ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<read_memory>:
  endbr64
  push   %r12
  lea    D3(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    $0xa,%ebx
  nopw   0x0(%rax,%rax,1)
L9:
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    0x0(%rbp),%esi
  mov    %r12,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L9
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: tag: D3
type_guessed: string

tool: D3: Child: read value = %2d


assistant: ```c
void read_memory(void *shared_memory) {
  int i;
  int value;
  for (i = 0; i < 10; i++) {
    sleep(1);
    value = *((int *)shared_memory);
    printf("Child: read value = %2d\n", value);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_memory>:
  endbr64
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    $0xa,%ebx
  nopw   0x0(%rax,%rax,1)
L6:
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    0x0(%rbp),%esi
  mov    %r12,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_label: D4
type_infered: string

tool: D4: Child: read value = %2d


assistant: ```c
void read_memory(void *shared_memory) {
  int i;
  int value;
  for (i = 0; i < 10; i++) {
    sleep(1);
    value = *((int *)shared_memory);
    printf("Child: read value = %2d\n", value);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<test_003>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    $0xa,%edi
  call   <mem_alloc@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <mem_size@plt>
  cmp    $0xa,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <mem_size@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <mem_end@plt>
  mov    -0x10(%rbp),%rdx
  add    $0xa,%rdx
  cmp    %rdx,%rax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <mem_end@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <mem_free@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <mem_free@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  nop
  leave
  ret
```

assistant: ```c
void test_003(void) {
  char *str1 = NULL;
  char *str2 = NULL;
  str1 = (char *)mem_alloc(10);
  assert(str1 != NULL);
  str2 = (char *)malloc(10);
  assert(str2 != NULL);
  assert(mem_size(str1) == 10);
  assert(mem_size(str2) == 0);
  assert(mem_end(str1) == str1 + 10);
  assert(mem_end(str2) == NULL);
  str1 = mem_free(str1);
  assert(str1 == NULL);
  str2 = mem_free(str2);
  assert(str2 == NULL);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test_003>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <mem_alloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <mem_size@plt>
  cmp    $0xa,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <mem_size@plt>
  test   %eax,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <mem_end@plt>
  lea    0xa(%rbx),%rdx
  cmp    %rax,%rdx
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <mem_end@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <mem_free@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <mem_free@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void test_003(void) {
  char *str1 = NULL;
  char *str2 = NULL;
  str1 = (char *)mem_alloc(10);
  assert(str1 != NULL);
  str2 = (char *)malloc(10);
  assert(str2 != NULL);
  assert(mem_size(str1) == 10);
  assert(mem_size(str2) == 0);
  assert(mem_end(str1) == str1 + 10);
  assert(mem_end(str2) == NULL);
  str1 = mem_free(str1);
  assert(str1 == NULL);
  str2 = mem_free(str2);
  assert(str2 == NULL);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<test_003>:
  endbr64
  push   %r12
  mov    $0xa,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <mem_alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%rbp
  setne  %dil
  call   <assert@plt>
  mov    $0xa,%edi
  call   <malloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r12
  setne  %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <mem_size@plt>
  xor    %edi,%edi
  cmp    $0xa,%eax
  sete   %dil
  call   <assert@plt>
  mov    %r12,%rdi
  call   <mem_size@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <mem_end@plt>
  lea    0xa(%rbp),%rdx
  xor    %edi,%edi
  cmp    %rax,%rdx
  sete   %dil
  call   <assert@plt>
  mov    %r12,%rdi
  call   <mem_end@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <mem_free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %r12,%rdi
  call   <mem_free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    10b0 <assert@plt>
```

