1);
        curr_swaps++;
      }
    }
    total_swaps += curr_swaps;
    if (!curr_swaps) {
      break;
    }
  }
  *count = total_swaps;
}
```