ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r15
  lea    D26(%rip),%r15
  push   %r14
  lea    D24(%rip),%r14
  push   %r13
  lea    D29(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  lea    D23(%rip),%r12
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  movl   $0x16,0x4(%rsp)
  mov    %rbp,%rsi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r15,%rdi
  xor    %eax,%eax
  movl   $0xb,0x4(%rsp)
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  movl   $0x2,0x4(%rsp)
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D26
type_guessed: string
name: retrieve_data
arguments: data_tag: D24
type_guessed: string
name: retrieve_data
arguments: data_tag: D29
type_guessed: string
name: retrieve_data
arguments: data_tag: D23
type_guessed: string

tool: D26: Address of pointer pc: %d


tool: D24: Content of pointer pc: %d



tool: D29: Address of c: %d


tool: D23: Value of c: %d



assistant: ```c
int main(void) {
  int *pc;
  int c;
  c = 22;
  printf("Address of c: %d\n", &c);
  printf("Value of c: %d\n\n", c);
  pc = &c;
  printf("Address of pointer pc: %d\n", pc);
  printf("Content of pointer pc: %d\n\n", *pc);
  c = 11;
  printf("Address of pointer pc: %d\n", pc);
  printf("Content of pointer pc: %d\n\n", *pc);
  *pc = 2;
  printf("Address of c: %d\n", &c);
  printf("Value of c: %d\n\n", c);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  lea    D19(%rip),%r15
  push   %r14
  lea    D14(%rip),%r14
  push   %r13
  lea    D20(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  lea    D23(%rip),%r12
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  movl   $0x16,0x4(%rsp)
  mov    %rbp,%rsi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r15,%rdi
  xor    %eax,%eax
  movl   $0xb,0x4(%rsp)
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  movl   $0x2,0x4(%rsp)
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D19
type_guessed: string
name: retrieve_data
arguments: data_label: D14
type_guessed: string
name: retrieve_data
arguments: data_label: D20
type_guessed: string
name: retrieve_data
arguments: data_label: D23
type_guessed: string

tool: D19: Address of pointer pc: %d


tool: D14: Content of pointer pc: %d



tool: D20: Address of c: %d


tool: D23: Value of c: %d



assistant: ```c
int main(void) {
  int *pc;
  int c;
  c = 22;
  printf("Address of c: %d\n", &c);
  printf("Value of c: %d\n\n", c);
  pc = &c;
  printf("Address of pointer pc: %d\n", pc);
  printf("Content of pointer pc: %d\n\n", *pc);
  c = 11;
  printf("Address of pointer pc: %d\n", pc);
  printf("Content of pointer pc: %d\n\n", *pc);
  *pc = 2;
  printf("Address of c: %d\n", &c);
  printf("Value of c: %d\n\n", c);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0xfa0,-0xa(%rbp)
  movzwl -0xa(%rbp),%eax
  cwtl
  lea    -0xa(%rbp),%rdx
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D9
type_infered: string

tool: D9: x value is %d and x addr is %p


assistant: ```c
int main() {
  short x = 4000;
  printf("x value is %d and x addr is %p\n", x, &x);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movw   $0xfa0,0x6(%rsp)
  lea    0x6(%rsp),%rdx
  mov    $0xfa0,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D1
type: string

tool: D1: x value is %d and x addr is %p


assistant: ```c
int main() {
  short x = 4000;
  printf("x value is %d and x addr is %p\n", x, &x);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xfa0,%esi
  lea    D4(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rdx
  mov    $0xfa0,%eax
  mov    %ax,0x6(%rsp)
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D4
type_infered: string

tool: D4: x value is %d and x addr is %p


assistant: ```c
int main() {
  short x = 4000;
  printf("x value is %d and x addr is %p\n", x, &x);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0xfa0,%esi
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rdx
  mov    $0xfa0,%eax
  mov    %ax,0x6(%rsp)
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D5
type_infered: string

