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

```asm
<RandomFunc>:
  endbr64
  push   %rbx
  mov    %rsi,%r8
  sub    $0x10,%rsp
  movzwl (%rdi),%ecx
  movzwl 0x2(%rdi),%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  shl    $0x7,%ecx
  and    $0xb500,%cx
  mov    %cx,0x6(%rsp)
  test   %di,%di
  je     L33
  lea    0x6(%rsp),%rdx
  jmp    L31
  nopl   0x0(%rax)
L3:
  xor    %ecx,%ecx
  mov    0x1(%rdx),%cl
  mov    (%rdx),%ch
  add    $0x1,%eax
  add    $0x2,%rdx
  mov    %cx,-0x2(%rdx)
  movzwl 0x6(%rsp),%ecx
  cmp    %di,%ax
  je     L33
L31:
  cmp    %cx,%ax
  jb     L3
  movzwl %cx,%esi
  shr    $0x2,%cx
  add    $0x1,%eax
  add    $0x2,%rdx
  and    $0x6,%ecx
  or     $0x1,%ecx
  sar    %cl,%esi
  mov    %si,0x6(%rsp)
  mov    %esi,%ecx
  cmp    %di,%ax
  jne    L31
L33:
  shr    $0x5,%cx
  xor    $0x9c9f,%cx
  mov    %cx,(%r8)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x10,%rsp
  pop    %rbx
  ret
L40:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  {
    state[0UL] = (input[0UL] & 914778474UL) << (unsigned short)7;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          copy11 = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy11;
        } else {
          state[local1] >>= ((state[0UL] >> (unsigned short)3) & (unsigned short)7) | 1UL;
          state[local1] >>= ((state[local1] >> (unsigned short)3) & (unsigned short)7) | 1UL;
        }
      } else {
        state[0UL] = state[local1] >> (((state[local1] >> (unsigned short)2) & (unsigned short)7) | 1UL);
      }
      local1++;
    }
    output[0UL] = (state[0UL] ^ 380867558UL) >> (unsigned short)5;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<stoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0xa,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x30,%al
  jne    L58
  mov    -0x28(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x78,%al
  je     L24
  mov    -0x28(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x58,%al
  jne    L0
L24:
  addq   $0x2,-0x28(%rbp)
  movl   $0x10,-0x18(%rbp)
  jmp    L54
L0:
  movl   $0x8,-0x18(%rbp)
  addq   $0x1,-0x28(%rbp)
L54:
  jmp    L58
L33:
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,%ebx
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <xdtoi@plt>
  add    %ebx,%eax
  mov    %eax,-0x14(%rbp)
L58:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L33
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int stoi(s)
char *s;
{
  int base = 10;
  int n = 0;
  if (*s == '0') {
    if (s[1] == 'x' || s[1] == 'X') {
      s += 2;
      base = 16;
    } else {
      base = 8;
      s += 1;
    }
  }
  while (*s)
    n = n * base + xdtoi(*s++);
  return n;
}
```
user: Translate the below assembly code into C programming language:

```asm
<stoi>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0xa,%r12d
  cmpb   $0x30,(%rdi)
  je     L31
L26:
  movzbl 0x0(%rbp),%edi
  mov    $0x0,%ebx
  test   %dil,%dil
  je     L46
L17:
  imul   %r12d,%ebx
  add    $0x1,%rbp
  movsbl %dil,%edi
  mov    $0x0,%eax
  call   <xdtoi@plt>
  add    %eax,%ebx
  movzbl 0x0(%rbp),%edi
  test   %dil,%dil
  jne    L17
L46:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L31:
  movzbl 0x1(%rdi),%eax
  and    $0xffffffdf,%eax
  cmp    $0x58,%al
  jne    L29
  add    $0x2,%rbp
  mov    $0x10,%r12d
  jmp    L26
L29:
  add    $0x1,%rbp
  mov    $0x8,%r12d
  jmp    L26
```

