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
<aligned_calloc>:
  endbr64
  sub    $0x8,%rsp
  add    $0x48,%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  test   %rax,%rax
  je     L9
  lea    0x48(%rax),%r8
  and    $0xffffffffffffffc0,%r8
  mov    %rax,-0x8(%r8)
  mov    %r8,%rax
  add    $0x8,%rsp
  ret
  xchg   %ax,%ax
L9:
  xor    %r8d,%r8d
  add    $0x8,%rsp
  mov    %r8,%rax
  ret
```

assistant: ```c
void *aligned_calloc(size_t size) {
  void *mem = calloc(size + 64 + sizeof(void *), 1);
  void **ptr = NULL;
  if (mem != NULL) {
    ptr = (void **)((uintptr_t)(mem + 64 + sizeof(void *)) & ~(uintptr_t)(64 - 1));
    ptr[-1] = mem;
  }
  return ptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<aligned_calloc>:
  endbr64
  sub    $0x8,%rsp
  add    $0x48,%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  test   %rax,%rax
  je     L0
  lea    0x48(%rax),%r8
  and    $0xffffffffffffffc0,%r8
  mov    %rax,-0x8(%r8)
  mov    %r8,%rax
  add    $0x8,%rsp
  ret
  xchg   %ax,%ax
L0:
  xor    %r8d,%r8d
  add    $0x8,%rsp
  mov    %r8,%rax
  ret
```

assistant: ```c
void *aligned_calloc(size_t size) {
  void *mem = calloc(size + 64 + sizeof(void *), 1);
  void **ptr = NULL;
  if (mem != NULL) {
    ptr = (void **)((uintptr_t)(mem + 64 + sizeof(void *)) & ~(uintptr_t)(64 - 1));
    ptr[-1] = mem;
  }
  return ptr;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0x1c(%rbp)
  movl   $0xffffffff,-0x18(%rbp)
  movq   $0x0,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0xffffffff,-0x14(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rdi
  call   <produce_y@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <produce_z@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <produce_v@plt>
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %eax,%edx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  mov    %edx,%edi
  call   <consumer@plt>
  mov    -0x1c(%rbp),%eax
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x18(%rbp),%eax
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x10(%rbp),%eax
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int main() {
  int y = -1, z = -1;
  int v[3] = {-1};
  produce_y(&y);
  produce_z(&z);
  produce_v(v);
  consumer(z + y, v);
#pragma omp taskwait
  assert(y == 1);
  assert(z == 2);
  assert(v[0] == 0);
  assert(v[1] == 1);
  assert(v[2] == 2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0xffffffff,0x4(%rsp)
  movl   $0xffffffff,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movl   $0xffffffff,0xc(%rsp)
  lea    0x4(%rsp),%rdi
  call   <produce_y@plt>
  lea    0x8(%rsp),%rdi
  call   <produce_z@plt>
  lea    0xc(%rsp),%rbx
  mov    %rbx,%rdi
  call   <produce_v@plt>
  mov    0x4(%rsp),%edi
  add    0x8(%rsp),%edi
  mov    %rbx,%rsi
  call   <consumer@plt>
  cmpl   $0x1,0x4(%rsp)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmpl   $0x2,0x8(%rsp)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmpl   $0x0,0xc(%rsp)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmpl   $0x1,0x10(%rsp)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmpl   $0x2,0x14(%rsp)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int y = -1, z = -1;
  int v[3] = {-1};
  produce_y(&y);
  produce_z(&z);
  produce_v(v);
  consumer(z + y, v);
#pragma omp taskwait
  assert(y == 1);
  assert(z == 2);
  assert(v[0] == 0);
  assert(v[1] == 1);
  assert(v[2] == 2);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  lea    0xc(%rsp),%rbp
  movl   $0xffffffff,0x4(%rsp)
  movl   $0xffffffff,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movl   $0xffffffff,0xc(%rsp)
  call   <produce_y@plt>
  lea    0x8(%rsp),%rdi
  call   <produce_z@plt>
  mov    %rbp,%rdi
  call   <produce_v@plt>
  mov    %rbp,%rsi
  mov    0x4(%rsp),%edi
  add    0x8(%rsp),%edi
  call   <consumer@plt>
  xor    %edi,%edi
  cmpl   $0x1,0x4(%rsp)
  sete   %dil
  call   <assert@plt>
  xor    %edi,%edi
  cmpl   $0x2,0x8(%rsp)
  sete   %dil
  call   <assert@plt>
  mov    0xc(%rsp),%eax
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  xor    %edi,%edi
  cmpl   $0x1,0x10(%rsp)
  sete   %dil
  call   <assert@plt>
  xor    %edi,%edi
  cmpl   $0x2,0x14(%rsp)
  sete   %dil
  call   <assert@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int y = -1, z = -1;
  int v[3] = {-1};
  produce_y(&y);
  produce_z(&z);
  produce_v(v);
  consumer(z + y, v);
#pragma omp taskwait
  assert(y == 1);
  assert(z == 2);
  assert(v[0] == 0);
  assert(v[1] == 1);
  assert(v[2] == 2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  lea    0xc(%rsp),%rbp
  movl   $0xffffffff,0x4(%rsp)
  movl   $0xffffffff,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movl   $0xffffffff,0xc(%rsp)
  call   <produce_y@plt>
  lea    0x8(%rsp),%rdi
  call   <produce_z@plt>
  mov    %rbp,%rdi
  call   <produce_v@plt>
  mov    %rbp,%rsi
  mov    0x4(%rsp),%edi
  add    0x8(%rsp),%edi
  call   <consumer@plt>
  xor    %edi,%edi
  cmpl   $0x1,0x4(%rsp)
  sete   %dil
  call   <assert@plt>
  xor    %edi,%edi
  cmpl   $0x2,0x8(%rsp)
  sete   %dil
  call   <assert@plt>
  mov    0xc(%rsp),%eax
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  xor    %edi,%edi
  cmpl   $0x1,0x10(%rsp)
  sete   %dil
  call   <assert@plt>
  xor    %edi,%edi
  cmpl   $0x2,0x14(%rsp)
  sete   %dil
  call   <assert@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int y = -1, z = -1;
  int v[3] = {-1};
  produce_y(&y);
  produce_z(&z);
  produce_v(v);
  consumer(z + y, v);
#pragma omp taskwait
  assert(y == 1);
  assert(z == 2);
  assert(v[0] == 0);
  assert(v[1] == 1);
  assert(v[2] == 2);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<getField>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D15(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L42
L22:
  subl   $0x1,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L17
  mov    -0x10(%rbp),%rax
  jmp    L41
L17:
  lea    D19(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x10(%rbp)
L42:
  cmpq   $0x0,-0x10(%rbp)
  je     L29
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L22
L29:
  mov    $0x0,%eax
L41:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D15
datatype: string
name: disassemble_data
arguments: data_label: D19
datatype: string

