ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<nla_memcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <nla_len@plt>
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L0
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <nla_data@plt>
  mov    %eax,%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %ecx,%edi
  call   <memcmp@plt>
  mov    %eax,-0x4(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int nla_memcmp(const struct nlattr *nla, const void *data, size_t size) {
  int d = nla_len(nla) - size;
  if (d == 0)
    d = memcmp(nla_data(nla), data, size);
  return d;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<nla_memcmp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%rbx
  call   <nla_len@plt>
  sub    %ebx,%eax
  je     L14
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L14:
  mov    %rbp,%rdi
  call   <nla_data@plt>
  mov    %eax,%edi
  mov    %rbx,%rdx
  mov    %r12,%rsi
  call   <memcmp@plt>
  jmp    L15
```

assistant: ```c
int nla_memcmp(const struct nlattr *nla, const void *data, size_t size) {
  int d = nla_len(nla) - size;
  if (d == 0)
    d = memcmp(nla_data(nla), data, size);
  return d;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<nla_memcmp>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <nla_len@plt>
  sub    %r12d,%eax
  je     L1
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %rbp,%rdi
  call   <nla_data@plt>
  pop    %rbp
  mov    %r12,%rdx
  mov    %r13,%rsi
  pop    %r12
  mov    %eax,%edi
  pop    %r13
  jmp    1080 <memcmp@plt>
```

assistant: ```c
int nla_memcmp(const struct nlattr *nla, const void *data, size_t size) {
  int d = nla_len(nla) - size;
  if (d == 0)
    d = memcmp(nla_data(nla), data, size);
  return d;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<nla_memcmp>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <nla_len@plt>
  sub    %r12d,%eax
  je     L1
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %rbp,%rdi
  call   <nla_data@plt>
  pop    %rbp
  mov    %r12,%rdx
  mov    %r13,%rsi
  pop    %r12
  mov    %eax,%edi
  pop    %r13
  jmp    1080 <memcmp@plt>
```

assistant: ```c
int nla_memcmp(const struct nlattr *nla, const void *data, size_t size) {
  int d = nla_len(nla) - size;
  if (d == 0)
    d = memcmp(nla_data(nla), data, size);
  return d;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<getpathbuffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L6
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <die@plt>
L6:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D4
type: string

tool: D4: out of memory

assistant: ```c
char *getpathbuffer(void) {
  char *buf;
  if (!(buf = malloc(1023 + 1)))
    (die("out of memory"));
  return buf;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getpathbuffer>:
  endbr64
  push   %rbx
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L15
L17:
  mov    %rbx,%rax
  pop    %rbx
  ret
L15:
  lea    D2(%rip),%rdi
  call   <die@plt>
  jmp    L17
```

tool_calls: name: read_data
arguments: data_label: D2
datatype: string

tool: D2: out of memory

assistant: ```c
char *getpathbuffer(void) {
  char *buf;
  if (!(buf = malloc(1023 + 1)))
    (die("out of memory"));
  return buf;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getpathbuffer>:
  endbr64
  push   %r12
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L6
  mov    %r12,%rax
  pop    %r12
  ret
  xchg   %ax,%ax
L6:
  lea    D9(%rip),%rdi
  call   <die@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D9
datatype: string

tool: D9: out of memory

assistant: ```c
char *getpathbuffer(void) {
  char *buf;
  if (!(buf = malloc(1023 + 1)))
    (die("out of memory"));
  return buf;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getpathbuffer>:
  endbr64
  push   %r12
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L8
  mov    %r12,%rax
  pop    %r12
  ret
  xchg   %ax,%ax
L8:
  lea    D1(%rip),%rdi
  call   <die@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

