ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<adc_chunk_offset>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  call   <adc_chunk_type@plt>
  cmp    $0x2,%eax
  je     L19
  cmp    $0x3,%eax
  je     L17
  cmp    $0x1,%eax
  pop    %rbx
  setne  %al
  movzbl %al,%eax
  neg    %eax
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  movzbl (%rbx),%eax
  movzbl 0x1(%rbx),%edx
  pop    %rbx
  shl    $0x8,%eax
  and    $0x300,%eax
  add    %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L17:
  movzbl 0x1(%rbx),%eax
  movzbl 0x2(%rbx),%edx
  pop    %rbx
  shl    $0x8,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
int adc_chunk_offset(unsigned char *chunk_start) {
  unsigned char *c = chunk_start;
  switch (adc_chunk_type(*c)) {
  case 0x01:
    return 0;
  case 0x02:
    return ((((unsigned char)*c & 0x03)) << 8) + (unsigned char)*(c + 1);
  case 0x03:
    return (((unsigned char)*(c + 1)) << 8) + (unsigned char)*(c + 2);
  }
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<adc_chunk_offset>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  call   <adc_chunk_type@plt>
  cmp    $0x2,%eax
  je     L8
  cmp    $0x3,%eax
  je     L10
  cmp    $0x1,%eax
  pop    %rbx
  setne  %al
  movzbl %al,%eax
  neg    %eax
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  movzbl (%rbx),%eax
  movzbl 0x1(%rbx),%edx
  pop    %rbx
  shl    $0x8,%eax
  and    $0x300,%eax
  add    %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  movzbl 0x1(%rbx),%eax
  movzbl 0x2(%rbx),%edx
  pop    %rbx
  shl    $0x8,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
int adc_chunk_offset(unsigned char *chunk_start) {
  unsigned char *c = chunk_start;
  switch (adc_chunk_type(*c)) {
  case 0x01:
    return 0;
  case 0x02:
    return ((((unsigned char)*c & 0x03)) << 8) + (unsigned char)*(c + 1);
  case 0x03:
    return (((unsigned char)*(c + 1)) << 8) + (unsigned char)*(c + 2);
  }
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x3e8,-0x8(%rbp)
  jmp    L90
L21:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x80,-0x4(%rbp)
  je     L115
  cmpl   $0x80,-0x4(%rbp)
  jg     L45
  cmpl   $0xa,-0x4(%rbp)
  je     L90
  cmpl   $0xd,-0x4(%rbp)
  je     L90
  jmp    L45
L115:
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <ferror@plt>
  test   %eax,%eax
  je     L45
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    D51(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L69
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L45
L69:
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <clearerr@plt>
  subl   $0x1,-0x8(%rbp)
  jmp    L90
L45:
  mov    -0x4(%rbp),%eax
  jmp    L40
L90:
  cmpl   $0x0,-0x8(%rbp)
  jne    L21
  mov    $0x80,%eax
L40:
  leave
  ret
```

assistant: ```c
int readc() {
  unsigned tries = 1000;
  int c;
  while (tries) {
    c = getchar();
    switch (c) {
    case '\n':
      continue;
    case '\r':
      continue;
    case EOF:
      if (ferror(stdin)) {
        if ((errno == EINTR) || (errno == EIO)) {
          clearerr(stdin);
          tries--;
          continue;
        }
      }
    default:
      return c;
    }
  }
  return EOF;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<readc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x3e8,%ebp
L69:
  test   %ebp,%ebp
  je     L23
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    $0xd,%eax
  je     L69
  cmp    $0x80,%eax
  je     L83
  cmp    $0xa,%eax
  je     L69
L36:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L83:
  mov    D15(%rip),%rax
  mov    (%rax),%edi
  call   <ferror@plt>
  test   %eax,%eax
  je     L36
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  mov    D25(%rip),%rdx