tool: D5: x value is %d and x addr is %p


assistant: ```c
int main() {
  short x = 4000;
  printf("x value is %d and x addr is %p\n", x, &x);
  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
<ping>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %rdx,-0x58(%rbp)
  mov    %rcx,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x3b9aca00,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0x38(%rbp),%rcx
  mov    -0x44(%rbp),%eax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <recebe_mesmo@plt>
  mov    -0x38(%rbp),%edx
  mov    -0x60(%rbp),%rcx
  mov    -0x44(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <recebe_mesmo@plt>
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x60(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  movl   $0x0,-0x34(%rbp)
  jmp    L3
L1:
  lea    -0x30(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  lea    -0x28(%rbp),%rsi
  mov    -0x44(%rbp),%eax
  mov    %eax,%edi
  call   <ping1@plt>
  mov    -0x30(%rbp),%rax
  cmp    %rax,-0x18(%rbp)
  jl     L8
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x50(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x20(%rbp),%rdx
  mov    -0x58(%rbp),%rax
  mov    %rdx,(%rax)
L8:
  addl   $0x1,-0x34(%rbp)
L3:
  cmpl   $0x3e7,-0x34(%rbp)
  jle    L1
  lea    -0x10(%rbp),%rsi
  mov    -0x44(%rbp),%eax
  mov    $0x0,%ecx
  mov    $0x8,%edx
  mov    %eax,%edi
  call   <send@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

assistant: ```c
void ping(int socket, long long *mlocal, long long *mremote, char *remotename) {
  long delta;
  long long local, remote;
  long mdelta = 1e9;
  int i;
  struct timeval t0 = {0, 0};
  int remotenamelen;
  recebe_mesmo(socket, (char *)&remotenamelen, sizeof(remotenamelen));
  recebe_mesmo(socket, remotename, remotenamelen);
  remotename[remotenamelen] = '\0';
  for (i = 0; i < 1000; i++) {
    ping1(socket, &local, &remote, &delta);
    if (delta <= mdelta) {
      mdelta = delta;
      *mlocal = local;
      *mremote = remote;
    }
  }
  send(socket, (void *)&t0, sizeof(t0), 0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ping>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,%r12d
  mov    %rsi,%r15
  mov    %rdx,%r14
  mov    %rcx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x20(%rsp)
  movl   $0x0,0x24(%rsp)
  lea    0x4(%rsp),%rsi
  mov    $0x4,%edx
  call   <recebe_mesmo@plt>
  mov    0x4(%rsp),%edx
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <recebe_mesmo@plt>
  movslq 0x4(%rsp),%rax
  movb   $0x0,(%rbx,%rax,1)
  mov    $0x3e8,%ebx
  mov    $0x3b9aca00,%ebp
  lea    0x8(%rsp),%r13
  jmp    L47
L24:
  sub    $0x1,%ebx
  je     L28
L47:
  lea    0x18(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %r13,%rcx
  mov    %r12d,%edi
  call   <ping1@plt>
  mov    0x8(%rsp),%rax
  cmp    %rbp,%rax
  jg     L24
  mov    0x10(%rsp),%rdx
  mov    %rdx,(%r15)
  mov    0x18(%rsp),%rdx
  mov    %rdx,(%r14)
  mov    %rax,%rbp
  jmp    L24
L28:
  lea    0x20(%rsp),%rsi
  mov    $0x0,%ecx
  mov    $0x8,%edx
  mov    %r12d,%edi
  call   <send@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ping(int socket, long long *mlocal, long long *mremote, char *remotename) {
  long delta;
  long long local, remote;
  long mdelta = 1e9;
  int i;
  struct timeval t0 = {0, 0};
  int remotenamelen;
  recebe_mesmo(socket, (char *)&remotenamelen, sizeof(remotenamelen));
  recebe_mesmo(socket, remotename, remotenamelen);
  remotename[remotenamelen] = '\0';
  for (i = 0; i < 1000; i++) {
    ping1(socket, &local, &remote, &delta);
    if (delta <= mdelta) {
      mdelta = delta;
      *mlocal = local;
      *mremote = remote;
    }
  }
  send(socket, (void *)&t0, sizeof(t0), 0);
}
```
user: What is the c source code of the assembly code below:

