ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<strbeginswith>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jae    L5
  mov    $0x0,%eax
  jmp    L17
L5:
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L17:
  leave
  ret
```

assistant: ```c
int strbeginswith(const char *string, const char *what) {
  size_t string_len = strlen(string), what_len = strlen(what);
  if (string_len < what_len) {
    return 0;
  }
  return !strncmp(string, what, what_len);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strbeginswith>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %rax,%r12
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  cmp    %rdx,%r12
  jb     L4
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strbeginswith(const char *string, const char *what) {
  size_t string_len = strlen(string), what_len = strlen(what);
  if (string_len < what_len) {
    return 0;
  }
  return !strncmp(string, what, what_len);
}
```
user: What is the c source code of the assembly code below:

```asm
<strbeginswith>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  mov    %rax,%rdx
  xor    %eax,%eax
  cmp    %rdx,%rbx
  jb     L4
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strbeginswith(const char *string, const char *what) {
  size_t string_len = strlen(string), what_len = strlen(what);
  if (string_len < what_len) {
    return 0;
  }
  return !strncmp(string, what, what_len);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strbeginswith>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  mov    %rax,%rdx
  xor    %eax,%eax
  cmp    %rdx,%rbx
  jb     L6
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strbeginswith(const char *string, const char *what) {
  size_t string_len = strlen(string), what_len = strlen(what);
  if (string_len < what_len) {
    return 0;
  }
  return !strncmp(string, what, what_len);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<convert_upper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L30
L14:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L39
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jg     L39
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    -0x20(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L39:
  addl   $0x1,-0x4(%rbp)
L30:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L14
  nop
  pop    %rbp
  ret
```

assistant: ```c
void *convert_upper(char *str) {
  int i = 0;
  while (str[i]) {
    if (str[i] >= 'a' && str[i] <= 'z')
      str[i] -= ('a' - 'A');
    i++;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<convert_upper>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L14
L19:
  ret
L26:
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  je     L19
L14:
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L26
  sub    $0x20,%eax
  mov    %al,(%rdi)
  jmp    L26
```

assistant: ```c
void *convert_upper(char *str) {
  int i = 0;
  while (str[i]) {
    if (str[i] >= 'a' && str[i] <= 'z')
      str[i] -= ('a' - 'A');
    i++;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<convert_upper>:
  endbr64
  jmp    L20
  cs nopw 0x0(%rax,%rax,1)
L14:
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L4
  sub    $0x20,%eax
  mov    %al,(%rdi)
L4:
  add    $0x1,%rdi
L20:
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L14
  ret
```

assistant: ```c
void *convert_upper(char *str) {
  int i = 0;
  while (str[i]) {
    if (str[i] >= 'a' && str[i] <= 'z')
      str[i] -= ('a' - 'A');
    i++;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<convert_upper>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L32
  nopl   0x0(%rax,%rax,1)
L30:
  lea    -0x61(%rax),%edx
  cmp    $0x19,%dl
  ja     L15
  sub    $0x20,%eax
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L30
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L30
L32:
  ret
```

assistant: ```c
void *convert_upper(char *str) {
  int i = 0;
  while (str[i]) {
    if (str[i] >= 'a' && str[i] <= 'z')
      str[i] -= ('a' - 'A');
    i++;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,%rax
  mov    %esi,-0xc(%rbp)
  mov    $0x0,%ebx
  jmp    L19
L16:
  mov    %rax,%rdx
  lea    0x1(%rdx),%rax
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  mov    %ebx,%ecx
  add    %ecx,%edx
  mov    %edx,%ebx
L19:
  movzbl (%rax),%edx
  test   %dl,%dl
  jne    L16
  mov    %ebx,%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int Hash(s, mod)
register char *s;
int mod;
{
  register int h = 0;
  while (*s)
    h += (unsigned)*s++;
  return h % mod;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<Hash>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L6
  mov    $0x0,%eax
L18:
  add    $0x1,%rdi
  movsbl %dl,%edx
  add    %edx,%eax
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L18
L22:
  cltd
  idiv   %esi
  mov    %edx,%eax
  ret
L6:
  mov    $0x0,%eax
  jmp    L22
```