tool: D15: ;

tool: D19: ;


assistant: ```c
char *getField(char *line, int num) {
  char *tmp = strdup(line);
  char *tok;
  for (tok = strtok(tmp, ";"); tok && *tok; tok = strtok(NULL, ";\n")) {
    if (!--num)
      return tok;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<getField>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebx
  call   <strdup@plt>
  mov    %rax,%rdi
  lea    D5(%rip),%rsi
  call   <strtok@plt>
  lea    D15(%rip),%rbp
  test   %rax,%rax
  je     L34
L16:
  cmpb   $0x0,(%rax)
  je     L38
  sub    $0x1,%ebx
  je     L34
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  test   %rax,%rax
  jne    L16
  jmp    L34
L38:
  mov    $0x0,%eax
L34:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D5
datatype: string
name: retrieve_data
arguments: label: D15
datatype: string

tool: D5: ;

tool: D15: ;


assistant: ```c
char *getField(char *line, int num) {
  char *tmp = strdup(line);
  char *tok;
  for (tok = strtok(tmp, ";"); tok && *tok; tok = strtok(NULL, ";\n")) {
    if (!--num)
      return tok;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getField>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <strdup@plt>
  lea    D19(%rip),%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  jne    L24
  jmp    L41
  xchg   %ax,%ax
L32:
  sub    $0x1,%ebx
  je     L20
  xor    %edi,%edi
  mov    %rbp,%rsi
  call   <strtok@plt>
  test   %rax,%rax
  je     L41
L24:
  cmpb   $0x0,(%rax)
  jne    L32
L41:
  xor    %eax,%eax
L20:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D2
data_type: string
name: retrieve_data
arguments: data_label: D19
data_type: string

tool: D2: ;


tool: D19: ;

