ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<hexdump_file>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r15
  mov    %rdx,0x8(%rsp)
  mov    %rcx,%r13
  mov    %rcx,%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L106
  mov    %rax,%r14
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  jmp    L68
L106:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L41
L7:
  mov    %al,(%r14,%rbp,1)
  add    $0x1,%rbp
L59:
  cmp    %r13,%rbp
  jae    L25
L13:
  add    $0x1,%rbx
L68:
  mov    %r12,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L55
  mov    %r12,%rdi
  call   <getc@plt>
  cmp    %r15,%rbx
  jb     L59
  cmp    0x8(%rsp),%rbx
  jbe    L7
  jmp    L59
L25:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <hexdump@plt>
  mov    $0x0,%ebp
  jmp    L13
L55:
  test   %rbp,%rbp
  jne    L81
L105:
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L41:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L81:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <hexdump@plt>
  jmp    L105
```

tool_calls: name: disassemble_data
arguments: data_label: D8
type_guessed: string

tool: D8: Could not allocate memory for byte_buffer


assistant: ```c
int hexdump_file(FILE *infile, unsigned long start, unsigned long stop, unsigned long width) {
  char ch;
  unsigned long f_index = 0;
  unsigned long bb_index = 0;
  unsigned char *byte_buffer = malloc(width);
  if (byte_buffer == NULL) {
    printf("Could not allocate memory for byte_buffer\n");
    return -1;
  }
  while (!feof(infile)) {
    ch = getc(infile);
    if ((f_index >= start) && (f_index <= stop)) {
      byte_buffer[bb_index] = ch;
      bb_index++;
    }
    if (bb_index >= width) {
      hexdump(byte_buffer, bb_index, width);
      bb_index = 0;
    }
    f_index++;
  }
  if (bb_index)
    hexdump(byte_buffer, bb_index, width);
  fclose(infile);
  free(byte_buffer);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<hexdump_file>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdi,%r12
  mov    %rcx,%rdi
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdx,0x8(%rsp)
  call   <malloc@plt>
  test   %rax,%rax
  je     L68
  mov    %rax,%r14
  xor    %ebp,%ebp
  xor    %ebx,%ebx
  jmp    L83
  nopw   0x0(%rax,%rax,1)
L67:
  add    $0x1,%rbx
L83:
  mov    %r12,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L97
  mov    %r12,%rdi
  call   <getc@plt>
  cmp    %r15,%rbx
  jb     L59
  cmp    0x8(%rsp),%rbx
  ja     L59
  mov    %al,(%r14,%rbp,1)
  add    $0x1,%rbp
L59:
  cmp    %r13,%rbp
  jb     L67
  mov    %rbp,%rsi
  mov    %r13,%rdx
  mov    %r14,%rdi
  xor    %ebp,%ebp
  call   <hexdump@plt>
  jmp    L67
  nopl   0x0(%rax,%rax,1)
L97:
  test   %rbp,%rbp
  jne    L40
L51:
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    %r14,%rdi
  call   <free@plt>
  xor    %eax,%eax
L19:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L40:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <hexdump@plt>
  jmp    L51
L68:
  lea    D5(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L19
```