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

tool: D1: out of memory

assistant: ```c
char *getpathbuffer(void) {
  char *buf;
  if (!(buf = malloc(1023 + 1)))
    (die("out of memory"));
  return buf;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<chpl_decRefCountsForDomainsInArrayEltTypes2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <compilerAssert@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void chpl_decRefCountsForDomainsInArrayEltTypes2(void) {
  compilerAssert();
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<chpl_decRefCountsForDomainsInArrayEltTypes2>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <compilerAssert@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void chpl_decRefCountsForDomainsInArrayEltTypes2(void) {
  compilerAssert();
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<chpl_decRefCountsForDomainsInArrayEltTypes2>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <compilerAssert@plt>
```

assistant: ```c
void chpl_decRefCountsForDomainsInArrayEltTypes2(void) {
  compilerAssert();
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<chpl_decRefCountsForDomainsInArrayEltTypes2>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <compilerAssert@plt>
```

assistant: ```c
void chpl_decRefCountsForDomainsInArrayEltTypes2(void) {
  compilerAssert();
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x810,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x810(%rbp)
  movq   $0x0,-0x808(%rbp)
  lea    -0x800(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x7e,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movq   $0x0,-0x410(%rbp)
  movq   $0x0,-0x408(%rbp)
  lea    -0x400(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x7e,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  lea    -0x410(%rbp),%rdx
  lea    -0x810(%rbp),%rax
  mov    $0x400,%ecx
  mov    $0x400,%esi
  mov    %rax,%rdi
  call   <getcurtauth@plt>
  lea    -0x810(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x410(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D19
type_infered: string
name: disassemble_data
arguments: data_label: D8
type_infered: string

tool: D19: user : [%s]


tool: D8: pass : [%s]


assistant: ```c
int main() {
  char user[1024] = {0};
  char pass[1024] = {0};
  getcurtauth(user, sizeof(user), pass, sizeof(pass));
  printf("user : [%s]\n", user);
  printf("pass : [%s]\n", pass);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  lea    0x10(%rsp),%rdi
  mov    $0x7e,%ecx
  rep stos %rax,%es:(%rdi)
  movq   $0x0,0x400(%rsp)
  movq   $0x0,0x408(%rsp)
  lea    0x410(%rsp),%rdi
  mov    $0x7e,%ecx
  rep stos %rax,%es:(%rdi)
  lea    0x400(%rsp),%rbx
  mov    %rsp,%rbp
  mov    $0x400,%ecx
  mov    %rbx,%rdx
  mov    $0x400,%esi
  mov    %rbp,%rdi
  call   <getcurtauth@plt>
  mov    %rbp,%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x818,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D5
type_guessed: string
name: read_data
arguments: data_label: D1
type_guessed: string

tool: D5: user : [%s]


tool: D1: pass : [%s]


assistant: ```c
int main() {
  char user[1024] = {0};
  char pass[1024] = {0};
  getcurtauth(user, sizeof(user), pass, sizeof(pass));
  printf("user : [%s]\n", user);
  printf("pass : [%s]\n", pass);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x7e,%ecx
  pxor   %xmm0,%xmm0
  mov    $0x400,%esi
  push   %rbp
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    %rsp,%r12
  movaps %xmm0,(%rsp)
  lea    0x400(%rsp),%rbp
  movaps %xmm0,0x400(%rsp)
  mov    %rbp,%rdx
  rep stos %rax,%es:(%rdi)
  lea    0x410(%rsp),%rdi
  mov    $0x7e,%ecx
  rep stos %rax,%es:(%rdi)
  mov    $0x400,%ecx
  mov    %r12,%rdi
  call   <getcurtauth@plt>
  mov    %r12,%rsi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x818,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D17
datatype: string
name: parse_data
arguments: label: D18
datatype: string

tool: D17: user : [%s]


tool: D18: pass : [%s]