assistant: ```c
void test_003(void) {
  char *str1 = NULL;
  char *str2 = NULL;
  str1 = (char *)mem_alloc(10);
  assert(str1 != NULL);
  str2 = (char *)malloc(10);
  assert(str2 != NULL);
  assert(mem_size(str1) == 10);
  assert(mem_size(str2) == 0);
  assert(mem_end(str1) == str1 + 10);
  assert(mem_end(str2) == NULL);
  str1 = mem_free(str1);
  assert(str1 == NULL);
  str2 = mem_free(str2);
  assert(str2 == NULL);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test_003>:
  endbr64
  push   %r12
  mov    $0xa,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <mem_alloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%rbp
  setne  %dil
  call   <assert@plt>
  mov    $0xa,%edi
  call   <malloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%r12
  setne  %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <mem_size@plt>
  xor    %edi,%edi
  cmp    $0xa,%eax
  sete   %dil
  call   <assert@plt>
  mov    %r12,%rdi
  call   <mem_size@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <mem_end@plt>
  lea    0xa(%rbp),%rdx
  xor    %edi,%edi
  cmp    %rax,%rdx
  sete   %dil
  call   <assert@plt>
  mov    %r12,%rdi
  call   <mem_end@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <mem_free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %r12,%rdi
  call   <mem_free@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    10b0 <assert@plt>
```

assistant: ```c
void test_003(void) {
  char *str1 = NULL;
  char *str2 = NULL;
  str1 = (char *)mem_alloc(10);
  assert(str1 != NULL);
  str2 = (char *)malloc(10);
  assert(str2 != NULL);
  assert(mem_size(str1) == 10);
  assert(mem_size(str2) == 0);
  assert(mem_end(str1) == str1 + 10);
  assert(mem_end(str2) == NULL);
  str1 = mem_free(str1);
  assert(str1 == NULL);
  str2 = mem_free(str2);
  assert(str2 == NULL);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xe(%rbp),%rax
  mov    $0x6,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L3
L36:
  mov    -0x18(%rbp),%eax
  cltq
  movzbl -0xe(%rbp,%rax,1),%ecx
  mov    $0xffffffab,%edx
  mov    %edx,%eax
  mul    %cl
  shr    $0x8,%ax
  mov    %eax,%edx
  shr    $0x2,%dl
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  movzbl %dl,%eax
  cltq
  movzbl -0xe(%rbp,%rax,1),%edx
  mov    -0x18(%rbp),%eax
  cltq
  movzbl -0xe(%rbp,%rax,1),%eax
  xor    %eax,%edx
  mov    -0x18(%rbp),%eax
  cltq
  mov    %dl,-0xe(%rbp,%rax,1)
  addl   $0x1,-0x18(%rbp)
L3:
  cmpl   $0x5,-0x18(%rbp)
  jle    L36
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L9
L13:
  mov    -0x18(%rbp),%eax
  cltq
  movzbl -0xe(%rbp,%rax,1),%eax
  movzbl %al,%eax
  add    %eax,-0x14(%rbp)
  addl   $0x1,-0x18(%rbp)
L9:
  cmpl   $0x5,-0x18(%rbp)
  jle    L13
  cmpl   $0x123,-0x14(%rbp)
  jne    L19
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L19:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type: string

tool: D0: You won!

assistant: ```c
main() {
  int i, count;
  unsigned char buffer[6];
  read(0, buffer, 6);
  for (i = 0; i < 6; i++)
    buffer[i] = buffer[buffer[i] % 6] ^ buffer[i];
  count = 0;
  for (i = 0; i < 6; i++)
    count += buffer[i];
  if (count == 0x123)
    printf("You won!");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x2(%rsp),%rbx
  mov    $0x6,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    %rbx,%rsi
  lea    0x8(%rsp),%r8
  mov    %rbx,%rcx
  mov    $0xffffffab,%r9d
L28:
  movzbl (%rcx),%edx
  mov    %edx,%eax
  mul    %r9b
  shr    $0xa,%ax
  lea    (%rax,%rax,2),%eax
  add    %eax,%eax
  mov    %edx,%edi
  sub    %eax,%edi
  movzbl %dil,%eax
  xor    0x2(%rsp,%rax,1),%dl
  mov    %dl,(%rcx)
  add    $0x1,%rcx
  cmp    %r8,%rcx
  jne    L28
  mov    $0x0,%eax
L13:
  movzbl (%rsi),%edx
  add    %edx,%eax
  add    $0x1,%rsi
  cmp    %r8,%rsi
  jne    L13
  cmp    $0x123,%eax
  je     L44
L3:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L44:
  lea    D5(%rip),%rdi
  call   <printf@plt>
  jmp    L3
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D5
type: string

tool: D5: You won!

assistant: ```c
main() {
  int i, count;
  unsigned char buffer[6];
  read(0, buffer, 6);
  for (i = 0; i < 6; i++)
    buffer[i] = buffer[buffer[i] % 6] ^ buffer[i];
  count = 0;
  for (i = 0; i < 6; i++)
    count += buffer[i];
  if (count == 0x123)