assistant: ```c
int Hash(s, mod)
register char *s;
int mod;
{
  register int h = 0;
  while (*s)
    h += (unsigned)*s++;
  return h % mod;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Hash>:
  endbr64
  movsbl (%rdi),%edx
  test   %dl,%dl
  je     L19
  xor    %eax,%eax
  nopl   (%rax)
L12:
  add    $0x1,%rdi
  add    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L12
  cltd
  idiv   %esi
  mov    %edx,%r8d
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int Hash(s, mod)
register char *s;
int mod;
{
  register int h = 0;
  while (*s)
    h += (unsigned)*s++;
  return h % mod;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Hash>:
  endbr64
  movsbl (%rdi),%edx
  test   %dl,%dl
  je     L7
  xor    %eax,%eax
  nopl   (%rax)
L10:
  add    $0x1,%rdi
  add    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L10
  cltd
  idiv   %esi
  mov    %edx,%r8d
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int Hash(s, mod)
register char *s;
int mod;
{
  register int h = 0;
  while (*s)
    h += (unsigned)*s++;
  return h % mod;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<MD5_Final>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    0x30(%rsi),%r12
  mov    (%rsi),%rdi
  movb   $0x80,(%r12,%rdi,1)
  add    $0x1,%rdi
  mov    D12(%rip),%rax
  mov    (%rax),%edx
  lea    -0x8(%rdx),%eax
  cltq
  cmp    %rdi,%rax
  jb     L11
L4:
  mov    D12(%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   <md5_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    $0x1,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  sub    %edi,%edx
  add    %r12,%rdi
  mov    $0x0,%esi
  call   <memset@plt>
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <md5_block_data_order@plt>
  mov    $0x0,%edi
  jmp    L4
```

assistant: ```c
int MD5_Final(unsigned char *md, MD5_CTX *c) {
  unsigned char *p = (unsigned char *)c->data;
  size_t n = c->num;
  p[n] = 0x80;
  n++;
  if (n > (MD5_CBLOCK - 8)) {
    memset(p + n, 0, MD5_CBLOCK - n);
    n = 0;
    md5_block_data_order(c, p, 1);
  }
  memset(p + n, 0, MD5_CBLOCK - 8 - n);
  p += MD5_CBLOCK - 8;
  (*((unsigned int *)(p)) = (c->Nl), (p) += 4);
  (*((unsigned int *)(p)) = (c->Nh), (p) += 4);
  p -= MD5_CBLOCK;
  md5_block_data_order(c, p, 1);
  c->num = 0;
  memset(p, 0, MD5_CBLOCK);
  do {
    unsigned long ll;
    ll = (c)->A;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->B;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->C;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->D;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
  } while (0);
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<MD5_Final>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  mov    0x30(%rsi),%rbp
  mov    (%rsi),%rax
  mov    D3(%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     L4
L18:
  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   <md5_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)
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L4:
  sub    %eax,%edx
  xor    %esi,%esi
  call   <memset@plt>
  mov    $0x1,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <md5_block_data_order@plt>
  mov    (%r12),%edx
  mov    %rbp,%rdi
  xor    %esi,%esi
  jmp    L18
```

assistant: ```c
int MD5_Final(unsigned char *md, MD5_CTX *c) {
  unsigned char *p = (unsigned char *)c->data;
  size_t n = c->num;
  p[n] = 0x80;
  n++;
  if (n > (MD5_CBLOCK - 8)) {
    memset(p + n, 0, MD5_CBLOCK - n);
    n = 0;
    md5_block_data_order(c, p, 1);
  }
  memset(p + n, 0, MD5_CBLOCK - 8 - n);
  p += MD5_CBLOCK - 8;
  (*((unsigned int *)(p)) = (c->Nl), (p) += 4);
  (*((unsigned int *)(p)) = (c->Nh), (p) += 4);
  p -= MD5_CBLOCK;
  md5_block_data_order(c, p, 1);
  c->num = 0;
  memset(p, 0, MD5_CBLOCK);
  do {
    unsigned long ll;
    ll = (c)->A;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->B;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->C;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->D;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
  } while (0);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<MD5_Final>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  mov    0x30(%rsi),%rbp
  mov    (%rsi),%rax
  mov    D5(%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     L12
L4:
  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   <md5_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    $0x1,%eax
  shufps $0x88,%xmm1,%xmm0
  movups %xmm0,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L12:
  sub    %eax,%edx
  xor    %esi,%esi
  call   <memset@plt>
  mov    $0x1,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <md5_block_data_order@plt>
  mov    (%r12),%edx
  mov    %rbp,%rdi
  xor    %esi,%esi
  jmp    L4
```

assistant: ```c
int MD5_Final(unsigned char *md, MD5_CTX *c) {
  unsigned char *p = (unsigned char *)c->data;
  size_t n = c->num;
  p[n] = 0x80;
  n++;
  if (n > (MD5_CBLOCK - 8)) {
    memset(p + n, 0, MD5_CBLOCK - n);
    n = 0;
    md5_block_data_order(c, p, 1);
  }
  memset(p + n, 0, MD5_CBLOCK - 8 - n);
  p += MD5_CBLOCK - 8;
  (*((unsigned int *)(p)) = (c->Nl), (p) += 4);
  (*((unsigned int *)(p)) = (c->Nh), (p) += 4);
  p -= MD5_CBLOCK;
  md5_block_data_order(c, p, 1);
  c->num = 0;
  memset(p, 0, MD5_CBLOCK);
  do {
    unsigned long ll;
    ll = (c)->A;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->B;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->C;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
    ll = (c)->D;
    (*((unsigned int *)((md))) = (ll), ((md)) += 4);
  } while (0);
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<p264_add8x8_idct_mmxext>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <p264_add4x4_idct_mmxext@plt>
  mov    -0x18(%rbp),%rax
  lea    0x40(%rax),%rdx
  mov    -0x8(%rbp),%rax
  lea    0x10(%rax),%rcx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  mov    %rcx,%rdi
  call   <p264_add4x4_idct_mmxext@plt>
  mov    -0x18(%rbp),%rax
  lea    0x80(%rax),%rdx
  mov    -0xc(%rbp),%eax
  shl    $0x2,%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rax,%rcx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  mov    %rcx,%rdi
  call   <p264_add4x4_idct_mmxext@plt>
  mov    -0x18(%rbp),%rax
  lea    0xc0(%rax),%rdx
  mov    -0xc(%rbp),%eax
  shl    $0x2,%eax
  cltq
  add    $0x4,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rax,%rcx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  mov    %rcx,%rdi
  call   <p264_add4x4_idct_mmxext@plt>
  nop
  leave
  ret
```

