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

```asm
<get_commands>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  lea    D10(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L13
L37:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <trim_word@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L18
  cmpq   $0x0,-0x18(%rbp)
  jne    L6
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  jmp    L45
L6:
  mov    -0x18(%rbp),%rax
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x18(%rbp)
L45:
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,(%rdx)
L18:
  lea    D10(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x10(%rbp)
L13:
  cmpq   $0x0,-0x10(%rbp)
  jne    L37
  mov    -0x30(%rbp),%rax
  mov    -0x1c(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D10
type_infered: string

tool: D10: ;

assistant: ```c
char **get_commands(char *cmd_text, int *num_commands) {
  char **commands = NULL;
  char *cmd, *trimmed_cmd;
  int i = 0;
  cmd = strtok(cmd_text, ";");
  while (cmd != NULL) {
    trimmed_cmd = trim_word(cmd);
    if (trimmed_cmd) {
      if (!commands)
        commands = (char **)malloc(sizeof(char *));
      else
        commands = (char **)realloc(commands, sizeof(char *));
      commands[i++] = trimmed_cmd;
    }
    cmd = strtok(NULL, ";");
  }
  *num_commands = i;
  return commands;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_commands>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r12
  lea    D2(%rip),%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L22
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  lea    D2(%rip),%r13
  jmp    L0
L49:
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
L4:
  movslq %ebp,%rax
  mov    %r14,(%rbx,%rax,8)
  lea    0x1(%rbp),%ebp
L26:
  mov    %r13,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L53
L0:
  call   <trim_word@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L26
  test   %rbx,%rbx
  je     L49
  mov    $0x8,%esi
  mov    %rbx,%rdi
  call   <realloc@plt>
  mov    %rax,%rbx
  jmp    L4
L22:
  mov    %rax,%rbx
  mov    $0x0,%ebp
L53:
  mov    %ebp,(%r12)
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
type: string

tool: D2: ;

assistant: ```c
char **get_commands(char *cmd_text, int *num_commands) {
  char **commands = NULL;
  char *cmd, *trimmed_cmd;
  int i = 0;
  cmd = strtok(cmd_text, ";");
  while (cmd != NULL) {
    trimmed_cmd = trim_word(cmd);
    if (trimmed_cmd) {
      if (!commands)
        commands = (char **)malloc(sizeof(char *));
      else
        commands = (char **)realloc(commands, sizeof(char *));
      commands[i++] = trimmed_cmd;
    }
    cmd = strtok(NULL, ";");
  }
  *num_commands = i;
  return commands;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_commands>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  lea    D2(%rip),%rbx
  mov    %rbx,%rsi
  call   <strtok@plt>
  test   %rax,%rax
  je     L36
  mov    %rax,%rdi
  xor    %r13d,%r13d
  xor    %ebp,%ebp
  jmp    L0
  nopl   (%rax)
L11:
  mov    %rbp,%rdi
  mov    $0x8,%esi
  call   <realloc@plt>
  mov    %rax,%rbp
L44:
  movslq %r13d,%rax
  add    $0x1,%r13d
  mov    %r14,0x0(%rbp,%rax,8)
L10:
  xor    %edi,%edi
  mov    %rbx,%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L14
L0:
  call   <trim_word@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L10
  test   %rbp,%rbp
  jne    L11
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  jmp    L44
  nop
L36:
  xor    %ebp,%ebp
  xor    %r13d,%r13d
L14:
  pop    %rbx
  mov    %r13d,(%r12)
  mov    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
type_infered: string

tool: D2: ;

assistant: ```c
char **get_commands(char *cmd_text, int *num_commands) {
  char **commands = NULL;
  char *cmd, *trimmed_cmd;
  int i = 0;
  cmd = strtok(cmd_text, ";");
  while (cmd != NULL) {
    trimmed_cmd = trim_word(cmd);
    if (trimmed_cmd) {
      if (!commands)
        commands = (char **)malloc(sizeof(char *));
      else
        commands = (char **)realloc(commands, sizeof(char *));
      commands[i++] = trimmed_cmd;
    }
    cmd = strtok(NULL, ";");
  }
  *num_commands = i;
  return commands;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<get_commands>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  lea    D7(%rip),%rbx
  mov    %rbx,%rsi
  call   <strtok@plt>
  test   %rax,%rax
  je     L29
  mov    %rax,%rdi
  xor    %r13d,%r13d
  xor    %ebp,%ebp
  jmp    L16
  nopl   (%rax)
L4:
  mov    %rbp,%rdi
  mov    $0x8,%esi
  call   <realloc@plt>
  mov    %rax,%rbp
L57:
  movslq %r13d,%rax
  add    $0x1,%r13d
  mov    %r14,0x0(%rbp,%rax,8)
L21:
  xor    %edi,%edi
  mov    %rbx,%rsi
  call   <strtok@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  je     L5
L16:
  call   <trim_word@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L21
  test   %rbp,%rbp
  jne    L4
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  jmp    L57
  nop
L29:
  xor    %ebp,%ebp
  xor    %r13d,%r13d
L5:
  pop    %rbx
  mov    %r13d,(%r12)
  mov    %rbp,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
datatype: string

tool: D7: ;

assistant: ```c
char **get_commands(char *cmd_text, int *num_commands) {
  char **commands = NULL;
  char *cmd, *trimmed_cmd;
  int i = 0;
  cmd = strtok(cmd_text, ";");
  while (cmd != NULL) {
    trimmed_cmd = trim_word(cmd);
    if (trimmed_cmd) {
      if (!commands)
        commands = (char **)malloc(sizeof(char *));
      else
