ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RIPEMD160_Final>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    0x38(%rsi),%r12
  mov    (%rsi),%rdi
  movb   $0x80,(%r12,%rdi,1)
  add    $0x1,%rdi
  mov    D5(%rip),%rax
  mov    (%rax),%edx
  lea    -0x8(%rdx),%eax
  cltq
  cmp    %rdi,%rax
  jb     L3
L4:
  mov    D5(%rip),%r13
  mov    0x0(%r13),%eax
  lea    -0x8(%rax),%edx
  sub    %edi,%edx
  add    %r12,%rdi
  mov    $0x0,%esi
  call   <memset@plt>
  movslq 0x0(%r13),%rax
  lea    -0x8(%r12,%rax,1),%rax
  mov    0x8(%rbx),%edx
  mov    %edx,(%rax)
  mov    0xc(%rbx),%edx
  mov    %edx,0x4(%rax)
  lea    0x8(%rax),%r12
  movslq 0x0(%r13),%rax
  sub    %rax,%r12
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <ripemd160_block_data_order@plt>
  movq   $0x0,(%rbx)
  mov    0x0(%r13),%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <memset@plt>
  mov    0x10(%rbx),%rax
  mov    %eax,0x0(%rbp)
  mov    0x18(%rbx),%rax
  mov    %eax,0x4(%rbp)
  mov    0x20(%rbx),%rax
  mov    %eax,0x8(%rbp)
  mov    0x28(%rbx),%rax
  mov    %eax,0xc(%rbp)
  mov    0x30(%rbx),%rax
  mov    %eax,0x10(%rbp)
  mov    $0x1,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  sub    %edi,%edx
  add    %r12,%rdi
  mov    $0x0,%esi
  call   <memset@plt>
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <ripemd160_block_data_order@plt>
  mov    $0x0,%edi
  jmp    L4
```

assistant: ```c
int RIPEMD160_Final(unsigned char *md, RIPEMD160_CTX *c) {
  unsigned char *p = (unsigned char *)c->data;
  size_t n = c->num;
  p[n] = 0x80;
  n++;
  if (n > (RIPEMD160_CBLOCK - 8)) {
    memset(p + n, 0, RIPEMD160_CBLOCK - n);
    n = 0;
    ripemd160_block_data_order(c, p, 1);
  }
  memset(p + n, 0, RIPEMD160_CBLOCK - 8 - n);
  p += RIPEMD160_CBLOCK - 8;
  (*((unsigned int *)(p)) = (c->Nl), (p) += 4);
  (*((unsigned int *)(p)) = (c->Nh), (p) += 4);
  p -= RIPEMD160_CBLOCK;
  ripemd160_block_data_order(c, p, 1);
  c->num = 0;
  memset(p, 0, RIPEMD160_CBLOCK);
  do {
    unsigned long ll;
    ll = (c)->A;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->B;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->C;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->D;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->E;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
  } while (0);
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<RIPEMD160_Final>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  mov    0x38(%rsi),%rbp
  mov    (%rsi),%rax
  mov    D4(%rip),%r12
  movb   $0x80,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  mov    (%r12),%edx
  lea    0x0(%rbp,%rax,1),%rdi
  mov    %eax,%esi
  lea    -0x8(%rdx),%ecx
  movslq %ecx,%rcx
  cmp    %rax,%rcx
  jb     L15
L3:
  sub    %esi,%edx
  xor    %esi,%esi
  sub    $0x8,%edx
  call   <memset@plt>
  movslq (%r12),%rax
  mov    0x8(%rbx),%edx
  mov    %rbx,%rdi
  lea    -0x8(%rbp,%rax,1),%rax
  mov    %edx,(%rax)
  mov    0xc(%rbx),%edx
  lea    0x8(%rax),%rbp
  mov    %edx,0x4(%rax)
  movslq (%r12),%rax
  mov    $0x1,%edx
  sub    %rax,%rbp
  mov    %rbp,%rsi
  call   <ripemd160_block_data_order@plt>
  mov    (%r12),%edx
  mov    %rbp,%rdi
  xor    %esi,%esi
  movq   $0x0,(%rbx)
  call   <memset@plt>
  mov    0x10(%rbx),%rax
  mov    %eax,0x0(%r13)
  mov    0x18(%rbx),%rax
  mov    %eax,0x4(%r13)
  mov    0x20(%rbx),%rax
  mov    %eax,0x8(%r13)
  mov    0x28(%rbx),%rax
  mov    %eax,0xc(%r13)
  mov    0x30(%rbx),%rax
  mov    %eax,0x10(%r13)
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L15:
  sub    %eax,%edx
  xor    %esi,%esi
  call   <memset@plt>
  mov    $0x1,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <ripemd160_block_data_order@plt>
  mov    (%r12),%edx
  mov    %rbp,%rdi
  xor    %esi,%esi
  jmp    L3
```

assistant: ```c
int RIPEMD160_Final(unsigned char *md, RIPEMD160_CTX *c) {
  unsigned char *p = (unsigned char *)c->data;
  size_t n = c->num;
