ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<combine_hangul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1100,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0xac00,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1161,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x11a7,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  js     L1
  cmpl   $0x12,-0x10(%rbp)
  jg     L1
  cmpl   $0x0,-0x8(%rbp)
  js     L1
  cmpl   $0x14,-0x8(%rbp)
  jg     L1
  mov    -0x10(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  imul   $0x1c,%eax,%eax
  lea    0xac00(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  jmp    L52
L1:
  cmpl   $0x0,-0xc(%rbp)
  js     L4
  cmpl   $0x2ba3,-0xc(%rbp)
  jg     L4
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x4,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1c,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  test   %edx,%edx
  jne    L4
  cmpl   $0x0,-0x4(%rbp)
  js     L4
  cmpl   $0x1c,-0x4(%rbp)
  jg     L4
  mov    -0x14(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  jmp    L52
L4:
  mov    $0x0,%eax
L52:
  pop    %rbp
  ret
```

assistant: ```c
int combine_hangul(uint32_t a, uint32_t b, uint32_t *result) {
  int LIndex = a - 0x1100;
  int SIndex = a - 0xAC00;
  int VIndex = b - 0x1161;
  int TIndex = b - 0x11A7;
  if (0 <= LIndex && LIndex < 19 && 0 <= VIndex && VIndex < 21) {
    *result = 0xAC00 + (LIndex * 21 + VIndex) * 28;
    return 1;
  } else if (0 <= SIndex && SIndex < (19 * (21 * 28)) && (SIndex % 28) == 0 && 0 <= TIndex && TIndex <= 28) {
    *result = a + TIndex;
    return 1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<combine_hangul>:
  endbr64
  mov    %edi,%ecx
  mov    %rdx,%r8
  lea    -0x1100(%rdi),%eax
  lea    -0xac00(%rdi),%edx
  lea    -0x1161(%rsi),%edi
  cmp    $0x14,%edi
  ja     L41
  cmp    $0x12,%eax
  jbe    L44
L41:
  mov    $0x0,%eax
  cmp    $0x2ba3,%edx
  ja     L27
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x4,%eax
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  imul   $0x1c,%eax,%edi
  mov    %edx,%eax
  sub    %edi,%eax
  jne    L18
  sub    $0x11a7,%esi
  cmp    $0x1c,%esi
  ja     L27
  add    %esi,%ecx
  mov    %ecx,(%r8)
  mov    $0x1,%eax
  ret
L44:
  lea    (%rax,%rax,4),%edx
  lea    (%rax,%rdx,4),%eax
  lea    -0x1161(%rsi,%rax,1),%eax
  imul   $0x1c,%eax,%eax
  add    $0xac00,%eax
  mov    %eax,(%r8)
  mov    $0x1,%eax
  ret
L18:
  mov    $0x0,%eax
L27:
  ret
```

assistant: ```c
int combine_hangul(uint32_t a, uint32_t b, uint32_t *result) {
  int LIndex = a - 0x1100;
  int SIndex = a - 0xAC00;
  int VIndex = b - 0x1161;
  int TIndex = b - 0x11A7;
  if (0 <= LIndex && LIndex < 19 && 0 <= VIndex && VIndex < 21) {
    *result = 0xAC00 + (LIndex * 21 + VIndex) * 28;
    return 1;
  } else if (0 <= SIndex && SIndex < (19 * (21 * 28)) && (SIndex % 28) == 0 && 0 <= TIndex && TIndex <= 28) {
    *result = a + TIndex;
    return 1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<combine_hangul>:
  endbr64
  mov    %edi,%ecx
  lea    -0x1161(%rsi),%eax
  mov    %rdx,%r8
  lea    -0x1100(%rdi),%edi
  lea    -0xac00(%rcx),%edx
  cmp    $0x14,%eax
  ja     L0
  cmp    $0x12,%edi
  jbe    L13
L0:
  xor    %eax,%eax
  cmp    $0x2ba3,%edx
  ja     L7
  movslq %edx,%rax
  mov    %edx,%edi
  imul   $0xffffffff92492493,%rax,%rax
  sar    $0x1f,%edi
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x4,%eax
  sub    %edi,%eax
