ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<unmangle_path>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x5c,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L55
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <read_oct@plt>
  cmp    $0x20,%al
  je     L24
L55:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L24:
  movb   $0x20,(%rbx)
  mov    $0x3,%r12d
  jmp    L43
L25:
  movzbl 0x0(%rbp),%eax
  mov    %al,(%rbx)
  test   %al,%al
  je     L55
L43:
  add    $0x1,%rbx
  movslq %r12d,%rbp
  add    %rbx,%rbp
  cmpb   $0x5c,0x0(%rbp)
  jne    L25
  mov    %rbp,%rdi
  call   <read_oct@plt>
  cmp    $0x20,%al
  jne    L25
  add    $0x3,%r12d
  movb   $0x20,(%rbx)
  jmp    L43
```

assistant: ```c
void unmangle_path(char *path) {
  char *p = strchr(path, '\\');
  if (p && read_oct(p) == ' ') {
    *p = ' ';
    int i = 3;
    do {
      p++;
      if (*(p + i) == '\\' && read_oct(p + i) == ' ') {
        *p = ' ';
        i += 3;
      } else
        *p = *(p + i);
    } while (*p);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<unmangle_path>:
  endbr64
  push   %r13
  mov    $0x5c,%esi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strchr@plt>
  test   %rax,%rax
  je     L3
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <read_oct@plt>
  cmp    $0x20,%al
  je     L68
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L68:
  movb   $0x20,(%rbx)
  mov    $0x3,%r13d
  mov    $0x3,%r12d
  jmp    L78
  nopl   0x0(%rax)
L2:
  mov    %al,(%rbx)
  test   %al,%al
  je     L3
L78:
  add    $0x1,%rbx
  lea    (%rbx,%r12,1),%rbp
  movzbl 0x0(%rbp),%eax
  cmp    $0x5c,%al
  jne    L2
  mov    %rbp,%rdi
  call   <read_oct@plt>
  cmp    $0x20,%al
  je     L57
  movzbl 0x0(%rbp),%eax
  jmp    L2
L57:
  add    $0x3,%r13d
  movb   $0x20,(%rbx)
  movslq %r13d,%r12
  jmp    L78
```

assistant: ```c
void unmangle_path(char *path) {
  char *p = strchr(path, '\\');
  if (p && read_oct(p) == ' ') {
    *p = ' ';
    int i = 3;
    do {
      p++;
      if (*(p + i) == '\\' && read_oct(p + i) == ' ') {
        *p = ' ';
        i += 3;
      } else
        *p = *(p + i);
    } while (*p);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<unmangle_path>:
  endbr64
  push   %r13
  mov    $0x5c,%esi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strchr@plt>
  test   %rax,%rax
  je     L6
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <read_oct@plt>
  cmp    $0x20,%al
  je     L25
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L25:
  movb   $0x20,(%rbx)
  mov    $0x3,%r13d
  mov    $0x3,%r12d
  jmp    L53
  nopl   0x0(%rax)
L60:
  mov    %al,(%rbx)
  test   %al,%al
  je     L6
L53:
  add    $0x1,%rbx
  lea    (%rbx,%r12,1),%rbp
  movzbl 0x0(%rbp),%eax
  cmp    $0x5c,%al
  jne    L60
  mov    %rbp,%rdi
  call   <read_oct@plt>
  cmp    $0x20,%al
  je     L68
  movzbl 0x0(%rbp),%eax
  jmp    L60
L68:
  add    $0x3,%r13d
  movb   $0x20,(%rbx)
  movslq %r13d,%r12
  jmp    L53
```

assistant: ```c
void unmangle_path(char *path) {
  char *p = strchr(path, '\\');
  if (p && read_oct(p) == ' ') {
    *p = ' ';
    int i = 3;
    do {
      p++;
      if (*(p + i) == '\\' && read_oct(p + i) == ' ') {
        *p = ' ';
        i += 3;
      } else
        *p = *(p + i);
    } while (*p);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_strmapi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strlen@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L28
  mov    $0x0,%eax
  jmp    L21
L8:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%eax
  mov    -0x20(%rbp),%ecx
  movslq %ecx,%rsi
  mov    -0x18(%rbp),%rcx
  lea    (%rsi,%rcx,1),%rbx
  mov    -0x30(%rbp),%rcx
  mov    %edx,%esi
  mov    %eax,%edi
  call   *%rcx
  mov    %al,(%rbx)
  addl   $0x1,-0x20(%rbp)
L28:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L8
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
L21:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *ft_strmapi(char const *s1, char (*f)(unsigned int, char)) {
  char *sfresh;
  int i;
  int j;
  i = 0;
  j = ft_strlen(s1);
  sfresh = malloc(sizeof(char) * j + 1);
  if (sfresh == NULL)
    return (NULL);
  while (s1[i]) {
    sfresh[i] = f(i, s1[i]);
    i++;
  }
  sfresh[i] = '\0';
  return (sfresh);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ft_strmapi>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  call   <ft_strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L7
  movzbl (%r12),%esi
  test   %sil,%sil
  je     L11
  mov    $0x0,%ebx
L0:
  movsbl %sil,%esi
  mov    %ebx,%edi
  call   *%r13
  mov    %al,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
  movzbl (%r12,%rbx,1),%esi
  test   %sil,%sil
  jne    L0
L15:
  movb   $0x0,0x0(%rbp,%rbx,1)
L7:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  mov    $0x0,%ebx
  jmp    L15
```