assistant: ```c
char *getField(char *line, int num) {
  char *tmp = strdup(line);
  char *tok;
  for (tok = strtok(tmp, ";"); tok && *tok; tok = strtok(NULL, ";\n")) {
    if (!--num)
      return tok;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getField>:
  endbr64
  push   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <strdup@plt>
  lea    D15(%rip),%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  jne    L27
  jmp    L36
  xchg   %ax,%ax
L25:
  sub    $0x1,%ebx
  je     L20
  xor    %edi,%edi
  mov    %rbp,%rsi
  call   <strtok@plt>
  test   %rax,%rax
  je     L36
L27:
  cmpb   $0x0,(%rax)
  jne    L25
L36:
  xor    %eax,%eax
L20:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type: string
name: disassemble_data
arguments: data_label: D15
type: string

tool: D0: ;


tool: D15: ;

assistant: ```c
char *getField(char *line, int num) {
  char *tmp = strdup(line);
  char *tok;
  for (tok = strtok(tmp, ";"); tok && *tok; tok = strtok(NULL, ";\n")) {
    if (!--num)
      return tok;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<img_basename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  jmp    L38
L26:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L2
  subl   $0x1,-0x18(%rbp)
L38:
  cmpl   $0x0,-0x18(%rbp)
  jns    L26
  jmp    L8
L2:
  nop
L8:
  addl   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    %eax,%eax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *img_basename(filename)
char *filename;
{
  char *new, *part;
  int len, dex;
  len = strlen(filename);
  dex = len - 1;
  while (dex > -1) {
    if (filename[dex] == '/') {
      break;
    } else {
      dex--;
    }
  }
  dex++;
  part = &(filename[dex]);
  len = strlen(part);
  new = (char *)malloc((unsigned)((len + 1) * sizeof(char)));
  strcpy(new, part);
  return (new);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<img_basename>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <strlen@plt>
  sub    $0x1,%eax
  js     L27
  movslq %eax,%rdx
  add    %rbx,%rdx
L19:
  cmpb   $0x2f,(%rdx)
  je     L27
  sub    $0x1,%eax
  sub    $0x1,%rdx
  cmp    $0xffffffff,%eax
  jne    L19
L27:
  add    $0x1,%eax
  cltq
  add    %rax,%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *img_basename(filename)
char *filename;
{
  char *new, *part;
  int len, dex;
  len = strlen(filename);
  dex = len - 1;
  while (dex > -1) {
    if (filename[dex] == '/') {
      break;
    } else {
      dex--;
    }
  }
  dex++;
  part = &(filename[dex]);
  len = strlen(part);
  new = (char *)malloc((unsigned)((len + 1) * sizeof(char)));
  strcpy(new, part);
  return (new);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<img_basename>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%edx
  sub    $0x1,%edx
  js     L0
  movslq %edx,%rax
  add    %rbx,%rax
  jmp    L16
  nopl   0x0(%rax)
L9:
  sub    $0x1,%edx
  sub    $0x1,%rax
  cmp    $0xffffffff,%edx
  je     L18
L16:
  cmpb   $0x2f,(%rax)
  mov    %rax,%rbp
  jne    L9
  add    $0x1,%edx
  movslq %edx,%rdx
  lea    (%rbx,%rdx,1),%rbp
L18:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  mov    %rax,%rbx
  call   <malloc@plt>
  lea    0x1(%rbx),%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  cltq
  lea    (%rbx,%rax,1),%rbp
  jmp    L18
```

assistant: ```c
char *img_basename(filename)
char *filename;
{
  char *new, *part;
  int len, dex;
  len = strlen(filename);
  dex = len - 1;
  while (dex > -1) {
    if (filename[dex] == '/') {
      break;
    } else {
      dex--;
    }
  }
  dex++;
  part = &(filename[dex]);
  len = strlen(part);
  new = (char *)malloc((unsigned)((len + 1) * sizeof(char)));
  strcpy(new, part);
  return (new);
}
```
user: What is the c source code of the assembly code below:

```asm
<img_basename>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%edx
  sub    $0x1,%edx
  js     L30
  movslq %edx,%rax
  add    %rbx,%rax
  jmp    L16
  nopl   0x0(%rax)
L21:
  sub    $0x1,%edx
  sub    $0x1,%rax
  cmp    $0xffffffff,%edx
  je     L29
L16:
  cmpb   $0x2f,(%rax)
  mov    %rax,%rbp
  jne    L21
  add    $0x1,%edx
  movslq %edx,%rdx
  lea    (%rbx,%rdx,1),%rbp
L29:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  mov    %rax,%rbx
  call   <malloc@plt>
  lea    0x1(%rbx),%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L30:
  cltq
  lea    (%rbx,%rax,1),%rbp
  jmp    L29
```