assistant: ```c
int main() {
  char user[1024] = {0};
  char pass[1024] = {0};
  getcurtauth(user, sizeof(user), pass, sizeof(pass));
  printf("user : [%s]\n", user);
  printf("pass : [%s]\n", pass);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x7e,%ecx
  pxor   %xmm0,%xmm0
  mov    $0x400,%esi
  push   %rbp
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    %rsp,%r12
  movaps %xmm0,(%rsp)
  lea    0x400(%rsp),%rbp
  movaps %xmm0,0x400(%rsp)
  mov    %rbp,%rdx
  rep stos %rax,%es:(%rdi)
  lea    0x410(%rsp),%rdi
  mov    $0x7e,%ecx
  rep stos %rax,%es:(%rdi)
  mov    $0x400,%ecx
  mov    %r12,%rdi
  call   <getcurtauth@plt>
  mov    %r12,%rsi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x818,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D15
type_guessed: string
name: read_data
arguments: tag: D13
type_guessed: string

tool: D15: user : [%s]


tool: D13: pass : [%s]


assistant: ```c
int main() {
  char user[1024] = {0};
  char pass[1024] = {0};
  getcurtauth(user, sizeof(user), pass, sizeof(pass));
  printf("user : [%s]\n", user);
  printf("pass : [%s]\n", pass);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<bf_new>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rax
  shr    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x14(%rbp),%eax
  mov    $0x7,%esi
  mov    %eax,%ecx
  shr    %cl,%rsi
  mov    %rsi,%rax
  and    $0x1,%eax
  add    %rdx,%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *bf_new(unsigned b) {
  char *bf = calloc(1, (((1UL << b) / 8) + (((1UL << b) % 8) ? 1 : 0)));
  return bf;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<bf_new>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%ecx
  mov    $0x1,%esi
  shl    %cl,%rsi
  shr    $0x3,%rsi
  mov    $0x7,%eax
  shr    %cl,%rax
  and    $0x1,%eax
  add    %rax,%rsi
  mov    $0x1,%edi
  call   <calloc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *bf_new(unsigned b) {
  char *bf = calloc(1, (((1UL << b) / 8) + (((1UL << b) % 8) ? 1 : 0)));
  return bf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bf_new>:
  endbr64
  mov    %edi,%ecx
  mov    $0x1,%esi
  mov    $0x7,%eax
  mov    $0x1,%edi
  shl    %cl,%rsi
  shr    %cl,%rax
  shr    $0x3,%rsi
  and    $0x1,%eax
  add    %rax,%rsi
  jmp    1050 <calloc@plt>
```

assistant: ```c
char *bf_new(unsigned b) {
  char *bf = calloc(1, (((1UL << b) / 8) + (((1UL << b) % 8) ? 1 : 0)));
  return bf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bf_new>:
  endbr64
  mov    %edi,%ecx
  mov    $0x1,%esi
  mov    $0x7,%eax
  mov    $0x1,%edi
  shl    %cl,%rsi
  shr    %cl,%rax
  shr    $0x3,%rsi
  and    $0x1,%eax
  add    %rax,%rsi
  jmp    1050 <calloc@plt>
```

assistant: ```c
char *bf_new(unsigned b) {
  char *bf = calloc(1, (((1UL << b) / 8) + (((1UL << b) % 8) ? 1 : 0)));
  return bf;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<mapif_guild_broken>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  movw   $0x3836,(%rax)
  lea    -0x20(%rbp),%rax
  add    $0x2,%rax
  mov    -0x24(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%eax
  mov    %al,-0x1a(%rbp)
  lea    -0x20(%rbp),%rax
  mov    $0x7,%esi
  mov    %rax,%rdi
  call   <mapif_sendall@plt>
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type_infered: string

tool: D3: int_guild: broken %d


assistant: ```c
int mapif_guild_broken(int guild_id, int flag) {
  unsigned char buf[16];
  (*(unsigned short *)(((unsigned char *)(buf)) + (0))) = 0x3836;
  (*(unsigned int *)(((unsigned char *)(buf)) + (2))) = guild_id;
  (*(unsigned char *)(((unsigned char *)(buf)) + (6))) = flag;
  mapif_sendall(buf, 7);
  printf("int_guild: broken %d\n", guild_id);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<mapif_guild_broken>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movw   $0x3836,(%rsp)
  mov    %edi,0x2(%rsp)
  mov    %sil,0x6(%rsp)
  mov    %rsp,%rdi
  mov    $0x7,%esi
  call   <mapif_sendall@plt>
  mov    %ebx,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D2
type: string

