ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<autohelperowl_attackpat185>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebp
  mov    %edx,%ebx
  mov    %ecx,%r14d
  mov    %esi,%ecx
  mov    $0x0,%esi
  mov    $0xffffffff,%edi
  call   <offset@plt>
  mov    %eax,%r12d
  mov    %ebp,%ecx
  mov    %ebx,%edx
  mov    $0xffffffff,%esi
  mov    $0x0,%edi
  call   <offset@plt>
  mov    %eax,%r13d
  mov    %ebp,%ecx
  mov    %ebx,%edx
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <offset@plt>
  mov    $0x3,%edi
  sub    %r14d,%edi
  sub    $0x8,%rsp
  push   %rax
  mov    %r12d,%r9d
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    $0x2,%esi
  call   <play_break_through_n@plt>
  add    $0x10,%rsp
  cmp    $0x3,%eax
  sete   %al
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int autohelperowl_attackpat185(struct pattern *patt, int transformation, int move, int color, int action) {
  int a, b, c;
  (void)patt;
  (void)color;
  (void)action;
  a = offset(-1, 0, move, transformation);
  b = offset(0, -1, move, transformation);
  c = offset(0, 1, move, transformation);
  return play_break_through_n((1 + 2 - (color)), 2, move, a, b, a, c) == 3;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<autohelperowl_attackpat185>:
  endbr64
  push   %r14
  mov    $0xffffffff,%edi
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %ecx,%ebp
  mov    %esi,%ecx
  xor    %esi,%esi
  push   %rbx
  call   <offset@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  xor    %edi,%edi
  mov    $0xffffffff,%esi
  mov    %eax,%r14d
  call   <offset@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  xor    %edi,%edi
  mov    $0x1,%esi
  mov    %eax,%ebx
  call   <offset@plt>
  sub    $0x8,%rsp
  mov    %r14d,%ecx
  mov    %r12d,%edx
  push   %rax
  mov    $0x3,%edi
  mov    %r14d,%r9d
  mov    %ebx,%r8d
  sub    %ebp,%edi
  mov    $0x2,%esi
  call   <play_break_through_n@plt>
  pop    %rdx
  pop    %rcx
  cmp    $0x3,%eax
  pop    %rbx
  pop    %rbp
  sete   %al
  pop    %r12
  pop    %r13
  movzbl %al,%eax
  pop    %r14
  ret
```

assistant: ```c
int autohelperowl_attackpat185(struct pattern *patt, int transformation, int move, int color, int action) {
  int a, b, c;
  (void)patt;
  (void)color;
  (void)action;
  a = offset(-1, 0, move, transformation);
  b = offset(0, -1, move, transformation);
  c = offset(0, 1, move, transformation);
  return play_break_through_n((1 + 2 - (color)), 2, move, a, b, a, c) == 3;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<autohelperowl_attackpat185>:
  endbr64
  push   %r14
  mov    $0xffffffff,%edi
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %ecx,%ebp
  mov    %esi,%ecx
  xor    %esi,%esi
  push   %rbx
  call   <offset@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  xor    %edi,%edi
  mov    $0xffffffff,%esi
  mov    %eax,%r14d
  call   <offset@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  xor    %edi,%edi
  mov    $0x1,%esi
  mov    %eax,%ebx
  call   <offset@plt>
  sub    $0x8,%rsp
  mov    %r14d,%ecx
  mov    %r12d,%edx
  push   %rax
  mov    $0x3,%edi
  mov    %r14d,%r9d
  mov    %ebx,%r8d
  sub    %ebp,%edi
  mov    $0x2,%esi
  call   <play_break_through_n@plt>
  pop    %rdx
  pop    %rcx
  cmp    $0x3,%eax
  pop    %rbx
  pop    %rbp
  sete   %al
  pop    %r12
  pop    %r13
  movzbl %al,%eax
  pop    %r14
  ret
```

assistant: ```c
int autohelperowl_attackpat185(struct pattern *patt, int transformation, int move, int color, int action) {
  int a, b, c;
  (void)patt;
  (void)color;
  (void)action;
  a = offset(-1, 0, move, transformation);
  b = offset(0, -1, move, transformation);
  c = offset(0, 1, move, transformation);
  return play_break_through_n((1 + 2 - (color)), 2, move, a, b, a, c) == 3;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<process>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <str_accumulate@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <str_accumulate@plt>
  mov    %rax,-0x8(%rbp)
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <str_accumulate@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <pthread_self@plt>
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    %ecx,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
type_infered: string
name: retrieve_data
arguments: label: D10
type_infered: string
name: retrieve_data
arguments: label: D20
type_infered: string

