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

```asm
<utf8_validate_char>:
  endbr64
  movzbl (%rdi),%eax
  mov    $0x1,%r8d
  test   %al,%al
  jns    L59
  xor    %r8d,%r8d
  cmp    $0xc1,%al
  jbe    L59
  lea    0x1(%rdi),%rdx
  cmp    $0xdf,%al
  jbe    L190
  cmp    $0xef,%al
  ja     L183
  sub    %rdx,%rsi
  cmp    $0x1,%rsi
  jle    L59
  movzbl 0x1(%rdi),%edx
  cmp    $0xe0,%al
  je     L130
  cmp    $0xed,%al
  je     L31
L136:
  and    $0xffffffc0,%edx
  xor    %r8d,%r8d
  cmp    $0x80,%dl
  jne    L59
  movzbl 0x2(%rdi),%eax
  xor    %r8d,%r8d
  and    $0xffffffc0,%eax
  cmp    $0x80,%al
  sete   %r8b
  lea    (%r8,%r8,2),%r8d
L59:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L190:
  sub    %rdx,%rsi
  test   %rsi,%rsi
  jle    L59
  movzbl 0x1(%rdi),%eax
  xor    %r8d,%r8d
  and    $0xffffffc0,%eax
  cmp    $0x80,%al
  sete   %r8b
  add    %r8d,%r8d
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L183:
  cmp    $0xf4,%al
  ja     L59
  sub    %rdx,%rsi
  cmp    $0x2,%rsi
  jle    L59
  movzbl 0x1(%rdi),%edx
  cmp    $0xf0,%al
  je     L229
  cmp    $0xf4,%al
  je     L117
L207:
  and    $0xffffffc0,%edx
  xor    %r8d,%r8d
  cmp    $0x80,%dl
  jne    L59
  movzbl 0x2(%rdi),%eax
  and    $0xffffffc0,%eax
  cmp    $0x80,%al
  jne    L59
  movzbl 0x3(%rdi),%eax
  xor    %r8d,%r8d
  and    $0xffffffc0,%eax
  cmp    $0x80,%al
  sete   %r8b
  shl    $0x2,%r8d
  jmp    L59
  nopl   (%rax)
L130:
  cmp    $0x9f,%dl
  ja     L136
  jmp    L59
  cs nopw 0x0(%rax,%rax,1)
L31:
  cmp    $0x9f,%dl
  jbe    L136
  jmp    L59
  xchg   %ax,%ax
L229:
  cmp    $0x8f,%dl
  ja     L207
  jmp    L59
  nopw   0x0(%rax,%rax,1)
L117:
  cmp    $0x8f,%dl
  jbe    L207
  jmp    L59
```

assistant: ```c
int utf8_validate_char(const char *s, const char *e) {
  unsigned char c = *s++;
  if (c <= 0x7F) {
    return 1;
  } else if (c <= 0xC1) {
    return 0;
  } else if (c <= 0xDF) {
    if (e - s < 1)
      return 0;
    if (((unsigned char)*s++ & 0xC0) != 0x80)
      return 0;
    return 2;
  } else if (c <= 0xEF) {
    if (e - s < 2)
      return 0;
    if (c == 0xE0 && (unsigned char)*s < 0xA0)
      return 0;
    if (c == 0xED && (unsigned char)*s > 0x9F)
      return 0;
    if (((unsigned char)*s++ & 0xC0) != 0x80)
      return 0;
    if (((unsigned char)*s++ & 0xC0) != 0x80)
      return 0;
    return 3;
  } else if (c <= 0xF4) {
    if (e - s < 3)
      return 0;
    if (c == 0xF0 && (unsigned char)*s < 0x90)
      return 0;
    if (c == 0xF4 && (unsigned char)*s > 0x8F)
      return 0;
    if (((unsigned char)*s++ & 0xC0) != 0x80)
      return 0;
    if (((unsigned char)*s++ & 0xC0) != 0x80)
      return 0;
    if (((unsigned char)*s++ & 0xC0) != 0x80)
      return 0;
    return 4;
  } else {
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cap_tree>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D3(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L2
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L2:
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D3
datatype: string
name: parse_data
arguments: label: D2
datatype: string

tool: 

tool: D2: 