assistant: ```c
int stoi(s)
char *s;
{
  int base = 10;
  int n = 0;
  if (*s == '0') {
    if (s[1] == 'x' || s[1] == 'X') {
      s += 2;
      base = 16;
    } else {
      base = 8;
      s += 1;
    }
  }
  while (*s)
    n = n * base + xdtoi(*s++);
  return n;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<stoi>:
  endbr64
  push   %r12
  mov    $0xa,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movsbl (%rdi),%edi
  cmp    $0x30,%dil
  je     L36
L38:
  xor    %ebx,%ebx
  test   %dil,%dil
  je     L44
  nopl   0x0(%rax)
L27:
  imul   %r12d,%ebx
  add    $0x1,%rbp
  xor    %eax,%eax
  call   <xdtoi@plt>
  movsbl 0x0(%rbp),%edi
  add    %eax,%ebx
  test   %dil,%dil
  jne    L27
L44:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L36:
  movsbl 0x1(%rbp),%edi
  mov    %edi,%eax
  and    $0xffffffdf,%eax
  cmp    $0x58,%al
  jne    L4
  movsbl 0x2(%rbp),%edi
  mov    $0x10,%r12d
  add    $0x2,%rbp
  jmp    L38
  nopl   (%rax)
L4:
  add    $0x1,%rbp
  mov    $0x8,%r12d
  jmp    L38
```

assistant: ```c
int stoi(s)
char *s;
{
  int base = 10;
  int n = 0;
  if (*s == '0') {
    if (s[1] == 'x' || s[1] == 'X') {
      s += 2;
      base = 16;
    } else {
      base = 8;
      s += 1;
    }
  }
  while (*s)
    n = n * base + xdtoi(*s++);
  return n;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<stoi>:
  endbr64
  push   %r12
  mov    $0xa,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movsbl (%rdi),%edi
  cmp    $0x30,%dil
  je     L5
L23:
  xor    %ebx,%ebx
  test   %dil,%dil
  je     L18
  nopl   0x0(%rax)
L28:
  imul   %r12d,%ebx
  add    $0x1,%rbp
  xor    %eax,%eax
  call   <xdtoi@plt>
  movsbl 0x0(%rbp),%edi
  add    %eax,%ebx
  test   %dil,%dil
  jne    L28
L18:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L5:
  movsbl 0x1(%rbp),%edi
  mov    %edi,%eax
  and    $0xffffffdf,%eax
  cmp    $0x58,%al
  jne    L30
  movsbl 0x2(%rbp),%edi
  mov    $0x10,%r12d
  add    $0x2,%rbp
  jmp    L23
  nopl   (%rax)
L30:
  add    $0x1,%rbp
  mov    $0x8,%r12d
  jmp    L23
```

assistant: ```c
int stoi(s)
char *s;
{
  int base = 10;
  int n = 0;
  if (*s == '0') {
    if (s[1] == 'x' || s[1] == 'X') {
      s += 2;
      base = 16;
    } else {
      base = 8;
      s += 1;
    }
  }
  while (*s)
    n = n * base + xdtoi(*s++);
  return n;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  lea    -0x2(%rax),%edx
  mov    -0x20(%rbp),%eax
  sub    $0x2,%eax
  imul   %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x1,-0x1c(%rbp)
  jmp    L38
L42:
  mov    -0x20(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x18(%rbp)
L76:
  mov    $0x20,%edi
  call   <putchar@plt>
  cmpl   $0x1,-0x1c(%rbp)
  jne    L14
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L37
L14:
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    $0x23,%edi
  call   <putchar@plt>
  movl   $0x2,-0x14(%rbp)
L39:
  mov    $0x23,%edi
  call   <putchar@plt>
  mov    $0x23,%edi
  call   <putchar@plt>
  addl   $0x1,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L39
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
L37:
  addl   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L76
  movl   $0x0,-0x18(%rbp)
  jmp    L66
L55:
  mov    $0x2a,%edi
  call   <putchar@plt>
  addl   $0x1,-0x18(%rbp)
L66:
  mov    -0x20(%rbp),%eax
  add    %eax,%eax
  sub    $0x1,%eax
  cmp    %eax,-0x18(%rbp)
  jl     L55
  addl   $0x1,-0x1c(%rbp)
L38:
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jl     L42
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D12
type_infered: string
name: disassemble_data
arguments: tag: D1
type_infered: string