        commands = (char **)realloc(commands, sizeof(char *));
      commands[i++] = trimmed_cmd;
    }
    cmd = strtok(NULL, ";");
  }
  *num_commands = i;
  return commands;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdx
  mov    $0xffff,%esi
  mov    $0x1c40000,%edi
  call   <mmap_perith@plt>
  mov    %rax,-0x10(%rbp)
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  add    $0x800,%rax
  mov    %rax,%rdi
  call   <print_pllc@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  add    $0xc00,%rax
  mov    %rax,%rdi
  call   <print_pllc@plt>
  lea    -0x14(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    $0xffff,%esi
  mov    %rax,%rdi
  call   <munmap_perith@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D8
type: string
name: disassemble_data
arguments: data_label: D2
type: string

tool: D8: --------------------
PLLC1:


tool: D2: --------------------
PLLC2:


assistant: ```c
int main(int argc, char *argv[]) {
  int fd;
  volatile void *sys_mod = mmap_perith(((0x01C40000) + 0), (0x01C4FFFF - 0x01C40000), &fd);
  printf("--------------------\nPLLC1:\n");
  print_pllc(sys_mod + 0x800);
  printf("--------------------\nPLLC2:\n");
  print_pllc(sys_mod + 0xC00);
  munmap_perith(sys_mod, (0x01C4FFFF - 0x01C40000), &fd);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rdx
  mov    $0xffff,%esi
  mov    $0x1c40000,%edi
  call   <mmap_perith@plt>
  mov    %rax,%rbx
  lea    D19(%rip),%rdi
  call   <printf@plt>
  lea    0x800(%rbx),%rdi
  call   <print_pllc@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    0xc00(%rbx),%rdi
  call   <print_pllc@plt>
  mov    %rbp,%rdx
  mov    $0xffff,%esi
  mov    %rbx,%rdi
  call   <munmap_perith@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D19
type_infered: string
name: read_data
arguments: data_tag: D13
type_infered: string

tool: D19: --------------------
PLLC1:


tool: D13: --------------------
PLLC2:


assistant: ```c
int main(int argc, char *argv[]) {
  int fd;
  volatile void *sys_mod = mmap_perith(((0x01C40000) + 0), (0x01C4FFFF - 0x01C40000), &fd);
  printf("--------------------\nPLLC1:\n");
  print_pllc(sys_mod + 0x800);
  printf("--------------------\nPLLC2:\n");
  print_pllc(sys_mod + 0xC00);
  munmap_perith(sys_mod, (0x01C4FFFF - 0x01C40000), &fd);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xffff,%esi
  mov    $0x1c40000,%edi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %r12,%rdx
  call   <mmap_perith@plt>
  lea    D17(%rip),%rdi
  mov    %rax,%rbp
  call   <printf@plt>
  lea    0x800(%rbp),%rdi
  call   <print_pllc@plt>
  lea    D14(%rip),%rdi
  call   <printf@plt>
  lea    0xc00(%rbp),%rdi
  call   <print_pllc@plt>
  mov    %r12,%rdx
  mov    $0xffff,%esi
  mov    %rbp,%rdi
  call   <munmap_perith@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D17
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D14
type_infered: string

tool: D17: --------------------
PLLC1:


tool: D14: --------------------
PLLC2:


assistant: ```c
int main(int argc, char *argv[]) {
  int fd;
  volatile void *sys_mod = mmap_perith(((0x01C40000) + 0), (0x01C4FFFF - 0x01C40000), &fd);
  printf("--------------------\nPLLC1:\n");
  print_pllc(sys_mod + 0x800);
  printf("--------------------\nPLLC2:\n");
  print_pllc(sys_mod + 0xC00);
  munmap_perith(sys_mod, (0x01C4FFFF - 0x01C40000), &fd);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xffff,%esi
  mov    $0x1c40000,%edi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %r12,%rdx
  call   <mmap_perith@plt>
  lea    D8(%rip),%rdi
  mov    %rax,%rbp
  call   <printf@plt>
  lea    0x800(%rbp),%rdi
  call   <print_pllc@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    0xc00(%rbp),%rdi
  call   <print_pllc@plt>
  mov    %r12,%rdx
  mov    $0xffff,%esi
  mov    %rbp,%rdi
  call   <munmap_perith@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D8
type_infered: string
name: retrieve_data
arguments: label: D1
type_infered: string

tool: D8: --------------------
PLLC1:


tool: D1: --------------------
PLLC2:


assistant: ```c
int main(int argc, char *argv[]) {
  int fd;
  volatile void *sys_mod = mmap_perith(((0x01C40000) + 0), (0x01C4FFFF - 0x01C40000), &fd);
  printf("--------------------\nPLLC1:\n");
  print_pllc(sys_mod + 0x800);
  printf("--------------------\nPLLC2:\n");
  print_pllc(sys_mod + 0xC00);
  munmap_perith(sys_mod, (0x01C4FFFF - 0x01C40000), &fd);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <showdir@plt>
  test   %rax,%rax
  je     L38
  mov    $0x1,%eax
  jmp    L1
L38:
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <showdir@plt>
  test   %rax,%rax
  je     L11
  mov    $0x1,%eax
  jmp    L1
L11:
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
L1:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D4
type: string
name: unpack_data
arguments: tag: D2
type: string
name: unpack_data
arguments: tag: D29
type: string

tool: D4: /

tool: D2: /var/

tool: D29: Test was successful

assistant: ```c
int main(int argc, char **argv) {
  if (showdir("/"))
    return 1;
  if (showdir("/var/"))
    return 1;
  puts("Test was successful");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D23(%rip),%rdi
  call   <showdir@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  je     L6
L22:
  add    $0x8,%rsp
  ret
L6:
  lea    D27(%rip),%rdi
  call   <showdir@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L22
  lea    D12(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  jmp    L22
```

