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
<longestUniqueSubsttr>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    $0x400,%edi
  mov    %eax,%r12d
  call   <malloc@plt>
  lea    0x8(%rax),%rdi
  mov    %rax,%rcx
  movq   $0xffffffffffffffff,(%rax)
  mov    %rax,%r8
  movq   $0xffffffffffffffff,0x3f8(%rax)
  and    $0xfffffffffffffff8,%rdi
  mov    $0xffffffffffffffff,%rax
  sub    %rdi,%rcx
  add    $0x400,%ecx
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
  movsbq (%rbx),%rax
  movl   $0x0,(%r8,%rax,4)
  cmp    $0x1,%r12d
  jle    L48
  movslq %r12d,%r9
  mov    $0x1,%eax
  mov    $0x1,%r11d
  mov    $0x1,%edx
  jmp    L14
  nopl   (%rax)
L29:
  mov    %eax,%r10d
  sub    %edx,%r10d
  cmp    %esi,%r10d
  jg     L41
  cmp    %edx,%r11d
  mov    %edi,(%rcx)
  cmovl  %edx,%r11d
  mov    %eax,%edx
  add    $0x1,%rax
  sub    %esi,%edx
  cmp    %rax,%r9
  je     L36
L14:
  movsbq (%rbx,%rax,1),%rcx
  mov    %eax,%edi
  lea    (%r8,%rcx,4),%rcx
  mov    (%rcx),%esi
  cmp    $0xffffffff,%esi
  jne    L29
L41:
  add    $0x1,%rax
  mov    %edi,(%rcx)
  add    $0x1,%edx
  cmp    %rax,%r9
  jne    L14
L36:
  cmp    %r11d,%edx
  cmovge %edx,%r11d
  mov    %r11d,%r12d
L11:
  mov    %r8,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
L48:
  mov    $0x1,%r12d
  jmp    L11
```

assistant: ```c
int longestUniqueSubsttr(char *str) {
  int n = strlen(str);
  int cur_len = 1;
  int max_len = 1;
  int prev_index;
  int i;
  int *visited = (int *)malloc(sizeof(int) * 256);
  for (i = 0; i < 256; i++)
    visited[i] = -1;
  visited[str[0]] = 0;
  for (i = 1; i < n; i++) {
    prev_index = visited[str[i]];
    if (prev_index == -1 || i - cur_len > prev_index)
      cur_len++;
    else {
      if (cur_len > max_len)
        max_len = cur_len;
      cur_len = i - prev_index;
    }
    visited[str[i]] = i;
  }
  if (cur_len > max_len)
    max_len = cur_len;
  free(visited);
  return max_len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<parse_double>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtod@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  cmp    %rax,-0x30(%rbp)
  je     L25
  mov    $0x0,%eax
  jmp    L27
L25:
  mov    -0x38(%rbp),%rax
  movsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    $0x1,%eax
L27:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
int parse_double(const char *start, const char *end, double *out) {
  char *endptr;
  double val = strtod(start, &endptr);
  if (endptr != end)
    return 0;
  *out = val;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_double>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtod@plt>
  mov    $0x0,%eax
  cmp    %rbx,(%rsp)
  je     L14
L23:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L26
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L14:
  movsd  %xmm0,0x0(%rbp)
  mov    $0x1,%eax
  jmp    L23
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int parse_double(const char *start, const char *end, double *out) {
  char *endptr;
  double val = strtod(start, &endptr);
  if (endptr != end)
    return 0;