    printf("You won!");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x6,%edx
  xor    %edi,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x2(%rsp),%rbx
  mov    %rbx,%rsi
  call   <read@plt>
  mov    %rbx,%rsi
  lea    0x8(%rsp),%r8
  mov    %rbx,%rcx
  mov    $0xffffffab,%r9d
  xchg   %ax,%ax
L46:
  movzbl (%rcx),%edx
  add    $0x1,%rcx
  mov    %edx,%eax
  mov    %edx,%edi
  mul    %r9b
  shr    $0xa,%ax
  lea    (%rax,%rax,2),%eax
  add    %eax,%eax
  sub    %eax,%edi
  movzbl %dil,%eax
  xor    0x2(%rsp,%rax,1),%dl
  mov    %dl,-0x1(%rcx)
  cmp    %r8,%rcx
  jne    L46
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L8:
  movzbl (%rsi),%edx
  add    $0x1,%rsi
  add    %edx,%eax
  cmp    %r8,%rsi
  jne    L8
  cmp    $0x123,%eax
  je     L45
L28:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L45:
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L28
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D3
data_type: string

tool: D3: You won!

assistant: ```c
main() {
  int i, count;
  unsigned char buffer[6];
  read(0, buffer, 6);
  for (i = 0; i < 6; i++)
    buffer[i] = buffer[buffer[i] % 6] ^ buffer[i];
  count = 0;
  for (i = 0; i < 6; i++)
    count += buffer[i];
  if (count == 0x123)
    printf("You won!");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  xor    %edi,%edi
  mov    $0x6,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x2(%rsp),%rsi
  call   <read@plt>
  movzbl 0x2(%rsp),%r9d
  mov    $0xffffffab,%edx
  movzbl 0x3(%rsp),%r10d
  movzbl 0x4(%rsp),%r8d
  movzbl 0x5(%rsp),%edi
  mov    %r9d,%eax
  mov    %r9d,%ecx
  movzbl 0x6(%rsp),%esi
  mul    %dl
  shr    $0xa,%ax
  lea    (%rax,%rax,2),%eax
  add    %eax,%eax
  sub    %eax,%ecx
  movzbl %cl,%eax
  mov    %r10d,%ecx
  xor    0x2(%rsp,%rax,1),%r9b
  mov    %r10d,%eax
  mul    %dl
  mov    %r9b,0x2(%rsp)
  movzbl %r9b,%r9d
  shr    $0xa,%ax
  lea    (%rax,%rax,2),%eax
  add    %eax,%eax
  sub    %eax,%ecx
  movzbl %cl,%eax
  mov    %r8d,%ecx
  xor    0x2(%rsp,%rax,1),%r10b
  mov    %r8d,%eax
  mul    %dl
  mov    %r10b,0x3(%rsp)
  movzbl %r10b,%r10d
  shr    $0xa,%ax
  lea    (%rax,%rax,2),%eax
  add    %eax,%eax
  sub    %eax,%ecx
  movzbl %cl,%eax
  mov    %edi,%ecx
  xor    0x2(%rsp,%rax,1),%r8b
  mov    %edi,%eax
  mul    %dl
  mov    %r8b,0x4(%rsp)
  movzbl %r8b,%r8d
  shr    $0xa,%ax
  lea    (%rax,%rax,2),%eax
  add    %eax,%eax
  sub    %eax,%ecx
  movzbl %cl,%eax
  mov    %esi,%ecx
  xor    0x2(%rsp,%rax,1),%dil
  mov    %esi,%eax
  mul    %dl
  mov    %dil,0x5(%rsp)
  movzbl %dil,%edi
  shr    $0xa,%ax
  lea    (%rax,%rax,2),%eax
  add    %eax,%eax
  sub    %eax,%ecx
  movzbl %cl,%eax
  movzbl 0x7(%rsp),%ecx
  xor    0x2(%rsp,%rax,1),%sil
  mov    %edx,%eax
  mul    %cl
  mov    %sil,0x6(%rsp)
  movzbl %sil,%esi
  mov    %eax,%edx
  mov    %ecx,%eax
  shr    $0xa,%dx
  lea    (%rdx,%rdx,2),%edx
  add    %edx,%edx
  sub    %edx,%eax
  add    %r10d,%r9d
  movzbl %al,%eax
  add    %r9d,%r8d
  xor    0x2(%rsp,%rax,1),%cl
  add    %r8d,%edi
  mov    %cl,0x7(%rsp)
  add    %edi,%esi
  movzbl %cl,%ecx
  add    %esi,%ecx
  cmp    $0x123,%ecx
  je     L25
L29:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L25:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L29
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D4
type: string

tool: D4: You won!

assistant: ```c
main() {
  int i, count;
  unsigned char buffer[6];
  read(0, buffer, 6);
  for (i = 0; i < 6; i++)
    buffer[i] = buffer[buffer[i] % 6] ^ buffer[i];
  count = 0;
  for (i = 0; i < 6; i++)
    count += buffer[i];
  if (count == 0x123)
    printf("You won!");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_bisect>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  sub    -0x18(%rbp),%eax
  cmp    $0x3,%eax
  jg     L53
  jmp    L69
L10:
  subl   $0x1,-0x24(%rbp)
L69:
  cmpl   $0x0,-0x24(%rbp)
  jle    L4
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jle    L10
L4:
  mov    -0x24(%rbp),%eax
  jmp    L14
L53:
  mov    -0x18(%rbp),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jge    L41
  mov    -0x4(%rbp),%ecx
  mov    -0x18(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <_bisect>
  jmp    L14
L41:
  mov    -0x24(%rbp),%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <_bisect>
L14:
  leave
  ret
```