assistant: ```c
char *ft_strmapi(char const *s1, char (*f)(unsigned int, char)) {
  char *sfresh;
  int i;
  int j;
  i = 0;
  j = ft_strlen(s1);
  sfresh = malloc(sizeof(char) * j + 1);
  if (sfresh == NULL)
    return (NULL);
  while (s1[i]) {
    sfresh[i] = f(i, s1[i]);
    i++;
  }
  sfresh[i] = '\0';
  return (sfresh);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ft_strmapi>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <ft_strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L3
  movsbl 0x0(%rbp),%esi
  test   %sil,%sil
  je     L5
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L24:
  mov    %ebx,%edi
  call   *%r13
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  movsbl 0x0(%rbp,%rbx,1),%esi
  test   %sil,%sil
  jne    L24
  add    %r12,%rbx
L21:
  movb   $0x0,(%rbx)
L3:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %rax,%rbx
  jmp    L21
```

assistant: ```c
char *ft_strmapi(char const *s1, char (*f)(unsigned int, char)) {
  char *sfresh;
  int i;
  int j;
  i = 0;
  j = ft_strlen(s1);
  sfresh = malloc(sizeof(char) * j + 1);
  if (sfresh == NULL)
    return (NULL);
  while (s1[i]) {
    sfresh[i] = f(i, s1[i]);
    i++;
  }
  sfresh[i] = '\0';
  return (sfresh);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ft_strmapi>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <ft_strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L37
  movsbl 0x0(%rbp),%esi
  test   %sil,%sil
  je     L2
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L20:
  mov    %ebx,%edi
  call   *%r13
  mov    %al,(%r12,%rbx,1)
  add    $0x1,%rbx
  movsbl 0x0(%rbp,%rbx,1),%esi
  test   %sil,%sil
  jne    L20
  add    %r12,%rbx
L21:
  movb   $0x0,(%rbx)
L37:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %rax,%rbx
  jmp    L21
```