tool: D2: int_guild: broken %d


assistant: ```c
int mapif_guild_broken(int guild_id, int flag) {
  unsigned char buf[16];
  (*(unsigned short *)(((unsigned char *)(buf)) + (0))) = 0x3836;
  (*(unsigned int *)(((unsigned char *)(buf)) + (2))) = guild_id;
  (*(unsigned char *)(((unsigned char *)(buf)) + (6))) = flag;
  mapif_sendall(buf, 7);
  printf("int_guild: broken %d\n", guild_id);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<mapif_guild_broken>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edi,0x2(%rsp)
  mov    $0x3836,%eax
  mov    %rsp,%rdi
  mov    %sil,0x6(%rsp)
  mov    $0x7,%esi
  mov    %ax,(%rsp)
  call   <mapif_sendall@plt>
  mov    %ebp,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type: string

tool: D9: int_guild: broken %d


assistant: ```c
int mapif_guild_broken(int guild_id, int flag) {
  unsigned char buf[16];
  (*(unsigned short *)(((unsigned char *)(buf)) + (0))) = 0x3836;
  (*(unsigned int *)(((unsigned char *)(buf)) + (2))) = guild_id;
  (*(unsigned char *)(((unsigned char *)(buf)) + (6))) = flag;
  mapif_sendall(buf, 7);
  printf("int_guild: broken %d\n", guild_id);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mapif_guild_broken>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edi,0x2(%rsp)
  mov    $0x3836,%eax
  mov    %rsp,%rdi
  mov    %sil,0x6(%rsp)
  mov    $0x7,%esi
  mov    %ax,(%rsp)
  call   <mapif_sendall@plt>
  mov    %ebp,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D8
type_infered: string

tool: D8: int_guild: broken %d


assistant: ```c
int mapif_guild_broken(int guild_id, int flag) {
  unsigned char buf[16];
  (*(unsigned short *)(((unsigned char *)(buf)) + (0))) = 0x3836;
  (*(unsigned int *)(((unsigned char *)(buf)) + (2))) = guild_id;
  (*(unsigned char *)(((unsigned char *)(buf)) + (6))) = flag;
  mapif_sendall(buf, 7);
  printf("int_guild: broken %d\n", guild_id);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <check_obstack_printf_int@plt>
  mov    $0x0,%eax
  call   <check_obstack_printf_typical@plt>
  mov    $0x0,%eax
  call   <check_obstack_printf_two_times@plt>
  mov    D52(%rip),%rax
  movq   %rax,%xmm0
  mov    $0x2,%ecx
  lea    D12(%rip),%rax
  mov    %rax,%rdx
  lea    D7(%rip),%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <check_obstack_vprintf@plt>
  lea    D40(%rip),%rax
  mov    %rax,%rdx
  lea    D11(%rip),%rax
  mov    %rax,%rsi
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <check_obstack_vprintf@plt>
  mov    $0x0,%eax
  call   <check_open_obstack_stream@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D52
type: f64
name: unpack_data
arguments: data_tag: D12
type: string
name: unpack_data
arguments: data_tag: D7
type: string
name: unpack_data
arguments: data_tag: D28
type: string
name: unpack_data
arguments: data_tag: D40
type: string
name: unpack_data
arguments: data_tag: D11
type: string
name: unpack_data
arguments: data_tag: D47
type: string

