ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<read>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x50(%rdi),%r12
  lea    D1(%rip),%rbp
L4:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

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

tool: D1: %d

assistant: ```c
void read(int *array) {
  int i;
  for (i = 0; i < 20; i++)
    scanf("%d", array + i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<read>:
  endbr64
  push   %r12
  lea    0x50(%rdi),%r12
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  cs nopw 0x0(%rax,%rax,1)
L0:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %r12,%rbx
  jne    L0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

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

tool: D5: %d

assistant: ```c
void read(int *array) {
  int i;
  for (i = 0; i < 20; i++)
    scanf("%d", array + i);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read>:
  endbr64
  push   %r12
  lea    0x50(%rdi),%r12
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %r12,%rbx
  jne    L7
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
data_type: string

tool: D3: %d

assistant: ```c
void read(int *array) {
  int i;
  for (i = 0; i < 20; i++)
    scanf("%d", array + i);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ccache_init_thread_rwlock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <pthread_rwlockattr_init@plt>
  test   %rax,%rax
  je     L41
  mov    $0xffffffff,%eax
  jmp    L10
L41:
  mov    D0(%rip),%rax
  mov    (%rax),%edx
  lea    -0xc(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <pthread_rwlockattr_setpshared@plt>
  test   %rax,%rax
  je     L2
  mov    $0xffffffff,%eax
  jmp    L10
L2:
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <pthread_rwlock_init@plt>
  test   %rax,%rax
  je     L39
  mov    $0xffffffff,%eax
  jmp    L10
L39:
  mov    $0x0,%eax
L10:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L36
  call   <__stack_chk_fail@plt>
L36:
  leave
  ret
```

assistant: ```c
int ccache_init_thread_rwlock(pthread_rwlock_t *lock) {
  pthread_rwlockattr_t attr;
  if (0 != pthread_rwlockattr_init(&attr)) {
    return -1;
  }
  if (0 != pthread_rwlockattr_setpshared(&attr, PTHREAD_PROCESS_SHARED)) {
    return -1;
  }
  if (0 != pthread_rwlock_init(lock, &attr)) {
    return -1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ccache_init_thread_rwlock>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <pthread_rwlockattr_init@plt>
  test   %rax,%rax
  jne    L6
  lea    0x4(%rsp),%rdi
  mov    D4(%rip),%rax
  mov    (%rax),%esi
  call   <pthread_rwlockattr_setpshared@plt>
  test   %rax,%rax
  jne    L13
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <pthread_rwlock_init@plt>
  neg    %rax
  sbb    %eax,%eax
L31:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L29
  add    $0x10,%rsp
  pop    %rbx
  ret
L6:
  mov    $0xffffffff,%eax
  jmp    L31
L13:
  mov    $0xffffffff,%eax
  jmp    L31
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ccache_init_thread_rwlock(pthread_rwlock_t *lock) {
  pthread_rwlockattr_t attr;
  if (0 != pthread_rwlockattr_init(&attr)) {
    return -1;
  }
  if (0 != pthread_rwlockattr_setpshared(&attr, PTHREAD_PROCESS_SHARED)) {
    return -1;
  }
  if (0 != pthread_rwlock_init(lock, &attr)) {
    return -1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ccache_init_thread_rwlock>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %r12,%rdi
  call   <pthread_rwlockattr_init@plt>
  test   %rax,%rax
  jne    L29
  mov    D5(%rip),%rax
  mov    %r12,%rdi
  mov    (%rax),%esi
  call   <pthread_rwlockattr_setpshared@plt>
  test   %rax,%rax
  jne    L29