  p[n] = 0x80;
  n++;
  if (n > (RIPEMD160_CBLOCK - 8)) {
    memset(p + n, 0, RIPEMD160_CBLOCK - n);
    n = 0;
    ripemd160_block_data_order(c, p, 1);
  }
  memset(p + n, 0, RIPEMD160_CBLOCK - 8 - n);
  p += RIPEMD160_CBLOCK - 8;
  (*((unsigned int *)(p)) = (c->Nl), (p) += 4);
  (*((unsigned int *)(p)) = (c->Nh), (p) += 4);
  p -= RIPEMD160_CBLOCK;
  ripemd160_block_data_order(c, p, 1);
  c->num = 0;
  memset(p, 0, RIPEMD160_CBLOCK);
  do {
    unsigned long ll;
    ll = (c)->A;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->B;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->C;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->D;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->E;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
  } while (0);
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RIPEMD160_Final>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  mov    0x38(%rsi),%rbp
  mov    (%rsi),%rax
  mov    D1(%rip),%r12
  movb   $0x80,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  mov    (%r12),%edx
  lea    0x0(%rbp,%rax,1),%rdi
  mov    %eax,%esi
  lea    -0x8(%rdx),%ecx
  movslq %ecx,%rcx
  cmp    %rax,%rcx
  jb     L13
L8:
  sub    %esi,%edx
  xor    %esi,%esi
  sub    $0x8,%edx
  call   <memset@plt>
  movslq (%r12),%rax
  mov    0x8(%rbx),%edx
  mov    %rbx,%rdi
  lea    -0x8(%rbp,%rax,1),%rax
  mov    %edx,(%rax)
  mov    0xc(%rbx),%edx
  lea    0x8(%rax),%rbp
  mov    %edx,0x4(%rax)
  movslq (%r12),%rax
  mov    $0x1,%edx
  sub    %rax,%rbp
  mov    %rbp,%rsi
  call   <ripemd160_block_data_order@plt>
  mov    (%r12),%edx
  mov    %rbp,%rdi
  xor    %esi,%esi
  movq   $0x0,(%rbx)
  call   <memset@plt>
  movdqu 0x10(%rbx),%xmm0
  movdqu 0x20(%rbx),%xmm1
  mov    0x30(%rbx),%rax
  shufps $0x88,%xmm1,%xmm0
  movups %xmm0,0x0(%r13)
  mov    %eax,0x10(%r13)
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L13:
  sub    %eax,%edx
  xor    %esi,%esi
  call   <memset@plt>
  mov    $0x1,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <ripemd160_block_data_order@plt>
  mov    (%r12),%edx
  mov    %rbp,%rdi
  xor    %esi,%esi
  jmp    L8
```

assistant: ```c
int RIPEMD160_Final(unsigned char *md, RIPEMD160_CTX *c) {
  unsigned char *p = (unsigned char *)c->data;
  size_t n = c->num;
  p[n] = 0x80;
  n++;
  if (n > (RIPEMD160_CBLOCK - 8)) {
    memset(p + n, 0, RIPEMD160_CBLOCK - n);
    n = 0;
    ripemd160_block_data_order(c, p, 1);
  }
  memset(p + n, 0, RIPEMD160_CBLOCK - 8 - n);
  p += RIPEMD160_CBLOCK - 8;
  (*((unsigned int *)(p)) = (c->Nl), (p) += 4);
  (*((unsigned int *)(p)) = (c->Nh), (p) += 4);
  p -= RIPEMD160_CBLOCK;
  ripemd160_block_data_order(c, p, 1);
  c->num = 0;
  memset(p, 0, RIPEMD160_CBLOCK);
  do {
    unsigned long ll;
    ll = (c)->A;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->B;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->C;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->D;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->E;
    (void)(*((unsigned int *)((md))) = (ll), ((md)) += 4);
  } while (0);
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strmcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strmlen@plt>
  mov    %eax,-0x28(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strmlen@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L45
  mov    -0x28(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x38(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L43
  mov    -0x28(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x40(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  jmp    L43
L45:
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x28(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x38(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L43
  mov    -0x20(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x24(%rbp),%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x28(%rbp),%eax