assistant: ```c
char *ft_strmapi(char const *s1, char (*f)(unsigned int, char)) {
  char *sfresh;
  int i;
  int j;
  i = 0;
  j = ft_strlen(s1);
  sfresh = malloc(sizeof(char) * j + 1);
  if (sfresh == NULL)
    return (NULL);
  while (s1[i]) {
    sfresh[i] = f(i, s1[i]);
    i++;
  }
  sfresh[i] = '\0';
  return (sfresh);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memized_cut_rod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L8
L18:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0xffff0001,(%rax)
  addl   $0x1,-0x10(%rbp)
L8:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L18
  mov    -0x8(%rbp),%rdx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <memized_cut_rod_aux@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int memized_cut_rod(int *p, int n) {
  int i, *r, max;
  r = (int *)malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    r[i] = -65535;
  }
  max = memized_cut_rod_aux(p, n, r);
  free(r);
  return (max);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<memized_cut_rod>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebx
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %ebx,%ebx
  jle    L13
  mov    %ebx,%edx
  lea    (%rax,%rdx,4),%rdx
L16:
  movl   $0xffff0001,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L16
L13:
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <memized_cut_rod_aux@plt>
  mov    %eax,%ebx
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int memized_cut_rod(int *p, int n) {
  int i, *r, max;
  r = (int *)malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    r[i] = -65535;
  }
  max = memized_cut_rod_aux(p, n, r);
  free(r);
  return (max);
}
```
user: What is the c source code of the assembly code below:

```asm
<memized_cut_rod>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  lea    0x0(,%rsi,4),%edi
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L13
  mov    %ebp,%edx
  lea    (%rax,%rdx,4),%rdx
  nopw   0x0(%rax,%rax,1)
L9:
  movl   $0xffff0001,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L9
L13:
  mov    %r12,%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <memized_cut_rod_aux@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <free@plt>
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int memized_cut_rod(int *p, int n) {
  int i, *r, max;
  r = (int *)malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    r[i] = -65535;
  }
  max = memized_cut_rod_aux(p, n, r);
  free(r);
  return (max);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memized_cut_rod>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  lea    0x0(,%rsi,4),%edi
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L10
  lea    -0x1(%rbp),%eax
  cmp    $0x2,%eax
  jbe    L28
  mov    %ebp,%edx
  movdqa D5(%rip),%xmm0
  mov    %r12,%rax
  shr    $0x2,%edx
  shl    $0x4,%rdx
  add    %r12,%rdx
  nopl   0x0(%rax)
L68:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L68
  mov    %ebp,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%bpl
  je     L10
L47:
  movslq %eax,%rdx
  movl   $0xffff0001,(%r12,%rdx,4)
  lea    0x0(,%rdx,4),%rcx
  lea    0x1(%rax),%edx
  cmp    %edx,%ebp
  jle    L10
  movl   $0xffff0001,0x4(%r12,%rcx,1)
  add    $0x2,%eax
  cmp    %eax,%ebp
  jle    L10
  movl   $0xffff0001,0x8(%r12,%rcx,1)
L10:
  mov    %r12,%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <memized_cut_rod_aux@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <free@plt>
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  xor    %eax,%eax
  jmp    L47
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
type_infered: byte[16]

tool: {"D5": [0x01,0x00,0xff,0xff,0x01,0x00,0xff,0xff,0x01,0x00,0xff,0xff,0x01,0x00,0xff,0xff]}

assistant: ```c
int memized_cut_rod(int *p, int n) {
  int i, *r, max;
  r = (int *)malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    r[i] = -65535;
  }
  max = memized_cut_rod_aux(p, n, r);
  free(r);
  return (max);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  lea    -0xc(%rbp),%rdx
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <waitid@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
int main(void) {
  siginfo_t info;
  return waitid(P_PID, getpid(), &info, 0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %eax,%esi
  lea    0x4(%rsp),%rdx
  mov    $0x0,%ecx
  mov    D9(%rip),%rax
  mov    (%rax),%edi
  call   <waitid@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  siginfo_t info;
  return waitid(P_PID, getpid(), &info, 0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <getpid@plt>
  lea    0x4(%rsp),%rdx
  xor    %ecx,%ecx
  mov    %eax,%esi
  mov    D3(%rip),%rax
  mov    (%rax),%edi
  call   <waitid@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  siginfo_t info;
  return waitid(P_PID, getpid(), &info, 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <getpid@plt>
  lea    0x4(%rsp),%rdx
  xor    %ecx,%ecx
  mov    %eax,%esi
  mov    D2(%rip),%rax
  mov    (%rax),%edi
  call   <waitid@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  siginfo_t info;
  return waitid(P_PID, getpid(), &info, 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    -0x40(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <palindromo@plt>
  test   %rax,%rax
  je     L21
  lea    D33(%rip),%rax
  jmp    L19
L21:
  lea    D29(%rip),%rax
L19:
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D34
type: string
name: retrieve_data
arguments: label: D39
type: string
name: retrieve_data
arguments: label: D33
type: string
name: retrieve_data
arguments: label: D29
type: string