```asm
<ping>:
  endbr64
  push   %r15
  mov    %edi,%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x3b9aca00,%ebp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x48,%rsp
  mov    %rsi,(%rsp)
  lea    0x14(%rsp),%rsi
  lea    0x18(%rsp),%r14
  mov    %rdx,0x8(%rsp)
  mov    $0x4,%edx
  lea    0x28(%rsp),%r13
  lea    0x20(%rsp),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x30(%rsp)
  call   <recebe_mesmo@plt>
  mov    0x14(%rsp),%edx
  mov    %rbx,%rsi
  mov    %r15d,%edi
  call   <recebe_mesmo@plt>
  movslq 0x14(%rsp),%rax
  movb   $0x0,(%rbx,%rax,1)
  mov    $0x3e8,%ebx
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %r15d,%edi
  call   <ping1@plt>
  mov    0x18(%rsp),%rax
  cmp    %rbp,%rax
  jg     L19
  mov    0x20(%rsp),%rdx
  mov    (%rsp),%rcx
  mov    %rax,%rbp
  mov    0x8(%rsp),%rdi
  mov    %rdx,(%rcx)
  mov    0x28(%rsp),%rdx
  mov    %rdx,(%rdi)
L19:
  sub    $0x1,%ebx
  jne    L6
  xor    %ecx,%ecx
  lea    0x30(%rsp),%rsi
  mov    $0x8,%edx
  mov    %r15d,%edi
  call   <send@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ping(int socket, long long *mlocal, long long *mremote, char *remotename) {
  long delta;
  long long local, remote;
  long mdelta = 1e9;
  int i;
  struct timeval t0 = {0, 0};
  int remotenamelen;
  recebe_mesmo(socket, (char *)&remotenamelen, sizeof(remotenamelen));
  recebe_mesmo(socket, remotename, remotenamelen);
  remotename[remotenamelen] = '\0';
  for (i = 0; i < 1000; i++) {
    ping1(socket, &local, &remote, &delta);
    if (delta <= mdelta) {
      mdelta = delta;
      *mlocal = local;
      *mremote = remote;
    }
  }
  send(socket, (void *)&t0, sizeof(t0), 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ping>:
  endbr64
  push   %r15
  mov    %edi,%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x3b9aca00,%ebp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x48,%rsp
  mov    %rsi,(%rsp)
  lea    0x14(%rsp),%rsi
  lea    0x18(%rsp),%r14
  mov    %rdx,0x8(%rsp)
  mov    $0x4,%edx
  lea    0x28(%rsp),%r13
  lea    0x20(%rsp),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x30(%rsp)
  call   <recebe_mesmo@plt>
  mov    0x14(%rsp),%edx
  mov    %rbx,%rsi
  mov    %r15d,%edi
  call   <recebe_mesmo@plt>
  movslq 0x14(%rsp),%rax
  movb   $0x0,(%rbx,%rax,1)
  mov    $0x3e8,%ebx
  nopw   0x0(%rax,%rax,1)
L15:
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %r15d,%edi
  call   <ping1@plt>
  mov    0x18(%rsp),%rax
  cmp    %rbp,%rax
  jg     L12
  mov    0x20(%rsp),%rdx
  mov    (%rsp),%rcx
  mov    %rax,%rbp
  mov    0x8(%rsp),%rdi
  mov    %rdx,(%rcx)
  mov    0x28(%rsp),%rdx
  mov    %rdx,(%rdi)
L12:
  sub    $0x1,%ebx
  jne    L15
  xor    %ecx,%ecx
  lea    0x30(%rsp),%rsi
  mov    $0x8,%edx
  mov    %r15d,%edi