tool: D3: Result of 

tool: D10:  thread

tool: D20: Thread %lx: "%s"


assistant: ```c
void *process(void *arg) {
  char *res;
  res = str_accumulate("Result of ");
  res = str_accumulate((char *)arg);
  res = str_accumulate(" thread");
  printf("Thread %lx: \"%s\"\n", pthread_self(), res);
  return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<process>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D14(%rip),%rdi
  call   <str_accumulate@plt>
  mov    %rbx,%rdi
  call   <str_accumulate@plt>
  lea    D8(%rip),%rdi
  call   <str_accumulate@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <pthread_self@plt>
  mov    %eax,%esi
  mov    %rbx,%rdx
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: tag: D14
type_guessed: string
name: read_data
arguments: tag: D8
type_guessed: string
name: read_data
arguments: tag: D19
type_guessed: string

tool: D14: Result of 

tool: D8:  thread

tool: D19: Thread %lx: "%s"


assistant: ```c
void *process(void *arg) {
  char *res;
  res = str_accumulate("Result of ");
  res = str_accumulate((char *)arg);
  res = str_accumulate(" thread");
  printf("Thread %lx: \"%s\"\n", pthread_self(), res);
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<process>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  lea    D7(%rip),%rdi
  call   <str_accumulate@plt>
  mov    %r12,%rdi
  call   <str_accumulate@plt>
  lea    D4(%rip),%rdi
  call   <str_accumulate@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <pthread_self@plt>
  mov    %r12,%rdx
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
type_infered: string
name: disassemble_data
arguments: data_label: D4
type_infered: string
name: disassemble_data
arguments: data_label: D0
type_infered: string

tool: D7: Result of 

tool: D4:  thread

tool: D0: Thread %lx: "%s"


assistant: ```c
void *process(void *arg) {
  char *res;
  res = str_accumulate("Result of ");
  res = str_accumulate((char *)arg);
  res = str_accumulate(" thread");
  printf("Thread %lx: \"%s\"\n", pthread_self(), res);
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<process>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  lea    D26(%rip),%rdi
  call   <str_accumulate@plt>
  mov    %r12,%rdi
  call   <str_accumulate@plt>
  lea    D4(%rip),%rdi
  call   <str_accumulate@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <pthread_self@plt>
  mov    %r12,%rdx
  lea    D11(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D26
data_type: string
name: unpack_data
arguments: label: D4
data_type: string
name: unpack_data
arguments: label: D11
data_type: string

tool: D26: Result of 

tool: D4:  thread

tool: D11: Thread %lx: "%s"