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <pthread_rwlock_init@plt>
  neg    %rax
  sbb    %eax,%eax
L36:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L29:
  mov    $0xffffffff,%eax
  jmp    L36
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ccache_init_thread_rwlock(pthread_rwlock_t *lock) {
  pthread_rwlockattr_t attr;
  if (0 != pthread_rwlockattr_init(&attr)) {
    return -1;
  }
  if (0 != pthread_rwlockattr_setpshared(&attr, PTHREAD_PROCESS_SHARED)) {
    return -1;
  }
  if (0 != pthread_rwlock_init(lock, &attr)) {
    return -1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ccache_init_thread_rwlock>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %r12,%rdi
  call   <pthread_rwlockattr_init@plt>
  test   %rax,%rax
  jne    L1
  mov    D9(%rip),%rax
  mov    %r12,%rdi
  mov    (%rax),%esi
  call   <pthread_rwlockattr_setpshared@plt>
  test   %rax,%rax
  jne    L1
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <pthread_rwlock_init@plt>
  neg    %rax
  sbb    %eax,%eax
L26:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L1:
  mov    $0xffffffff,%eax
  jmp    L26
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ccache_init_thread_rwlock(pthread_rwlock_t *lock) {
  pthread_rwlockattr_t attr;
  if (0 != pthread_rwlockattr_init(&attr)) {
    return -1;
  }
  if (0 != pthread_rwlockattr_setpshared(&attr, PTHREAD_PROCESS_SHARED)) {
    return -1;
  }
  if (0 != pthread_rwlock_init(lock, &attr)) {
    return -1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<revertVector>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L17
L1:
  subq   $0x4,-0x18(%rbp)
  addq   $0x4,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  addl   $0x1,-0xc(%rbp)
L17:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L1
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    $0x1,%eax
  sub    %rdx,%rax
  shl    $0x2,%rax
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *revertVector(int *vec, int N) {
  int i;
  int *rvec = (int *)malloc(sizeof(int) * N);
  vec += N;
  for (i = 0; i < N; i++) {
    vec--;
    rvec++;
    *rvec = *vec;
  }
  rvec -= N - 1;
  return rvec;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<revertVector>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  movslq %esi,%rbx
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  lea    (%r12,%rbx,4),%rdi
  test   %ebp,%ebp
  jle    L8
  mov    %ebp,%ebp
  mov    %rbp,%rdx
  neg    %rdx
  lea    (%rdi,%rdx,4),%r8
  mov    %rdi,%rdx
L15:
  sub    $0x4,%rdx
  mov    (%rdx),%esi
  mov    %rdi,%rcx
  sub    %rdx,%rcx
  mov    %esi,(%rcx,%rax,1)
  cmp    %r8,%rdx
  jne    L15
  lea    (%rax,%rbp,4),%rdx
L8:
  mov    $0x1,%eax
  sub    %rbx,%rax
  lea    (%rdx,%rax,4),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *revertVector(int *vec, int N) {
  int i;
  int *rvec = (int *)malloc(sizeof(int) * N);
  vec += N;
  for (i = 0; i < N; i++) {
    vec--;
    rvec++;
    *rvec = *vec;
  }
  rvec -= N - 1;
  return rvec;
}
```
user: What is the c source code of the assembly code below:

```asm
<revertVector>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  lea    (%r12,%rbp,4),%rdi
  test   %ebp,%ebp
  jle    L20
  mov    %rbp,%r9
  mov    %rdi,%rdx
  neg    %r9
  shl    $0x2,%r9
  lea    (%rdi,%r9,1),%r8
  nopl   0x0(%rax,%rax,1)
L2:
  sub    $0x4,%rdx
  mov    %rdi,%rcx
  mov    (%rdx),%esi
  sub    %rdx,%rcx
  mov    %esi,(%rcx,%rax,1)
  cmp    %r8,%rdx
  jne    L2
  sub    %r9,%rax
  mov    %rax,%rdx
L0:
  mov    $0x1,%eax
  pop    %rbx
  sub    %rbp,%rax
  pop    %rbp
  pop    %r12
  lea    (%rdx,%rax,4),%rax
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  mov    %rax,%rdx
  jmp    L0
```

