ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<hdoip_cli_get_edid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <hoic_get_edid@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int hdoip_cli_get_edid(int fd, int fdr, char **argv, int argc) {
  hoic_get_edid(fd, argv[0]);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<hdoip_cli_get_edid>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rdx),%rsi
  call   <hoic_get_edid@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int hdoip_cli_get_edid(int fd, int fdr, char **argv, int argc) {
  hoic_get_edid(fd, argv[0]);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<hdoip_cli_get_edid>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rdx),%rsi
  call   <hoic_get_edid@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int hdoip_cli_get_edid(int fd, int fdr, char **argv, int argc) {
  hoic_get_edid(fd, argv[0]);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hdoip_cli_get_edid>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rdx),%rsi
  call   <hoic_get_edid@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int hdoip_cli_get_edid(int fd, int fdr, char **argv, int argc) {
  hoic_get_edid(fd, argv[0]);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<generate_passwd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,-0xc(%rbp)
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    $0x0,%edx
  divl   -0xc(%rbp)
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L23
L21:
  mov    $0x0,%eax
  call   <rand@plt>
  addl   $0x1,-0x14(%rbp)
L23:
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  cmp    %eax,-0x14(%rbp)
  jl     L21
  movl   $0x0,-0x10(%rbp)
  jmp    L37
L7:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  lea    0x41(%rax),%ecx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x10(%rbp)
L37:
  mov    -0x10(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L7
  nop
  nop
  leave
  ret
```

assistant: ```c
void generate_passwd(char password[], int length) {
  pid_t pid = getpid();
  time_t current_time = time(NULL);
  unsigned int seed = current_time / pid;
  srand(seed);
  for (int i = 0; i < abs(pid); i++)
    rand();
  for (int i = 0; i < length; i++)
    password[i] = 'A' + (rand() % 26);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<generate_passwd>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%r12d
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%ebx
  mov    $0x0,%edi
  call   <time@plt>
  mov    $0x0,%edx
  div    %ebx
  mov    %eax,%edi
  call   <srand@plt>
  mov    %ebx,%ebp
  neg    %ebp
  cmovs  %ebx,%ebp
  test   %ebp,%ebp
  jle    L0
  mov    $0x0,%ebx
L32:
  mov    $0x0,%eax
  call   <rand@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jne    L32
L0:
  test   %r12d,%r12d
  jle    L28
  mov    %r13,%rbx
  lea    -0x1(%r12),%eax
  lea    0x1(%r13,%rax,1),%rbp
L1:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x23,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    $0x41,%eax
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L1
L28:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void generate_passwd(char password[], int length) {
  pid_t pid = getpid();
  time_t current_time = time(NULL);
  unsigned int seed = current_time / pid;
  srand(seed);
  for (int i = 0; i < abs(pid); i++)
    rand();
  for (int i = 0; i < length; i++)
    password[i] = 'A' + (rand() % 26);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<generate_passwd>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <getpid@plt>
  xor    %edi,%edi
  mov    %eax,%ebx
  call   <time@plt>
  xor    %edx,%edx
  mov    %ebx,%r12d
  div    %ebx
  mov    %eax,%edi
  call   <srand@plt>
  neg    %r12d
  cmovs  %ebx,%r12d
  test   %ebx,%ebx
  je     L7
  xor    %ebx,%ebx
  nop
L20:
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <rand@plt>
  cmp    %r12d,%ebx
  jne    L20
L7:
  test   %r13d,%r13d
  jle    L9
  lea    -0x1(%r13),%eax
  lea    0x1(%rbp,%rax,1),%rbx
  nopl   (%rax)
L11:
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    $0x41,%eax
  mov    %al,-0x1(%rbp)
  cmp    %rbp,%rbx
  jne    L11
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void generate_passwd(char password[], int length) {
