ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<lw_protocol_chat_talk>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x3f0,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    $0x3e8,%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <memset@plt>
  mov    %rbp,%r8
  lea    D6(%rip),%rcx
  lea    D19(%rip),%rdx
  mov    $0x3e7,%esi
  mov    %r12,%rdi
  call   <snprintf@plt>
  movb   $0x0,0x3e7(%rsp)
  mov    $0x0,%eax
  cmpl   $0x0,(%rbx)
  jns    L4
L6:
  mov    0x3e8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L31
  add    $0x3f0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  call   <lw_sock_send_str@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L6
  mov    %rbx,%rdi
  call   <recv_ok@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L6
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D19
type_infered: string

tool: D6: CHATTALK

tool: D19: %s '%s'

assistant: ```c
int lw_protocol_chat_talk(int *sock, char *text) {
  int result = 0;
  char message[1000];
  {
    {
      memset(((message)), 0, (sizeof(message)));
      snprintf(((message)), (sizeof(message)) - 1, (("%s '%s'")), (("CHATTALK")), ((text)));
      ((message))[(sizeof(message)) - 1] = '\0';
    };
  };
  if (*sock >= 0 && lw_sock_send_str(sock, message)) {
    if (recv_ok(sock)) {
      result = 1;
    }
  }
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<lw_protocol_chat_talk>:
  endbr64
  push   %r12
  mov    $0x3e8,%edx
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <memset@plt>
  mov    %rbx,%r8
  mov    $0x3e7,%esi
  mov    %r12,%rdi
  lea    D14(%rip),%rcx
  lea    D17(%rip),%rdx
  call   <snprintf@plt>
  mov    0x0(%rbp),%eax
  movb   $0x0,0x3e7(%rsp)
  test   %eax,%eax
  jns    L3
L16:
  xor    %eax,%eax
L35:
  mov    0x3e8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x3f0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <lw_sock_send_str@plt>
  test   %rax,%rax
  je     L16
  mov    %rbp,%rdi
  call   <recv_ok@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L35
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D14
data_type: string
name: unpack_data
arguments: label: D17
data_type: string

tool: D14: CHATTALK

tool: D17: %s '%s'

assistant: ```c
int lw_protocol_chat_talk(int *sock, char *text) {
  int result = 0;
  char message[1000];
  {
    {
      memset(((message)), 0, (sizeof(message)));
      snprintf(((message)), (sizeof(message)) - 1, (("%s '%s'")), (("CHATTALK")), ((text)));
      ((message))[(sizeof(message)) - 1] = '\0';
    };
  };
  if (*sock >= 0 && lw_sock_send_str(sock, message)) {
    if (recv_ok(sock)) {
      result = 1;
    }
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<lw_protocol_chat_talk>:
  endbr64
  push   %r12
  mov    $0x3e8,%edx
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <memset@plt>
  mov    %rbx,%r8
  mov    $0x3e7,%esi
  mov    %r12,%rdi
  lea    D6(%rip),%rcx
  lea    D7(%rip),%rdx
  call   <snprintf@plt>
  mov    0x0(%rbp),%eax
  movb   $0x0,0x3e7(%rsp)
  test   %eax,%eax
  jns    L22
L5:
  xor    %eax,%eax
L19:
  mov    0x3e8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x3f0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L22:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <lw_sock_send_str@plt>
  test   %rax,%rax
  je     L5
  mov    %rbp,%rdi
  call   <recv_ok@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L19
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D6
data_type: string
name: read_data
arguments: label: D7
data_type: string

tool: D6: CHATTALK

tool: D7: %s '%s'

