ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<read_word>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r14
  mov    %edx,%r13d
L80:
  mov    %rbp,%rdi
  call   <fgetc@plt>
  mov    %eax,%ebx
  cmp    $0x20,%eax
  je     L80
  and    $0xfffffffb,%eax
  cmp    $0x9,%eax
  je     L80
  mov    D6(%rip),%rax
  cmp    %ebx,(%rax)
  je     L99
  cmp    $0xa,%ebx
  je     L129
  mov    %bl,(%r14)
  cmp    $0x2,%r13d
  jle    L27
  lea    0x1(%r14),%r12
  sub    $0x1,%r13d
  mov    $0x1,%ebx
L95:
  mov    %rbp,%rdi
  call   <fgetc@plt>
  mov    %eax,%edx
  and    $0xfffffffb,%edx
  cmp    $0x9,%edx
  je     L125
  cmp    $0x20,%eax
  je     L125
  mov    D6(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L125
  cmp    $0xa,%eax
  je     L122
  mov    %al,(%r12)
  add    $0x1,%ebx
  add    $0x1,%r12
  cmp    %r13d,%ebx
  jne    L95
  mov    %r13d,%ebx
  jmp    L125
L122:
  mov    %rbp,%rsi
  mov    $0xa,%edi
  call   <ungetc@plt>
L125:
  movslq %ebx,%rax
  movb   $0x0,(%r14,%rax,1)
L99:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  mov    $0x1,%ebx
  jmp    L125
L129:
  mov    $0x0,%ebx
  jmp    L99
```

assistant: ```c
int read_word(FILE *f, char *buf, int L) {
  int i = 0, c = 0;
  while ((c = fgetc(f)) == ' ' || c == '\t' || c == '\r')
    ;
  if (c == EOF)
    return EOF;
  if (c == '\n')
    return 0;
  buf[i] = c;
  for (i = 1; i < L - 1; i++) {
    c = fgetc(f);
    if (c == ' ' || c == '\t' || c == '\r' || c == EOF)
      break;
    if (c == '\n') {
      ungetc(c, f);
      break;
    }
    buf[i] = c;
  }
  buf[i] = '\0';
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<read_word>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L142:
  mov    %r15,%rdi
  call   <fgetc@plt>
  mov    %eax,%r12d
  cmp    $0x20,%eax
  je     L142
  and    $0xfffffffb,%eax
  cmp    $0x9,%eax
  je     L142
  mov    D8(%rip),%rbp
  cmp    %r12d,0x0(%rbp)
  je     L148
  cmp    $0xa,%r12d
  je     L40
  mov    %r12b,(%rbx)
  cmp    $0x2,%r13d
  jle    L22
  sub    $0x3,%r13d
  mov    $0x1,%r14d
  add    $0x2,%r13
  jmp    L28
  nopl   0x0(%rax)
L47:
  cmp    $0x20,%eax
  je     L21
  cmp    %eax,0x0(%rbp)
  je     L21
  cmp    $0xa,%eax
  je     L138
  mov    %al,(%rbx,%r14,1)
  add    $0x1,%r14
  add    $0x1,%r12d
  cmp    %r14,%r13
  je     L86
L28:
  mov    %r15,%rdi
  mov    %r14d,%r12d
  call   <fgetc@plt>
  mov    %eax,%edx
  and    $0xfffffffb,%edx
  cmp    $0x9,%edx
  jne    L47
L21:
  movb   $0x0,(%rbx,%r14,1)
L148:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L40:
  xor    %r12d,%r12d
  jmp    L148
L138:
  mov    %r15,%rsi
  mov    $0xa,%edi
  call   <ungetc@plt>
  jmp    L21
L86:
  movslq %r12d,%r14
  jmp    L21
L22:
  mov    $0x1,%r14d
  mov    $0x1,%r12d
  jmp    L21
```

assistant: ```c
int read_word(FILE *f, char *buf, int L) {
  int i = 0, c = 0;
  while ((c = fgetc(f)) == ' ' || c == '\t' || c == '\r')
    ;
  if (c == EOF)
    return EOF;
  if (c == '\n')
    return 0;
  buf[i] = c;
  for (i = 1; i < L - 1; i++) {
    c = fgetc(f);
    if (c == ' ' || c == '\t' || c == '\r' || c == EOF)
      break;
    if (c == '\n') {
      ungetc(c, f);
      break;
    }
    buf[i] = c;
  }
  buf[i] = '\0';
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_word>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %r15,%rdi
  call   <fgetc@plt>
  mov    %eax,%r12d
  cmp    $0x20,%eax
  je     L14
  and    $0xfffffffb,%eax
  cmp    $0x9,%eax
  je     L14
  mov    D5(%rip),%rbp
  cmp    %r12d,0x0(%rbp)
  je     L15
  cmp    $0xa,%r12d
  je     L72
  mov    %r12b,(%rbx)
  cmp    $0x2,%r13d
  jle    L91
  sub    $0x3,%r13d
  mov    $0x1,%r14d
  add    $0x2,%r13
  jmp    L53
  nopl   0x0(%rax)
L74:
  cmp    $0x20,%eax
  je     L116
  cmp    %eax,0x0(%rbp)
  je     L116
  cmp    $0xa,%eax
  je     L33
  mov    %al,(%rbx,%r14,1)
  add    $0x1,%r14
  add    $0x1,%r12d
  cmp    %r13,%r14
  je     L89
L53:
  mov    %r15,%rdi
  mov    %r14d,%r12d
  call   <fgetc@plt>
  mov    %eax,%edx
  and    $0xfffffffb,%edx