  *out = val;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<parse_double>:
  endbr64
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtod@plt>
  xor    %eax,%eax
  cmp    %rbx,(%rsp)
  jne    L8
  movsd  %xmm0,0x0(%rbp)
  mov    $0x1,%eax
L8:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int parse_double(const char *start, const char *end, double *out) {
  char *endptr;
  double val = strtod(start, &endptr);
  if (endptr != end)
    return 0;
  *out = val;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<parse_double>:
  endbr64
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtod@plt>
  xor    %eax,%eax
  cmp    %rbx,(%rsp)
  jne    L11
  movsd  %xmm0,0x0(%rbp)
  mov    $0x1,%eax
L11:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int parse_double(const char *start, const char *end, double *out) {
  char *endptr;
  double val = strtod(start, &endptr);
  if (endptr != end)
    return 0;
  *out = val;
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Qround>:
  endbr64
  mov    %esi,%eax
  sar    $0x18,%eax
  and    $0xf,%eax
  lea    (%rdi,%rax,4),%r10
  mov    %esi,%eax
  sar    $0x10,%eax
  and    $0xf,%eax
  lea    (%rdi,%rax,4),%r9
  mov    (%r9),%ecx
  mov    %esi,%eax
  sar    $0x8,%eax
  and    $0xf,%eax
  lea    (%rdi,%rax,4),%r8
  and    $0xf,%esi
  lea    (%rdi,%rsi,4),%rsi
  mov    %ecx,%r11d
  add    (%r10),%r11d
  mov    %r11d,%eax
  xor    (%rsi),%eax
  mov    %eax,%edx
  shl    $0x10,%edx
  sar    $0x10,%eax
  or     %eax,%edx
  mov    %edx,%edi
  add    (%r8),%edi
  xor    %edi,%ecx
  mov    %ecx,%eax
  shl    $0xc,%eax
  sar    $0x14,%ecx
  or     %ecx,%eax
  lea    (%r11,%rax,1),%ecx
  xor    %ecx,%edx
  mov    %edx,%r11d
  shl    $0x8,%r11d
  sar    $0x18,%edx
  or     %edx,%r11d
  lea    (%rdi,%r11,1),%edx
  xor    %edx,%eax
  mov    %eax,%edi
  shl    $0x7,%edi
  sar    $0x19,%eax
  mov    %ecx,(%r10)
  or     %edi,%eax
  mov    %eax,(%r9)
  mov    %edx,(%r8)
  mov    %r11d,(%rsi)
  ret
```

assistant: ```c
void Qround(uint32_t *s, uint32_t abcd) {
  uint32_t A = s[(abcd >> 24) & 0x0F];
  uint32_t B = s[(abcd >> 16) & 0x0F];
  uint32_t C = s[(abcd >> 8) & 0x0F];
  uint32_t D = s[(abcd >> 0) & 0x0F];
  A += B;
  D ^= A;
  D = (((D) << ((16) & 0x1F)) | ((D) >> (0x20 - ((16) & 0x1F))));
  C += D;
  B ^= C;
  B = (((B) << ((12) & 0x1F)) | ((B) >> (0x20 - ((12) & 0x1F))));
  A += B;
  D ^= A;
  D = (((D) << ((8) & 0x1F)) | ((D) >> (0x20 - ((8) & 0x1F))));
  C += D;
  B ^= C;
  B = (((B) << ((7) & 0x1F)) | ((B) >> (0x20 - ((7) & 0x1F))));
  s[(abcd >> 24) & 0x0F] = A;
  s[(abcd >> 16) & 0x0F] = B;
  s[(abcd >> 8) & 0x0F] = C;
  s[(abcd >> 0) & 0x0F] = D;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Qround>:
  endbr64
  mov    %esi,%eax
  sar    $0x18,%eax
  and    $0xf,%eax
  lea    (%rdi,%rax,4),%r10
  mov    %esi,%eax
  sar    $0x10,%eax
  mov    (%r10),%r11d
  and    $0xf,%eax
  lea    (%rdi,%rax,4),%r9
  mov    %esi,%eax
  and    $0xf,%esi
  sar    $0x8,%eax
  mov    (%r9),%ecx
  lea    (%rdi,%rsi,4),%rsi
  and    $0xf,%eax
  lea    (%rdi,%rax,4),%r8
  mov    (%rsi),%eax
  add    %ecx,%r11d
  mov    (%r8),%edi