tool: D12: %d

tool: D1: %d


assistant: ```c
int main() {
  int n, c, s, i, j;
  scanf("%d", &n);
  int br = 0;
  br = (n - 2) * (n - 2);
  printf("%d\n", br);
  for (c = 1; c < n; c++) {
    s = n - c;
    i = 0;
    do {
      printf(" ");
      if (c == 1) {
        printf("*");
        printf("\n");
      } else {
        printf("*");
        printf("#");
        j = 2;
        do {
          printf("#");
          printf("#");
          j++;
        } while (j < c);
        printf("*");
        printf("\n");
      }
      i++;
    } while (i < s);
    for (i = 0; i < (n * 2) - 1; i++)
      printf("*");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x2(%rax),%esi
  imul   %esi,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%r13d
  cmp    $0x1,%r13d
  jle    L108
  mov    $0x1,%ebp
  jmp    L98
L66:
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    $0x23,%edi
  call   <putchar@plt>
  mov    $0x2,%ebx
L57:
  mov    $0x23,%edi
  call   <putchar@plt>
  mov    $0x23,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jl     L57
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
L42:
  add    $0x1,%r12d
  cmp    %r12d,%r13d
  jle    L36
L91:
  mov    $0x20,%edi
  call   <putchar@plt>
  cmp    $0x1,%ebp
  jne    L66
  mov    $0x2a,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L42
L36:
  mov    0x4(%rsp),%eax
  add    %eax,%eax
  cmp    $0x1,%eax
  jle    L76
  mov    $0x0,%ebx
L39:
  mov    $0x2a,%edi
  call   <putchar@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax,%rax,1),%eax
  cmp    %ebx,%eax
  jg     L39
L76:
  add    $0x1,%ebp
  mov    0x4(%rsp),%r13d
  cmp    %ebp,%r13d
  jle    L108
L98:
  sub    %ebp,%r13d
  mov    $0x0,%r12d
  jmp    L91
L108:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D10
type_guessed: string
name: parse_data
arguments: tag: D1
type_guessed: string

tool: D10: %d

tool: D1: %d


assistant: ```c
int main() {
  int n, c, s, i, j;
  scanf("%d", &n);
  int br = 0;
  br = (n - 2) * (n - 2);
  printf("%d\n", br);
  for (c = 1; c < n; c++) {
    s = n - c;
    i = 0;
    do {
      printf(" ");
      if (c == 1) {
        printf("*");
        printf("\n");
      } else {
        printf("*");
        printf("#");
        j = 2;
        do {
          printf("#");
          printf("#");
          j++;
        } while (j < c);
        printf("*");
        printf("\n");
      }
      i++;
    } while (i < s);
    for (i = 0; i < (n * 2) - 1; i++)
      printf("*");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  lea    D0(%rip),%rdi
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    D3(%rip),%rdi
  lea    -0x2(%rax),%esi
  xor    %eax,%eax
  imul   %esi,%esi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x1,%eax
  jle    L41
  nopl   0x0(%rax)
L32:
  sub    %ebp,%eax
  xor    %r13d,%r13d
  mov    %eax,%r12d
  jmp    L66
  nopw   0x0(%rax,%rax,1)
L25:
  mov    $0x2a,%edi
  add    $0x1,%r13d
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    %r13d,%r12d
  jle    L27
L66:
  mov    $0x20,%edi
  call   <putchar@plt>
  cmp    $0x1,%ebp
  je     L25
  mov    $0x2a,%edi
  mov    $0x2,%ebx
  call   <putchar@plt>
  mov    $0x23,%edi
  call   <putchar@plt>
  nopl   (%rax)
L98:
  mov    $0x23,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  mov    $0x23,%edi
  call   <putchar@plt>
  cmp    %ebp,%ebx
  jl     L98
  jmp    L25
  nopl   (%rax)
L27:
  mov    0x4(%rsp),%eax
  lea    (%rax,%rax,1),%edx
  cmp    $0x1,%edx
  jle    L94
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L30:
  mov    $0x2a,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax,%rax,1),%edx
  cmp    %ebx,%edx
  jg     L30
L94:
  add    $0x1,%ebp
  cmp    %ebp,%eax
  jg     L32
L41:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D0
type: string
name: unpack_data
arguments: tag: D3
type: string