  cmp    $0x9,%edx
  jne    L74
L116:
  movb   $0x0,(%rbx,%r14,1)
L15:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L72:
  xor    %r12d,%r12d
  jmp    L15
L33:
  mov    %r15,%rsi
  mov    $0xa,%edi
  call   <ungetc@plt>
  jmp    L116
L89:
  movslq %r12d,%r14
  jmp    L116
L91:
  mov    $0x1,%r14d
  mov    $0x1,%r12d
  jmp    L116
```

assistant: ```c
int read_word(FILE *f, char *buf, int L) {
  int i = 0, c = 0;
  while ((c = fgetc(f)) == ' ' || c == '\t' || c == '\r')
    ;
  if (c == EOF)
    return EOF;
  if (c == '\n')
    return 0;
  buf[i] = c;
  for (i = 1; i < L - 1; i++) {
    c = fgetc(f);
    if (c == ' ' || c == '\t' || c == '\r' || c == EOF)
      break;
    if (c == '\n') {
      ungetc(c, f);
      break;
    }
    buf[i] = c;
  }
  buf[i] = '\0';
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
L0:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x1(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
  cmpb   $0x0,-0x1(%rbp)
  jne    L0
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void strcpy(char *restrict dst, const char *restrict src) {
  char c;
  do {
    c = *(src++);
    *(dst++) = c;
  } while (c != '\0');
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcpy>:
  endbr64
  mov    $0x0,%eax
L7:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L7
  ret
```

assistant: ```c
void strcpy(char *restrict dst, const char *restrict src) {
  char c;
  do {
    c = *(src++);
    *(dst++) = c;
  } while (c != '\0');
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strcpy>:
  endbr64
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L3:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L3
  ret
```

assistant: ```c
void strcpy(char *restrict dst, const char *restrict src) {
  char c;
  do {
    c = *(src++);
    *(dst++) = c;
  } while (c != '\0');
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strcpy>:
  endbr64
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L2:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L2
  ret
```

assistant: ```c
void strcpy(char *restrict dst, const char *restrict src) {
  char c;
  do {
    c = *(src++);
    *(dst++) = c;
  } while (c != '\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
<cube_120convert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L13:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <whichpos@plt>
  mov    %rax,%rdx
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <whichrot@plt>
  mov    %rax,%rdx
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  cltq
  lea    0x1(%rax),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L3:
  cmpl   $0x13,-0x4(%rbp)
  jle    L13
  mov    $0x1,%eax
  leave
  ret
```

assistant: ```c
int cube_120convert(const char *input, char *output) {
  int i;
  for (i = 0; i < 20; ++i) {
    ((char *)output + (i << 1))[0] = (char)whichpos(input + i * 6);
    ((char *)output + (i << 1))[1] = (char)whichrot(input + i * 6);
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cube_120convert>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  lea    0x78(%rdi),%r12
L9:
  mov    %rbx,%rdi
  call   <whichpos@plt>
  mov    %al,0x0(%rbp)
  mov    %rbx,%rdi
  call   <whichrot@plt>
  mov    %al,0x1(%rbp)
  add    $0x6,%rbx
  add    $0x2,%rbp
  cmp    %r12,%rbx
  jne    L9
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int cube_120convert(const char *input, char *output) {
  int i;
  for (i = 0; i < 20; ++i) {
    ((char *)output + (i << 1))[0] = (char)whichpos(input + i * 6);
    ((char *)output + (i << 1))[1] = (char)whichrot(input + i * 6);
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cube_120convert>:
  endbr64
  push   %r12
  lea    0x78(%rdi),%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %rbx,%rdi
  add    $0x2,%rbp
  call   <whichpos@plt>
  mov    %rbx,%rdi
  add    $0x6,%rbx
  mov    %al,-0x2(%rbp)
  call   <whichrot@plt>
  mov    %al,-0x1(%rbp)
  cmp    %r12,%rbx
  jne    L7
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int cube_120convert(const char *input, char *output) {