  call   <send@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ping(int socket, long long *mlocal, long long *mremote, char *remotename) {
  long delta;
  long long local, remote;
  long mdelta = 1e9;
  int i;
  struct timeval t0 = {0, 0};
  int remotenamelen;
  recebe_mesmo(socket, (char *)&remotenamelen, sizeof(remotenamelen));
  recebe_mesmo(socket, remotename, remotenamelen);
  remotename[remotenamelen] = '\0';
  for (i = 0; i < 1000; i++) {
    ping1(socket, &local, &remote, &delta);
    if (delta <= mdelta) {
      mdelta = delta;
      *mlocal = local;
      *mremote = remote;
    }
  }
  send(socket, (void *)&t0, sizeof(t0), 0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ssl_x509_subject_cn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <X509_get_subject_name@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L52
  mov    $0x0,%eax
  jmp    L24
L52:
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  mov    -0x18(%rbp),%rax
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %rax,%rdi
  call   <X509_NAME_get_text_by_NID@plt>
  add    $0x1,%eax
  cltq
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L31
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L17
L31:
  mov    $0x0,%eax
  jmp    L24
L17:
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <X509_NAME_get_text_by_NID@plt>
  cmp    $0xffffffff,%eax
  jne    L59
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L24
L59:
  mov    -0x30(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
L24:
  leave
  ret
```

assistant: ```c
char *ssl_x509_subject_cn(X509 *crt, size_t *psz) {
  X509_NAME *ptr;
  char *cn;
  size_t sz;
  ptr = X509_get_subject_name(crt);
  if (!ptr)
    return NULL;
  sz = X509_NAME_get_text_by_NID(ptr, NID_commonName, NULL, 0) + 1;
  if ((sz == 0) || !(cn = malloc(sz)))
    return NULL;
  if (X509_NAME_get_text_by_NID(ptr, NID_commonName, cn, sz) == -1) {
    free(cn);
    return NULL;
  }
  *psz = sz;
  return cn;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ssl_x509_subject_cn>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%r12
  call   <X509_get_subject_name@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L58
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    D13(%rip),%rax
  mov    (%rax),%esi
  mov    %r13,%rdi
  call   <X509_NAME_get_text_by_NID@plt>
  lea    0x1(%rax),%ebx
  movslq %ebx,%rbx
  mov    $0x0,%ebp
  test   %rbx,%rbx
  je     L16
  mov    %rbx,%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L16
  mov    %rbx,%rcx
  mov    %rax,%rdx
  mov    D13(%rip),%rax
  mov    (%rax),%esi
  mov    %r13,%rdi
  call   <X509_NAME_get_text_by_NID@plt>
  cmp    $0xffffffff,%eax
  je     L44
  mov    %rbx,(%r12)
L16:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L44:
  mov    %rbp,%rdi
  call   <free@plt>
  mov    $0x0,%ebp
  jmp    L16
L58:
  mov    %rax,%rbp
  jmp    L16
```