tool_calls: name: retrieve_data
arguments: data_label: D5
datatype: string

tool: D5: Could not allocate memory for byte_buffer


assistant: ```c
int hexdump_file(FILE *infile, unsigned long start, unsigned long stop, unsigned long width) {
  char ch;
  unsigned long f_index = 0;
  unsigned long bb_index = 0;
  unsigned char *byte_buffer = malloc(width);
  if (byte_buffer == NULL) {
    printf("Could not allocate memory for byte_buffer\n");
    return -1;
  }
  while (!feof(infile)) {
    ch = getc(infile);
    if ((f_index >= start) && (f_index <= stop)) {
      byte_buffer[bb_index] = ch;
      bb_index++;
    }
    if (bb_index >= width) {
      hexdump(byte_buffer, bb_index, width);
      bb_index = 0;
    }
    f_index++;
  }
  if (bb_index)
    hexdump(byte_buffer, bb_index, width);
  fclose(infile);
  free(byte_buffer);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<hexdump_file>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdi,%r12
  mov    %rcx,%rdi
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdx,0x8(%rsp)
  call   <malloc@plt>
  test   %rax,%rax
  je     L81
  mov    %rax,%r14
  xor    %ebp,%ebp
  xor    %ebx,%ebx
  jmp    L84
  nopw   0x0(%rax,%rax,1)
L28:
  add    $0x1,%rbx
L84:
  mov    %r12,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L68
  mov    %r12,%rdi
  call   <getc@plt>
  cmp    %r15,%rbx
  jb     L56
  cmp    0x8(%rsp),%rbx
  ja     L56
  mov    %al,(%r14,%rbp,1)
  add    $0x1,%rbp
L56:
  cmp    %r13,%rbp
  jb     L28
  mov    %rbp,%rsi
  mov    %r13,%rdx
  mov    %r14,%rdi
  xor    %ebp,%ebp
  call   <hexdump@plt>
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L68:
  test   %rbp,%rbp
  jne    L94
L60:
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    %r14,%rdi
  call   <free@plt>
  xor    %eax,%eax
L57:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L94:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <hexdump@plt>
  jmp    L60
L81:
  lea    D3(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L57
```

tool_calls: name: unpack_data
arguments: label: D3
type_infered: string

tool: D3: Could not allocate memory for byte_buffer


assistant: ```c
int hexdump_file(FILE *infile, unsigned long start, unsigned long stop, unsigned long width) {
  char ch;
  unsigned long f_index = 0;
  unsigned long bb_index = 0;
  unsigned char *byte_buffer = malloc(width);
  if (byte_buffer == NULL) {
    printf("Could not allocate memory for byte_buffer\n");
    return -1;
  }
  while (!feof(infile)) {
    ch = getc(infile);
    if ((f_index >= start) && (f_index <= stop)) {
      byte_buffer[bb_index] = ch;
      bb_index++;
    }
    if (bb_index >= width) {
      hexdump(byte_buffer, bb_index, width);
      bb_index = 0;
    }
    f_index++;
  }
  if (bb_index)
    hexdump(byte_buffer, bb_index, width);
  fclose(infile);
  free(byte_buffer);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<cutPoly3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <initEST@plt>
  test   %eax,%eax
  jne    L1
  mov    $0x0,%eax
  jmp    L55
L1:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <initPST@plt>
  test   %eax,%eax
  jne    L83
  mov    $0x0,%eax
  jmp    L55
L83:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <cutST@plt>
  test   %eax,%eax
  jne    L45
  mov    $0x0,%eax
  jmp    L55
L45:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <finishST@plt>
  test   %eax,%eax
  jne    L69
  mov    $0x0,%eax
  jmp    L55
L69:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fetchST@plt>
  test   %eax,%eax
  jne    L98
  mov    $0x0,%eax
  jmp    L55
L98:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <freeST@plt>
  test   %eax,%eax
  jne    L82
  mov    $0x0,%eax
  jmp    L55
L82:
  mov    $0x1,%eax
L55:
  leave
  ret
```