tool: D0: %d

tool: D3: %d


assistant: ```c
int main() {
  int n, c, s, i, j;
  scanf("%d", &n);
  int br = 0;
  br = (n - 2) * (n - 2);
  printf("%d\n", br);
  for (c = 1; c < n; c++) {
    s = n - c;
    i = 0;
    do {
      printf(" ");
      if (c == 1) {
        printf("*");
        printf("\n");
      } else {
        printf("*");
        printf("#");
        j = 2;
        do {
          printf("#");
          printf("#");
          j++;
        } while (j < c);
        printf("*");
        printf("\n");
      }
      i++;
    } while (i < s);
    for (i = 0; i < (n * 2) - 1; i++)
      printf("*");
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D2(%rip),%rdi
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    D10(%rip),%rdi
  lea    -0x2(%rax),%esi
  xor    %eax,%eax
  imul   %esi,%esi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x1,%eax
  jle    L88
  nopl   0x0(%rax)
L5:
  sub    %ebp,%eax
  xor    %r13d,%r13d
  mov    %eax,%r12d
  jmp    L99
  nopw   0x0(%rax,%rax,1)
L52:
  mov    $0x2a,%edi
  add    $0x1,%r13d
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    %r13d,%r12d
  jle    L20
L99:
  mov    $0x20,%edi
  call   <putchar@plt>
  cmp    $0x1,%ebp
  je     L52
  mov    $0x2a,%edi
  mov    $0x2,%ebx
  call   <putchar@plt>
  mov    $0x23,%edi
  call   <putchar@plt>
  nopl   (%rax)
L83:
  mov    $0x23,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  mov    $0x23,%edi
  call   <putchar@plt>
  cmp    %ebp,%ebx
  jl     L83
  jmp    L52
  nopl   (%rax)
L20:
  mov    0x4(%rsp),%eax
  lea    (%rax,%rax,1),%edx
  cmp    $0x1,%edx
  jle    L17
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L94:
  mov    $0x2a,%edi
  add    $0x1,%ebx
  call   <putchar@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax,%rax,1),%edx
  cmp    %ebx,%edx
  jg     L94
L17:
  add    $0x1,%ebp
  cmp    %eax,%ebp
  jl     L5
L88:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L74
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L74:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type: string
name: retrieve_data
arguments: data_label: D10
type: string

tool: D2: %d

tool: D10: %d


assistant: ```c
int main() {
  int n, c, s, i, j;
  scanf("%d", &n);
  int br = 0;
  br = (n - 2) * (n - 2);
  printf("%d\n", br);
  for (c = 1; c < n; c++) {
    s = n - c;
    i = 0;
    do {
      printf(" ");
      if (c == 1) {
        printf("*");
        printf("\n");
      } else {
        printf("*");
        printf("#");
        j = 2;
        do {
          printf("#");
          printf("#");
          j++;
        } while (j < c);
        printf("*");
        printf("\n");
      }
      i++;
    } while (i < s);
    for (i = 0; i < (n * 2) - 1; i++)
      printf("*");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rev_word>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L20
L10:
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L20
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <rev_str@plt>
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
L20:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <rev_str@plt>
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <rev_str@plt>
  nop
  leave
  ret
```

assistant: ```c
void rev_word(char *str) {
  char *start = str;
  char *curr = str;
  while (*curr) {
    curr++;
    if (*curr == ' ') {
      rev_str(start, curr - 1);
      start = curr + 1;
    }
  }
  rev_str(start, curr - 1);
  rev_str(str, curr - 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<rev_word>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  cmpb   $0x0,(%rdi)
  je     L18
  mov    %rdi,%rbx
  jmp    L19
L2:
  cmpb   $0x0,(%rbx)
  je     L43
L19:
  mov    %rbx,%rsi
  add    $0x1,%rbx
  cmpb   $0x20,(%rbx)
  jne    L2
  call   <rev_str@plt>
  lea    0x1(%rbx),%rdi
  jmp    L2
L18:
  mov    %rdi,%rbx
L43:
  sub    $0x1,%rbx
  mov    %rbx,%rsi
  call   <rev_str@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <rev_str@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void rev_word(char *str) {
  char *start = str;
  char *curr = str;
  while (*curr) {
    curr++;
    if (*curr == ' ') {
      rev_str(start, curr - 1);
      start = curr + 1;
    }
  }
  rev_str(start, curr - 1);
  rev_str(str, curr - 1);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rev_word>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  cmpb   $0x0,(%rdi)
  je     L35
  mov    %rdi,%rbx
  jmp    L34
  nopl   (%rax)
L49:
  cmpb   $0x0,(%rbx)
  je     L51
L34:
  mov    %rbx,%rbp
  add    $0x1,%rbx
  cmpb   $0x20,(%rbx)
  jne    L49
  mov    %rbp,%rsi
  call   <rev_str@plt>
  cmpb   $0x0,(%rbx)
  lea    0x1(%rbx),%rdi
  jne    L34
L51:
  mov    %rbp,%rsi
  call   <rev_str@plt>
  pop    %rbx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1050 <rev_str@plt>
  nopl   0x0(%rax)
L35:
  lea    -0x1(%rdi),%rbp
  jmp    L51
```