  lea    0x1(%rax),%ebx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%r12
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strmlen@plt>
  cltq
  lea    (%r12,%rax,1),%rcx
  mov    -0x40(%rbp),%rax
  mov    %ebx,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
L43:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <freeChar@plt>
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L35
  call   <__stack_chk_fail@plt>
L35:
  add    $0x30,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
void strmcat(char **base_str, char *append_str) {
  char *tmp = NULL;
  int const applen = strmlen(append_str);
  int const baselen = strmlen(*base_str);
  if (*base_str == NULL) {
    *base_str = (char *)malloc(applen + 1);
    if (*base_str != NULL)
      memcpy(*base_str, append_str, applen + 1);
  } else {
    tmp = *base_str;
    *base_str = malloc(baselen + applen + 1);
    if (*base_str != NULL) {
      memcpy(*base_str, tmp, baselen);
      memcpy(*base_str + strmlen(tmp), append_str, applen + 1);
    }
  }
  freeChar(&tmp);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strmcat>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsi,%rdi
  call   <strmlen@plt>
  mov    %eax,%ebp
  mov    (%rbx),%rdi
  call   <strmlen@plt>
  mov    %eax,%r13d
  mov    (%rbx),%rax
  test   %rax,%rax
  je     L44
  mov    %rax,(%rsp)
  lea    0x1(%rbp,%r13,1),%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L21
  mov    %r13d,%edx
  mov    (%rsp),%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    (%rbx),%rbx
  mov    (%rsp),%rdi
  call   <strmlen@plt>
  lea    0x1(%rbp),%edx
  cltq
  lea    (%rbx,%rax,1),%rdi
  mov    %r12,%rsi
  call   <memcpy@plt>
L21:
  mov    %rsp,%rdi
  call   <freeChar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L44:
  add    $0x1,%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L21
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  jmp    L21
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strmcat(char **base_str, char *append_str) {
  char *tmp = NULL;
  int const applen = strmlen(append_str);
  int const baselen = strmlen(*base_str);
  if (*base_str == NULL) {
    *base_str = (char *)malloc(applen + 1);
    if (*base_str != NULL)
      memcpy(*base_str, append_str, applen + 1);
  } else {
    tmp = *base_str;
    *base_str = malloc(baselen + applen + 1);
    if (*base_str != NULL) {
      memcpy(*base_str, tmp, baselen);
      memcpy(*base_str + strmlen(tmp), append_str, applen + 1);
    }
  }
  freeChar(&tmp);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strmcat>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  call   <strmlen@plt>
  mov    (%rbx),%rdi
  mov    %eax,%ebp
  call   <strmlen@plt>
  mov    %eax,%r13d
  mov    (%rbx),%rax
  test   %rax,%rax
  je     L35
  lea    0x1(%rbp,%r13,1),%edi
  mov    %rax,(%rsp)
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L26
  mov    (%rsp),%rsi
  mov    %r13d,%edx
  call   <memcpy@plt>
  mov    (%rsp),%rdi
  mov    (%rbx),%rbx
  call   <strmlen@plt>
  lea    0x1(%rbp),%edx
  mov    %r12,%rsi
  cltq
  lea    (%rbx,%rax,1),%rdi
  call   <memcpy@plt>
L26:
  mov    %rsp,%rdi
  call   <freeChar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L35:
  lea    0x1(%rbp),%r13d
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L26
  mov    %r13d,%edx
  mov    %r12,%rsi
  call   <memcpy@plt>
  jmp    L26
L48:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strmcat(char **base_str, char *append_str) {
  char *tmp = NULL;
  int const applen = strmlen(append_str);
  int const baselen = strmlen(*base_str);
  if (*base_str == NULL) {
    *base_str = (char *)malloc(applen + 1);
    if (*base_str != NULL)
      memcpy(*base_str, append_str, applen + 1);
  } else {
    tmp = *base_str;
    *base_str = malloc(baselen + applen + 1);
    if (*base_str != NULL) {
      memcpy(*base_str, tmp, baselen);