assistant: ```c
int _bisect(int val, int *vec, int l, int r) {
  if ((r - l) <= 3) {
    while (((r - 1) >= 0) && (vec[r - 1] >= val)) {
      r--;
    }
    return r;
  } else {
    int i = (l + r) / 2;
    if (vec[i] > val) {
      return _bisect(val, vec, l, i);
    } else {
      return _bisect(val, vec, i, r);
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<_bisect>:
  endbr64
  mov    %ecx,%eax
  sub    %edx,%eax
  cmp    $0x3,%eax
  jle    L0
  sub    $0x8,%rsp
  lea    (%rcx,%rdx,1),%r8d
  mov    %r8d,%eax
  shr    $0x1f,%eax
  add    %r8d,%eax
  sar    %eax
  movslq %eax,%r8
  cmp    %edi,(%rsi,%r8,4)
  jle    L39
  mov    %eax,%ecx
  call   <_bisect>
L40:
  add    $0x8,%rsp
  ret
L0:
  test   %ecx,%ecx
  jle    L25
  movslq %ecx,%rax
  lea    (%rsi,%rax,4),%rdx
  mov    %ecx,%eax
L9:
  cmp    %edi,-0x4(%rdx)
  jl     L53
  sub    $0x4,%rdx
  sub    $0x1,%eax
  jne    L9
  ret
L53:
  ret
L39:
  mov    %eax,%edx
  call   <_bisect>
  jmp    L40
L25:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int _bisect(int val, int *vec, int l, int r) {
  if ((r - l) <= 3) {
    while (((r - 1) >= 0) && (vec[r - 1] >= val)) {
      r--;
    }
    return r;
  } else {
    int i = (l + r) / 2;
    if (vec[i] > val) {
      return _bisect(val, vec, l, i);
    } else {
      return _bisect(val, vec, i, r);
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_bisect>:
  endbr64
  mov    %edi,%r8d
  mov    %ecx,%eax
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L45:
  lea    (%rdx,%rax,1),%ecx
  mov    %ecx,%edi
  shr    $0x1f,%edi
  add    %ecx,%edi
  sar    %edi
  movslq %edi,%rcx
  cmp    %r8d,(%rsi,%rcx,4)
  cmovle %edi,%edx
  cmovg  %edi,%eax
L25:
  mov    %eax,%ecx
  sub    %edx,%ecx
  cmp    $0x3,%ecx
  jg     L45
  test   %eax,%eax
  jle    L55
  movslq %eax,%rdx
  lea    (%rsi,%rdx,4),%rdx
  jmp    L52
  nop
L1:
  sub    $0x4,%rdx
  sub    $0x1,%eax
  je     L55
L52:
  cmp    %r8d,-0x4(%rdx)
  jge    L1
  ret
L55:
  ret
```

assistant: ```c
int _bisect(int val, int *vec, int l, int r) {
  if ((r - l) <= 3) {
    while (((r - 1) >= 0) && (vec[r - 1] >= val)) {
      r--;
    }
    return r;
  } else {
    int i = (l + r) / 2;
    if (vec[i] > val) {
      return _bisect(val, vec, l, i);
    } else {
      return _bisect(val, vec, i, r);
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<_bisect>:
  endbr64
  mov    %edi,%r8d
  mov    %ecx,%eax
  jmp    L51
  nopl   0x0(%rax,%rax,1)
L54:
  lea    (%rdx,%rax,1),%ecx
  mov    %ecx,%edi
  shr    $0x1f,%edi
  add    %ecx,%edi
  sar    %edi
  movslq %edi,%rcx
  cmp    %r8d,(%rsi,%rcx,4)
  cmovle %edi,%edx
  cmovg  %edi,%eax
L51:
  mov    %eax,%ecx
  sub    %edx,%ecx
  cmp    $0x3,%ecx
  jg     L54
  test   %eax,%eax
  jle    L7
  movslq %eax,%rdx
  lea    (%rsi,%rdx,4),%rdx
  jmp    L37
  nop
L48:
  sub    $0x4,%rdx
  sub    $0x1,%eax
  je     L7
L37:
  cmp    %r8d,-0x4(%rdx)
  jge    L48
  ret
L7:
  ret
```