assistant: ```c
int lw_protocol_chat_talk(int *sock, char *text) {
  int result = 0;
  char message[1000];
  {
    {
      memset(((message)), 0, (sizeof(message)));
      snprintf(((message)), (sizeof(message)) - 1, (("%s '%s'")), (("CHATTALK")), ((text)));
      ((message))[(sizeof(message)) - 1] = '\0';
    };
  };
  if (*sock >= 0 && lw_sock_send_str(sock, message)) {
    if (recv_ok(sock)) {
      result = 1;
    }
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<cmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L17
  cmpl   $0x0,-0x8(%rbp)
  jne    L17
  mov    $0x0,%eax
  jmp    L18
L17:
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    -0x8(%rbp),%esi
  movslq %esi,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %esi,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %esi,%edx
  cmp    %edx,%ecx
  je     L24
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %ecx,%edx
  sub    %eax,%edx
  mov    -0x8(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%esi
  mov    %esi,%eax
  shl    $0x2,%eax
  add    %esi,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%esi
  mov    %edx,%eax
  sub    %esi,%eax
  jmp    L18
L24:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  sub    %eax,%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  imul   $0x66666667,%rcx,%rcx
  shr    $0x20,%rcx
  sar    $0x2,%ecx
  sar    $0x1f,%eax
  mov    %eax,%esi
  mov    %ecx,%eax
  sub    %esi,%eax
  mov    %edx,%esi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <cmp>
L18:
  leave
  ret
```

assistant: ```c
cmp(a, b) {
  if (!a && !b)
    return 0;
  if (a % 10 != b % 10)
    return a % 10 - b % 10;
  return cmp(a / 10, b / 10);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cmp>:
  endbr64
  mov    %edi,%eax
  or     %esi,%eax
  je     L1
  mov    %edi,%edx
  mov    %esi,%ecx
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  mov    %edi,%eax
  sub    %esi,%eax
  movslq %ecx,%rsi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x22,%rsi
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%esi
  lea    (%rsi,%rsi,4),%edi
  add    %edi,%edi
  mov    %ecx,%esi
  sub    %edi,%esi
  cmp    %esi,%eax
  je     L18
  sub    %esi,%eax
  ret
L18:
  sub    $0x8,%rsp
  movslq %ecx,%rsi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x22,%rsi
  sar    $0x1f,%ecx
  sub    %ecx,%esi
  movslq %edx,%rdi
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  sar    $0x1f,%edx
  sub    %edx,%edi
  mov    $0x0,%eax
  call   <cmp>
  add    $0x8,%rsp
  ret
L1:
  ret
```

assistant: ```c
cmp(a, b) {
  if (!a && !b)
    return 0;
  if (a % 10 != b % 10)
    return a % 10 - b % 10;
  return cmp(a / 10, b / 10);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<cmp>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%r8d
  or     %esi,%eax
  je     L7
  movslq %edi,%rsi
  mov    %edi,%edx
  movslq %r8d,%rcx
  mov    %r8d,%r9d
  imul   $0x66666667,%rsi,%rax
  sar    $0x1f,%edx
  sar    $0x1f,%r9d
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  mov    %edi,%eax
  add    %edx,%edx
  sub    %edx,%eax
  imul   $0x66666667,%rcx,%rdx
  sar    $0x22,%rdx
  sub    %r9d,%edx
  lea    (%rdx,%rdx,4),%r9d
  mov    %r8d,%edx
  add    %r9d,%r9d
  sub    %r9d,%edx
  cmp    %eax,%edx
  je     L40
  jmp    L4
  nopl   (%rax)
L46:
  movslq %edi,%rsi
  mov    %edi,%edx
  movslq %r8d,%rcx
  mov    %r8d,%r9d
  imul   $0x66666667,%rsi,%rax
  sar    $0x1f,%edx
  sar    $0x1f,%r9d
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  mov    %edi,%eax
  add    %edx,%edx
  sub    %edx,%eax
  imul   $0x66666667,%rcx,%rdx
  sar    $0x22,%rdx
  sub    %r9d,%edx
  lea    (%rdx,%rdx,4),%r9d
  mov    %r8d,%edx
  add    %r9d,%r9d
  sub    %r9d,%edx
  cmp    %edx,%eax
  jne    L4
L40:
  imul   $0x66666667,%rcx,%rcx
  mov    %r8d,%eax
  imul   $0x66666667,%rsi,%rsi
  sar    $0x1f,%eax
  sar    $0x22,%rcx
  sub    %eax,%ecx
  sar    $0x22,%rsi
  mov    %edi,%eax
  sar    $0x1f,%eax
  mov    %esi,%edi
  mov    %ecx,%r8d
  sub    %eax,%edi
  mov    %ecx,%eax
  or     %edi,%eax
  jne    L46
L7:
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  sub    %edx,%eax
  ret
```

assistant: ```c
cmp(a, b) {
  if (!a && !b)
    return 0;
  if (a % 10 != b % 10)
    return a % 10 - b % 10;
  return cmp(a / 10, b / 10);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cmp>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%r8d
  or     %esi,%eax
  je     L8
  movslq %edi,%rsi
  mov    %edi,%edx
  movslq %r8d,%rcx
  mov    %r8d,%r9d
  imul   $0x66666667,%rsi,%rax
  sar    $0x1f,%edx
  sar    $0x1f,%r9d
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  mov    %edi,%eax
  add    %edx,%edx
  sub    %edx,%eax
  imul   $0x66666667,%rcx,%rdx
  sar    $0x22,%rdx
  sub    %r9d,%edx
  lea    (%rdx,%rdx,4),%r9d
  mov    %r8d,%edx
  add    %r9d,%r9d
  sub    %r9d,%edx
  cmp    %eax,%edx
  je     L17
  jmp    L36
  nopl   (%rax)
L41:
  movslq %edi,%rsi
  mov    %edi,%edx
  movslq %r8d,%rcx
  mov    %r8d,%r9d
  imul   $0x66666667,%rsi,%rax
  sar    $0x1f,%edx
  sar    $0x1f,%r9d
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  mov    %edi,%eax
  add    %edx,%edx
  sub    %edx,%eax
  imul   $0x66666667,%rcx,%rdx
  sar    $0x22,%rdx
  sub    %r9d,%edx
  lea    (%rdx,%rdx,4),%r9d
  mov    %r8d,%edx
  add    %r9d,%r9d
  sub    %r9d,%edx
  cmp    %edx,%eax
  jne    L36
L17:
  imul   $0x66666667,%rcx,%rcx
  mov    %r8d,%eax
  imul   $0x66666667,%rsi,%rsi
  sar    $0x1f,%eax
  sar    $0x22,%rcx
  sub    %eax,%ecx
  sar    $0x22,%rsi
  mov    %edi,%eax
  sar    $0x1f,%eax
  mov    %esi,%edi
  mov    %ecx,%r8d
  sub    %eax,%edi
  mov    %ecx,%eax
  or     %edi,%eax
  jne    L41
L8:
  ret
  nopl   0x0(%rax,%rax,1)
L36:
  sub    %edx,%eax
  ret
```