      memcpy(*base_str + strmlen(tmp), append_str, applen + 1);
    }
  }
  freeChar(&tmp);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strmcat>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  call   <strmlen@plt>
  mov    (%rbx),%rdi
  mov    %eax,%ebp
  call   <strmlen@plt>
  mov    %eax,%r13d
  mov    (%rbx),%rax
  test   %rax,%rax
  je     L35
  lea    0x1(%rbp,%r13,1),%edi
  mov    %rax,(%rsp)
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L28
  mov    (%rsp),%rsi
  mov    %r13d,%edx
  call   <memcpy@plt>
  mov    (%rsp),%rdi
  mov    (%rbx),%rbx
  call   <strmlen@plt>
  lea    0x1(%rbp),%edx
  mov    %r12,%rsi
  cltq
  lea    (%rbx,%rax,1),%rdi
  call   <memcpy@plt>
L28:
  mov    %rsp,%rdi
  call   <freeChar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L35:
  lea    0x1(%rbp),%r13d
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L28
  mov    %r13d,%edx
  mov    %r12,%rsi
  call   <memcpy@plt>
  jmp    L28
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void strmcat(char **base_str, char *append_str) {
  char *tmp = NULL;
  int const applen = strmlen(append_str);
  int const baselen = strmlen(*base_str);
  if (*base_str == NULL) {
    *base_str = (char *)malloc(applen + 1);
    if (*base_str != NULL)
      memcpy(*base_str, append_str, applen + 1);
  } else {
    tmp = *base_str;
    *base_str = malloc(baselen + applen + 1);
    if (*base_str != NULL) {
      memcpy(*base_str, tmp, baselen);
      memcpy(*base_str + strmlen(tmp), append_str, applen + 1);
    }
  }
  freeChar(&tmp);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<band_negative_Good>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <unknown_function@plt>
  mov    %eax,-0x18(%rbp)
  mov    $0x0,%eax
  call   <unknown_function@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0xfffffffe,-0x18(%rbp)
  jge    L33
  cmpl   $0xffffffff,-0x14(%rbp)
  jge    L33
  mov    -0x18(%rbp),%eax
  and    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  addl   $0x5,-0x10(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  js     L33
  mov    -0x10(%rbp),%eax
  cltq
  movb   $0x0,-0xb(%rbp,%rax,1)
L33:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
void band_negative_Good() {
  char a[3];
  int x = unknown_function();
  int y = unknown_function();
  if (x <= -3 && y <= -2) {
    int z = x & y;
    z = z + 5;
    if (z >= 0) {
      a[z] = 0;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

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

assistant: ```c
void band_negative_Good() {
  char a[3];
  int x = unknown_function();
  int y = unknown_function();
  if (x <= -3 && y <= -2) {
    int z = x & y;
    z = z + 5;
    if (z >= 0) {
      a[z] = 0;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<band_negative_Good>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <unknown_function@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1050 <unknown_function@plt>
```

assistant: ```c
void band_negative_Good() {
  char a[3];
  int x = unknown_function();
  int y = unknown_function();
  if (x <= -3 && y <= -2) {
    int z = x & y;
    z = z + 5;
    if (z >= 0) {
      a[z] = 0;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<band_negative_Good>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <unknown_function@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1050 <unknown_function@plt>
```

assistant: ```c
void band_negative_Good() {
  char a[3];
  int x = unknown_function();
  int y = unknown_function();
  if (x <= -3 && y <= -2) {
    int z = x & y;
    z = z + 5;
    if (z >= 0) {
      a[z] = 0;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
L6:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
  nop
  pop    %rbp
  ret
```

assistant: ```c
char *strcpy(char *dest, const char *src) {
  do {
    *dest++ = *src++;
  } while (*src != 0);
}
```
user: What is the c source code of the assembly code below?