  pid_t pid = getpid();
  time_t current_time = time(NULL);
  unsigned int seed = current_time / pid;
  srand(seed);
  for (int i = 0; i < abs(pid); i++)
    rand();
  for (int i = 0; i < length; i++)
    password[i] = 'A' + (rand() % 26);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<generate_passwd>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <getpid@plt>
  xor    %edi,%edi
  mov    %eax,%ebx
  call   <time@plt>
  xor    %edx,%edx
  mov    %ebx,%r12d
  div    %ebx
  mov    %eax,%edi
  call   <srand@plt>
  neg    %r12d
  cmovs  %ebx,%r12d
  test   %ebx,%ebx
  je     L29
  xor    %ebx,%ebx
  nop
L7:
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <rand@plt>
  cmp    %r12d,%ebx
  jne    L7
L29:
  test   %r13d,%r13d
  jle    L11
  lea    -0x1(%r13),%eax
  lea    0x1(%rbp,%rax,1),%rbx
  nopl   (%rax)
L39:
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    $0x41,%eax
  mov    %al,-0x1(%rbp)
  cmp    %rbp,%rbx
  jne    L39
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void generate_passwd(char password[], int length) {
  pid_t pid = getpid();
  time_t current_time = time(NULL);
  unsigned int seed = current_time / pid;
  srand(seed);
  for (int i = 0; i < abs(pid); i++)
    rand();
  for (int i = 0; i < length; i++)
    password[i] = 'A' + (rand() % 26);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<convert_base10>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  movl   $0x1,-0x1c(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  jmp    L63
L59:
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0x28(%rbp)
  mov    %eax,-0x20(%rbp)
  addl   $0x1,-0x1c(%rbp)
L63:
  cmpl   $0x0,-0x20(%rbp)
  jg     L59
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L12
  mov    $0x0,%eax
  jmp    L45
L12:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  subl   $0x1,-0x1c(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  jmp    L56
L10:
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0x28(%rbp)
  mov    %edx,%eax
  cmp    $0x9,%eax
  jle    L3
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0x28(%rbp)
  subl   $0x1,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %edx,%edi
  call   <ifhex@plt>
  mov    %al,(%rbx)
  jmp    L72
L3:
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0x28(%rbp)
  mov    %edx,%eax
  lea    0x30(%rax),%ecx
  subl   $0x1,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L72:
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0x28(%rbp)
  mov    %eax,-0x20(%rbp)
L56:
  cmpl   $0x0,-0x20(%rbp)
  jg     L10
  mov    -0x18(%rbp),%rax
L45:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *convert_base10(int nbr, int base_to) {
  int resdiv;
  int i;
  char *result;
  i = 1;
  resdiv = nbr;
  while (resdiv > 0) {
    resdiv = resdiv / base_to;
    ++i;
  }
  if ((result = malloc(sizeof(char) * i)) == NULL)
    return (0);
  result[i] = '\0';
  --i;
  resdiv = nbr;
  while (resdiv > 0) {
    if ((resdiv % base_to) > 9)
      result[--i] = ifhex(resdiv % base_to);
    else
      result[--i] = (resdiv % base_to) + 48;
    resdiv /= base_to;
  }
  return (result);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<convert_base10>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L29
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %edi,%edx
  mov    $0x1,%r14d
L76:
  mov    %edx,%eax
  cltd
  idiv   %ebx
  mov    %eax,%edx
  mov    %r14d,%r12d
  add    $0x1,%r14d
  test   %eax,%eax
  jg     L76
  mov    %r14d,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L40
  movslq %r14d,%r14
  movb   $0x0,0x0(%r13,%r14,1)
  sub    $0x1,%r12d
  jmp    L67
L29:
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L40
  movb   $0x0,0x1(%rax)
L40:
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L61:
  movslq %r12d,%r14
  add    %r13,%r14
  mov    %edx,%edi
  call   <ifhex@plt>
  mov    %al,(%r14)
L34:
  mov    %ebp,%eax
  cltd
  idiv   %ebx
  mov    %eax,%ebp
  sub    $0x1,%r12d
  test   %eax,%eax
  jle    L40
L67:
  mov    %ebp,%eax
  cltd
  idiv   %ebx
  cmp    $0x9,%edx
  jg     L61
  movslq %r12d,%rax
  add    $0x30,%edx
  mov    %dl,0x0(%r13,%rax,1)
  jmp    L34
```