assistant: ```c
void p264_add8x8_idct_mmxext(uint8_t *p_dst, int i_dst, int16_t dct[4][4][4]) {
  p264_add4x4_idct_mmxext(p_dst, i_dst, dct[0]);
  p264_add4x4_idct_mmxext(&p_dst[4], i_dst, dct[1]);
  p264_add4x4_idct_mmxext(&p_dst[4 * i_dst + 0], i_dst, dct[2]);
  p264_add4x4_idct_mmxext(&p_dst[4 * i_dst + 4], i_dst, dct[3]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<p264_add8x8_idct_mmxext>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%ebp
  mov    %rdx,%rbx
  call   <p264_add4x4_idct_mmxext@plt>
  lea    0x40(%rbx),%rdx
  lea    0x10(%r13),%rdi
  mov    %ebp,%esi
  call   <p264_add4x4_idct_mmxext@plt>
  lea    0x0(,%rbp,4),%r12d
  movslq %r12d,%r12
  shl    $0x2,%r12
  lea    0x80(%rbx),%rdx
  lea    0x0(%r13,%r12,1),%rdi
  mov    %ebp,%esi
  call   <p264_add4x4_idct_mmxext@plt>
  lea    0xc0(%rbx),%rdx
  lea    0x10(%r13,%r12,1),%rdi
  mov    %ebp,%esi
  call   <p264_add4x4_idct_mmxext@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void p264_add8x8_idct_mmxext(uint8_t *p_dst, int i_dst, int16_t dct[4][4][4]) {
  p264_add4x4_idct_mmxext(p_dst, i_dst, dct[0]);
  p264_add4x4_idct_mmxext(&p_dst[4], i_dst, dct[1]);
  p264_add4x4_idct_mmxext(&p_dst[4 * i_dst + 0], i_dst, dct[2]);
  p264_add4x4_idct_mmxext(&p_dst[4 * i_dst + 4], i_dst, dct[3]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<p264_add8x8_idct_mmxext>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  lea    0x0(,%rbp,4),%r12d
  mov    %rdx,%rbx
  movslq %r12d,%r12
  shl    $0x2,%r12
  sub    $0x8,%rsp
  call   <p264_add4x4_idct_mmxext@plt>
  lea    0x40(%rbx),%rdx
  lea    0x10(%r13),%rdi
  mov    %ebp,%esi
  call   <p264_add4x4_idct_mmxext@plt>
  lea    0x0(%r13,%r12,1),%rdi
  lea    0x80(%rbx),%rdx
  mov    %ebp,%esi
  call   <p264_add4x4_idct_mmxext@plt>
  add    $0x8,%rsp
  lea    0x10(%r13,%r12,1),%rdi
  mov    %ebp,%esi
  lea    0xc0(%rbx),%rdx
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1050 <p264_add4x4_idct_mmxext@plt>
```

assistant: ```c
void p264_add8x8_idct_mmxext(uint8_t *p_dst, int i_dst, int16_t dct[4][4][4]) {
  p264_add4x4_idct_mmxext(p_dst, i_dst, dct[0]);
  p264_add4x4_idct_mmxext(&p_dst[4], i_dst, dct[1]);
  p264_add4x4_idct_mmxext(&p_dst[4 * i_dst + 0], i_dst, dct[2]);
  p264_add4x4_idct_mmxext(&p_dst[4 * i_dst + 4], i_dst, dct[3]);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<p264_add8x8_idct_mmxext>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  lea    0x0(,%rbp,4),%r12d
  mov    %rdx,%rbx
  movslq %r12d,%r12
  shl    $0x2,%r12
  sub    $0x8,%rsp
  call   <p264_add4x4_idct_mmxext@plt>
  lea    0x40(%rbx),%rdx
  lea    0x10(%r13),%rdi
  mov    %ebp,%esi
  call   <p264_add4x4_idct_mmxext@plt>
  lea    0x0(%r13,%r12,1),%rdi
  lea    0x80(%rbx),%rdx
  mov    %ebp,%esi
  call   <p264_add4x4_idct_mmxext@plt>
  add    $0x8,%rsp
  lea    0x10(%r13,%r12,1),%rdi
  mov    %ebp,%esi
  lea    0xc0(%rbx),%rdx
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1050 <p264_add4x4_idct_mmxext@plt>
```