assistant: ```c
cmp(a, b) {
  if (!a && !b)
    return 0;
  if (a % 10 != b % 10)
    return a % 10 - b % 10;
  return cmp(a / 10, b / 10);
}
```
user: Translate the below assembly code into C programming language:

```asm
<write_next_dobldobl_solution_with_diagnostics>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %rcx,-0x48(%rbp)
  mov    %r8d,-0x4c(%rbp)
  mov    %r9d,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x3c(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x40(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x4c(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x50(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    0x10(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x48(%rbp),%rcx
  lea    -0x28(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0xb1,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x2c(%rbp)
  cmpl   $0x0,-0x2c(%rbp)
  jne    L6
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
L6:
  mov    -0x2c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
int write_next_dobldobl_solution_with_diagnostics(int *k, int n, int m, double *sol, int nbstep, int nbfail, int nbiter, int nbsyst) {
  int a[5], b[2], fail;
  b[0] = n;
  b[1] = m;
  a[0] = nbstep;
  a[1] = nbfail;
  a[2] = nbiter;
  a[3] = nbsyst;
  a[4] = *k;
  fail = _ada_use_c2phc4c(177, a, b, sol, 0);
  if (fail == 0)
    (*k)++;
  return fail;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<write_next_dobldobl_solution_with_diagnostics>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %esi,0x8(%rsp)
  mov    %edx,0xc(%rsp)
  mov    %r8d,0x10(%rsp)
  mov    %r9d,0x14(%rsp)
  mov    0x40(%rsp),%eax
  mov    %eax,0x18(%rsp)
  mov    0x48(%rsp),%eax
  mov    %eax,0x1c(%rsp)
  mov    (%rdi),%eax
  mov    %eax,0x20(%rsp)
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    $0x0,%r8d
  mov    $0xb1,%edi
  call   <_ada_use_c2phc4c@plt>
  test   %eax,%eax
  jne    L2
  addl   $0x1,(%rbx)
L2:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x30,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int write_next_dobldobl_solution_with_diagnostics(int *k, int n, int m, double *sol, int nbstep, int nbfail, int nbiter, int nbsyst) {
  int a[5], b[2], fail;
  b[0] = n;
  b[1] = m;
  a[0] = nbstep;
  a[1] = nbfail;
  a[2] = nbiter;
  a[3] = nbsyst;
  a[4] = *k;
  fail = _ada_use_c2phc4c(177, a, b, sol, 0);
  if (fail == 0)
    (*k)++;
  return fail;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<write_next_dobldobl_solution_with_diagnostics>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    0x40(%rsp),%eax
  mov    %esi,0x8(%rsp)
  lea    0x10(%rsp),%rsi
  mov    %edx,0xc(%rsp)
  lea    0x8(%rsp),%rdx
  mov    %eax,0x18(%rsp)
  mov    0x48(%rsp),%eax
  mov    %r8d,0x10(%rsp)
  xor    %r8d,%r8d
  mov    %eax,0x1c(%rsp)
  mov    (%rdi),%eax
  mov    $0xb1,%edi
  mov    %r9d,0x14(%rsp)
  mov    %eax,0x20(%rsp)
  call   <_ada_use_c2phc4c@plt>
  test   %eax,%eax
  jne    L13
  addl   $0x1,(%rbx)
L13:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x30,%rsp
  pop    %rbx
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int write_next_dobldobl_solution_with_diagnostics(int *k, int n, int m, double *sol, int nbstep, int nbfail, int nbiter, int nbsyst) {