assistant: ```c
char *img_basename(filename)
char *filename;
{
  char *new, *part;
  int len, dex;
  len = strlen(filename);
  dex = len - 1;
  while (dex > -1) {
    if (filename[dex] == '/') {
      break;
    } else {
      dex--;
    }
  }
  dex++;
  part = &(filename[dex]);
  len = strlen(part);
  new = (char *)malloc((unsigned)((len + 1) * sizeof(char)));
  strcpy(new, part);
  return (new);
}
```
user: Translate the below assembly code into C programming language:

```asm
<utf32_is_valid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0xd7ff,-0x4(%rbp)
  jle    L7
  cmpl   $0xdfff,-0x4(%rbp)
  jg     L7
  mov    $0x0,%eax
  jmp    L13
L7:
  cmpl   $0x10ffff,-0x4(%rbp)
  setle  %al
  movzbl %al,%eax
L13:
  pop    %rbp
  ret
```

assistant: ```c
bool utf32_is_valid(uint32_t cp) {
  if ((cp >= 0xd800) && (cp <= 0xdfff))
    return false;
  return cp <= 0x10ffff;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<utf32_is_valid>:
  endbr64
  lea    -0xd800(%rdi),%edx
  mov    $0x0,%eax
  cmp    $0x7ff,%edx
  jbe    L8
  cmp    $0x10ffff,%edi
  setle  %al
  movzbl %al,%eax
L8:
  ret
```

assistant: ```c
bool utf32_is_valid(uint32_t cp) {
  if ((cp >= 0xd800) && (cp <= 0xdfff))
    return false;
  return cp <= 0x10ffff;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<utf32_is_valid>:
  endbr64
  lea    -0xd800(%rdi),%edx
  xor    %eax,%eax
  cmp    $0x7ff,%edx
  jbe    L4
  xor    %eax,%eax
  cmp    $0x10ffff,%edi
  setle  %al
L4:
  ret
```

assistant: ```c
bool utf32_is_valid(uint32_t cp) {
  if ((cp >= 0xd800) && (cp <= 0xdfff))
    return false;
  return cp <= 0x10ffff;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<utf32_is_valid>:
  endbr64
  lea    -0xd800(%rdi),%edx
  xor    %eax,%eax
  cmp    $0x7ff,%edx
  jbe    L8
  xor    %eax,%eax
  cmp    $0x10ffff,%edi
  setle  %al
L8:
  ret
```

assistant: ```c
bool utf32_is_valid(uint32_t cp) {
  if ((cp >= 0xd800) && (cp <= 0xdfff))
    return false;
  return cp <= 0x10ffff;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<iio_create_xml_context_mem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%edx
  mov    D8(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
struct iio_context *iio_create_xml_context_mem(const char *xml, size_t len) {
  errno = ENOSYS;
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<iio_create_xml_context_mem>:
  endbr64
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    D12(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  ret
```

assistant: ```c
struct iio_context *iio_create_xml_context_mem(const char *xml, size_t len) {
  errno = ENOSYS;
  return NULL;
}
```
user: What is the c source code of the assembly code below:

```asm
<iio_create_xml_context_mem>:
  endbr64
  mov    D15(%rip),%rax
  mov    (%rax),%edx
  mov    D7(%rip),%rax
  mov    %edx,(%rax)
  xor    %eax,%eax
  ret
```