  xor    %r11d,%eax
  mov    %eax,%edx
  sar    $0x10,%eax
  shl    $0x10,%edx
  or     %eax,%edx
  add    %edx,%edi
  xor    %edi,%ecx
  mov    %ecx,%eax
  sar    $0x14,%ecx
  shl    $0xc,%eax
  or     %ecx,%eax
  lea    (%r11,%rax,1),%ecx
  xor    %ecx,%edx
  mov    %ecx,(%r10)
  mov    %edx,%r11d
  sar    $0x18,%edx
  shl    $0x8,%r11d
  or     %edx,%r11d
  lea    (%rdi,%r11,1),%edx
  xor    %edx,%eax
  mov    %eax,%edi
  sar    $0x19,%eax
  shl    $0x7,%edi
  or     %edi,%eax
  mov    %eax,(%r9)
  mov    %edx,(%r8)
  mov    %r11d,(%rsi)
  ret
```

assistant: ```c
void Qround(uint32_t *s, uint32_t abcd) {
  uint32_t A = s[(abcd >> 24) & 0x0F];
  uint32_t B = s[(abcd >> 16) & 0x0F];
  uint32_t C = s[(abcd >> 8) & 0x0F];
  uint32_t D = s[(abcd >> 0) & 0x0F];
  A += B;
  D ^= A;
  D = (((D) << ((16) & 0x1F)) | ((D) >> (0x20 - ((16) & 0x1F))));
  C += D;
  B ^= C;
  B = (((B) << ((12) & 0x1F)) | ((B) >> (0x20 - ((12) & 0x1F))));
  A += B;
  D ^= A;
  D = (((D) << ((8) & 0x1F)) | ((D) >> (0x20 - ((8) & 0x1F))));
  C += D;
  B ^= C;
  B = (((B) << ((7) & 0x1F)) | ((B) >> (0x20 - ((7) & 0x1F))));
  s[(abcd >> 24) & 0x0F] = A;
  s[(abcd >> 16) & 0x0F] = B;
  s[(abcd >> 8) & 0x0F] = C;
  s[(abcd >> 0) & 0x0F] = D;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Qround>:
  endbr64
  mov    %esi,%eax
  sar    $0x18,%eax
  and    $0xf,%eax
  lea    (%rdi,%rax,4),%r10
  mov    %esi,%eax
  sar    $0x10,%eax
  mov    (%r10),%r11d
  and    $0xf,%eax
  lea    (%rdi,%rax,4),%r9
  mov    %esi,%eax
  and    $0xf,%esi
  sar    $0x8,%eax
  mov    (%r9),%ecx
  lea    (%rdi,%rsi,4),%rsi
  and    $0xf,%eax
  lea    (%rdi,%rax,4),%r8
  mov    (%rsi),%eax
  add    %ecx,%r11d
  mov    (%r8),%edi
  xor    %r11d,%eax
  mov    %eax,%edx
  sar    $0x10,%eax
  shl    $0x10,%edx
  or     %eax,%edx
  add    %edx,%edi
  xor    %edi,%ecx
  mov    %ecx,%eax
  sar    $0x14,%ecx
  shl    $0xc,%eax
  or     %ecx,%eax
  lea    (%r11,%rax,1),%ecx
  xor    %ecx,%edx
  mov    %ecx,(%r10)
  mov    %edx,%r11d
  sar    $0x18,%edx
  shl    $0x8,%r11d
  or     %edx,%r11d
  lea    (%rdi,%r11,1),%edx
  xor    %edx,%eax
  mov    %eax,%edi
  sar    $0x19,%eax
  shl    $0x7,%edi
  or     %edi,%eax
  mov    %eax,(%r9)
  mov    %edx,(%r8)
  mov    %r11d,(%rsi)
  ret
```

assistant: ```c
void Qround(uint32_t *s, uint32_t abcd) {
  uint32_t A = s[(abcd >> 24) & 0x0F];
  uint32_t B = s[(abcd >> 16) & 0x0F];
  uint32_t C = s[(abcd >> 8) & 0x0F];
  uint32_t D = s[(abcd >> 0) & 0x0F];
  A += B;
  D ^= A;
  D = (((D) << ((16) & 0x1F)) | ((D) >> (0x20 - ((16) & 0x1F))));
  C += D;
  B ^= C;
  B = (((B) << ((12) & 0x1F)) | ((B) >> (0x20 - ((12) & 0x1F))));
  A += B;
  D ^= A;
  D = (((D) << ((8) & 0x1F)) | ((D) >> (0x20 - ((8) & 0x1F))));
  C += D;
  B ^= C;
  B = (((B) << ((7) & 0x1F)) | ((B) >> (0x20 - ((7) & 0x1F))));
  s[(abcd >> 24) & 0x0F] = A;
  s[(abcd >> 16) & 0x0F] = B;
  s[(abcd >> 8) & 0x0F] = C;
  s[(abcd >> 0) & 0x0F] = D;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<cbm_lock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <cbmarch_lock@plt>
  nop
  leave
  ret
```

assistant: ```c