assistant: ```c
int cutPoly3(struct Plex *plex) {
  if (!initEST(plex))
    return (0);
  if (!initPST(plex))
    return (0);
  if (!cutST(plex))
    return (0);
  if (!finishST(plex))
    return (0);
  if (!fetchST(plex))
    return (0);
  if (!freeST(plex))
    return (0);
  return (1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cutPoly3>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <initEST@plt>
  test   %eax,%eax
  jne    L30
L41:
  pop    %rbx
  ret
L30:
  mov    %rbx,%rdi
  call   <initPST@plt>
  test   %eax,%eax
  je     L41
  mov    %rbx,%rdi
  call   <cutST@plt>
  test   %eax,%eax
  je     L41
  mov    %rbx,%rdi
  call   <finishST@plt>
  test   %eax,%eax
  je     L41
  mov    %rbx,%rdi
  call   <fetchST@plt>
  test   %eax,%eax
  je     L41
  mov    %rbx,%rdi
  call   <freeST@plt>
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  jmp    L41
```

assistant: ```c
int cutPoly3(struct Plex *plex) {
  if (!initEST(plex))
    return (0);
  if (!initPST(plex))
    return (0);
  if (!cutST(plex))
    return (0);
  if (!finishST(plex))
    return (0);
  if (!fetchST(plex))
    return (0);
  if (!freeST(plex))
    return (0);
  return (1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<cutPoly3>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <initEST@plt>
  test   %eax,%eax
  jne    L1
L34:
  xor    %eax,%eax
  pop    %rbp
  ret
  nopl   (%rax)
L1:
  mov    %rbp,%rdi
  call   <initPST@plt>
  test   %eax,%eax
  je     L34
  mov    %rbp,%rdi
  call   <cutST@plt>
  test   %eax,%eax
  je     L34
  mov    %rbp,%rdi
  call   <finishST@plt>
  test   %eax,%eax
  je     L34
  mov    %rbp,%rdi
  call   <fetchST@plt>
  test   %eax,%eax
  je     L34
  mov    %rbp,%rdi
  call   <freeST@plt>
  pop    %rbp
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int cutPoly3(struct Plex *plex) {
  if (!initEST(plex))
    return (0);
  if (!initPST(plex))
    return (0);
  if (!cutST(plex))
    return (0);
  if (!finishST(plex))
    return (0);
  if (!fetchST(plex))
    return (0);
  if (!freeST(plex))
    return (0);
  return (1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cutPoly3>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <initEST@plt>
  test   %eax,%eax
  jne    L37
L6:
  xor    %eax,%eax
  pop    %rbp
  ret
  nopl   (%rax)
L37:
  mov    %rbp,%rdi
  call   <initPST@plt>
  test   %eax,%eax
  je     L6
  mov    %rbp,%rdi
  call   <cutST@plt>
  test   %eax,%eax
  je     L6
  mov    %rbp,%rdi
  call   <finishST@plt>
  test   %eax,%eax
  je     L6
  mov    %rbp,%rdi
  call   <fetchST@plt>
  test   %eax,%eax
  je     L6
  mov    %rbp,%rdi
  call   <freeST@plt>
  pop    %rbp
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int cutPoly3(struct Plex *plex) {
  if (!initEST(plex))
    return (0);
  if (!initPST(plex))
    return (0);
  if (!cutST(plex))
    return (0);
  if (!finishST(plex))
    return (0);
  if (!fetchST(plex))
    return (0);
  if (!freeST(plex))
    return (0);
  return (1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<socket_connected>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x4,-0xc(%rbp)
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <getpeername@plt>
  cmp    $0xffffffff,%eax
  jne    L2
  lea    -0x11(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    $0x0,%eax
  jmp    L17
L2:
  mov    $0x1,%eax
L17:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int socket_connected(int s) {
  struct sockaddr_in sa;
  int dummy;
  char ch;
  dummy = sizeof sa;
  if (getpeername(s, (struct sockaddr *)&sa, &dummy) == -1) {
    read(s, &ch, 1);
    return 0;
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<socket_connected>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x4,0x14(%rsp)
  lea    0x14(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  call   <getpeername@plt>
  mov    %eax,%edx
  mov    $0x1,%eax
  cmp    $0xffffffff,%edx
  je     L12
L18:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L21
  add    $0x20,%rsp
  pop    %rbx
  ret
L12:
  lea    0xf(%rsp),%rsi
  mov    $0x1,%edx
  mov    %ebx,%edi
  call   <read@plt>
  mov    $0x0,%eax
  jmp    L18
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int socket_connected(int s) {
  struct sockaddr_in sa;
  int dummy;
  char ch;
  dummy = sizeof sa;
  if (getpeername(s, (struct sockaddr *)&sa, &dummy) == -1) {
    read(s, &ch, 1);
    return 0;
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<socket_connected>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  movl   $0x4,0x14(%rsp)
  call   <getpeername@plt>
  mov    %eax,%r8d
  mov    $0x1,%eax
  cmp    $0xffffffff,%r8d
  je     L16
L26:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  add    $0x20,%rsp
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L16:
  lea    0xf(%rsp),%rsi
  mov    $0x1,%edx
  mov    %ebp,%edi
  call   <read@plt>
  xor    %eax,%eax
  jmp    L26
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int socket_connected(int s) {
  struct sockaddr_in sa;
  int dummy;
  char ch;
  dummy = sizeof sa;
  if (getpeername(s, (struct sockaddr *)&sa, &dummy) == -1) {
    read(s, &ch, 1);
    return 0;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<socket_connected>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  movl   $0x4,0x14(%rsp)
  call   <getpeername@plt>
  mov    %eax,%r8d
  mov    $0x1,%eax
  cmp    $0xffffffff,%r8d
  je     L24
L2:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x20,%rsp
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L24:
  lea    0xf(%rsp),%rsi
  mov    $0x1,%edx
  mov    %ebp,%edi
  call   <read@plt>
  xor    %eax,%eax
  jmp    L2
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int socket_connected(int s) {
  struct sockaddr_in sa;
  int dummy;
  char ch;
  dummy = sizeof sa;
  if (getpeername(s, (struct sockaddr *)&sa, &dummy) == -1) {
    read(s, &ch, 1);
    return 0;
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0x7,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rsi
  mov    $0x0,%edi
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L9:
  cmp    %rdx,%rsp
  je     L32
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L9
L32:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L8
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L8:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x32,-0x30(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L56
L47:
  mov    -0x20(%rbp),%rax
  mov    -0x34(%rbp),%edx
  movslq %edx,%rdx
  mov    -0x34(%rbp),%ecx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0x34(%rbp)
L56:
  mov    -0x34(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L47
  movl   $0x0,-0x30(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L19
L11:
  mov    -0x20(%rbp),%rax
  mov    -0x34(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  add    %eax,-0x30(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    $0x0,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L19:
  mov    -0x34(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L11
  mov    -0x30(%rbp),%eax
  mov    %eax,%edx
  mov    $0x0,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x30(%rbp),%eax
  mov    %eax,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsp
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L34
  call   <__stack_chk_fail@plt>
L34:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D0
data_type: string
name: read_data
arguments: tag: D6
data_type: string
name: read_data
arguments: tag: D26
data_type: string

tool: D0: thread %d suma a[%d] / 

tool: D6: 
* thread %d suma= %d

tool: D26: suma=%d 


assistant: ```c
main() {
  int i, n = 7;
  int a[n], suma = 50;
  for (i = 0; i < n; i++)
    a[i] = i;
#pragma omp parallel
  {
    suma = 0;
#pragma omp for
    for (i = 0; i < n; i++) {
      suma = suma + a[i];
      printf("thread %d suma a[%d] / ", 0, i);
    }
    printf("\n* thread %d suma= %d", 0, suma);
  }
  printf("\n");
  printf("suma=%d \n", suma);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x1,0x4(%rsp)
  movl   $0x2,0x8(%rsp)
  movl   $0x3,0xc(%rsp)
  movl   $0x4,0x10(%rsp)
  movl   $0x5,0x14(%rsp)
  movl   $0x6,0x18(%rsp)
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  lea    D9(%rip),%r12
L10:
  add    (%rsp,%rbx,4),%ebp
  mov    %ebx,%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x7,%rbx
  jne    L10
  mov    %ebp,%edx
  mov    $0x0,%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %ebp,%esi
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D9
datatype: string
name: unpack_data
arguments: data_tag: D10
datatype: string
name: unpack_data
arguments: data_tag: D17
datatype: string

tool: D9: thread %d suma a[%d] / 

tool: D10: 
* thread %d suma= %d

tool: D17: suma=%d 


assistant: ```c
main() {
  int i, n = 7;
  int a[n], suma = 50;
  for (i = 0; i < n; i++)
    a[i] = i;
#pragma omp parallel
  {
    suma = 0;
#pragma omp for
    for (i = 0; i < n; i++) {
      suma = suma + a[i];
      printf("thread %d suma a[%d] / ", 0, i);
    }
    printf("\n* thread %d suma= %d", 0, suma);
  }
  printf("\n");
  printf("suma=%d \n", suma);
}
```
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   %r12
  lea    D10(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x200000001,%rax
  mov    %rax,0x4(%rsp)
  movabs $0x400000003,%rax
  mov    %rax,0xc(%rsp)
  movabs $0x600000005,%rax
  mov    %rax,0x14(%rsp)
  xor    %eax,%eax
  jmp    L20
  cs nopw 0x0(%rax,%rax,1)
L22:
  mov    (%rsp,%rbx,4),%eax
L20:
  mov    %ebx,%edx
  add    %eax,%ebp
  xor    %esi,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x7,%rbx
  jne    L22
  xor    %esi,%esi
  mov    %ebp,%edx
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D10
type_guessed: string
name: read_data
arguments: label: D4
type_guessed: string
name: read_data
arguments: label: D15
type_guessed: string