tool: D34: Informe uma string: 

tool: D39: %s

tool: D33: Eh palindromo

tool: D29: Nao eh palindromo

assistant: ```c
int main(int argc, const char *argv[]) {
  char string[50];
  printf("Informe uma string: ");
  scanf("%s", string);
  printf("%s\n", palindromo(string, strlen(string)) ? "Eh palindromo" : "Nao eh palindromo");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <palindromo@plt>
  test   %rax,%rax
  lea    D28(%rip),%rdi
  lea    D2(%rip),%rax
  cmove  %rax,%rdi
  call   <puts@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D10
type_infered: string
name: disassemble_data
arguments: data_label: D8
type_infered: string
name: disassemble_data
arguments: data_label: D28
type_infered: string
name: disassemble_data
arguments: data_label: D2
type_infered: string

tool: D10: Informe uma string: 

tool: D8: %s

tool: D28: Eh palindromo

tool: D2: Nao eh palindromo

assistant: ```c
int main(int argc, const char *argv[]) {
  char string[50];
  printf("Informe uma string: ");
  scanf("%s", string);
  printf("%s\n", palindromo(string, strlen(string)) ? "Eh palindromo" : "Nao eh palindromo");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D39(%rip),%rdi
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <palindromo@plt>
  lea    D28(%rip),%rdi
  test   %rax,%rax
  lea    D2(%rip),%rax
  cmove  %rax,%rdi
  call   <puts@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D39
type_infered: string
name: parse_data
arguments: data_tag: D29
type_infered: string
name: parse_data
arguments: data_tag: D28
type_infered: string
name: parse_data
arguments: data_tag: D2
type_infered: string

tool: D39: Informe uma string: 

tool: D29: %s

tool: D28: Eh palindromo

tool: D2: Nao eh palindromo

assistant: ```c
int main(int argc, const char *argv[]) {
  char string[50];
  printf("Informe uma string: ");
  scanf("%s", string);
  printf("%s\n", palindromo(string, strlen(string)) ? "Eh palindromo" : "Nao eh palindromo");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D11(%rip),%rdi
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <palindromo@plt>
  lea    D21(%rip),%rdi
  test   %rax,%rax
  lea    D32(%rip),%rax
  cmove  %rax,%rdi
  call   <puts@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D11
data_type: string
name: read_data
arguments: data_tag: D7
data_type: string
name: read_data
arguments: data_tag: D21
data_type: string
name: read_data
arguments: data_tag: D32
data_type: string

tool: D11: Informe uma string: 