assistant: ```c
void cap_tree(char *eol) {
  if (0 == strcmp(eol, ""))
    printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cap_tree>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L1
L6:
  add    $0x8,%rsp
  ret
L1:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  jmp    L6
```

tool_calls: name: retrieve_data
arguments: tag: D7
datatype: string
name: retrieve_data
arguments: tag: D9
datatype: string

tool: 

tool: D9: 


assistant: ```c
void cap_tree(char *eol) {
  if (0 == strcmp(eol, ""))
    printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cap_tree>:
  endbr64
  sub    $0x8,%rsp
  lea    D14(%rip),%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L1
  add    $0x8,%rsp
  ret
  xchg   %ax,%ax
L1:
  lea    D6(%rip),%rdi
  add    $0x8,%rsp
  jmp    1060 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D14
type_infered: string
name: extract_symbolic_data
arguments: tag: D6
type_infered: string

tool: 

tool: D6: 


assistant: ```c
void cap_tree(char *eol) {
  if (0 == strcmp(eol, ""))
    printf("\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cap_tree>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L9
  add    $0x8,%rsp
  ret
  xchg   %ax,%ax
L9:
  lea    D6(%rip),%rdi
  add    $0x8,%rsp
  jmp    1060 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type: string
name: disassemble_data
arguments: data_tag: D6
type: string

tool: 

tool: D6: 


assistant: ```c
void cap_tree(char *eol) {
  if (0 == strcmp(eol, ""))
    printf("\n");
}
```
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
  lea    -0x13000(%rsp),%r11
L36:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L36
  sub    $0x8b0,%rsp
  mov    %edi,-0x138a4(%rbp)
  mov    %rsi,-0x138b0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1389c(%rbp)
  jmp    L50
L18:
  movl   $0x0,-0x138a0(%rbp)
  jmp    L68
L49:
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,-0x13898(%rbp)
  cmpq   $0x0,-0x13898(%rbp)
  jne    L10
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <exit@plt>
L10:
  mov    -0x138a0(%rbp),%eax
  cltq
  mov    -0x13898(%rbp),%rdx
  mov    %rdx,-0x13890(%rbp,%rax,8)
  mov    -0x13898(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,-0x138a0(%rbp)
L68:
  cmpl   $0x270f,-0x138a0(%rbp)
  jle    L49
  addl   $0x1,-0x1389c(%rbp)
L50:
  cmpl   $0x4e1f,-0x1389c(%rbp)
  jle    L18
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  char *stack[10000];
  char *blob;
  int i, j;
  for (j = 0; j < 20000; j++) {
    for (i = 0; i < 10000; i++) {
      blob = malloc(1024);
      if (blob == NULL) {
        exit(EXIT_FAILURE);
      }
      stack[i] = blob;
      free(blob);
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x4e20,%r12d
  jmp    L36
L57:
  mov    %rbx,%rdi
  call   <free@plt>
  sub    $0x1,%ebp
  je     L6
L43:
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L57
  mov    D9(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  jmp    L57
L6:
  sub    $0x1,%r12d
  je     L44
L36:
  mov    $0x2710,%ebp
  jmp    L43
L44:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  char *stack[10000];
  char *blob;
  int i, j;
  for (j = 0; j < 20000; j++) {
    for (i = 0; i < 10000; i++) {
      blob = malloc(1024);
      if (blob == NULL) {
        exit(EXIT_FAILURE);
      }
      stack[i] = blob;
      free(blob);
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x4e20,%r12d
  push   %rbp
  push   %rbx
L32:
  mov    $0x2710,%ebx
  jmp    L13
  nopl   (%rax)
L7:
  mov    %rbp,%rdi
  call   <free@plt>
  sub    $0x1,%ebx
  je     L36
L13:
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L7
  mov    D4(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  sub    $0x1,%ebx
  jne    L13
L36:
  sub    $0x1,%r12d
  jne    L32
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  char *stack[10000];
  char *blob;
  int i, j;
  for (j = 0; j < 20000; j++) {
    for (i = 0; i < 10000; i++) {
      blob = malloc(1024);
      if (blob == NULL) {
        exit(EXIT_FAILURE);
      }
      stack[i] = blob;
      free(blob);
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x4e20,%ebp
  push   %rbx
  sub    $0x8,%rsp
L33:
  mov    $0x2710,%ebx
  jmp    L7
  cs nopw 0x0(%rax,%rax,1)
L39:
  call   <free@plt>
  sub    $0x1,%ebx
  je     L25
L7:
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L39
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  xor    %edi,%edi
  call   <free@plt>
  sub    $0x1,%ebx
  jne    L7
L25:
  sub    $0x1,%ebp
  jne    L33
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  char *stack[10000];
  char *blob;
  int i, j;
  for (j = 0; j < 20000; j++) {
    for (i = 0; i < 10000; i++) {
      blob = malloc(1024);
      if (blob == NULL) {
        exit(EXIT_FAILURE);
      }
      stack[i] = blob;
      free(blob);
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0xa,%esi
  mov    $0x2,%edi
  call   <afct@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D7
datatype: string