assistant: ```c
void rev_word(char *str) {
  char *start = str;
  char *curr = str;
  while (*curr) {
    curr++;
    if (*curr == ' ') {
      rev_str(start, curr - 1);
      start = curr + 1;
    }
  }
  rev_str(start, curr - 1);
  rev_str(str, curr - 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<rev_word>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  cmpb   $0x0,(%rdi)
  je     L39
  mov    %rdi,%rbp
  jmp    L43
  nopl   (%rax)
L33:
  test   %al,%al
  je     L15
L42:
  mov    %rbx,%rbp
L43:
  movzbl 0x1(%rbp),%eax
  lea    0x1(%rbp),%rbx
  cmp    $0x20,%al
  jne    L33
  mov    %rbp,%rsi
  call   <rev_str@plt>
  cmpb   $0x0,0x1(%rbp)
  lea    0x2(%rbp),%rdi
  jne    L42
L15:
  mov    %rbp,%rsi
  call   <rev_str@plt>
  pop    %rbx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1050 <rev_str@plt>
  nopl   0x0(%rax)
L39:
  lea    -0x1(%rdi),%rbp
  jmp    L15
```

assistant: ```c
void rev_word(char *str) {
  char *start = str;
  char *curr = str;
  while (*curr) {
    curr++;
    if (*curr == ' ') {
      rev_str(start, curr - 1);
      start = curr + 1;
    }
  }
  rev_str(start, curr - 1);
  rev_str(str, curr - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<set_pixel>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <GetVramAddr@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void set_pixel(int x, int y, unsigned long color) {
  unsigned long *vptrc = 0, *vptrl = 0;
  vptrc = (unsigned long *)GetVramAddr(x, y);
  vptrl = (unsigned long *)vptrc;
  *vptrl = (unsigned long)color;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<set_pixel>:
  endbr64
  push   %rbx
  mov    %rdx,%rbx
  call   <GetVramAddr@plt>
  mov    %rbx,(%rax)
  pop    %rbx
  ret
```

assistant: ```c
void set_pixel(int x, int y, unsigned long color) {
  unsigned long *vptrc = 0, *vptrl = 0;
  vptrc = (unsigned long *)GetVramAddr(x, y);
  vptrl = (unsigned long *)vptrc;
  *vptrl = (unsigned long)color;
}
```
user: What is the c source code of the assembly code below?

```asm
<set_pixel>:
  endbr64
  push   %rbx
  mov    %rdx,%rbx
  call   <GetVramAddr@plt>
  mov    %rbx,(%rax)
  pop    %rbx
  ret
```

assistant: ```c
void set_pixel(int x, int y, unsigned long color) {
  unsigned long *vptrc = 0, *vptrl = 0;
  vptrc = (unsigned long *)GetVramAddr(x, y);
  vptrl = (unsigned long *)vptrc;
  *vptrl = (unsigned long)color;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rsi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L2
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <abort@plt>
L2:
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
type_guessed: string
name: retrieve_data
arguments: data_tag: D19
type_guessed: string
name: retrieve_data
arguments: data_tag: D48
type_guessed: string
name: retrieve_data
arguments: data_tag: D32
type_guessed: string
name: retrieve_data
arguments: data_tag: D24
type_guessed: string

tool: D13: open nofile.txt


tool: D19: r

tool: D48: nofile.txt

tool: D32: Going to abort the program


tool: D24: Going to close nofile.txt


assistant: ```c
int main() {
  FILE *fp;
  printf("open nofile.txt\n");
  fp = fopen("nofile.txt", "r");
  if (fp == NULL) {
    printf("Going to abort the program\n");
    abort();
  }
  printf("Going to close nofile.txt\n");
  fclose(fp);
  return (0);
}
```