  int a[5], b[2], fail;
  b[0] = n;
  b[1] = m;
  a[0] = nbstep;
  a[1] = nbfail;
  a[2] = nbiter;
  a[3] = nbsyst;
  a[4] = *k;
  fail = _ada_use_c2phc4c(177, a, b, sol, 0);
  if (fail == 0)
    (*k)++;
  return fail;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<write_next_dobldobl_solution_with_diagnostics>:
  endbr64
  push   %rbx
  movd   %esi,%xmm0
  movd   %edx,%xmm2
  movd   %r9d,%xmm4
  punpckldq %xmm2,%xmm0
  mov    %rdi,%rbx
  sub    $0x30,%rsp
  movd   0x40(%rsp),%xmm1
  movd   0x48(%rsp),%xmm3
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   %xmm0,0x8(%rsp)
  movd   %r8d,%xmm0
  mov    (%rdi),%eax
  punpckldq %xmm3,%xmm1
  punpckldq %xmm4,%xmm0
  xor    %r8d,%r8d
  mov    $0xb1,%edi
  punpcklqdq %xmm1,%xmm0
  mov    %eax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <_ada_use_c2phc4c@plt>
  test   %eax,%eax
  jne    L5
  addl   $0x1,(%rbx)
L5:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x30,%rsp
  pop    %rbx
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int write_next_dobldobl_solution_with_diagnostics(int *k, int n, int m, double *sol, int nbstep, int nbfail, int nbiter, int nbsyst) {
  int a[5], b[2], fail;
  b[0] = n;
  b[1] = m;
  a[0] = nbstep;
  a[1] = nbfail;
  a[2] = nbiter;
  a[3] = nbsyst;
  a[4] = *k;
  fail = _ada_use_c2phc4c(177, a, b, sol, 0);
  if (fail == 0)
    (*k)++;
  return fail;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<transfer_YUV_PLANAR_to_YUVA8888>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x10(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x2(%rax),%rdx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  add    $0x3,%rax
  movb   $0xff,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x4(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void transfer_YUV_PLANAR_to_YUVA8888(unsigned char *(*output), unsigned char *input_y, unsigned char *input_u, unsigned char *input_v) {
  (*output)[0] = *input_y;
  (*output)[1] = *input_u;
  (*output)[2] = *input_v;
  (*output)[3] = 0xff;
  (*output) += 4;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<transfer_YUV_PLANAR_to_YUVA8888>:
  endbr64
  mov    (%rdi),%rax
  movzbl (%rsi),%esi
  mov    %sil,(%rax)
  mov    (%rdi),%rax
  movzbl (%rdx),%edx
  mov    %dl,0x1(%rax)
  mov    (%rdi),%rax
  movzbl (%rcx),%edx
  mov    %dl,0x2(%rax)
  mov    (%rdi),%rax
  movb   $0xff,0x3(%rax)
  addq   $0x4,(%rdi)
  ret
```

assistant: ```c
void transfer_YUV_PLANAR_to_YUVA8888(unsigned char *(*output), unsigned char *input_y, unsigned char *input_u, unsigned char *input_v) {
  (*output)[0] = *input_y;
  (*output)[1] = *input_u;
  (*output)[2] = *input_v;
  (*output)[3] = 0xff;
  (*output) += 4;
}
```
user: What is the c source code of the assembly code below:

```asm
<transfer_YUV_PLANAR_to_YUVA8888>:
  endbr64
  mov    (%rdi),%rax
  movzbl (%rsi),%esi
  mov    %sil,(%rax)
  mov    (%rdi),%rax
  movzbl (%rdx),%edx
  mov    %dl,0x1(%rax)
  mov    (%rdi),%rax
  movzbl (%rcx),%edx
  mov    %dl,0x2(%rax)
  mov    (%rdi),%rax
  movb   $0xff,0x3(%rax)
  addq   $0x4,(%rdi)
  ret
```