assistant: ```c
char *convert_base10(int nbr, int base_to) {
  int resdiv;
  int i;
  char *result;
  i = 1;
  resdiv = nbr;
  while (resdiv > 0) {
    resdiv = resdiv / base_to;
    ++i;
  }
  if ((result = malloc(sizeof(char) * i)) == NULL)
    return (0);
  result[i] = '\0';
  --i;
  resdiv = nbr;
  while (resdiv > 0) {
    if ((resdiv % base_to) > 9)
      result[--i] = ifhex(resdiv % base_to);
    else
      result[--i] = (resdiv % base_to) + 48;
    resdiv /= base_to;
  }
  return (result);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<convert_base10>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L7
  mov    %edi,%ebx
  mov    %esi,%r12d
  mov    %edi,%eax
  mov    $0x1,%ebp
L70:
  cltd
  movslq %ebp,%r13
  add    $0x1,%ebp
  idiv   %r12d
  test   %eax,%eax
  jg     L70
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L19
  movslq %ebp,%rbp
  movb   $0x0,(%r14,%rbp,1)
  lea    -0x1(%r14,%r13,1),%rbp
  jmp    L88
  nopl   0x0(%rax)
L55:
  mov    %al,0x0(%rbp)
  sub    $0x1,%rbp
  test   %ebx,%ebx
  jle    L86
L88:
  mov    %ebx,%eax
  cltd
  idiv   %r12d
  mov    %eax,%ebx
  lea    0x30(%rdx),%eax
  cmp    $0x9,%edx
  jle    L55
  mov    %edx,%edi
  sub    $0x1,%rbp
  call   <ifhex@plt>
  mov    %al,0x1(%rbp)
  test   %ebx,%ebx
  jg     L88
L86:
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L7:
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L19
  movb   $0x0,0x1(%rax)
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L19:
  xor    %r14d,%r14d
  jmp    L86
```

assistant: ```c
char *convert_base10(int nbr, int base_to) {
  int resdiv;
  int i;
  char *result;
  i = 1;
  resdiv = nbr;
  while (resdiv > 0) {
    resdiv = resdiv / base_to;
    ++i;
  }
  if ((result = malloc(sizeof(char) * i)) == NULL)
    return (0);
  result[i] = '\0';
  --i;
  resdiv = nbr;
  while (resdiv > 0) {
    if ((resdiv % base_to) > 9)
      result[--i] = ifhex(resdiv % base_to);
    else
      result[--i] = (resdiv % base_to) + 48;
    resdiv /= base_to;
  }
  return (result);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<convert_base10>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L19
  mov    %edi,%ebx
  mov    %esi,%r12d
  mov    %edi,%eax
  mov    $0x1,%ebp
L56:
  cltd
  movslq %ebp,%r13
  add    $0x1,%ebp
  idiv   %r12d
  test   %eax,%eax
  jg     L56
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L47
  movslq %ebp,%rbp
  movb   $0x0,(%r14,%rbp,1)
  lea    -0x1(%r14,%r13,1),%rbp
  xchg   %ax,%ax
L74:
  mov    %ebx,%eax
  cltd
  idiv   %r12d
  mov    %eax,%ebx
  cmp    $0x9,%edx
  jg     L30
  add    $0x30,%edx
  sub    $0x1,%rbp
  mov    %dl,0x1(%rbp)
  test   %eax,%eax
  jg     L74
L38:
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  mov    %edx,%edi
  sub    $0x1,%rbp
  call   <ifhex@plt>
  mov    %al,0x1(%rbp)
  test   %ebx,%ebx
  jg     L74
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L19:
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L47
  movb   $0x0,0x1(%rax)
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L47:
  xor    %r14d,%r14d
  jmp    L38
```