assistant: ```c
void *process(void *arg) {
  char *res;
  res = str_accumulate("Result of ");
  res = str_accumulate((char *)arg);
  res = str_accumulate(" thread");
  printf("Thread %lx: \"%s\"\n", pthread_self(), res);
  return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<read_mem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x48(%rbp),%eax
  mov    %eax,-0x20(%rbp)
L46:
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  mov    %edx,%ecx
  or     %eax,%ecx
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  or     %eax,%edx
  mov    -0x44(%rbp),%eax
  shl    $0x14,%eax
  mov    $0x0,%r9d
  mov    $0xffffffff,%r8d
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <mmap@plt>
  mov    %rax,-0x30(%rbp)
  mov    D32(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x30(%rbp)
  je     L46
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L13
L44:
  lea    -0x20(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0x10,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L49
  mov    $0x1,%eax
  jmp    L12
L49:
  addl   $0x4,-0x34(%rbp)
L13:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x44(%rbp),%edx
  shl    $0x14,%edx
  movslq %edx,%rdx
  shr    $0x4,%rdx
  cmp    %rdx,%rax
  jb     L44
  mov    $0x0,%eax
L12:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L40
  call   <__stack_chk_fail@plt>
L40:
  leave
  ret
```

assistant: ```c
int read_mem(int mb_count, int value) {
  int i;
  char *ptr = NULL;
  int *cur = NULL;
  int buf[4] = {value};
  do {
    ptr = mmap(NULL, mb_count * (1024 * 1024) * sizeof(char), PROT_READ | PROT_WRITE, MAP_ANONYMOUS | MAP_PRIVATE, -1, 0);
  } while (ptr == MAP_FAILED);
  cur = (int *)ptr;
  for (i = 0; i < (mb_count * (1024 * 1024) * sizeof(char)) / (4 * sizeof(int)); i += 4) {
    if (memcmp(cur, buf, 4 * sizeof(int)) == 0) {
      return 1;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_mem>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  mov    %esi,(%rsp)
  shl    $0x14,%edi
  mov    %edi,%ebp
  mov    D40(%rip),%r12
L69:
  mov    (%r12),%ecx
  mov    D31(%rip),%rax
  or     (%rax),%ecx
  mov    D0(%rip),%rax
  mov    (%rax),%edx
  mov    D36(%rip),%rax
  or     (%rax),%edx
  mov    $0x0,%r9d
  mov    $0xffffffff,%r8d
  mov    %ebp,%esi
  mov    $0x0,%edi
  call   <mmap@plt>
  mov    %rax,%rbx
  mov    D29(%rip),%rax
  cmp    %rbx,(%rax)
  je     L69
  movslq %ebp,%rbp
  mov    %rbp,%r13
  shr    $0x4,%r13
  cmp    $0xf,%rbp
  jbe    L68
  mov    $0x0,%ebp
  mov    %rsp,%r12
L43:
  mov    $0x10,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  je     L17
  add    $0x4,%rbp
  cmp    %r13,%rbp
  jb     L43
  mov    $0x0,%eax
  jmp    L53
L68:
  mov    $0x0,%eax
  jmp    L53
L17:
  mov    $0x1,%eax
L53:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L47
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L47:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_mem(int mb_count, int value) {
  int i;
  char *ptr = NULL;
  int *cur = NULL;
  int buf[4] = {value};
  do {
    ptr = mmap(NULL, mb_count * (1024 * 1024) * sizeof(char), PROT_READ | PROT_WRITE, MAP_ANONYMOUS | MAP_PRIVATE, -1, 0);
  } while (ptr == MAP_FAILED);
  cur = (int *)ptr;
  for (i = 0; i < (mb_count * (1024 * 1024) * sizeof(char)) / (4 * sizeof(int)); i += 4) {
    if (memcmp(cur, buf, 4 * sizeof(int)) == 0) {
      return 1;
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<read_mem>:
  endbr64
  push   %r15
  pxor   %xmm0,%xmm0
  shl    $0x14,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movslq %edi,%rbx
  sub    $0x28,%rsp
  mov    D4(%rip),%r15
  mov    D17(%rip),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  mov    D30(%rip),%r12
  mov    D44(%rip),%rbp
  mov    %esi,(%rsp)
  nop
L44:
  mov    D22(%rip),%rax
  mov    (%r15),%ecx
  xor    %r9d,%r9d
  mov    %ebx,%esi
  mov    0x0(%r13),%edx
  mov    $0xffffffff,%r8d
  or     (%r12),%edx
  xor    %edi,%edi
  or     (%rax),%ecx
  call   <mmap@plt>
  mov    %rax,%r14
  cmp    %rax,0x0(%rbp)
  je     L44
  shr    $0x4,%rbx
  je     L20
  xor    %ebp,%ebp
  mov    %rsp,%r12
  jmp    L26
  nopl   0x0(%rax)
L9:
  add    $0x4,%rbp
  cmp    %rbx,%rbp
  jae    L20
L26:
  mov    $0x10,%edx
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L9
  mov    $0x1,%eax
  jmp    L60
  nopl   (%rax)
L20:
  xor    %eax,%eax
L60:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L54
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L54:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_mem(int mb_count, int value) {
  int i;
  char *ptr = NULL;
  int *cur = NULL;
  int buf[4] = {value};
  do {
    ptr = mmap(NULL, mb_count * (1024 * 1024) * sizeof(char), PROT_READ | PROT_WRITE, MAP_ANONYMOUS | MAP_PRIVATE, -1, 0);