  cmp    (%rdx),%eax
  je     L43
  mov    D47(%rip),%rdx
  cmp    (%rdx),%eax
  jne    L36
L43:
  mov    D15(%rip),%rax
  mov    (%rax),%edi
  call   <clearerr@plt>
  sub    $0x1,%ebp
  jmp    L69
L23:
  mov    $0x80,%ebx
  jmp    L36
```

assistant: ```c
int readc() {
  unsigned tries = 1000;
  int c;
  while (tries) {
    c = getchar();
    switch (c) {
    case '\n':
      continue;
    case '\r':
      continue;
    case EOF:
      if (ferror(stdin)) {
        if ((errno == EINTR) || (errno == EIO)) {
          clearerr(stdin);
          tries--;
          continue;
        }
      }
    default:
      return c;
    }
  }
  return EOF;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<readc>:
  endbr64
  push   %rbp
  push   %rbx
  mov    $0x3e8,%ebx
  sub    $0x8,%rsp
L74:
  test   %ebx,%ebx
  je     L5
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xd,%eax
  je     L74
  cmp    $0x80,%eax
  je     L50
  cmp    $0xa,%eax
  je     L74
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L50:
  mov    D34(%rip),%rbp
  mov    0x0(%rbp),%edi
  call   <ferror@plt>
  test   %eax,%eax
  je     L5
  mov    D8(%rip),%rax
  mov    D12(%rip),%rdx
  mov    (%rax),%eax
  cmp    (%rdx),%eax
  je     L72
  mov    D20(%rip),%rdx
  cmp    (%rdx),%eax
  je     L72
L5:
  add    $0x8,%rsp
  mov    $0x80,%eax
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L72:
  mov    0x0(%rbp),%edi
  sub    $0x1,%ebx
  call   <clearerr@plt>
  jmp    L74
```

assistant: ```c
int readc() {
  unsigned tries = 1000;
  int c;
  while (tries) {
    c = getchar();
    switch (c) {
    case '\n':
      continue;
    case '\r':
      continue;
    case EOF:
      if (ferror(stdin)) {
        if ((errno == EINTR) || (errno == EIO)) {
          clearerr(stdin);
          tries--;
          continue;
        }
      }
    default:
      return c;
    }
  }
  return EOF;
}
```
user: What is the c source code of the assembly code below:

```asm
<readc>:
  endbr64
  push   %rbp
  push   %rbx
  mov    $0x3e9,%ebx
  sub    $0x8,%rsp
L3:
  sub    $0x1,%ebx
  je     L35
L17:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xd,%eax
  je     L17
  cmp    $0x80,%eax
  je     L18
  cmp    $0xa,%eax
  je     L17
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  mov    D24(%rip),%rbp
  mov    0x0(%rbp),%edi
  call   <ferror@plt>
  test   %eax,%eax
  je     L35
  mov    D31(%rip),%rax
  mov    D18(%rip),%rdx
  mov    (%rax),%eax
  cmp    (%rdx),%eax
  je     L32
  mov    D2(%rip),%rdx
  cmp    (%rdx),%eax
  je     L32
L35:
  add    $0x8,%rsp
  mov    $0x80,%eax
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L32:
  mov    0x0(%rbp),%edi
  call   <clearerr@plt>
  jmp    L3
```

assistant: ```c
int readc() {
  unsigned tries = 1000;
  int c;
  while (tries) {
    c = getchar();
    switch (c) {
    case '\n':
      continue;
    case '\r':
      continue;
    case EOF:
      if (ferror(stdin)) {
        if ((errno == EINTR) || (errno == EIO)) {
          clearerr(stdin);
          tries--;
          continue;
        }
      }
    default:
      return c;
    }
  }
  return EOF;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<tzbytecnt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L17
  xor    %eax,%eax
  tzcnt  -0x18(%rbp),%rax
  jmp    L8
L17:
  mov    $0x40,%eax
L8:
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    0x7(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x3,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t tzbytecnt(uint64_t val) {
  uint32_t cnt;
  cnt = (val == 0) ? 64 : __builtin_ctzll(val);
  cnt = cnt / 8;
  return cnt;
}
```
user: What is the c source code of the assembly code below?