```asm
<strcpy>:
  endbr64
L2:
  add    $0x1,%rsi
  add    $0x1,%rdi
  movzbl -0x1(%rsi),%eax
  mov    %al,-0x1(%rdi)
  cmpb   $0x0,(%rsi)
  jne    L2
  ret
```

assistant: ```c
char *strcpy(char *dest, const char *src) {
  do {
    *dest++ = *src++;
  } while (*src != 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcpy>:
  endbr64
  movzbl (%rsi),%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L7:
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L7
  ret
```

assistant: ```c
char *strcpy(char *dest, const char *src) {
  do {
    *dest++ = *src++;
  } while (*src != 0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcpy>:
  endbr64
  nopl   0x0(%rax)
L9:
  movsb  %ds:(%rsi),%es:(%rdi)
  cmpb   $0x0,(%rsi)
  jne    L9
  ret
```

assistant: ```c
char *strcpy(char *dest, const char *src) {
  do {
    *dest++ = *src++;
  } while (*src != 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<genlmsg_alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <NLMSG_LENGTH@plt>
  mov    -0x18(%rbp),%rdx
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <nlmsg_alloc@plt>
  leave
  ret
```

assistant: ```c
extern struct genlmsg *genlmsg_alloc(size_t size) {
  size_t len = NLMSG_LENGTH(GENL_HDRLEN) + size;
  return (struct genlmsg *)nlmsg_alloc(len);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<genlmsg_alloc>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    D5(%rip),%rax
  mov    (%rax),%edi
  call   <NLMSG_LENGTH@plt>
  lea    (%rax,%rbx,1),%rdi
  call   <nlmsg_alloc@plt>
  pop    %rbx
  ret
```

assistant: ```c
extern struct genlmsg *genlmsg_alloc(size_t size) {
  size_t len = NLMSG_LENGTH(GENL_HDRLEN) + size;
  return (struct genlmsg *)nlmsg_alloc(len);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<genlmsg_alloc>:
  endbr64
  mov    D2(%rip),%rax
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rax),%edi
  call   <NLMSG_LENGTH@plt>
  lea    (%rax,%rbx,1),%rdi
  pop    %rbx
  jmp    1060 <nlmsg_alloc@plt>
```

assistant: ```c
extern struct genlmsg *genlmsg_alloc(size_t size) {
  size_t len = NLMSG_LENGTH(GENL_HDRLEN) + size;
  return (struct genlmsg *)nlmsg_alloc(len);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<genlmsg_alloc>:
  endbr64
  mov    D2(%rip),%rax
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rax),%edi
  call   <NLMSG_LENGTH@plt>
  lea    (%rax,%rbx,1),%rdi
  pop    %rbx
  jmp    1060 <nlmsg_alloc@plt>
```

assistant: ```c
extern struct genlmsg *genlmsg_alloc(size_t size) {
  size_t len = NLMSG_LENGTH(GENL_HDRLEN) + size;
  return (struct genlmsg *)nlmsg_alloc(len);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<is_corner>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L161
  cmpl   $0x0,-0x18(%rbp)
  jne    L161
  movl   $0x1,-0x4(%rbp)
L161:
  mov    -0x20(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L93
  mov    -0x1c(%rbp),%eax
  sub    $0x2,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L141
  cmpl   $0x0,-0x18(%rbp)
  je     L48
L141:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L118
  cmpl   $0x1,-0x18(%rbp)
  jne    L118
L48:
  movl   $0x1,-0x4(%rbp)
  jmp    L118
L93:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L118
  cmpl   $0x0,-0x18(%rbp)
  jne    L118
  movl   $0x1,-0x4(%rbp)
L118:
  mov    -0x20(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L125
  cmpl   $0x0,-0x14(%rbp)
  jne    L183
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x18(%rbp)
  jne    L183
  movl   $0x1,-0x4(%rbp)
  jmp    L183
L125:
  cmpl   $0x0,-0x14(%rbp)
  jne    L51
  mov    -0x20(%rbp),%eax
  sub    $0x2,%eax
  cmp    %eax,-0x18(%rbp)
  je     L55
L51:
  cmpl   $0x1,-0x14(%rbp)
  jne    L183
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x18(%rbp)
  jne    L183
L55:
  movl   $0x1,-0x4(%rbp)
L183:
  mov    -0x1c(%rbp),%eax
  sub    $0x2,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L210
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x18(%rbp)
  je     L22
L210:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L10
  mov    -0x20(%rbp),%eax
  sub    $0x2,%eax
  cmp    %eax,-0x18(%rbp)
  jne    L10
L22:
  movl   $0x1,-0x4(%rbp)
L10:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int is_corner(int column, int row, int width, int height) {
  int corner = 0;
  if ((column == 0) && (row == 0)) {
    corner = 1;
  }
  if (height % 2) {
    if (((column == width - 2) && (row == 0)) || ((column == width - 1) && (row == 1))) {
      corner = 1;
    }
  } else {
    if ((column == width - 1) && (row == 0)) {
      corner = 1;
    }
  }
  if (height % 2) {
    if ((column == 0) && (row == height - 1)) {
      corner = 1;
    }
  } else {
    if (((column == 0) && (row == height - 2)) || ((column == 1) && (row == height - 1))) {
      corner = 1;
    }
  }
  if (((column == width - 2) && (row == height - 1)) || ((column == width - 1) && (row == height - 2))) {
    corner = 1;
  }
  return corner;
}
```