void cbm_lock(int HandleDevice) {
  ;
  cbmarch_lock(HandleDevice);
  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
<cbm_lock>:
  endbr64
  sub    $0x8,%rsp
  call   <cbmarch_lock@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void cbm_lock(int HandleDevice) {
  ;
  cbmarch_lock(HandleDevice);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<cbm_lock>:
  endbr64
  jmp    1050 <cbmarch_lock@plt>
```

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

```asm
<cbm_lock>:
  endbr64
  jmp    1050 <cbmarch_lock@plt>
```

assistant: ```c
void cbm_lock(int HandleDevice) {
  ;
  cbmarch_lock(HandleDevice);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<compress_BC5>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %edx,-0x74(%rbp)
  mov    %ecx,-0x78(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x54(%rbp)
  jmp    L12
L25:
  movl   $0x0,-0x58(%rbp)
  jmp    L28
L30:
  lea    -0x50(%rbp),%r8
  mov    -0x78(%rbp),%edi
  mov    -0x74(%rbp),%ecx
  mov    -0x54(%rbp),%edx
  mov    -0x58(%rbp),%esi
  mov    -0x70(%rbp),%rax
  mov    %r8,%r9
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <extract_block@plt>
  lea    -0x50(%rbp),%rcx
  mov    -0x68(%rbp),%rax
  mov    $0xfffffffe,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <encode_alpha_block_DXT5@plt>
  mov    -0x68(%rbp),%rax
  lea    0x8(%rax),%rcx
  lea    -0x50(%rbp),%rax
  mov    $0xffffffff,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <encode_alpha_block_DXT5@plt>
  addq   $0x10,-0x68(%rbp)
  addl   $0x4,-0x58(%rbp)
L28:
  mov    -0x58(%rbp),%eax
  cmp    -0x74(%rbp),%eax
  jl     L30
  addl   $0x4,-0x54(%rbp)
L12:
  mov    -0x54(%rbp),%eax
  cmp    -0x78(%rbp),%eax
  jl     L25
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

assistant: ```c
void compress_BC5(unsigned char *dst, const unsigned char *src, int w, int h) {
  unsigned char block[64];
  int x, y;
  for (y = 0; y < h; y += 4) {
    for (x = 0; x < w; x += 4) {
      extract_block(src, x, y, w, h, block);
      encode_alpha_block_DXT5(dst, block, -2);
      encode_alpha_block_DXT5(dst + 8, block, -1);
      dst += 16;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<compress_BC5>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %rsi,0x8(%rsp)
  mov    %ecx,0x14(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  test   %ecx,%ecx
  jle    L2
  mov    %edx,%r14d
  mov    %ecx,%eax
  sub    $0x1,%eax
  and    $0xfffffffc,%eax
  add    $0x4,%eax
  mov    %eax,0x28(%rsp)
  lea    -0x1(%rdx),%edx
  mov    %edx,%eax
  shr    $0x2,%eax
  add    $0x1,%eax
  mov    %eax,%eax
  shl    $0x4,%rax
  mov    %rax,0x20(%rsp)
  mov    $0x0,%r13d
  and    $0xfffffffc,%edx
  mov    %edx,0x2c(%rsp)
  jmp    L12
L53:
  mov    0x2c(%rsp),%eax
  lea    0x4(%rax),%r15d
  mov    0x18(%rsp),%rbp
  mov    $0x0,%ebx
  lea    0x30(%rsp),%r12
L69:
  mov    %r12,%r9
  mov    0x14(%rsp),%r8d
  mov    %r14d,%ecx
  mov    %r13d,%edx
  mov    %ebx,%esi
  mov    0x8(%rsp),%rdi
  call   <extract_block@plt>
  mov    $0xfffffffe,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <encode_alpha_block_DXT5@plt>