```asm
<tzbytecnt>:
  endbr64
  xor    %eax,%eax
  tzcnt  %rdi,%rax
  test   %rdi,%rdi
  mov    $0x40,%edx
  cmovne %eax,%edx
  lea    0x7(%rdx),%eax
  test   %edx,%edx
  cmovns %edx,%eax
  sar    $0x3,%eax
  ret
```

assistant: ```c
uint32_t tzbytecnt(uint64_t val) {
  uint32_t cnt;
  cnt = (val == 0) ? 64 : __builtin_ctzll(val);
  cnt = cnt / 8;
  return cnt;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tzbytecnt>:
  endbr64
  xor    %eax,%eax
  mov    $0x8,%edx
  tzcnt  %rdi,%rax
  sar    $0x3,%eax
  test   %rdi,%rdi
  cmove  %edx,%eax
  ret
```

assistant: ```c
uint32_t tzbytecnt(uint64_t val) {
  uint32_t cnt;
  cnt = (val == 0) ? 64 : __builtin_ctzll(val);
  cnt = cnt / 8;
  return cnt;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tzbytecnt>:
  endbr64
  xor    %eax,%eax
  mov    $0x8,%edx
  tzcnt  %rdi,%rax
  sar    $0x3,%eax
  test   %rdi,%rdi
  cmove  %edx,%eax
  ret
```

assistant: ```c
uint32_t tzbytecnt(uint64_t val) {
  uint32_t cnt;
  cnt = (val == 0) ? 64 : __builtin_ctzll(val);
  cnt = cnt / 8;
  return cnt;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rop3_153_1_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  xor    -0x1c(%rbp),%al
  mov    %al,-0x1(%rbp)
  notb   -0x1(%rbp)
  movzbl -0x1c(%rbp),%eax
  not    %eax
  mov    %eax,%edx
  movzbl -0x1(%rbp),%eax
  and    %eax,%edx
  movzbl -0x20(%rbp),%eax
  not    %eax
  and    %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x1c(%rbp),%al
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x20(%rbp),%al
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_153_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  stk1 = *D ^ S;
  stk1 = ~stk1;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rop3_153_1_1>:
  endbr64
  mov    %edx,%eax
  or     %esi,%eax
  mov    %eax,%edx
  or     (%rdi),%dl
  not    %edx
  and    (%rdi),%al
  or     %eax,%edx
  mov    %dl,(%rdi)
  ret
```

assistant: ```c
void rop3_153_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  stk1 = *D ^ S;
  stk1 = ~stk1;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: Translate the below assembly code into C programming language:

```asm
<rop3_153_1_1>:
  endbr64
  mov    %edx,%eax
  movzbl (%rdi),%edx
  or     %esi,%eax
  or     %eax,%edx
  and    (%rdi),%al
  not    %edx
  or     %eax,%edx
  mov    %dl,(%rdi)
  ret
```

assistant: ```c
void rop3_153_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  stk1 = *D ^ S;
  stk1 = ~stk1;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rop3_153_1_1>:
  endbr64
  mov    %edx,%eax
  movzbl (%rdi),%edx
  or     %esi,%eax
  or     %eax,%edx
  and    (%rdi),%al
  not    %edx
  or     %eax,%edx
  mov    %dl,(%rdi)
  ret
```

assistant: ```c
void rop3_153_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  stk1 = *D ^ S;
  stk1 = ~stk1;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: Translate the below assembly code into C programming language:

```asm
<expand>:
  endbr64
  mov    %rsi,%r11
  movzbl (%rdi),%eax
  test   %al,%al
  je     L36
  push   %rbx
  mov    %rdi,%r10
  mov    %rdi,%r9
  mov    $0x0,%esi
  mov    $0x0,%edi
  mov    $0x0,%r8d
  jmp    L17
L89:
  cmpb   $0x2d,0x1(%r10,%r8,1)
  je     L98
L105:
  add    $0x1,%edi