tool: D10: thread %d suma a[%d] / 

tool: D4: 
* thread %d suma= %d

tool: D15: suma=%d 


assistant: ```c
main() {
  int i, n = 7;
  int a[n], suma = 50;
  for (i = 0; i < n; i++)
    a[i] = i;
#pragma omp parallel
  {
    suma = 0;
#pragma omp for
    for (i = 0; i < n; i++) {
      suma = suma + a[i];
      printf("thread %d suma a[%d] / ", 0, i);
    }
    printf("\n* thread %d suma= %d", 0, suma);
  }
  printf("\n");
  printf("suma=%d \n", suma);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D16(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x38,%rsp
  movdqa D24(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D26(%rip),%rax
  movl   $0x6,0x18(%rsp)
  mov    %rsp,%r13
  movaps %xmm0,(%rsp)
  mov    %rax,0x10(%rsp)
  nopl   0x0(%rax)
L10:
  add    0x0(%r13,%rbx,4),%ebp
  mov    %ebx,%edx
  xor    %esi,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x7,%rbx
  jne    L10
  xor    %esi,%esi
  mov    %ebp,%edx
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D16
type_infered: string
name: unpack_data
arguments: data_label: D24
type_infered: byte[16]
name: unpack_data
arguments: data_label: D26
type_infered: byte[8]
name: unpack_data
arguments: data_label: D42
type_infered: string
name: unpack_data
arguments: data_label: D10
type_infered: string

tool: D16: thread %d suma a[%d] / 

tool: {"D24": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D26": [0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00]}

tool: D42: 
* thread %d suma= %d

tool: D10: suma=%d 


assistant: ```c
main() {
  int i, n = 7;
  int a[n], suma = 50;
  for (i = 0; i < n; i++)
    a[i] = i;
#pragma omp parallel
  {
    suma = 0;
#pragma omp for
    for (i = 0; i < n; i++) {
      suma = suma + a[i];
      printf("thread %d suma a[%d] / ", 0, i);
    }
    printf("\n* thread %d suma= %d", 0, suma);
  }
  printf("\n");
  printf("suma=%d \n", suma);
}
```