  } while (ptr == MAP_FAILED);
  cur = (int *)ptr;
  for (i = 0; i < (mb_count * (1024 * 1024) * sizeof(char)) / (4 * sizeof(int)); i += 4) {
    if (memcmp(cur, buf, 4 * sizeof(int)) == 0) {
      return 1;
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<read_mem>:
  endbr64
  push   %r15
  pxor   %xmm0,%xmm0
  shl    $0x14,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movslq %edi,%rbx
  sub    $0x28,%rsp
  mov    D2(%rip),%r15
  mov    D8(%rip),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  mov    D48(%rip),%r12
  mov    D47(%rip),%rbp
  mov    %esi,(%rsp)
  nop
L0:
  mov    D6(%rip),%rax
  mov    (%r15),%ecx
  xor    %r9d,%r9d
  mov    %ebx,%esi
  mov    0x0(%r13),%edx
  mov    $0xffffffff,%r8d
  or     (%r12),%edx
  xor    %edi,%edi
  or     (%rax),%ecx
  call   <mmap@plt>
  mov    %rax,%r14
  cmp    %rax,0x0(%rbp)
  je     L0
  shr    $0x4,%rbx
  je     L24
  xor    %ebp,%ebp
  mov    %rsp,%r12
  jmp    L50
  nopl   0x0(%rax)
L60:
  add    $0x4,%rbp
  cmp    %rbx,%rbp
  jae    L24
L50:
  mov    $0x10,%edx
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L60
  mov    $0x1,%eax
  jmp    L25
  nopl   (%rax)
L24:
  xor    %eax,%eax
L25:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_mem(int mb_count, int value) {
  int i;
  char *ptr = NULL;
  int *cur = NULL;
  int buf[4] = {value};
  do {
    ptr = mmap(NULL, mb_count * (1024 * 1024) * sizeof(char), PROT_READ | PROT_WRITE, MAP_ANONYMOUS | MAP_PRIVATE, -1, 0);
  } while (ptr == MAP_FAILED);
  cur = (int *)ptr;
  for (i = 0; i < (mb_count * (1024 * 1024) * sizeof(char)) / (4 * sizeof(int)); i += 4) {
    if (memcmp(cur, buf, 4 * sizeof(int)) == 0) {
      return 1;
    }
  }
  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
<data_read_miss>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x87654320,%eax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

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

tool: D4: %i


assistant: ```c
void data_read_miss(void) {
  int *i = (void *)0x87654320;
  printf("%i\n", *i);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<data_read_miss>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x87654320,%eax
  mov    (%rax),%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
datatype: string

tool: D4: %i


assistant: ```c
void data_read_miss(void) {
  int *i = (void *)0x87654320;
  printf("%i\n", *i);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<data_read_miss>:
  endbr64
  mov    $0x87654320,%eax
  mov    (%rax),%esi
  lea    D2(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type_infered: string

tool: D2: %i


assistant: ```c
void data_read_miss(void) {
  int *i = (void *)0x87654320;
  printf("%i\n", *i);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<data_read_miss>:
  endbr64
  mov    $0x87654320,%eax
  mov    (%rax),%esi
  lea    D9(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D9
type_infered: string

tool: D9: %i


assistant: ```c
void data_read_miss(void) {
  int *i = (void *)0x87654320;
  printf("%i\n", *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
<isqrt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x4000,-0x4(%rbp)
  jmp    L50
L23:
  sarl   $0x2,-0x4(%rbp)
L50:
  mov    -0x4(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L23
  jmp    L35
L55:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L18
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  sub    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  add    %eax,-0x8(%rbp)
L18:
  sarl   -0x8(%rbp)
  sarl   $0x2,-0x4(%rbp)
L35:
  cmpl   $0x0,-0x4(%rbp)
  jne    L55
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L44
  addl   $0x1,-0x8(%rbp)
L44:
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint16_t isqrt(uint16_t a_nInput) {
  uint16_t op = a_nInput;
  uint16_t res = 0;
  uint16_t one = 1uL << 14;
  while (one > op) {
    one >>= 2;
  }
  while (one != 0) {
    if (op >= res + one) {
      op = op - (res + one);
      res = res + 2 * one;
    }
    res >>= 1;
    one >>= 2;
  }
  if (op > res) {
    res++;
  }
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<isqrt>:
  endbr64
  cmp    $0x3fff,%edi
  jg     L7
  mov    $0x4000,%eax
L63:
  sar    $0x2,%eax
  cmp    %eax,%edi
  jl     L63
  test   %eax,%eax
  je     L38
L12:
  mov    $0x0,%edx
  jmp    L3
L7:
  mov    $0x4000,%eax
  jmp    L12
L62:
  sar    %edx
  sar    $0x2,%eax
  je     L30
L3:
  lea    (%rdx,%rax,1),%ecx
  cmp    %edi,%ecx
  jg     L62
  sub    %ecx,%edi
  lea    (%rcx,%rax,1),%edx
  jmp    L62
L38:
  mov    %eax,%edx
L30:
  cmp    %edx,%edi
  setg   %al
  movzbl %al,%eax
  add    %eax,%edx
  mov    %edx,%eax
  ret
```