tool: D7: afct(2,10) =  %d


assistant: ```c
int main(int argc, char *argv[]) {
  printf("afct(2,10) =  %d\n", afct(2, 10));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  mov    $0x2,%edi
  call   <afct@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_infered: string

tool: D7: afct(2,10) =  %d


assistant: ```c
int main(int argc, char *argv[]) {
  printf("afct(2,10) =  %d\n", afct(2, 10));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  mov    $0x2,%edi
  call   <afct@plt>
  lea    D2(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
datatype: string

tool: D2: afct(2,10) =  %d


assistant: ```c
int main(int argc, char *argv[]) {
  printf("afct(2,10) =  %d\n", afct(2, 10));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  mov    $0x2,%edi
  call   <afct@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
datatype: string

tool: D5: afct(2,10) =  %d


assistant: ```c
int main(int argc, char *argv[]) {
  printf("afct(2,10) =  %d\n", afct(2, 10));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<EPIDividerSet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void EPIDividerSet(uint32_t ui32Base, uint32_t ui32Divider) {
  ;
  (*((volatile uint32_t *)(ui32Base + 0x00000004))) = ui32Divider;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<EPIDividerSet>:
  endbr64
  mov    %rsi,0x4(%rdi)
  ret
```

assistant: ```c
void EPIDividerSet(uint32_t ui32Base, uint32_t ui32Divider) {
  ;
  (*((volatile uint32_t *)(ui32Base + 0x00000004))) = ui32Divider;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<EPIDividerSet>:
  endbr64
  mov    %rsi,0x4(%rdi)
  ret
```

assistant: ```c
void EPIDividerSet(uint32_t ui32Base, uint32_t ui32Divider) {
  ;
  (*((volatile uint32_t *)(ui32Base + 0x00000004))) = ui32Divider;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<EPIDividerSet>:
  endbr64
  mov    %rsi,0x4(%rdi)
  ret
```

assistant: ```c
void EPIDividerSet(uint32_t ui32Base, uint32_t ui32Divider) {
  ;
  (*((volatile uint32_t *)(ui32Base + 0x00000004))) = ui32Divider;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<power>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L14
L18:
  mov    -0x4(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L18
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int power(int base, int n) {
  int i, p;
  p = 1;
  for (i = 1; i <= n; i = i + 1) {
    p = p * base;
  }
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<power>:
  endbr64
  test   %esi,%esi
  jle    L16
  add    $0x1,%esi
  mov    $0x1,%edx
  mov    $0x1,%eax
L25:
  imul   %edi,%edx
  add    $0x1,%eax
  cmp    %esi,%eax
  jne    L25
L21:
  mov    %edx,%eax
  ret
L16:
  mov    $0x1,%edx
  jmp    L21
```

assistant: ```c
int power(int base, int n) {
  int i, p;
  p = 1;
  for (i = 1; i <= n; i = i + 1) {
    p = p * base;
  }
  return p;
}
```
user: What is the c source code of the assembly code below:

```asm
<power>:
  endbr64
  test   %esi,%esi
  jle    L0
  add    $0x1,%esi
  mov    $0x1,%r8d
  mov    $0x1,%eax
  cs nopw 0x0(%rax,%rax,1)
L8:
  add    $0x1,%eax
  imul   %edi,%r8d
  cmp    %esi,%eax
  jne    L8
  mov    %r8d,%eax
  ret
  nop
L0:
  mov    $0x1,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int power(int base, int n) {
  int i, p;
  p = 1;
  for (i = 1; i <= n; i = i + 1) {
    p = p * base;
  }
  return p;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<power>:
  endbr64
  test   %esi,%esi
  jle    L11
  add    $0x1,%esi
  mov    $0x1,%r8d
  mov    $0x1,%eax
  cs nopw 0x0(%rax,%rax,1)
L10:
  add    $0x1,%eax
  imul   %edi,%r8d
  cmp    %esi,%eax
  jne    L10
  mov    %r8d,%eax
  ret
  nop
L11:
  mov    $0x1,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int power(int base, int n) {
  int i, p;
  p = 1;
  for (i = 1; i <= n; i = i + 1) {
    p = p * base;
  }
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getop>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  nop
L21:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L21
  cmpl   $0x9,-0x4(%rbp)
  je     L21
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L87
  cmpl   $0x2e,-0x4(%rbp)
  je     L87
  mov    -0x4(%rbp),%eax
  jmp    L86
L87:
  movl   $0x0,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L34
  nop
L69:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L69
L34:
  cmpl   $0x2e,-0x4(%rbp)
  jne    L22
  nop
L16:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L16
L22:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L90
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <ungetch@plt>
L90:
  mov    $0x30,%eax
L86:
  leave
  ret
```

assistant: ```c
int getop(char s[]) {
  int i, c;
  while ((s[0] = c = getch()) == ' ' || c == '\t')
    ;
  s[1] = '\0';
  if (!isdigit(c) && c != '.')
    return c;
  i = 0;
  if (isdigit(c))
    while (isdigit(s[++i] = c = getch()))
      ;
  if (c == '.')
    while (isdigit(s[++i] = c = getch()))
      ;
  s[i] = '\0';
  if (c != EOF)
    ungetch(c);
  return '0';
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<getop>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
L68:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,%ebx
  mov    %al,0x0(%rbp)
  cmp    $0x9,%eax
  je     L68
  cmp    $0x20,%al
  je     L68
  movb   $0x0,0x1(%rbp)
  movsbl %al,%r12d
  mov    %r12d,%edi
  call   <isdigit@plt>
  cmp    $0x2e,%ebx
  je     L16
  test   %rax,%rax
  je     L42
L16:
  mov    %r12d,%edi
  call   <isdigit@plt>
  mov    $0x0,%r13d
  test   %rax,%rax
  je     L82
  mov    $0x1,%r12d
L99:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,%ebx
  mov    %r12d,%r13d
  mov    %al,0x0(%rbp,%r12,1)
  movsbl %al,%edi
  call   <isdigit@plt>
  add    $0x1,%r12
  test   %rax,%rax
  jne    L99
L82:
  cmp    $0x2e,%ebx
  je     L3
L81:
  movslq %r13d,%r13
  movb   $0x0,0x0(%rbp,%r13,1)
  mov    D8(%rip),%rax
  cmp    %ebx,(%rax)
  jne    L72
  mov    $0x30,%ebx
L42:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  lea    0x1(%r13),%r12d
  movslq %r12d,%r12
L19:
  mov    $0x0,%eax
  call   <getch@plt>
  mov    %eax,%ebx
  mov    %r12d,%r13d
  mov    %al,0x0(%rbp,%r12,1)
  movsbl %al,%edi
  call   <isdigit@plt>
  add    $0x1,%r12
  test   %rax,%rax
  jne    L19
  jmp    L81
L72:
  mov    %ebx,%edi
  call   <ungetch@plt>
  mov    $0x30,%ebx
  jmp    L42
```