assistant: ```c
int *revertVector(int *vec, int N) {
  int i;
  int *rvec = (int *)malloc(sizeof(int) * N);
  vec += N;
  for (i = 0; i < N; i++) {
    vec--;
    rvec++;
    *rvec = *vec;
  }
  rvec -= N - 1;
  return rvec;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<revertVector>:
  endbr64
  push   %r12
  movslq %esi,%r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x0(,%r12,4),%edi
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%rdx
  lea    0x0(,%r12,4),%rax
  lea    0x0(%rbp,%rax,1),%r8
  test   %r12d,%r12d
  jle    L86
  lea    0x4(%rdx,%rax,1),%rcx
  lea    0x4(%rdx),%rsi
  mov    %r12,%rbx
  cmp    %rbp,%rcx
  setbe  %cl
  cmp    %rsi,%r8
  setbe  %sil
  or     %sil,%cl
  je     L81
  lea    -0x1(%r12),%ecx
  cmp    $0x2,%ecx
  jbe    L81
  mov    %r12d,%ecx
  lea    -0x10(%rbp,%rax,1),%rax
  shr    $0x2,%ecx
  lea    -0x10(%rax),%rsi
  lea    0x4(%rdx,%rax,1),%rdi
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  sub    %rcx,%rsi
  nopw   0x0(%rax,%rax,1)
L77:
  movdqu (%rax),%xmm1
  mov    %rdi,%rcx
  sub    %rax,%rcx
  sub    $0x10,%rax
  pshufd $0x1b,%xmm1,%xmm0
  movups %xmm0,(%rcx)
  cmp    %rax,%rsi
  jne    L77
  mov    %ebx,%ecx
  mov    %ebx,%edi
  and    $0xfffffffc,%ecx
  mov    %ecx,%eax
  neg    %rax
  shl    $0x2,%rax
  cmp    %ecx,%ebx
  je     L37
  lea    -0x4(%r8,%rax,1),%rsi
  mov    %rdx,%r8
  sub    %rax,%r8
  lea    0x4(%r8),%rax
  mov    (%rsi),%r8d
  mov    %r8d,(%rax)
  lea    0x1(%rcx),%r8d
  cmp    %r8d,%ebx
  jle    L37
  mov    -0x4(%rsi),%r8d
  add    $0x2,%ecx
  mov    %r8d,0x4(%rax)
  cmp    %ecx,%ebx
  jle    L37
  mov    -0x8(%rsi),%ecx
  mov    %ecx,0x8(%rax)
L37:
  lea    (%rdx,%rdi,4),%rdx
L86:
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  sub    %r12,%rax
  pop    %r12
  lea    (%rdx,%rax,4),%rax
  ret
  nop
L81:
  mov    %ebx,%edi
  mov    %rdi,%rax
  neg    %rax
  lea    (%r8,%rax,4),%r9
  mov    %r8,%rax
  nop
L70:
  sub    $0x4,%rax
  mov    %r8,%rcx
  mov    (%rax),%esi
  sub    %rax,%rcx
  mov    %esi,(%rdx,%rcx,1)
  cmp    %r9,%rax
  jne    L70
  jmp    L37
```

assistant: ```c
int *revertVector(int *vec, int N) {
  int i;
  int *rvec = (int *)malloc(sizeof(int) * N);
  vec += N;
  for (i = 0; i < N; i++) {
    vec--;
    rvec++;
    *rvec = *vec;
  }
  rvec -= N - 1;
  return rvec;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<p64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x20,%edx
  mov    -0x44(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,-0x3c(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x38(%rbp)
  movl   $0x1,-0x30(%rbp)
  movl   $0x1d,-0x2c(%rbp)
  movl   $0x4,-0x28(%rbp)
  movl   $0x8,-0x24(%rbp)
  movl   $0x11,-0x20(%rbp)
  movl   $0xc,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0xe,-0x14(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L8
L14:
  mov    -0x38(%rbp),%eax
  add    %eax,-0x3c(%rbp)
  mov    -0x34(%rbp),%eax
  cltd
  shr    $0x1d,%edx
  add    %edx,%eax
  and    $0x7,%eax
  sub    %edx,%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%ecx
  roll   %cl,-0x38(%rbp)
  mov    -0x3c(%rbp),%eax
  xor    %eax,-0x38(%rbp)
  mov    -0x34(%rbp),%eax
  add    %eax,-0x38(%rbp)
  addl   $0x1,-0x34(%rbp)
L8:
  cmpl   $0x1f,-0x34(%rbp)
  jle    L14
  mov    -0x3c(%rbp),%eax
  mov    $0x20,%edx
  mov    %edx,%ecx
  shl    %cl,%eax
  xor    -0x38(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

assistant: ```c
uint64_t p64(uint64_t x) {
  uint32_t hi = x >> 32;
  uint32_t lo = ((x)&0x00000000FFFFFFFFULL);
  unsigned rcon[8] = {1, 29, 4, 8, 17, 12, 3, 14};
  for (int i = 0; i < 32; i++) {
    {
      (hi) += (lo);
      (lo) = ((((lo)) << ((rcon[i % 8]))) ^ (((lo)) >> (32 - ((rcon[i % 8])))));
      (lo) ^= (hi);
    };
    lo += i;
  }
  return ((((uint64_t)hi) << 32) ^ lo);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<p64>:
  endbr64
  sub    $0x38,%rsp
  mov    %edi,%eax
  mov    %fs:0x28,%rdx
  mov    %rdx,0x28(%rsp)
  xor    %edx,%edx
  mov    $0x20,%ecx
  sar    %cl,%edi
  movl   $0x1,(%rsp)
  movl   $0x1d,0x4(%rsp)
  movl   $0x4,0x8(%rsp)
  movl   $0x8,0xc(%rsp)