  lea    0x8(%rbp),%rdi
  mov    $0xffffffff,%edx
  mov    %r12,%rsi
  call   <encode_alpha_block_DXT5@plt>
  add    $0x10,%rbp
  add    $0x4,%ebx
  cmp    %r15d,%ebx
  jne    L69
  mov    0x20(%rsp),%rcx
  add    %rcx,0x18(%rsp)
L1:
  add    $0x4,%r13d
  cmp    0x28(%rsp),%r13d
  je     L2
L12:
  test   %r14d,%r14d
  jg     L53
  jmp    L1
L2:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L46:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void compress_BC5(unsigned char *dst, const unsigned char *src, int w, int h) {
  unsigned char block[64];
  int x, y;
  for (y = 0; y < h; y += 4) {
    for (x = 0; x < w; x += 4) {
      extract_block(src, x, y, w, h, block);
      encode_alpha_block_DXT5(dst, block, -2);
      encode_alpha_block_DXT5(dst + 8, block, -1);
      dst += 16;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<compress_BC5>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %rsi,0x10(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  test   %ecx,%ecx
  jle    L34
  lea    -0x1(%rcx),%eax
  mov    %edx,%r13d
  lea    -0x1(%rdx),%edx
  xor    %r12d,%r12d
  and    $0xfffffffc,%eax
  mov    %ecx,%r15d
  add    $0x4,%eax
  mov    %eax,0x28(%rsp)
  mov    %edx,%eax
  and    $0xfffffffc,%edx
  shr    $0x2,%eax
  mov    %edx,0x2c(%rsp)
  add    $0x1,%eax
  shl    $0x4,%rax
  mov    %rax,0x20(%rsp)
  nopl   0x0(%rax,%rax,1)
L18:
  test   %r13d,%r13d
  jle    L20
  mov    0x2c(%rsp),%eax
  mov    0x18(%rsp),%rbx
  xor    %r14d,%r14d
  lea    0x30(%rsp),%rbp
  add    $0x4,%eax
  mov    %eax,0xc(%rsp)
  mov    %r15d,%eax
  mov    %r13d,%r15d
  mov    %eax,%r13d
  nopw   0x0(%rax,%rax,1)
L35:
  mov    0x10(%rsp),%rdi
  mov    %rbp,%r9
  mov    %r13d,%r8d
  mov    %r15d,%ecx
  mov    %r14d,%esi
  mov    %r12d,%edx
  add    $0x4,%r14d
  call   <extract_block@plt>
  mov    %rbx,%rdi
  mov    $0xfffffffe,%edx
  mov    %rbp,%rsi
  call   <encode_alpha_block_DXT5@plt>
  lea    0x8(%rbx),%rdi
  mov    $0xffffffff,%edx
  mov    %rbp,%rsi
  call   <encode_alpha_block_DXT5@plt>
  add    $0x10,%rbx
  cmp    0xc(%rsp),%r14d
  jne    L35
  mov    0x20(%rsp),%rcx
  add    %rcx,0x18(%rsp)
  mov    %r13d,%eax
  mov    %r15d,%r13d
  mov    %eax,%r15d
L20:
  add    $0x4,%r12d
  cmp    0x28(%rsp),%r12d
  jne    L18
L34:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void compress_BC5(unsigned char *dst, const unsigned char *src, int w, int h) {
  unsigned char block[64];
  int x, y;
  for (y = 0; y < h; y += 4) {
    for (x = 0; x < w; x += 4) {
      extract_block(src, x, y, w, h, block);
      encode_alpha_block_DXT5(dst, block, -2);
      encode_alpha_block_DXT5(dst + 8, block, -1);
      dst += 16;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<compress_BC5>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %rsi,(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  test   %ecx,%ecx
  jle    L11
  mov    %edx,%r13d
  test   %edx,%edx
  jle    L11
  lea    -0x1(%rcx),%eax
  lea    -0x1(%rdx),%edx
  xor    %r12d,%r12d
  mov    %ecx,%r15d
  and    $0xfffffffc,%eax
  lea    0x20(%rsp),%rbp
  mov    %r13d,%r14d
  add    $0x4,%eax