tool: D52: 3.4

tool: D12: one

tool: D7: %s and %d then %.1f

tool: D28: one and 2 then 3.4

tool: D40: two
three

tool: D11: one
%s

tool: D47: one
two
three

assistant: ```c
int main(int argc, char *argv[]) {
  check_obstack_printf_int();
  check_obstack_printf_typical();
  check_obstack_printf_two_times();
  check_obstack_vprintf("one and 2 then 3.4", "%s and %d then %.1f", "one", 2, 3.4);
  check_obstack_vprintf("one\ntwo\nthree", "one\n%s", "two\nthree");
  check_open_obstack_stream();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <check_obstack_printf_int@plt>
  mov    $0x0,%eax
  call   <check_obstack_printf_typical@plt>
  mov    $0x0,%eax
  call   <check_obstack_printf_two_times@plt>
  movsd  D64(%rip),%xmm0
  mov    $0x2,%ecx
  lea    D57(%rip),%rdx
  lea    D31(%rip),%rsi
  lea    D1(%rip),%rdi
  mov    $0x1,%eax
  call   <check_obstack_vprintf@plt>
  lea    D60(%rip),%rdx
  lea    D40(%rip),%rsi
  lea    D45(%rip),%rdi
  mov    $0x0,%eax
  call   <check_obstack_vprintf@plt>
  mov    $0x0,%eax
  call   <check_open_obstack_stream@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D64
type_guessed: f64
name: retrieve_data
arguments: data_label: D57
type_guessed: string
name: retrieve_data
arguments: data_label: D31
type_guessed: string
name: retrieve_data
arguments: data_label: D1
type_guessed: string
name: retrieve_data
arguments: data_label: D60
type_guessed: string
name: retrieve_data
arguments: data_label: D40
type_guessed: string
name: retrieve_data
arguments: data_label: D45
type_guessed: string

tool: D64: 3.4

tool: D57: one

tool: D31: %s and %d then %.1f

tool: D1: one and 2 then 3.4

tool: D60: two
three

tool: D40: one
%s

tool: D45: one
two
three

assistant: ```c
int main(int argc, char *argv[]) {
  check_obstack_printf_int();
  check_obstack_printf_typical();
  check_obstack_printf_two_times();
  check_obstack_vprintf("one and 2 then 3.4", "%s and %d then %.1f", "one", 2, 3.4);
  check_obstack_vprintf("one\ntwo\nthree", "one\n%s", "two\nthree");
  check_open_obstack_stream();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <check_obstack_printf_int@plt>
  xor    %eax,%eax
  call   <check_obstack_printf_typical@plt>
  xor    %eax,%eax
  call   <check_obstack_printf_two_times@plt>
  mov    $0x2,%ecx
  movsd  D28(%rip),%xmm0
  lea    D34(%rip),%rdx
  lea    D61(%rip),%rsi
  lea    D10(%rip),%rdi
  mov    $0x1,%eax
  call   <check_obstack_vprintf@plt>
  lea    D69(%rip),%rdx
  lea    D7(%rip),%rsi
  xor    %eax,%eax
  lea    D17(%rip),%rdi
  call   <check_obstack_vprintf@plt>
  xor    %eax,%eax
  call   <check_open_obstack_stream@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D28
type_infered: f64
name: extract_symbolic_data
arguments: label: D34
type_infered: string
name: extract_symbolic_data
arguments: label: D61
type_infered: string
name: extract_symbolic_data
arguments: label: D10
type_infered: string
name: extract_symbolic_data
arguments: label: D69
type_infered: string
name: extract_symbolic_data
arguments: label: D7
type_infered: string
name: extract_symbolic_data
arguments: label: D17
type_infered: string

tool: D28: 3.4

tool: D34: one

tool: D61: %s and %d then %.1f

tool: D10: one and 2 then 3.4

tool: D69: two
three

tool: D7: one
%s

tool: D17: one
two
three

assistant: ```c
int main(int argc, char *argv[]) {
  check_obstack_printf_int();
  check_obstack_printf_typical();
  check_obstack_printf_two_times();
  check_obstack_vprintf("one and 2 then 3.4", "%s and %d then %.1f", "one", 2, 3.4);
  check_obstack_vprintf("one\ntwo\nthree", "one\n%s", "two\nthree");
  check_open_obstack_stream();
  return 0;
}
```