assistant: ```c
uint16_t isqrt(uint16_t a_nInput) {
  uint16_t op = a_nInput;
  uint16_t res = 0;
  uint16_t one = 1uL << 14;
  while (one > op) {
    one >>= 2;
  }
  while (one != 0) {
    if (op >= res + one) {
      op = op - (res + one);
      res = res + 2 * one;
    }
    res >>= 1;
    one >>= 2;
  }
  if (op > res) {
    res++;
  }
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isqrt>:
  endbr64
  mov    $0x4000,%eax
  cmp    $0x3fff,%edi
  jg     L22
  nopl   0x0(%rax)
L43:
  sar    $0x2,%eax
  cmp    %eax,%edi
  jl     L43
  test   %eax,%eax
  je     L20
L22:
  xor    %r8d,%r8d
  cs nopw 0x0(%rax,%rax,1)
L10:
  lea    (%r8,%rax,1),%edx
  cmp    %edi,%edx
  jg     L21
  sub    %edx,%edi
  lea    (%rdx,%rax,1),%r8d
L21:
  sar    %r8d
  sar    $0x2,%eax
  jne    L10
  xor    %eax,%eax
  cmp    %r8d,%edi
  setg   %al
  add    %eax,%r8d
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L20:
  xor    %r8d,%r8d
  xor    %eax,%eax
  cmp    %r8d,%edi
  setg   %al
  add    %eax,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
uint16_t isqrt(uint16_t a_nInput) {
  uint16_t op = a_nInput;
  uint16_t res = 0;
  uint16_t one = 1uL << 14;
  while (one > op) {
    one >>= 2;
  }
  while (one != 0) {
    if (op >= res + one) {
      op = op - (res + one);
      res = res + 2 * one;
    }
    res >>= 1;
    one >>= 2;
  }
  if (op > res) {
    res++;
  }
  return res;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<isqrt>:
  endbr64
  mov    $0x4000,%edx
  cmp    $0x3fff,%edi
  jg     L20
  nopl   0x0(%rax)
L1:
  sar    $0x2,%edx
  cmp    %edx,%edi
  jl     L1
  test   %edx,%edx
  je     L46
L20:
  xor    %eax,%eax
  nopl   (%rax)
L9:
  lea    (%rdx,%rax,1),%ecx
  cmp    %ecx,%edi
  jl     L38
  lea    (%rdx,%rcx,1),%eax
  sub    %ecx,%edi
  sar    %eax
  sar    $0x2,%edx
  jne    L9
  xor    %edx,%edx
  cmp    %eax,%edi
  setg   %dl
  add    %edx,%eax
  ret
  nopl   (%rax)
L38:
  sar    %eax
  sar    $0x2,%edx
  jne    L9
  xor    %edx,%edx
  cmp    %eax,%edi
  setg   %dl
  add    %edx,%eax
  ret
  nopl   0x0(%rax)
L46:
  xor    %eax,%eax
  xor    %edx,%edx
  cmp    %eax,%edi
  setg   %dl
  add    %edx,%eax
  ret
```

assistant: ```c
uint16_t isqrt(uint16_t a_nInput) {
  uint16_t op = a_nInput;
  uint16_t res = 0;
  uint16_t one = 1uL << 14;
  while (one > op) {
    one >>= 2;
  }
  while (one != 0) {
    if (op >= res + one) {
      op = op - (res + one);
      res = res + 2 * one;
    }
    res >>= 1;
    one >>= 2;
  }
  if (op > res) {
    res++;
  }
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ofl_error_type_to_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <open_memstream@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <ofl_error_type_print@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
char *ofl_error_type_to_string(uint16_t type) {
  char *str;
  size_t str_size;
  FILE *stream = open_memstream(&str, &str_size);
  ofl_error_type_print(stream, type);
  fclose(stream);
  return str;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ofl_error_type_to_string>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  call   <open_memstream@plt>
  mov    %rax,%rbx
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <ofl_error_type_print@plt>
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ofl_error_type_to_string(uint16_t type) {
  char *str;
  size_t str_size;
  FILE *stream = open_memstream(&str, &str_size);
  ofl_error_type_print(stream, type);
  fclose(stream);
  return str;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ofl_error_type_to_string>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  call   <open_memstream@plt>
  mov    %r12d,%esi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <ofl_error_type_print@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ofl_error_type_to_string(uint16_t type) {
  char *str;
  size_t str_size;
  FILE *stream = open_memstream(&str, &str_size);
  ofl_error_type_print(stream, type);
  fclose(stream);
  return str;
}
```