  mov    %eax,0x1c(%rsp)
  mov    %edx,%eax
  shr    $0x2,%eax
  add    $0x1,%eax
  shl    $0x4,%rax
  mov    %rax,0x10(%rsp)
  mov    %edx,%eax
  and    $0xfffffffc,%eax
  add    $0x4,%eax
  mov    %eax,0x18(%rsp)
  nopl   0x0(%rax)
L9:
  mov    %r15d,%eax
  mov    0x8(%rsp),%rbx
  mov    %r14d,%r15d
  xor    %r13d,%r13d
  mov    %eax,%r14d
  nopl   0x0(%rax)
L24:
  mov    (%rsp),%rdi
  mov    %rbp,%r9
  mov    %r14d,%r8d
  mov    %r15d,%ecx
  mov    %r13d,%esi
  mov    %r12d,%edx
  add    $0x4,%r13d
  call   <extract_block@plt>
  mov    %rbx,%rdi
  mov    $0xfffffffe,%edx
  mov    %rbp,%rsi
  call   <encode_alpha_block_DXT5@plt>
  lea    0x8(%rbx),%rdi
  mov    $0xffffffff,%edx
  mov    %rbp,%rsi
  call   <encode_alpha_block_DXT5@plt>
  add    $0x10,%rbx
  cmp    0x18(%rsp),%r13d
  jne    L24
  mov    %r14d,%eax
  mov    0x10(%rsp),%rcx
  mov    %r15d,%r14d
  add    %rcx,0x8(%rsp)
  mov    %eax,%r15d
  add    $0x4,%r12d
  cmp    0x1c(%rsp),%r12d
  jne    L9
L11:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void compress_BC5(unsigned char *dst, const unsigned char *src, int w, int h) {
  unsigned char block[64];
  int x, y;
  for (y = 0; y < h; y += 4) {
    for (x = 0; x < w; x += 4) {
      extract_block(src, x, y, w, h, block);
      encode_alpha_block_DXT5(dst, block, -2);
      encode_alpha_block_DXT5(dst + 8, block, -1);
      dst += 16;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strsave>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L4
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <allocate@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L4:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strsave(char *s) {
  char *ptr = NULL;
  if (s != NULL) {
    ptr = (char *)allocate(strlen(s) + 1);
    strcpy(ptr, s);
  }
  return (ptr);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strsave>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L0
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <allocate@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L0:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strsave(char *s) {
  char *ptr = NULL;
  if (s != NULL) {
    ptr = (char *)allocate(strlen(s) + 1);
    strcpy(ptr, s);
  }
  return (ptr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strsave>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L5
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <allocate@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
L5:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strsave(char *s) {
  char *ptr = NULL;
  if (s != NULL) {
    ptr = (char *)allocate(strlen(s) + 1);
    strcpy(ptr, s);
  }
  return (ptr);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strsave>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L0
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <allocate@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
L0:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strsave(char *s) {
  char *ptr = NULL;
  if (s != NULL) {
    ptr = (char *)allocate(strlen(s) + 1);
    strcpy(ptr, s);
  }
  return (ptr);
}
```