assistant: ```c
char *ssl_x509_subject_cn(X509 *crt, size_t *psz) {
  X509_NAME *ptr;
  char *cn;
  size_t sz;
  ptr = X509_get_subject_name(crt);
  if (!ptr)
    return NULL;
  sz = X509_NAME_get_text_by_NID(ptr, NID_commonName, NULL, 0) + 1;
  if ((sz == 0) || !(cn = malloc(sz)))
    return NULL;
  if (X509_NAME_get_text_by_NID(ptr, NID_commonName, cn, sz) == -1) {
    free(cn);
    return NULL;
  }
  *psz = sz;
  return cn;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ssl_x509_subject_cn>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  call   <X509_get_subject_name@plt>
  test   %rax,%rax
  je     L12
  mov    D0(%rip),%r13
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %rax,%rdi
  mov    %rax,%rbp
  mov    0x0(%r13),%esi
  call   <X509_NAME_get_text_by_NID@plt>
  lea    0x1(%rax),%ebx
  movslq %ebx,%rbx
  test   %rbx,%rbx
  je     L12
  mov    %rbx,%rdi
  call   <malloc@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L12
  mov    0x0(%r13),%esi
  mov    %rbx,%rcx
  mov    %rax,%rdx
  mov    %rbp,%rdi
  call   <X509_NAME_get_text_by_NID@plt>
  cmp    $0xffffffff,%eax
  je     L2
  mov    %rbx,(%r12)
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L2:
  mov    %r14,%rdi
  call   <free@plt>
L12:
  xor    %r14d,%r14d
  pop    %rbx
  pop    %rbp
  mov    %r14,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *ssl_x509_subject_cn(X509 *crt, size_t *psz) {
  X509_NAME *ptr;
  char *cn;
  size_t sz;
  ptr = X509_get_subject_name(crt);
  if (!ptr)
    return NULL;
  sz = X509_NAME_get_text_by_NID(ptr, NID_commonName, NULL, 0) + 1;
  if ((sz == 0) || !(cn = malloc(sz)))
    return NULL;
  if (X509_NAME_get_text_by_NID(ptr, NID_commonName, cn, sz) == -1) {
    free(cn);
    return NULL;
  }
  *psz = sz;
  return cn;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ssl_x509_subject_cn>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  call   <X509_get_subject_name@plt>
  test   %rax,%rax
  je     L8
  mov    D0(%rip),%r13
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %rax,%rdi
  mov    %rax,%rbp
  mov    0x0(%r13),%esi
  call   <X509_NAME_get_text_by_NID@plt>
  lea    0x1(%rax),%ebx
  movslq %ebx,%rbx
  test   %rbx,%rbx
  je     L8
  mov    %rbx,%rdi
  call   <malloc@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L8
  mov    0x0(%r13),%esi
  mov    %rbx,%rcx
  mov    %rax,%rdx
  mov    %rbp,%rdi
  call   <X509_NAME_get_text_by_NID@plt>
  cmp    $0xffffffff,%eax
  je     L22
  mov    %rbx,(%r12)
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L22:
  mov    %r14,%rdi
  call   <free@plt>
L8:
  xor    %r14d,%r14d
  pop    %rbx
  pop    %rbp
  mov    %r14,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *ssl_x509_subject_cn(X509 *crt, size_t *psz) {
  X509_NAME *ptr;
  char *cn;
  size_t sz;
  ptr = X509_get_subject_name(crt);
  if (!ptr)
    return NULL;
  sz = X509_NAME_get_text_by_NID(ptr, NID_commonName, NULL, 0) + 1;
  if ((sz == 0) || !(cn = malloc(sz)))
    return NULL;
  if (X509_NAME_get_text_by_NID(ptr, NID_commonName, cn, sz) == -1) {
    free(cn);
    return NULL;
  }
  *psz = sz;
  return cn;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<initCurses>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <initscr@plt>
  mov    $0x0,%eax
  call   <cbreak@plt>
  mov    $0x0,%eax
  call   <noecho@plt>
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <keypad@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void initCurses() {
  initscr();
  cbreak();
  noecho();
  keypad(stdscr, 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<initCurses>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <initscr@plt>
  mov    $0x0,%eax
  call   <cbreak@plt>
  mov    $0x0,%eax
  call   <noecho@plt>
  mov    $0x1,%esi
  mov    D6(%rip),%rax
  mov    (%rax),%edi
  call   <keypad@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void initCurses() {
  initscr();
  cbreak();
  noecho();
  keypad(stdscr, 1);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<initCurses>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <initscr@plt>
  xor    %eax,%eax
  call   <cbreak@plt>
  xor    %eax,%eax
  call   <noecho@plt>
  mov    D4(%rip),%rax
  mov    $0x1,%esi
  mov    (%rax),%edi
  add    $0x8,%rsp
  jmp    1090 <keypad@plt>
```

assistant: ```c
void initCurses() {
  initscr();
  cbreak();
  noecho();
  keypad(stdscr, 1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<initCurses>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <initscr@plt>
  xor    %eax,%eax
  call   <cbreak@plt>
  xor    %eax,%eax
  call   <noecho@plt>
  mov    D0(%rip),%rax
  mov    $0x1,%esi
  mov    (%rax),%edi
  add    $0x8,%rsp
  jmp    1090 <keypad@plt>
```

assistant: ```c
void initCurses() {
  initscr();
  cbreak();
  noecho();
  keypad(stdscr, 1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <funk@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
data_type: string

tool: D7: %d

assistant: ```c
int main() {
  int a = funk(2, 3);
  printf("%d", a);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <funk@plt>
  mov    %eax,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D5
type_guessed: string

tool: D5: %d

assistant: ```c
int main() {
  int a = funk(2, 3);
  printf("%d", a);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <funk@plt>
  lea    D9(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type_infered: string

tool: D9: %d

assistant: ```c
int main() {
  int a = funk(2, 3);
  printf("%d", a);
  return 0;
}
```