  movl   $0x11,0x10(%rsp)
  movl   $0xc,0x14(%rsp)
  movl   $0x3,0x18(%rsp)
  movl   $0xe,0x1c(%rsp)
  mov    $0x0,%esi
L10:
  add    %eax,%edi
  mov    %esi,%ecx
  sar    $0x1f,%ecx
  shr    $0x1d,%ecx
  lea    (%rsi,%rcx,1),%edx
  and    $0x7,%edx
  sub    %ecx,%edx
  movslq %edx,%rdx
  mov    (%rsp,%rdx,4),%ecx
  rol    %cl,%eax
  xor    %edi,%eax
  add    %esi,%eax
  add    $0x1,%esi
  cmp    $0x20,%esi
  jne    L10
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x38,%rsp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t p64(uint64_t x) {
  uint32_t hi = x >> 32;
  uint32_t lo = ((x)&0x00000000FFFFFFFFULL);
  unsigned rcon[8] = {1, 29, 4, 8, 17, 12, 3, 14};
  for (int i = 0; i < 32; i++) {
    {
      (hi) += (lo);
      (lo) = ((((lo)) << ((rcon[i % 8]))) ^ (((lo)) >> (32 - ((rcon[i % 8])))));
      (lo) ^= (hi);
    };
    lo += i;
  }
  return ((((uint64_t)hi) << 32) ^ lo);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<p64>:
  endbr64
  sub    $0x38,%rsp
  mov    %edi,%esi
  mov    %edi,%eax
  mov    $0x20,%ecx
  movabs $0x1d00000001,%rdi
  sar    %cl,%esi
  mov    $0x1,%ecx
  mov    %fs:0x28,%rdx
  mov    %rdx,0x28(%rsp)
  xor    %edx,%edx
  mov    %rdi,(%rsp)
  movabs $0x800000004,%rdi
  mov    %rdi,0x8(%rsp)
  movabs $0xc00000011,%rdi
  mov    %rdi,0x10(%rsp)
  movabs $0xe00000003,%rdi
  mov    %rdi,0x18(%rsp)
  jmp    L26
  nopl   (%rax)
L18:
  mov    %edx,%ecx
  and    $0x7,%ecx
  mov    (%rsp,%rcx,4),%ecx
L26:
  add    %eax,%esi
  rol    %cl,%eax
  xor    %esi,%eax
  add    %edx,%eax
  add    $0x1,%edx
  cmp    $0x20,%edx
  jne    L18
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x38,%rsp
  ret
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t p64(uint64_t x) {
  uint32_t hi = x >> 32;
  uint32_t lo = ((x)&0x00000000FFFFFFFFULL);
  unsigned rcon[8] = {1, 29, 4, 8, 17, 12, 3, 14};
  for (int i = 0; i < 32; i++) {
    {
      (hi) += (lo);
      (lo) = ((((lo)) << ((rcon[i % 8]))) ^ (((lo)) >> (32 - ((rcon[i % 8])))));
      (lo) ^= (hi);
    };
    lo += i;
  }
  return ((((uint64_t)hi) << 32) ^ lo);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<p64>:
  endbr64
  sub    $0x38,%rsp
  movdqa D19(%rip),%xmm0
  mov    %edi,%esi
  mov    %edi,%eax
  mov    %fs:0x28,%rdx
  mov    %rdx,0x28(%rsp)
  xor    %edx,%edx
  mov    $0x20,%ecx
  rol    %eax
  mov    $0x1,%edx
  sar    %cl,%esi
  movaps %xmm0,(%rsp)
  movdqa D18(%rip),%xmm0
  add    %edi,%esi
  movaps %xmm0,0x10(%rsp)
  xor    %esi,%eax
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %edx,%ecx
  add    %eax,%esi
  and    $0x7,%ecx
  mov    (%rsp,%rcx,4),%ecx
  rol    %cl,%eax
  xor    %esi,%eax
  add    %edx,%eax
  add    $0x1,%edx
  cmp    $0x20,%edx
  jne    L5
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x38,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D19
data_type: byte[16]
name: unpack_data
arguments: tag: D18
data_type: byte[16]

tool: {"D19": [0x01,0x00,0x00,0x00,0x1d,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: {"D18": [0x11,0x00,0x00,0x00,0x0c,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x0e,0x00,0x00,0x00]}

assistant: ```c
uint64_t p64(uint64_t x) {
  uint32_t hi = x >> 32;
  uint32_t lo = ((x)&0x00000000FFFFFFFFULL);
  unsigned rcon[8] = {1, 29, 4, 8, 17, 12, 3, 14};
  for (int i = 0; i < 32; i++) {
    {
      (hi) += (lo);
      (lo) = ((((lo)) << ((rcon[i % 8]))) ^ (((lo)) >> (32 - ((rcon[i % 8])))));
      (lo) ^= (hi);
    };
    lo += i;
  }
  return ((((uint64_t)hi) << 32) ^ lo);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<handler>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type_infered: string

tool: D0: reader: get signal


assistant: ```c
void handler(int signo) {
  printf("reader: get signal\n");
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<handler>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type_infered: string

tool: D0: reader: get signal


assistant: ```c
void handler(int signo) {
  printf("reader: get signal\n");
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<handler>:
  endbr64
  lea    D2(%rip),%rdi
  jmp    1050 <printf@plt>
```