assistant: ```c
char *convert_base10(int nbr, int base_to) {
  int resdiv;
  int i;
  char *result;
  i = 1;
  resdiv = nbr;
  while (resdiv > 0) {
    resdiv = resdiv / base_to;
    ++i;
  }
  if ((result = malloc(sizeof(char) * i)) == NULL)
    return (0);
  result[i] = '\0';
  --i;
  resdiv = nbr;
  while (resdiv > 0) {
    if ((resdiv % base_to) > 9)
      result[--i] = ifhex(resdiv % base_to);
    else
      result[--i] = (resdiv % base_to) + 48;
    resdiv /= base_to;
  }
  return (result);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <moreprolog@plt>
  cmp    $0x1,%eax
  jne    L4
  mov    $0x0,%eax
  call   <readpages@plt>
  mov    $0x0,%eax
  call   <writepages@plt>
  mov    $0x0,%eax
  call   <trailer@plt>
L4:
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
data_type: string

tool: D3: %%EndProlog


assistant: ```c
void reverse(void) {
  if (moreprolog("%%EndProlog\n") == 1) {
    readpages();
    writepages();
    trailer();
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<reverse>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <moreprolog@plt>
  cmp    $0x1,%eax
  je     L19
L13:
  add    $0x8,%rsp
  ret
L19:
  mov    $0x0,%eax
  call   <readpages@plt>
  mov    $0x0,%eax
  call   <writepages@plt>
  mov    $0x0,%eax
  call   <trailer@plt>
  jmp    L13
```

tool_calls: name: unpack_data
arguments: label: D9
data_type: string

tool: D9: %%EndProlog


assistant: ```c
void reverse(void) {
  if (moreprolog("%%EndProlog\n") == 1) {
    readpages();
    writepages();
    trailer();
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<reverse>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <moreprolog@plt>
  cmp    $0x1,%eax
  je     L0
  add    $0x8,%rsp
  ret
  xchg   %ax,%ax
L0:
  xor    %eax,%eax
  call   <readpages@plt>
  xor    %eax,%eax
  call   <writepages@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10b0 <trailer@plt>
```

tool_calls: name: parse_data
arguments: label: D1
type_infered: string

tool: D1: %%EndProlog


assistant: ```c
void reverse(void) {
  if (moreprolog("%%EndProlog\n") == 1) {
    readpages();
    writepages();
    trailer();
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<reverse>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <moreprolog@plt>
  cmp    $0x1,%eax
  je     L9
  add    $0x8,%rsp
  ret
  xchg   %ax,%ax
L9:
  xor    %eax,%eax
  call   <readpages@plt>
  xor    %eax,%eax
  call   <writepages@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10b0 <trailer@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D3
datatype: string

tool: D3: %%EndProlog


assistant: ```c
void reverse(void) {
  if (moreprolog("%%EndProlog\n") == 1) {
    readpages();
    writepages();
    trailer();
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x820,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x2a,-0x814(%rbp)
  jmp    L1
L25:
  lea    -0x410(%rbp),%rdx
  mov    D9(%rip),%rax
  mov    %rdx,%rsi
  mov    $0xa,%edi
  movq   %rax,%xmm0
  call   <gcvt@plt>
  mov    %rax,%rdx
  mov    D10(%rip),%rcx
  lea    -0x810(%rbp),%rax
  movq   %rcx,%xmm0
  lea    D5(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    %eax,%edx
  mov    -0x814(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x814(%rbp)
  lea    -0x810(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    -0x810(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <__wrap_write@plt>
L1:
  cmpl   $0x2a,-0x814(%rbp)
  je     L25
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
type: f64
name: read_data
arguments: data_tag: D10
type: byte[8]
name: read_data
arguments: data_tag: D5
type: string

tool: D9: 3.5

tool: {"D10": [0x55,0x55,0x55,0x55,0x55,0x55,0xd5,0x3f]}

tool: D5: hello world: %f %s


assistant: ```c
int main(void) {
  int ret = 42;
  char buf[1024];
  char buf2[1024];
  while (ret == 42) {
    ret += sprintf(buf, "hello world: %f %s\n", 1.0 / 3.0, gcvt(3.5, 10, buf2));
    __wrap_write(1, buf, strlen(buf));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  lea    0x400(%rsp),%r13
  lea    D6(%rip),%r12
L7:
  mov    %r13,%rsi
  mov    $0xa,%edi
  mov    D22(%rip),%rax
  movq   %rax,%xmm0
  call   <gcvt@plt>
  mov    %rax,%rdx
  mov    %rsp,%rbp
  mov    D0(%rip),%rax
  movq   %rax,%xmm0
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    $0x1,%edi
  call   <__wrap_write@plt>
  test   %ebx,%ebx
  je     L7
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x818,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D6
data_type: string
name: read_data
arguments: label: D22
data_type: f64
name: read_data
arguments: label: D0
data_type: byte[8]