assistant: ```c
struct iio_context *iio_create_xml_context_mem(const char *xml, size_t len) {
  errno = ENOSYS;
  return NULL;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<iio_create_xml_context_mem>:
  endbr64
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    D17(%rip),%rax
  mov    %edx,(%rax)
  xor    %eax,%eax
  ret
```

assistant: ```c
struct iio_context *iio_create_xml_context_mem(const char *xml, size_t len) {
  errno = ENOSYS;
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<USBDMAChannelCountSet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  cmpl   $0x40050000,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  cmpl   $0x7,-0x8(%rbp)
  setle  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x4(%rbp),%eax
  lea    0x20c(%rax),%edx
  mov    -0x8(%rbp),%eax
  shl    $0x4,%eax
  add    %edx,%eax
  cltq
  mov    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void USBDMAChannelCountSet(uint32_t ui32Base, uint32_t ui32Channel, uint32_t ui32Count) {
  ASSERT(ui32Base == 0x40050000);
  ASSERT(ui32Channel < 8);
  (*((volatile uint32_t *)(ui32Base + 0x0000020C + (0x10 * ui32Channel)))) = ui32Count;
}
```
user: Translate the below assembly code into C programming language:

```asm
<USBDMAChannelCountSet>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %edx,%r12d
  cmp    $0x40050000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  cmp    $0x7,%ebx
  setle  %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  shl    $0x4,%ebx
  lea    0x20c(%rbp,%rbx,1),%eax
  cltq
  mov    %r12d,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void USBDMAChannelCountSet(uint32_t ui32Base, uint32_t ui32Channel, uint32_t ui32Count) {
  ASSERT(ui32Base == 0x40050000);
  ASSERT(ui32Channel < 8);
  (*((volatile uint32_t *)(ui32Base + 0x0000020C + (0x10 * ui32Channel)))) = ui32Count;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<USBDMAChannelCountSet>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  push   %rbx
  cmp    $0x40050000,%ebp
  mov    %esi,%ebx
  sete   %dil
  call   <ASSERT@plt>
  xor    %edi,%edi
  cmp    $0x7,%ebx
  setle  %dil
  shl    $0x4,%ebx
  call   <ASSERT@plt>
  lea    0x20c(%rbp,%rbx,1),%eax
  cltq
  mov    %r12d,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void USBDMAChannelCountSet(uint32_t ui32Base, uint32_t ui32Channel, uint32_t ui32Count) {
  ASSERT(ui32Base == 0x40050000);
  ASSERT(ui32Channel < 8);
  (*((volatile uint32_t *)(ui32Base + 0x0000020C + (0x10 * ui32Channel)))) = ui32Count;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<USBDMAChannelCountSet>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  push   %rbx
  cmp    $0x40050000,%ebp
  mov    %esi,%ebx
  sete   %dil
  call   <ASSERT@plt>
  xor    %edi,%edi
  cmp    $0x7,%ebx
  setle  %dil
  shl    $0x4,%ebx
  call   <ASSERT@plt>
  lea    0x20c(%rbp,%rbx,1),%eax
  cltq
  mov    %r12d,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void USBDMAChannelCountSet(uint32_t ui32Base, uint32_t ui32Channel, uint32_t ui32Count) {
  ASSERT(ui32Base == 0x40050000);
  ASSERT(ui32Channel < 8);
  (*((volatile uint32_t *)(ui32Base + 0x0000020C + (0x10 * ui32Channel)))) = ui32Count;
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L36
  cmpl   $0x9,-0x4(%rbp)
  ja     L36
L37:
  subl   $0x1,-0x4(%rbp)
  shll   -0x8(%rbp)
  cmpl   $0x8,-0x4(%rbp)
  jbe    L29
  mov    $0x0,%eax
  call   <link_error@plt>
L29:
  cmpl   $0x0,-0x4(%rbp)
  jne    L37
L36:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
__attribute__((no, noclone)) int foo(unsigned int n, int r) {
  int i;
  if (n > 0) {
    asm("");
    if (n < 10) {
      asm("");
      do {
        --n;
        r *= 2;
        if (n >= 9)
          link_error();
      } while (n > 0);
    }
  }
  return r + n;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<foo>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  test   %edi,%edi
  je     L10
  cmp    $0x9,%edi
  ja     L10
  jmp    L11
L19:
  mov    $0x0,%eax
  call   <link_error@plt>
L11:
  sub    $0x1,%ebx
  add    %ebp,%ebp
  cmp    $0x8,%ebx
  ja     L19
  test   %ebx,%ebx
  jne    L11
L10:
  lea    0x0(%rbp,%rbx,1),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
__attribute__((no, noclone)) int foo(unsigned int n, int r) {
  int i;
  if (n > 0) {
    asm("");
    if (n < 10) {
      asm("");
      do {
        --n;
        r *= 2;
        if (n >= 9)
          link_error();
      } while (n > 0);
    }
  }
  return r + n;
}
```