assistant: ```c
void p264_add8x8_idct_mmxext(uint8_t *p_dst, int i_dst, int16_t dct[4][4][4]) {
  p264_add4x4_idct_mmxext(p_dst, i_dst, dct[0]);
  p264_add4x4_idct_mmxext(&p_dst[4], i_dst, dct[1]);
  p264_add4x4_idct_mmxext(&p_dst[4 * i_dst + 0], i_dst, dct[2]);
  p264_add4x4_idct_mmxext(&p_dst[4 * i_dst + 4], i_dst, dct[3]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test9>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L22
L1:
  mov    -0x1c(%rbp),%eax
  cltq
  movzbl -0x12(%rbp,%rax,1),%eax
  movsbl %al,%eax
  add    %eax,-0x18(%rbp)
  addl   $0x1,-0x1c(%rbp)
L22:
  cmpl   $0xa,-0x1c(%rbp)
  jle    L1
  mov    -0x18(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

assistant: ```c
int test9(void) {
  int i, sum = 0;
  char tab[10];
  for (i = 0; i < 11; i++) {
    sum += tab[i];
  }
  return sum;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test9>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rax
  lea    0x19(%rsp),%rsi
  mov    $0x0,%edx
L2:
  movsbl (%rax),%ecx
  add    %ecx,%edx
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L2
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    %edx,%eax
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test9(void) {
  int i, sum = 0;
  char tab[10];
  for (i = 0; i < 11; i++) {
    sum += tab[i];
  }
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test9>:
  endbr64
  sub    $0x28,%rsp
  xor    %r8d,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x19(%rsp),%rcx
  lea    0xe(%rsp),%rax
  nopl   (%rax)
L4:
  movsbl (%rax),%edx
  add    $0x1,%rax
  add    %edx,%r8d
  cmp    %rcx,%rax
  jne    L4
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    %r8d,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test9(void) {
  int i, sum = 0;
  char tab[10];
  for (i = 0; i < 11; i++) {
    sum += tab[i];
  }
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test9>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movsbl 0xe(%rsp),%edx
  movsbl 0xf(%rsp),%eax
  add    %eax,%edx
  movsbl 0x10(%rsp),%eax
  add    %edx,%eax
  movsbl 0x11(%rsp),%edx
  add    %eax,%edx
  movsbl 0x12(%rsp),%eax
  add    %edx,%eax
  movsbl 0x13(%rsp),%edx
  add    %eax,%edx
  movsbl 0x14(%rsp),%eax
  add    %edx,%eax
  movsbl 0x15(%rsp),%edx
  add    %eax,%edx
  movsbl 0x16(%rsp),%eax
  add    %edx,%eax
  movsbl 0x17(%rsp),%edx
  add    %eax,%edx
  movsbl 0x18(%rsp),%eax
  add    %edx,%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test9(void) {
  int i, sum = 0;
  char tab[10];
  for (i = 0; i < 11; i++) {
    sum += tab[i];
  }
  return sum;
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_trace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x60(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <backtrace@plt>
  mov    %rax,-0x70(%rbp)
  mov    -0x70(%rbp),%rdx
  lea    -0x60(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <backtrace_symbols@plt>
  mov    %rax,-0x68(%rbp)
  mov    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movq   $0x0,-0x78(%rbp)
  jmp    L18
L19:
  mov    -0x78(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addq   $0x1,-0x78(%rbp)
L18:
  mov    -0x78(%rbp),%rax
  cmp    -0x70(%rbp),%rax
  jb     L19
  mov    -0x68(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D6
datatype: string

tool: D6: Obtained %zd stack frames.


assistant: ```c
void print_trace(void) {
  void *array[10];
  size_t size;
  char **strings;
  size_t i;
  size = backtrace(array, 10);
  strings = backtrace_symbols(array, size);
  printf("Obtained %zd stack frames.\n", size);
  for (i = 0; i < size; i++)
    printf("%s\n", strings[i]);
  free(strings);
}
```