  int i;
  for (i = 0; i < 20; ++i) {
    ((char *)output + (i << 1))[0] = (char)whichpos(input + i * 6);
    ((char *)output + (i << 1))[1] = (char)whichrot(input + i * 6);
  }
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<cube_120convert>:
  endbr64
  push   %r12
  lea    0x78(%rdi),%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %rbx,%rdi
  add    $0x2,%rbp
  call   <whichpos@plt>
  mov    %rbx,%rdi
  add    $0x6,%rbx
  mov    %al,-0x2(%rbp)
  call   <whichrot@plt>
  mov    %al,-0x1(%rbp)
  cmp    %rbx,%r12
  jne    L9
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int cube_120convert(const char *input, char *output) {
  int i;
  for (i = 0; i < 20; ++i) {
    ((char *)output + (i << 1))[0] = (char)whichpos(input + i * 6);
    ((char *)output + (i << 1))[1] = (char)whichrot(input + i * 6);
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<var_value_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void var_value_free(void *value) {
  double *v = (double *)value;
  free(v);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<var_value_free>:
  endbr64
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void var_value_free(void *value) {
  double *v = (double *)value;
  free(v);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<var_value_free>:
  endbr64
  jmp    1050 <free@plt>
```

assistant: ```c
void var_value_free(void *value) {
  double *v = (double *)value;
  free(v);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<var_value_free>:
  endbr64
  jmp    1050 <free@plt>
```

assistant: ```c
void var_value_free(void *value) {
  double *v = (double *)value;
  free(v);
}
```
user: What is the c source code of the assembly code below:

```asm
<vminu4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <vcmpgeu4@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  and    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  not    %eax
  and    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  or     %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
unsigned int vminu4(unsigned int a, unsigned int b) {
  unsigned int r, s;
  s = vcmpgeu4(b, a);
  r = a & s;
  s = b & ~s;
  r = r | s;
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vminu4>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %edi,%esi
  mov    %ebp,%edi
  call   <vcmpgeu4@plt>
  xor    %ebp,%ebx
  and    %eax,%ebx
  mov    %ebx,%eax
  xor    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned int vminu4(unsigned int a, unsigned int b) {
  unsigned int r, s;
  s = vcmpgeu4(b, a);
  r = a & s;
  s = b & ~s;
  r = r | s;
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<vminu4>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  mov    %edi,%esi
  push   %rbx
  mov    %edi,%ebx
  mov    %ebp,%edi
  xor    %ebp,%ebx
  sub    $0x8,%rsp
  call   <vcmpgeu4@plt>
  add    $0x8,%rsp
  and    %eax,%ebx
  mov    %ebx,%eax
  pop    %rbx
  xor    %ebp,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int vminu4(unsigned int a, unsigned int b) {
  unsigned int r, s;
  s = vcmpgeu4(b, a);
  r = a & s;
  s = b & ~s;
  r = r | s;
  return r;
}
```
user: What is the c source code of the assembly code below?

```asm
<vminu4>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  mov    %edi,%esi
  push   %rbx
  mov    %edi,%ebx
  mov    %ebp,%edi
  xor    %ebp,%ebx
  sub    $0x8,%rsp
  call   <vcmpgeu4@plt>
  add    $0x8,%rsp
  and    %eax,%ebx
  mov    %ebx,%eax
  pop    %rbx
  xor    %ebp,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int vminu4(unsigned int a, unsigned int b) {
  unsigned int r, s;
  s = vcmpgeu4(b, a);
  r = a & s;
  s = b & ~s;
  r = r | s;
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<menu_principal>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    D50(%rip),%rdi
  call   <puts@plt>
  lea    D74(%rip),%rdi
  call   <puts@plt>
  lea    D109(%rip),%rdi
  call   <puts@plt>
  lea    D115(%rip),%rbx
L13:
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    $0x6,%eax
  ja     L13
  test   %eax,%eax
  je     L0
L10:
  lea    D40(%rip),%rax
  mov    %rax,0x10(%rsp)
  lea    D15(%rip),%rax
  mov    %rax,0x18(%rsp)
  lea    D54(%rip),%rax
  mov    %rax,0x20(%rsp)
  lea    D51(%rip),%rax
  mov    %rax,0x28(%rsp)
  lea    D48(%rip),%rax
  mov    %rax,0x30(%rsp)
  lea    D7(%rip),%rax
  mov    %rax,0x38(%rsp)
  lea    D88(%rip),%rax
  mov    %rax,0x40(%rsp)
  movslq 0xc(%rsp),%rax
  mov    0x10(%rsp,%rax,8),%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rsi
  lea    D81(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L10
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D50
data_type: string
name: disassemble_data
arguments: data_tag: D74
data_type: string
name: disassemble_data
arguments: data_tag: D109
data_type: string
name: disassemble_data
arguments: data_tag: D115
data_type: string
name: disassemble_data
arguments: data_tag: D40
data_type: string
name: disassemble_data
arguments: data_tag: D15
data_type: string
name: disassemble_data
arguments: data_tag: D54
data_type: string
name: disassemble_data
arguments: data_tag: D51
data_type: string
name: disassemble_data
arguments: data_tag: D48
data_type: string
name: disassemble_data
arguments: data_tag: D7
data_type: string
name: disassemble_data
arguments: data_tag: D88
data_type: string
name: disassemble_data
arguments: data_tag: D81
data_type: string

tool: D50: King of Tokyo


tool: D74: Jogador escolha um monstro:
---------------------------

tool: D109: 1-Godzilla
2-Feto Gigante Zumbi Nazista
3-Pac-Man
4-Blanca
5-Penha
6-Nemesis

0-SAIR


tool: D115: %d

tool: 

tool: D15: Godzilla

tool: D54: Feto Gigante Zumbi Nazista

tool: D51: Pac-Man

tool: D48: Blanca

tool: D7: Penha

tool: D88: Nemesis

tool: D81: VOCE ESCOLHEU: %s



assistant: ```c
void menu_principal(char *monstro1, char *monstro2) {
  printf("King of Tokyo\n\n");
  printf("Jogador escolha um monstro:\n---------------------------\n");
  printf("1-Godzilla\n2-Feto Gigante Zumbi Nazista\n3-Pac-Man\n4-Blanca\n5-Penha\n6-Nemesis\n\n0-SAIR\n\n");
  int escolha;
  do {
    scanf("%d", &escolha);
  } while (escolha < 0 || escolha > 6);
  if (escolha == 0)
    exit(0);
  char *monstros[] = {"", "Godzilla", "Feto Gigante Zumbi Nazista", "Pac-Man", "Blanca", "Penha", "Nemesis"};
  strcpy(monstro1, monstros[escolha]);
  printf("VOCE ESCOLHEU: %s\n\n", monstro1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<menu_principal>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  lea    D116(%rip),%rdi
  push   %rbp
  push   %rbx
  lea    D35(%rip),%rbx
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbp
  call   <puts@plt>
  lea    D106(%rip),%rdi
  call   <puts@plt>
  lea    D65(%rip),%rdi
  call   <puts@plt>
  nop
L23:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movslq 0xc(%rsp),%rax
  cmp    $0x6,%eax
  ja     L23
  test   %eax,%eax
  je     L3
L20:
  lea    D75(%rip),%rdx
  lea    D38(%rip),%rcx
  mov    %r12,%rdi
  movq   %rdx,%xmm1
  movq   %rcx,%xmm0
  lea    D48(%rip),%rdx
  punpcklqdq %xmm1,%xmm0
  lea    D10(%rip),%rcx
  movq   %rdx,%xmm2
  movaps %xmm0,0x10(%rsp)
  movq   %rcx,%xmm0
  lea    D91(%rip),%rdx
  lea    D19(%rip),%rcx
  punpcklqdq %xmm2,%xmm0
  movq   %rdx,%xmm3
  lea    D47(%rip),%rdx
  movaps %xmm0,0x20(%rsp)
  movq   %rcx,%xmm0
  punpcklqdq %xmm3,%xmm0
  mov    %rdx,0x40(%rsp)
  movaps %xmm0,0x30(%rsp)
  mov    0x10(%rsp,%rax,8),%rsi
  call   <strcpy@plt>
  xor    %eax,%eax
  mov    %r12,%rsi
  lea    D98(%rip),%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x50,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %edi,%edi
  call   <exit@plt>
  movslq 0xc(%rsp),%rax
  jmp    L20
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D116
data_type: string
name: read_data
arguments: label: D35
data_type: string
name: read_data
arguments: label: D106
data_type: string
name: read_data
arguments: label: D65
data_type: string
name: read_data
arguments: label: D75
data_type: string
name: read_data