assistant: ```c
int getop(char s[]) {
  int i, c;
  while ((s[0] = c = getch()) == ' ' || c == '\t')
    ;
  s[1] = '\0';
  if (!isdigit(c) && c != '.')
    return c;
  i = 0;
  if (isdigit(c))
    while (isdigit(s[++i] = c = getch()))
      ;
  if (c == '.')
    while (isdigit(s[++i] = c = getch()))
      ;
  s[i] = '\0';
  if (c != EOF)
    ungetch(c);
  return '0';
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getop>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L32:
  xor    %eax,%eax
  call   <getch@plt>
  mov    %al,(%rbx)
  mov    %eax,%r12d
  cmp    $0x9,%eax
  je     L32
  cmp    $0x20,%al
  je     L32
  movb   $0x0,0x1(%rbx)
  movsbl %al,%ebp
  mov    %ebp,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L76
  cmp    $0x2e,%r12d
  je     L59
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L59:
  mov    %ebp,%edi
  xor    %r13d,%r13d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L26
L96:
  lea    0x1(%rbx,%r13,1),%rbx
  nopl   0x0(%rax)
L51:
  xor    %eax,%eax
  mov    %rbx,%rbp
  add    $0x1,%rbx
  call   <getch@plt>
  mov    %al,-0x1(%rbx)
  movsbl %al,%edi
  mov    %eax,%r12d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L51
  mov    D2(%rip),%rax
  movb   $0x0,0x0(%rbp)
  cmp    %r12d,(%rax)
  jne    L99
L29:
  add    $0x8,%rsp
  mov    $0x30,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L76:
  mov    %ebp,%edi
  xor    %r13d,%r13d
  call   <isdigit@plt>
  test   %rax,%rax
  je     L10
L26:
  mov    $0x1,%ebp
  nopl   0x0(%rax)
L20:
  xor    %eax,%eax
  movslq %ebp,%r13
  call   <getch@plt>
  mov    %al,(%rbx,%rbp,1)
  movsbl %al,%edi
  mov    %eax,%r12d
  add    $0x1,%rbp
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L20
L10:
  movslq %r13d,%rbp
  add    %rbx,%rbp
  cmp    $0x2e,%r12d
  je     L96
  mov    D2(%rip),%rax
  movb   $0x0,0x0(%rbp)
  cmp    %r12d,(%rax)
  je     L29
L99:
  mov    %r12d,%edi
  mov    $0x30,%r12d
  call   <ungetch@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int getop(char s[]) {
  int i, c;
  while ((s[0] = c = getch()) == ' ' || c == '\t')
    ;
  s[1] = '\0';
  if (!isdigit(c) && c != '.')
    return c;
  i = 0;
  if (isdigit(c))
    while (isdigit(s[++i] = c = getch()))
      ;
  if (c == '.')
    while (isdigit(s[++i] = c = getch()))
      ;
  s[i] = '\0';
  if (c != EOF)
    ungetch(c);
  return '0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getop>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   (%rax)
L21:
  xor    %eax,%eax
  call   <getch@plt>
  mov    %al,(%rbx)
  mov    %eax,%r12d
  cmp    $0x9,%eax
  je     L21
  cmp    $0x20,%al
  je     L21
  movb   $0x0,0x1(%rbx)
  movsbl %al,%ebp
  mov    %ebp,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L103
  cmp    $0x2e,%r12d
  je     L92
L84:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L92:
  mov    %ebp,%edi
  xor    %r14d,%r14d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L129
L121:
  movslq %r14d,%r14
  lea    0x1(%rbx,%r14,1),%rbx
  nopw   0x0(%rax,%rax,1)
L116:
  xor    %eax,%eax
  mov    %rbx,%r13
  add    $0x1,%rbx
  call   <getch@plt>
  mov    %al,-0x1(%rbx)
  movsbl %al,%edi
  mov    %eax,%r12d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L116
  jmp    L112
  nopl   0x0(%rax,%rax,1)
L103:
  mov    %ebp,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L66
L129:
  mov    %ebx,%r15d
  lea    0x1(%rbx),%rbp
  neg    %r15d
  xchg   %ax,%ax
L124:
  xor    %eax,%eax
  lea    (%r15,%rbp,1),%r14d
  mov    %rbp,%r13
  add    $0x1,%rbp
  call   <getch@plt>
  mov    %al,-0x1(%rbp)
  movsbl %al,%edi
  mov    %eax,%r12d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L124
L100:
  cmp    $0x2e,%r12d
  je     L121
L112:
  mov    D0(%rip),%rax
  movb   $0x0,0x0(%r13)
  cmp    %r12d,(%rax)
  je     L77
  mov    %r12d,%edi
  mov    $0x30,%r12d
  call   <ungetch@plt>
  jmp    L84
  nop
L77:
  mov    $0x30,%r12d
  jmp    L84
  nopl   0x0(%rax,%rax,1)
L66:
  mov    %rbx,%r13
  xor    %r14d,%r14d
  jmp    L100
```

assistant: ```c
int getop(char s[]) {
  int i, c;
  while ((s[0] = c = getch()) == ' ' || c == '\t')
    ;
  s[1] = '\0';
  if (!isdigit(c) && c != '.')
    return c;
  i = 0;
  if (isdigit(c))
    while (isdigit(s[++i] = c = getch()))
      ;
  if (c == '.')
    while (isdigit(s[++i] = c = getch()))
      ;
  s[i] = '\0';
  if (c != EOF)
    ungetch(c);
  return '0';
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L9:
  mov    -0x24(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L9
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void get_string(char a[], char out[], int begin, int length) {
  for (int i = 0; i < length; i++) {
    out[i] = a[begin + i];
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_string>:
  endbr64
  test   %ecx,%ecx
  jle    L13
  mov    %ecx,%ecx
  mov    $0x0,%eax
  movslq %edx,%rdx
  add    %rdx,%rdi
L2:
  movzbl (%rdi,%rax,1),%edx
  mov    %dl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L2
L13:
  ret
```

assistant: ```c
void get_string(char a[], char out[], int begin, int length) {
  for (int i = 0; i < length; i++) {
    out[i] = a[begin + i];
  }
}
```