tool_calls: name: disassemble_data
arguments: label: D23
data_type: string
name: disassemble_data
arguments: label: D27
data_type: string
name: disassemble_data
arguments: label: D12
data_type: string

tool: D23: /

tool: D27: /var/

tool: D12: Test was successful

assistant: ```c
int main(int argc, char **argv) {
  if (showdir("/"))
    return 1;
  if (showdir("/var/"))
    return 1;
  puts("Test was successful");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D16(%rip),%rdi
  call   <showdir@plt>
  test   %rax,%rax
  je     L25
L22:
  mov    $0x1,%eax
L8:
  add    $0x8,%rsp
  ret
L25:
  lea    D20(%rip),%rdi
  call   <showdir@plt>
  test   %rax,%rax
  jne    L22
  lea    D8(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  jmp    L8
```

tool_calls: name: read_data
arguments: data_tag: D16
type: string
name: read_data
arguments: data_tag: D20
type: string
name: read_data
arguments: data_tag: D8
type: string

tool: D16: /

tool: D20: /var/

tool: D8: Test was successful

assistant: ```c
int main(int argc, char **argv) {
  if (showdir("/"))
    return 1;
  if (showdir("/var/"))
    return 1;
  puts("Test was successful");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D17(%rip),%rdi
  call   <showdir@plt>
  test   %rax,%rax
  je     L3
L18:
  mov    $0x1,%eax
L2:
  add    $0x8,%rsp
  ret
L3:
  lea    D18(%rip),%rdi
  call   <showdir@plt>
  test   %rax,%rax
  jne    L18
  lea    D10(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  jmp    L2
```