tool: D7: %s

tool: D21: Eh palindromo

tool: D32: Nao eh palindromo

assistant: ```c
int main(int argc, const char *argv[]) {
  char string[50];
  printf("Informe uma string: ");
  scanf("%s", string);
  printf("%s\n", palindromo(string, strlen(string)) ? "Eh palindromo" : "Nao eh palindromo");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sqlite_encode_binary>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %rdx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L123
  mov    %rdi,%r13
  mov    %esi,%r12d
  mov    %rsp,%rdi
  mov    $0x400,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  lea    -0x1(%r12),%r14d
  movslq %r14d,%rax
  add    %r13,%rax
  mov    %r13,%r11
  mov    %r14d,%r14d
  movslq %r12d,%rdx
  lea    -0x2(%r13,%rdx,1),%rcx
  sub    %r14,%rcx
L4:
  movzbl (%rax),%edx
  addl   $0x1,(%rsp,%rdx,4)
  sub    $0x1,%rax
  cmp    %rcx,%rax
  jne    L4
  lea    0x4(%rsp),%rdi
  mov    %r12d,%r10d
  mov    $0x2,%edx
  mov    $0x1,%ecx
  jmp    L127
L123:
  mov    $0x1,%ebp
  test   %rdx,%rdx
  je     L41
  movb   $0x78,(%rdx)
  movb   $0x0,0x1(%rdx)
  jmp    L41
L65:
  cmp    $0xff,%r8d
  jg     L24
L168:
  add    $0x1,%ecx
  add    $0x1,%edx
  add    $0x4,%rdi
L127:
  mov    %ecx,%r9d
  cmp    $0x27,%ecx
  je     L168
  mov    %edx,%r8d
  movzbl %dl,%esi
  mov    (%rdi),%eax
  add    (%rsp,%rsi,4),%eax
  lea    0x26(%rdx),%esi
  movzbl %sil,%esi
  add    (%rsp,%rsi,4),%eax
  cmp    %r10d,%eax
  jge    L65
  test   %eax,%eax
  je     L118
  mov    %eax,%r10d
  mov    %ecx,%ebp
  jmp    L65
L24:
  mov    %r10d,%eax
  mov    %ebp,%r9d
L118:
  lea    0x1(%rax,%r12,1),%ebp
  test   %rbx,%rbx
  je     L41
  mov    %r9d,%r8d
  mov    %r9b,(%rbx)
  lea    0x1(%r13,%r14,1),%rdi
  mov    $0x1,%ebp
  jmp    L16
L42:
  lea    0x1(%rbp),%ecx
  movslq %ebp,%rbp
  movb   $0x1,(%rbx,%rbp,1)
  add    $0x1,%edx
L101:
  lea    0x1(%rcx),%ebp
  movslq %ecx,%rsi
  mov    %dl,(%rbx,%rsi,1)
  add    $0x1,%r11
  cmp    %rdi,%r11
  je     L126
L16:
  movzbl (%r11),%edx
  sub    %r8d,%edx
  cmp    $0x1,%dl
  jbe    L42
  mov    %ebp,%ecx
  cmp    $0x27,%dl
  jne    L101
  jmp    L42
L126:
  movslq %ebp,%rdx
  movb   $0x0,(%rbx,%rdx,1)
  add    %r12d,%eax
  cmp    %ecx,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
L41:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  mov    %ebp,%eax
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sqlite_encode_binary(const unsigned char *in, int n, unsigned char *out) {
  int i, j, e, m;
  unsigned char x;
  int cnt[256];
  if (n <= 0) {
    if (out) {
      out[0] = 'x';
      out[1] = 0;
    }
    return 1;
  }
  memset(cnt, 0, sizeof(cnt));
  for (i = n - 1; i >= 0; i--) {
    cnt[in[i]]++;
  }
  m = n;
  for (i = 1; i < 256; i++) {
    int sum;
    if (i == '\'')
      continue;
    sum = cnt[i] + cnt[(i + 1) & 0xff] + cnt[(i + '\'') & 0xff];
    if (sum < m) {
      m = sum;
      e = i;
      if (m == 0)
        break;
    }
  }
  if (out == 0) {
    return n + m + 1;
  }
  out[0] = e;
  j = 1;
  for (i = 0; i < n; i++) {
    x = in[i] - e;
    if (x == 0 || x == 1 || x == '\'') {
      out[j++] = 1;
      x++;
    }
    out[j++] = x;
  }
  out[j] = 0;
  assert(j == n + m + 1);
  return j;
}
```