  movslq %edi,%r8
  lea    (%r10,%r8,1),%r9
  movzbl (%r9),%eax
  test   %al,%al
  je     L20
  mov    %edx,%esi
L17:
  lea    0x1(%rsi),%edx
  movslq %esi,%rcx
  mov    %al,(%r11,%rcx,1)
  mov    %eax,%ecx
  and    $0xffffffdf,%ecx
  sub    $0x41,%ecx
  cmp    $0x19,%cl
  jbe    L89
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L105
  jmp    L89
L98:
  lea    0x2(%r10,%r8,1),%r8
  movzbl (%r8),%ebx
  cmp    %al,%bl
  jle    L105
  movslq %edx,%rcx
  cmp    -0x1(%r11,%rcx,1),%bl
  jle    L2
  add    %r11,%rcx
L18:
  mov    %edx,%eax
  sub    %esi,%eax
  add    (%r9),%al
  add    $0x1,%edx
  mov    %al,(%rcx)
  add    $0x1,%rcx
  cmp    (%r8),%al
  jl     L18
L2:
  sub    $0x1,%edx
  add    $0x1,%edi
  jmp    L105
L36:
  mov    $0x0,%edx
  movslq %edx,%rdx
  movb   $0x0,(%rsi,%rdx,1)
  ret
L20:
  movslq %edx,%rdx
  movb   $0x0,(%r11,%rdx,1)
  pop    %rbx
  ret
```

assistant: ```c
void expand(char s1[], char s2[]) {
  int i, j, k;
  for (i = 0, j = 0; s1[i] != '\0'; i++) {
    s2[j++] = s1[i];
    if ((s1[i] >= 'a' && s1[i] <= 'z') || (s1[i] >= 'A' && s1[i] <= 'Z') || (s1[i] >= '0' && s1[i] <= '9'))
      if (s1[i + 1] == '-' && s1[i + 2] > s1[i]) {
        for (k = 1; s2[j - 1] < s1[i + 2]; k++)
          s2[j++] = s1[i] + k;
        j--;
        i++;
      }
  }
  s2[j] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<expand>:
  endbr64
  movzbl (%rdi),%eax
  mov    %rsi,%r11
  test   %al,%al
  je     L56
  mov    %rdi,%r10
  mov    %rdi,%r9
  push   %rbx
  xor    %esi,%esi
  xor    %edi,%edi
  xor    %r8d,%r8d
  jmp    L64
  nopw   0x0(%rax,%rax,1)
L84:
  add    $0x1,%edi
  movslq %edi,%r8
  lea    (%r10,%r8,1),%r9
  movzbl (%r9),%eax
  test   %al,%al
  je     L37
  mov    %edx,%esi
L64:
  movslq %esi,%rcx
  lea    0x1(%rsi),%edx
  mov    %al,(%r11,%rcx,1)
  mov    %eax,%ecx
  and    $0xffffffdf,%ecx
  sub    $0x41,%ecx
  cmp    $0x19,%cl
  jbe    L23
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L84
L23:
  cmpb   $0x2d,0x1(%r10,%r8,1)
  jne    L84
  lea    0x2(%r10,%r8,1),%r8
  cmp    %al,(%r8)
  jle    L84
  movslq %edx,%rcx
  add    %r11,%rcx
  nopl   0x0(%rax,%rax,1)
L72:
  mov    %edx,%eax
  mov    %edx,%ebx
  add    $0x1,%rcx
  add    $0x1,%edx
  sub    %esi,%eax
  add    (%r9),%al
  mov    %al,-0x1(%rcx)
  cmp    (%r8),%al
  jl     L72
  add    $0x1,%edi
  mov    %ebx,%edx
  jmp    L84
  nopw   0x0(%rax,%rax,1)
L37:
  movslq %edx,%rdx
  pop    %rbx
  add    %rdx,%r11
  movb   $0x0,(%r11)
  ret
  nopl   0x0(%rax)
L56:
  movb   $0x0,(%rsi)
  ret
```

assistant: ```c
void expand(char s1[], char s2[]) {
  int i, j, k;
  for (i = 0, j = 0; s1[i] != '\0'; i++) {
    s2[j++] = s1[i];
    if ((s1[i] >= 'a' && s1[i] <= 'z') || (s1[i] >= 'A' && s1[i] <= 'Z') || (s1[i] >= '0' && s1[i] <= '9'))
      if (s1[i + 1] == '-' && s1[i + 2] > s1[i]) {
        for (k = 1; s2[j - 1] < s1[i + 2]; k++)
          s2[j++] = s1[i] + k;
        j--;
        i++;
      }
  }
  s2[j] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<expand>:
  endbr64
  movzbl (%rdi),%eax
  mov    %rsi,%r11
  test   %al,%al
  je     L40
  mov    %rdi,%r10
  mov    %rdi,%r9
  push   %rbx
  xor    %esi,%esi
  xor    %edi,%edi
  xor    %r8d,%r8d
  jmp    L30
  nopw   0x0(%rax,%rax,1)
L21:
  add    $0x1,%edi
  movslq %edi,%r8
  lea    (%r10,%r8,1),%r9
  movzbl (%r9),%eax
  test   %al,%al
  je     L8
  mov    %edx,%esi
L30:
  movslq %esi,%rcx
  lea    0x1(%rsi),%edx
  mov    %al,(%r11,%rcx,1)
  mov    %eax,%ecx
  and    $0xffffffdf,%ecx
  sub    $0x41,%ecx
  cmp    $0x19,%cl
  jbe    L65
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L21
L65:
  cmpb   $0x2d,0x1(%r10,%r8,1)
  jne    L21
  lea    0x2(%r10,%r8,1),%r8
  cmp    %al,(%r8)
  jle    L21
  movslq %edx,%rcx
  add    %r11,%rcx
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %edx,%eax
  mov    %edx,%ebx
  add    $0x1,%rcx
  add    $0x1,%edx
  sub    %esi,%eax
  add    (%r9),%al
  mov    %al,-0x1(%rcx)
  cmp    (%r8),%al
  jl     L14
  add    $0x1,%edi
  mov    %ebx,%edx
  jmp    L21
  nopw   0x0(%rax,%rax,1)
L8:
  movslq %edx,%rdx
  pop    %rbx
  add    %rdx,%r11
  movb   $0x0,(%r11)
  ret
  nopl   0x0(%rax)
L40:
  movb   $0x0,(%rsi)
  ret
```

assistant: ```c
void expand(char s1[], char s2[]) {
  int i, j, k;
  for (i = 0, j = 0; s1[i] != '\0'; i++) {
    s2[j++] = s1[i];
    if ((s1[i] >= 'a' && s1[i] <= 'z') || (s1[i] >= 'A' && s1[i] <= 'Z') || (s1[i] >= '0' && s1[i] <= '9'))
      if (s1[i + 1] == '-' && s1[i + 2] > s1[i]) {
        for (k = 1; s2[j - 1] < s1[i + 2]; k++)
          s2[j++] = s1[i] + k;
        j--;
        i++;
      }
  }
  s2[j] = '\0';
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L10
  movl   $0x1,-0xc(%rbp)
  jmp    L20
L63:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L20:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0xc(%rbp)
  jle    L63
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xc(%rbp)
  jmp    L39
L11:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  subl   $0x1,-0xc(%rbp)
L39:
  cmpl   $0x0,-0xc(%rbp)
  jg     L11
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L58
L10:
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L58:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L56
  call   <__stack_chk_fail@plt>
L56:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D1
data_type: string
name: read_data
arguments: tag: D36
data_type: string

tool: D1: %d

tool: D36: Error

assistant: ```c
int main(void) {
  int number, i;
  scanf("%d", &number);
  if (number % 2 == 0) {
    for (i = 1; i <= (number / 2); i++) {
      printf("%d", i);
    }
    for (i = (number / 2); i >= 1; i--) {
      printf("%d", i);
    }
    printf("\n");
  } else {
    printf("Error\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   $0x1,%al
  jne    L61
  cmp    $0x1,%eax
  jle    L3
  mov    $0x1,%ebp
  lea    D18(%rip),%r12
L42:
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebp
  mov    0x4(%rsp),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,%ebx
  cmp    %ebp,%eax
  jge    L42
  cmp    $0x1,%edx
  jle    L3
  lea    D18(%rip),%rbp
L17:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x1,%ebx
  test   %ebx,%ebx
  jg     L17
L3:
  mov    $0xa,%edi
  call   <putchar@plt>
L22:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L61:
  lea    D2(%rip),%rdi
  call   <puts@plt>
  jmp    L22
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D18
datatype: string
name: disassemble_data
arguments: label: D2
datatype: string