tool_calls: name: retrieve_data
arguments: tag: D17
datatype: string
name: retrieve_data
arguments: tag: D18
datatype: string
name: retrieve_data
arguments: tag: D10
datatype: string

tool: D17: /

tool: D18: /var/

tool: D10: Test was successful

assistant: ```c
int main(int argc, char **argv) {
  if (showdir("/"))
    return 1;
  if (showdir("/var/"))
    return 1;
  puts("Test was successful");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ReadDecUint32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x20(%rbp),%rax
  not    %rax
  shr    $0x3f,%rax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  cmpq   $0x0,-0x28(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  movq   $0x0,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  movl   $0x0,(%rax)
  jmp    L47
L34:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %edx,%eax
  lea    -0x30(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jbe    L3
  mov    -0x28(%rbp),%rax
  movl   $0xffffffff,(%rax)
L3:
  addq   $0x1,-0x8(%rbp)
L47:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jge    L7
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L7
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L34
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int32_t ReadDecUint32(const char *buf, int32_t max_length, uint32_t *value) {
  assert(buf != NULL);
  assert(max_length >= 0);
  assert(value != NULL);
  int32_t length = 0;
  *value = 0U;
  while (length < max_length && '0' <= buf[length] && buf[length] <= '9') {
    uint32_t prev_value = *value;
    *value *= 10U;
    *value += (uint32_t)(buf[length] - '0');
    if (*value < prev_value) {
      *value = 0xFFFFFFFF;
    }
    ++length;
  }
  return length;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ReadDecUint32>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%rbx
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %r12,%rdi
  not    %rdi
  shr    $0x3f,%rdi
  call   <assert@plt>
  test   %rbx,%rbx
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  movl   $0x0,(%rbx)
  mov    $0x0,%eax
  mov    $0xffffffff,%edi
  test   %r12,%r12
  jle    L10
L1:
  movzbl 0x0(%rbp,%rax,1),%esi
  lea    -0x30(%rsi),%edx
  cmp    $0x9,%dl
  ja     L10
  mov    (%rbx),%ecx
  lea    (%rcx,%rcx,4),%edx
  add    %edx,%edx
  mov    %edx,(%rbx)
  movsbl 0x0(%rbp,%rax,1),%esi
  lea    -0x30(%rdx,%rsi,1),%edx
  cmp    %ecx,%edx
  cmovb  %edi,%edx
  mov    %edx,(%rbx)
  add    $0x1,%rax
  cmp    %rax,%r12
  jne    L1
  mov    %r12,%rax
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int32_t ReadDecUint32(const char *buf, int32_t max_length, uint32_t *value) {
  assert(buf != NULL);
  assert(max_length >= 0);
  assert(value != NULL);
  int32_t length = 0;
  *value = 0U;
  while (length < max_length && '0' <= buf[length] && buf[length] <= '9') {
    uint32_t prev_value = *value;
    *value *= 10U;
    *value += (uint32_t)(buf[length] - '0');
    if (*value < prev_value) {
      *value = 0xFFFFFFFF;
    }
    ++length;
  }
  return length;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ReadDecUint32>:
  endbr64
  push   %r12
  test   %rdi,%rdi
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  setne  %dil
  push   %rbx
  movzbl %dil,%edi
  mov    %rdx,%rbx
  call   <assert@plt>
  mov    %r12,%rdi
  not    %rdi
  shr    $0x3f,%rdi
  call   <assert@plt>
  xor    %edi,%edi
  test   %rbx,%rbx
  setne  %dil
  call   <assert@plt>
  movl   $0x0,(%rbx)
  xor    %eax,%eax
  mov    $0xffffffff,%edi