arguments: label: D38
data_type: string
name: read_data
arguments: label: D48
data_type: string
name: read_data
arguments: label: D10
data_type: string
name: read_data
arguments: label: D91
data_type: string
name: read_data
arguments: label: D19
data_type: string
name: read_data
arguments: label: D47
data_type: string
name: read_data
arguments: label: D98
data_type: string

tool: D116: King of Tokyo


tool: D35: %d

tool: D106: Jogador escolha um monstro:
---------------------------

tool: D65: 1-Godzilla
2-Feto Gigante Zumbi Nazista
3-Pac-Man
4-Blanca
5-Penha
6-Nemesis

0-SAIR


tool: D75: Godzilla

tool: 

tool: D48: Pac-Man

tool: D10: Feto Gigante Zumbi Nazista

tool: D91: Penha

tool: D19: Blanca

tool: D47: Nemesis

tool: D98: VOCE ESCOLHEU: %s



assistant: ```c
void menu_principal(char *monstro1, char *monstro2) {
  printf("King of Tokyo\n\n");
  printf("Jogador escolha um monstro:\n---------------------------\n");
  printf("1-Godzilla\n2-Feto Gigante Zumbi Nazista\n3-Pac-Man\n4-Blanca\n5-Penha\n6-Nemesis\n\n0-SAIR\n\n");
  int escolha;
  do {
    scanf("%d", &escolha);
  } while (escolha < 0 || escolha > 6);
  if (escolha == 0)
    exit(0);
  char *monstros[] = {"", "Godzilla", "Feto Gigante Zumbi Nazista", "Pac-Man", "Blanca", "Penha", "Nemesis"};
  strcpy(monstro1, monstros[escolha]);
  printf("VOCE ESCOLHEU: %s\n\n", monstro1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %edi,%edx
  mov    %dl,-0x4(%rbp)
  mov    %ecx,%edx
  mov    %dl,-0x8(%rbp)
  mov    %al,-0xc(%rbp)
  movsbl -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <ft_putchar@plt>
  movsbl -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <ft_putchar@plt>
  movsbl -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <ft_putchar@plt>
  cmpb   $0x37,-0x4(%rbp)
  je     L4
  mov    $0x2c,%edi
  call   <ft_putchar@plt>
  mov    $0x20,%edi
  call   <ft_putchar@plt>
L4:
  nop
  leave
  ret
```

assistant: ```c
void print(char a, char b, char c) {
  ft_putchar(a);
  ft_putchar(b);
  ft_putchar(c);
  if (a != 55) {
    ft_putchar(',');
    ft_putchar(' ');
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %esi,%ebp
  mov    %edx,%ebx
  movsbl %dil,%edi
  call   <ft_putchar@plt>
  movsbl %bpl,%edi
  call   <ft_putchar@plt>
  movsbl %bl,%edi
  call   <ft_putchar@plt>
  cmp    $0x37,%r12b
  jne    L5
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  mov    $0x2c,%edi
  call   <ft_putchar@plt>
  mov    $0x20,%edi
  call   <ft_putchar@plt>
  jmp    L18
```

assistant: ```c
void print(char a, char b, char c) {
  ft_putchar(a);
  ft_putchar(b);
  ft_putchar(c);
  if (a != 55) {
    ft_putchar(',');
    ft_putchar(' ');
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<print>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  movsbl %dil,%edi
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  call   <ft_putchar@plt>
  movsbl %bpl,%edi
  call   <ft_putchar@plt>
  movsbl %bl,%edi
  call   <ft_putchar@plt>
  cmp    $0x37,%r12b
  jne    L2
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L2:
  mov    $0x2c,%edi
  call   <ft_putchar@plt>
  pop    %rbx
  mov    $0x20,%edi
  pop    %rbp
  pop    %r12
  jmp    1050 <ft_putchar@plt>
```

assistant: ```c
void print(char a, char b, char c) {
  ft_putchar(a);
  ft_putchar(b);
  ft_putchar(c);
  if (a != 55) {
    ft_putchar(',');
    ft_putchar(' ');
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  movsbl %dil,%edi
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  call   <ft_putchar@plt>
  movsbl %bpl,%edi
  call   <ft_putchar@plt>
  movsbl %bl,%edi
  call   <ft_putchar@plt>
  cmp    $0x37,%r12b
  jne    L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L1:
  mov    $0x2c,%edi
  call   <ft_putchar@plt>
  pop    %rbx
  mov    $0x20,%edi
  pop    %rbp
  pop    %r12
  jmp    1050 <ft_putchar@plt>
```

assistant: ```c
void print(char a, char b, char c) {
  ft_putchar(a);
  ft_putchar(b);
  ft_putchar(c);
  if (a != 55) {
    ft_putchar(',');
    ft_putchar(' ');
  }
}
```