assistant: ```c
int _bisect(int val, int *vec, int l, int r) {
  if ((r - l) <= 3) {
    while (((r - 1) >= 0) && (vec[r - 1] >= val)) {
      r--;
    }
    return r;
  } else {
    int i = (l + r) / 2;
    if (vec[i] > val) {
      return _bisect(val, vec, l, i);
    } else {
      return _bisect(val, vec, i, r);
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<kbWaitForWriteReady>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x2000,-0x4(%rbp)
  jmp    L23
L6:
  subl   $0x1,-0x4(%rbp)
L23:
  cmpl   $0x0,-0x4(%rbp)
  jle    L14
  mov    $0x64,%edi
  call   <PM_inpb@plt>
  and    $0x2,%eax
  test   %eax,%eax
  jne    L6
L14:
  cmpl   $0x0,-0x4(%rbp)
  setg   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int kbWaitForWriteReady(void) {
  int timeout = 8192;
  while ((timeout > 0) && (PM_inpb(0x64) & 0x02))
    timeout--;
  return (timeout > 0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<kbWaitForWriteReady>:
  endbr64
  push   %rbx
  mov    $0x2000,%ebx
L3:
  mov    $0x64,%edi
  call   <PM_inpb@plt>
  test   $0x2,%al
  je     L4
  sub    $0x1,%ebx
  jne    L3
L4:
  test   %ebx,%ebx
  setg   %al
  movzbl %al,%eax
  pop    %rbx
  ret
```

assistant: ```c
int kbWaitForWriteReady(void) {
  int timeout = 8192;
  while ((timeout > 0) && (PM_inpb(0x64) & 0x02))
    timeout--;
  return (timeout > 0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<kbWaitForWriteReady>:
  endbr64
  push   %r12
  mov    $0x2000,%r12d
  jmp    L5
  xchg   %ax,%ax
L8:
  sub    $0x1,%r12d
  je     L21
L5:
  mov    $0x64,%edi
  call   <PM_inpb@plt>
  test   $0x2,%al
  jne    L8
  mov    $0x1,%r12d
L21:
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int kbWaitForWriteReady(void) {
  int timeout = 8192;
  while ((timeout > 0) && (PM_inpb(0x64) & 0x02))
    timeout--;
  return (timeout > 0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<kbWaitForWriteReady>:
  endbr64
  push   %r12
  mov    $0x2000,%r12d
  jmp    L27
  xchg   %ax,%ax
L22:
  sub    $0x1,%r12d
  je     L5
L27:
  mov    $0x64,%edi
  call   <PM_inpb@plt>
  test   $0x2,%al
  jne    L22
  mov    $0x1,%r12d
L5:
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int kbWaitForWriteReady(void) {
  int timeout = 8192;
  while ((timeout > 0) && (PM_inpb(0x64) & 0x02))
    timeout--;
  return (timeout > 0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L0:
  jmp    L0
```

assistant: ```c
int main(void) {
  for (;;) {
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
L6:
  jmp    L6
```

assistant: ```c
int main(void) {
  for (;;) {
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  nopl   0x0(%rax)
L5:
  jmp    L5
```

assistant: ```c
int main(void) {
  for (;;) {
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  nopl   0x0(%rax)
L3:
  jmp    L3
```

assistant: ```c
int main(void) {
  for (;;) {
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<my_putstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L19:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <my_putchar@plt>
  addl   $0x1,-0x4(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L19
  nop
  nop
  leave
  ret
```

assistant: ```c
void my_putstr(char *str) {
  int count;
  count = 0;
  while (*(str + count)) {
    my_putchar(*(str + count));
    count++;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<my_putstr>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L14
  push   %rbx
  lea    0x1(%rdi),%rbx
L9:
  movsbl %al,%edi
  call   <my_putchar@plt>
  add    $0x1,%rbx
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  jne    L9
  pop    %rbx
  ret
L14:
  ret
```

assistant: ```c
void my_putstr(char *str) {
  int count;
  count = 0;
  while (*(str + count)) {
    my_putchar(*(str + count));
    count++;
  }
}
```