  test   %r12,%r12
  jg     L24
  jmp    L2
  nopw   0x0(%rax,%rax,1)
L5:
  mov    (%rbx),%ecx
  lea    (%rcx,%rcx,4),%edx
  add    %edx,%edx
  mov    %edx,(%rbx)
  movsbl 0x0(%rbp,%rax,1),%esi
  lea    -0x30(%rdx,%rsi,1),%edx
  cmp    %ecx,%edx
  cmovb  %edi,%edx
  add    $0x1,%rax
  mov    %edx,(%rbx)
  cmp    %rax,%r12
  je     L2
L24:
  movzbl 0x0(%rbp,%rax,1),%esi
  lea    -0x30(%rsi),%edx
  cmp    $0x9,%dl
  jbe    L5
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int32_t ReadDecUint32(const char *buf, int32_t max_length, uint32_t *value) {
  assert(buf != NULL);
  assert(max_length >= 0);
  assert(value != NULL);
  int32_t length = 0;
  *value = 0U;
  while (length < max_length && '0' <= buf[length] && buf[length] <= '9') {
    uint32_t prev_value = *value;
    *value *= 10U;
    *value += (uint32_t)(buf[length] - '0');
    if (*value < prev_value) {
      *value = 0xFFFFFFFF;
    }
    ++length;
  }
  return length;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ReadDecUint32>:
  endbr64
  push   %r12
  test   %rdi,%rdi
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  setne  %dil
  push   %rbx
  movzbl %dil,%edi
  mov    %rdx,%rbx
  call   <assert@plt>
  mov    %r12,%rdi
  not    %rdi
  shr    $0x3f,%rdi
  call   <assert@plt>
  xor    %edi,%edi
  test   %rbx,%rbx
  setne  %dil
  call   <assert@plt>
  movl   $0x0,(%rbx)
  xor    %eax,%eax
  xor    %ecx,%ecx
  test   %r12,%r12
  jle    L64
  movzbl 0x0(%rbp,%rax,1),%edi
  lea    -0x30(%rdi),%edx
  cmp    $0x9,%dl
  ja     L64
  nopw   0x0(%rax,%rax,1)
L63:
  lea    (%rcx,%rcx,4),%edx
  add    %edx,%edx
  mov    %edx,(%rbx)
  movsbl 0x0(%rbp,%rax,1),%esi
  lea    -0x30(%rdx,%rsi,1),%edx
  cmp    %ecx,%edx
  jae    L13
  add    $0x1,%rax
  movl   $0xffffffff,(%rbx)
  cmp    %r12,%rax
  je     L64
  mov    $0xffffffff,%edx
  mov    %edx,%ecx
L52:
  movzbl 0x0(%rbp,%rax,1),%edi
  lea    -0x30(%rdi),%edx
  cmp    $0x9,%dl
  jbe    L63
L64:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L13:
  add    $0x1,%rax
  mov    %edx,(%rbx)
  cmp    %rax,%r12
  je     L64
  mov    %edx,%ecx
  jmp    L52
```

assistant: ```c
int32_t ReadDecUint32(const char *buf, int32_t max_length, uint32_t *value) {
  assert(buf != NULL);
  assert(max_length >= 0);
  assert(value != NULL);
  int32_t length = 0;
  *value = 0U;
  while (length < max_length && '0' <= buf[length] && buf[length] <= '9') {
    uint32_t prev_value = *value;
    *value *= 10U;
    *value += (uint32_t)(buf[length] - '0');
    if (*value < prev_value) {
      *value = 0xFFFFFFFF;
    }
    ++length;
  }
  return length;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x7de,%ecx
  mov    $0x1c,%edx
  mov    $0x1,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D36
type_guessed: string
name: extract_symbolic_data
arguments: tag: D26
type_guessed: string
name: extract_symbolic_data
arguments: tag: D15
type_guessed: string
name: extract_symbolic_data
arguments: tag: D21
type_guessed: string

tool: D36: Course: E G 13 02 A -                  Spring 2014 

tool: D26: Name:                                 Miles Lilly 

tool: D15: Date Cre         ated: %d/%d/%d 


tool: D21: Assignment: Which assignment is this? 

assistant: ```c
int main() {
  printf("Course: E G 13 02 A -                  Spring 2014 \n");
  printf("Name:                                 Miles Lilly \n");
  printf("Date Cre         ated: %d/%d/%d \n", 1, 28, 2014);
  printf("Assignment: Which assignment is this? \n");
  return 0;
}
```