tool: D18: %d

tool: D2: Error

assistant: ```c
int main(void) {
  int number, i;
  scanf("%d", &number);
  if (number % 2 == 0) {
    for (i = 1; i <= (number / 2); i++) {
      printf("%d", i);
    }
    for (i = (number / 2); i >= 1; i--) {
      printf("%d", i);
    }
    printf("\n");
  } else {
    printf("Error\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   %r12
  lea    D0(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   $0x1,%al
  jne    L6
  cmp    $0x1,%eax
  jle    L15
  mov    $0x1,%ebp
  nopw   0x0(%rax,%rax,1)
L30:
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  mov    %eax,%ebx
  shr    $0x1f,%ebx
  add    %eax,%ebx
  sar    %ebx
  cmp    %ebp,%ebx
  jge    L30
  sub    $0x1,%eax
  jle    L15
  nopl   (%rax)
L29:
  mov    %ebx,%esi
  xor    %eax,%eax
  mov    %r12,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L29
L15:
  mov    $0xa,%edi
  call   <putchar@plt>
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  lea    D9(%rip),%rdi
  call   <puts@plt>
  jmp    L9
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D0
data_type: string
name: read_data
arguments: tag: D9
data_type: string

tool: D0: %d

tool: D9: Error

assistant: ```c
int main(void) {
  int number, i;
  scanf("%d", &number);
  if (number % 2 == 0) {
    for (i = 1; i <= (number / 2); i++) {
      printf("%d", i);
    }
    for (i = (number / 2); i >= 1; i--) {
      printf("%d", i);
    }
    printf("\n");
  } else {
    printf("Error\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   $0x1,%al
  jne    L1
  cmp    $0x1,%eax
  jle    L8
  mov    $0x1,%ebp
  nopw   0x0(%rax,%rax,1)
L36:
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  mov    %eax,%ebx
  shr    $0x1f,%ebx
  add    %eax,%ebx
  sar    %ebx
  cmp    %ebp,%ebx
  jge    L36
  sub    $0x1,%eax
  jle    L8
  nopl   (%rax)
L19:
  mov    %ebx,%esi
  xor    %eax,%eax
  mov    %r12,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L19
L8:
  mov    $0xa,%edi
  call   <putchar@plt>
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  lea    D5(%rip),%rdi
  call   <puts@plt>
  jmp    L21
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D10
type_infered: string
name: read_data
arguments: label: D5
type_infered: string

tool: D10: %d

tool: D5: Error

assistant: ```c
int main(void) {
  int number, i;
  scanf("%d", &number);
  if (number % 2 == 0) {
    for (i = 1; i <= (number / 2); i++) {
      printf("%d", i);
    }
    for (i = (number / 2); i >= 1; i--) {
      printf("%d", i);
    }
    printf("\n");
  } else {
    printf("Error\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L14
L19:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  add    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdx
  shl    $0x4,%rdx
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L19
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x10(%rbp),%rax
  cqto
  idiv   %rcx
  mov    %rdx,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ltoi@plt>
  leave
  ret
```

assistant: ```c
int hash(char *message, int hash_size) {
  long digest = 0l;
  char *msg = message;
  while (*msg != '\0') {
    digest += *msg;
    digest *= 17;
    msg++;
  }
  digest %= hash_size;
  return ltoi(digest);
}
```