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

tool: D2: reader: get signal


assistant: ```c
void handler(int signo) {
  printf("reader: get signal\n");
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<handler>:
  endbr64
  lea    D7(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: label: D7
type: string

tool: D7: reader: get signal


assistant: ```c
void handler(int signo) {
  printf("reader: get signal\n");
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getCurrentPos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    $0x24,%esi
  mov    %eax,%edi
  call   <dxl_read_word@plt>
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <PrintErrorCode@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getCurrentPos(int Id) {
  int CurrentPos;
  CurrentPos = dxl_read_word(Id, 36);
  PrintErrorCode();
  return CurrentPos;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getCurrentPos>:
  endbr64
  push   %rbx
  mov    $0x24,%esi
  call   <dxl_read_word@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <PrintErrorCode@plt>
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int getCurrentPos(int Id) {
  int CurrentPos;
  CurrentPos = dxl_read_word(Id, 36);
  PrintErrorCode();
  return CurrentPos;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getCurrentPos>:
  endbr64
  push   %r12
  mov    $0x24,%esi
  call   <dxl_read_word@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <PrintErrorCode@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int getCurrentPos(int Id) {
  int CurrentPos;
  CurrentPos = dxl_read_word(Id, 36);
  PrintErrorCode();
  return CurrentPos;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<getCurrentPos>:
  endbr64
  push   %r12
  mov    $0x24,%esi
  call   <dxl_read_word@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <PrintErrorCode@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int getCurrentPos(int Id) {
  int CurrentPos;
  CurrentPos = dxl_read_word(Id, 36);
  PrintErrorCode();
  return CurrentPos;
}
```
user: What is the c source code of the assembly code below?

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L16:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  jne    L17
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  jmp    L2
L17:
  mov    -0x4(%rbp),%eax
  not    %eax
L2:
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L5:
  cmpl   $0xa,-0x4(%rbp)
  jle    L16
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void foo(int *x, int *y) {
  int i;
  for (i = 0; i < 11; i++)
    y[i] = (x[i] == 1) ? i + 1 : -(i + 1);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<foo>:
  endbr64
  mov    $0x1,%eax
L5:
  mov    %eax,%edx
  neg    %edx
  cmpl   $0x1,-0x4(%rdi,%rax,4)
  cmove  %eax,%edx
  mov    %edx,-0x4(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    $0xc,%rax
  jne    L5
  ret
```

assistant: ```c
void foo(int *x, int *y) {
  int i;
  for (i = 0; i < 11; i++)
    y[i] = (x[i] == 1) ? i + 1 : -(i + 1);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<foo>:
  endbr64
  mov    $0x1,%eax
  nopl   0x0(%rax)
L18:
  cmpl   $0x1,-0x4(%rdi,%rax,4)
  mov    %eax,%edx
  je     L0
  neg    %edx
L0:
  mov    %edx,-0x4(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    $0xc,%rax
  jne    L18
  ret
```

assistant: ```c
void foo(int *x, int *y) {
  int i;
  for (i = 0; i < 11; i++)
    y[i] = (x[i] == 1) ? i + 1 : -(i + 1);
}
```