tool: D6: hello world: %f %s


tool: D22: 3.5

tool: {"D0": [0x55,0x55,0x55,0x55,0x55,0x55,0xd5,0x3f]}

assistant: ```c
int main(void) {
  int ret = 42;
  char buf[1024];
  char buf2[1024];
  while (ret == 42) {
    ret += sprintf(buf, "hello world: %f %s\n", 1.0 / 3.0, gcvt(3.5, 10, buf2));
    __wrap_write(1, buf, strlen(buf));
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D15(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  lea    0x400(%rsp),%r13
  mov    %rsp,%rbp
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    D4(%rip),%rax
  mov    %r13,%rsi
  mov    $0xa,%edi
  movq   %rax,%xmm0
  call   <gcvt@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %rax,%rdx
  mov    D6(%rip),%rax
  movq   %rax,%xmm0
  call   <sprintf@plt>
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    $0x1,%edi
  mov    %eax,%edx
  call   <__wrap_write@plt>
  test   %ebx,%ebx
  je     L15
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x818,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D15
type_infered: string
name: parse_data
arguments: data_tag: D4
type_infered: f64
name: parse_data
arguments: data_tag: D6
type_infered: byte[8]

tool: D15: hello world: %f %s


tool: D4: 3.5

tool: {"D6": [0x55,0x55,0x55,0x55,0x55,0x55,0xd5,0x3f]}

assistant: ```c
int main(void) {
  int ret = 42;
  char buf[1024];
  char buf2[1024];
  while (ret == 42) {
    ret += sprintf(buf, "hello world: %f %s\n", 1.0 / 3.0, gcvt(3.5, 10, buf2));
    __wrap_write(1, buf, strlen(buf));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D12(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  lea    0x400(%rsp),%r13
  mov    %rsp,%rbp
  cs nopw 0x0(%rax,%rax,1)
L8:
  mov    D16(%rip),%rax
  mov    %r13,%rsi
  mov    $0xa,%edi
  movq   %rax,%xmm0
  call   <gcvt@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    %rax,%rdx
  mov    D5(%rip),%rax
  movq   %rax,%xmm0
  call   <sprintf@plt>
  mov    %rbp,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    $0x1,%edi
  mov    %eax,%edx
  call   <__wrap_write@plt>
  test   %ebx,%ebx
  je     L8
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x818,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D12
type_guessed: string
name: read_data
arguments: data_tag: D16
type_guessed: f64
name: read_data
arguments: data_tag: D5
type_guessed: byte[8]

tool: D12: hello world: %f %s


tool: D16: 3.5

tool: {"D5": [0x55,0x55,0x55,0x55,0x55,0x55,0xd5,0x3f]}

assistant: ```c
int main(void) {
  int ret = 42;
  char buf[1024];
  char buf2[1024];
  while (ret == 42) {
    ret += sprintf(buf, "hello world: %f %s\n", 1.0 / 3.0, gcvt(3.5, 10, buf2));
    __wrap_write(1, buf, strlen(buf));
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tpSDKCreate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    %r9d,-0x34(%rbp)
  mov    -0x34(%rbp),%r9d
  mov    -0x30(%rbp),%r8
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  push   0x28(%rbp)
  push   $0x1
  push   $0x0
  push   0x20(%rbp)
  mov    0x18(%rbp),%edi
  push   %rdi
  push   0x10(%rbp)
  mov    %rax,%rdi
  call   <MQTTCreate@plt>
  add    $0x30,%rsp
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int tpSDKCreate(char *host, int port, int keepalive, char *userName, char *password, int enableServerCertAuth, char *sudirribeTopic[], int sudirribeTopicSize, char *publishTopic, char *clientID) {
  int rc = MQTTCreate(host, port, keepalive, userName, password, enableServerCertAuth, sudirribeTopic, sudirribeTopicSize, publishTopic, NULL, 1, clientID);
  return rc;
}
```