assistant: ```c
void transfer_YUV_PLANAR_to_YUVA8888(unsigned char *(*output), unsigned char *input_y, unsigned char *input_u, unsigned char *input_v) {
  (*output)[0] = *input_y;
  (*output)[1] = *input_u;
  (*output)[2] = *input_v;
  (*output)[3] = 0xff;
  (*output) += 4;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<transfer_YUV_PLANAR_to_YUVA8888>:
  endbr64
  mov    (%rdi),%rax
  movzbl (%rsi),%esi
  mov    %sil,(%rax)
  mov    (%rdi),%rax
  movzbl (%rdx),%edx
  mov    %dl,0x1(%rax)
  mov    (%rdi),%rax
  movzbl (%rcx),%edx
  mov    %dl,0x2(%rax)
  mov    (%rdi),%rax
  movb   $0xff,0x3(%rax)
  addq   $0x4,(%rdi)
  ret
```

assistant: ```c
void transfer_YUV_PLANAR_to_YUVA8888(unsigned char *(*output), unsigned char *input_y, unsigned char *input_u, unsigned char *input_v) {
  (*output)[0] = *input_y;
  (*output)[1] = *input_u;
  (*output)[2] = *input_v;
  (*output)[3] = 0xff;
  (*output) += 4;
}
```
user: What is the c source code of the assembly code below:

```asm
<getPathsAppendedWithCommand>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    (%rsi),%rbx
  test   %rbx,%rbx
  je     L52
  mov    %rsi,%rbp
  lea    D11(%rip),%r13
L106:
  mov    $0x4,%edx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L16
  mov    0x0(%rbp),%rdi
  call   <strlen@plt>
  lea    -0x4(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    0x0(%rbp),%rsi
  add    $0x5,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L52:
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  lea    0x400(%rax),%rdx
L69:
  movq   $0x0,(%rax)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L69
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp)
  movzbl (%rbx),%eax
  test   %al,%al
  je     L5
  add    $0x1,%rbx
  mov    $0x0,%edx
  mov    $0x0,%r13d
  jmp    L59
L16:
  add    $0x8,%rbp
  mov    0x0(%rbp),%rbx
  test   %rbx,%rbx
  jne    L106
  jmp    L52
L100:
  movslq %r13d,%rax
  movslq %edx,%rdx
  mov    0x0(%rbp,%rax,8),%rcx
  movb   $0x0,(%rcx,%rdx,1)
  add    $0x1,%r13d
  lea    0x8(%rbp,%rax,8),%r14
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    %rax,(%r14)
  mov    $0x0,%edx
L4:
  add    $0x1,%rbx
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  je     L80
L59:
  cmp    $0x3a,%al
  je     L100
  movslq %r13d,%rcx
  movslq %edx,%rsi
  mov    0x0(%rbp,%rcx,8),%rcx
  mov    %al,(%rcx,%rsi,1)
  add    $0x1,%edx
  jmp    L4
L5:
  mov    $0x0,%edx
  mov    $0x0,%r13d
L80:
  movslq %r13d,%r13
  movslq %edx,%rdx
  mov    0x0(%rbp,%r13,8),%rax
  movb   $0x0,(%rax,%rdx,1)
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  je     L2
  mov    %rbp,%rbx
  lea    D5(%rip),%r13
L61:
  mov    %r13,%rsi
  call   <mystrcat@plt>
  mov    %r12,%rsi
  mov    (%rbx),%rdi
  call   <mystrcat@plt>
  add    $0x8,%rbx
  mov    (%rbx),%rdi
  test   %rdi,%rdi
  jne    L61
L2:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: disassemble_data
arguments: label: D11
data_type: string
name: disassemble_data
arguments: label: D5
data_type: string

tool: D11: PATH

tool: D5: /

assistant: ```c
char **getPathsAppendedWithCommand(char *command, char *envp[]) {
  int i, j = 0, k = 0;
  char *pathvar = NULL;
  char **paths = NULL;
  for (i = 0; envp[i]; i++) {
    if (strncmp(envp[i], "PATH", 4) == 0) {
      pathvar = malloc(strlen(envp[i]) - 4);
      strcpy(pathvar, envp[i] + 5);
      break;
    } else
      continue;
  }
  paths = malloc(sizeof(char *) * 128);
  for (i = 0; i < 128; i++) {
    paths[i] = NULL;
  }
  paths[0] = malloc(sizeof(char) * 128);
  for (i = 0; pathvar[i] != '\0'; i++) {
    if (pathvar[i] != ':') {
      paths[j][k] = pathvar[i];
      k++;
    } else {
      paths[j][k] = '\0';
      j++;
      paths[j] = malloc(sizeof(char) * 128);
      k = 0;
    }
  }
  paths[j][k] = '\0';
  for (i = 0; paths[i]; i++) {
    mystrcat(paths[i], "/");
    mystrcat(paths[i], command);
  }
  return paths;
}
```