  imul   $0x1c,%eax,%edi
  mov    %edx,%eax
  sub    %edi,%eax
  jne    L6
  sub    $0x11a7,%esi
  cmp    $0x1c,%esi
  jbe    L28
L7:
  ret
  xchg   %ax,%ax
L6:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  lea    (%rdi,%rdi,4),%edx
  lea    (%rdi,%rdx,4),%edx
  add    %edx,%eax
  imul   $0x1c,%eax,%eax
  add    $0xac00,%eax
  mov    %eax,(%r8)
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L28:
  add    %esi,%ecx
  mov    $0x1,%eax
  mov    %ecx,(%r8)
  ret
```

assistant: ```c
int combine_hangul(uint32_t a, uint32_t b, uint32_t *result) {
  int LIndex = a - 0x1100;
  int SIndex = a - 0xAC00;
  int VIndex = b - 0x1161;
  int TIndex = b - 0x11A7;
  if (0 <= LIndex && LIndex < 19 && 0 <= VIndex && VIndex < 21) {
    *result = 0xAC00 + (LIndex * 21 + VIndex) * 28;
    return 1;
  } else if (0 <= SIndex && SIndex < (19 * (21 * 28)) && (SIndex % 28) == 0 && 0 <= TIndex && TIndex <= 28) {
    *result = a + TIndex;
    return 1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<combine_hangul>:
  endbr64
  mov    %edi,%ecx
  lea    -0x1161(%rsi),%eax
  mov    %rdx,%r8
  lea    -0x1100(%rdi),%edi
  lea    -0xac00(%rcx),%edx
  cmp    $0x14,%eax
  ja     L29
  cmp    $0x12,%edi
  jbe    L20
L29:
  xor    %eax,%eax
  cmp    $0x2ba3,%edx
  ja     L9
  movslq %edx,%rax
  mov    %edx,%edi
  imul   $0xffffffff92492493,%rax,%rax
  sar    $0x1f,%edi
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x4,%eax
  sub    %edi,%eax
  imul   $0x1c,%eax,%edi
  mov    %edx,%eax
  sub    %edi,%eax
  jne    L7
  sub    $0x11a7,%esi
  cmp    $0x1c,%esi
  jbe    L18
L9:
  ret
  xchg   %ax,%ax
L7:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  lea    (%rdi,%rdi,4),%edx
  lea    (%rdi,%rdx,4),%edx
  add    %edx,%eax
  imul   $0x1c,%eax,%eax
  add    $0xac00,%eax
  mov    %eax,(%r8)
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L18:
  add    %esi,%ecx
  mov    $0x1,%eax
  mov    %ecx,(%r8)
  ret
```

assistant: ```c
int combine_hangul(uint32_t a, uint32_t b, uint32_t *result) {
  int LIndex = a - 0x1100;
  int SIndex = a - 0xAC00;
  int VIndex = b - 0x1161;
  int TIndex = b - 0x11A7;
  if (0 <= LIndex && LIndex < 19 && 0 <= VIndex && VIndex < 21) {
    *result = 0xAC00 + (LIndex * 21 + VIndex) * 28;
    return 1;
  } else if (0 <= SIndex && SIndex < (19 * (21 * 28)) && (SIndex % 28) == 0 && 0 <= TIndex && TIndex <= 28) {
    *result = a + TIndex;
    return 1;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r15
  lea    D31(%rip),%rsi
  lea    D48(%rip),%rdi
  push   %r14
  lea    D36(%rip),%r15
  push   %r13
  mov    $0xffffffff,%r13d
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x800(%rsp),%r14
  call   <fopen@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L5
  nopl   0x0(%rax)
L40:
  mov    %r12,%rcx
  mov    $0x1,%edx
  mov    $0x800,%esi
  mov    %rbx,%rdi
  call   <fread@plt>
  test   %rax,%rax
  je     L117
  cmpl   $0xff,(%rsp)
  je     L80
L118:
  test   %rbp,%rbp
  je     L40
L105:
  mov    %rbp,%rcx
  mov    $0x1,%edx
  mov    $0x800,%esi
  mov    %rbx,%rdi
  call   <fwrite@plt>
  jmp    L40
  nopl   0x0(%rax,%rax,1)
L80:
  cmpl   $0xd8,0x4(%rsp)
  jne    L118
  cmpl   $0xff,0x8(%rsp)
  jne    L118
  mov    0xc(%rsp),%eax
  sub    $0xe0,%eax
  cmp    $0x1,%eax
  ja     L118
  test   %rbp,%rbp
  je     L17
  mov    %rbp,%rdi
  call   <fclose@plt>
L17:
  add    $0x1,%r13d
  mov    %r15,%rsi
  mov    %r14,%rdi
  mov    %r13d,%edx
  call   <sprintf@plt>
  lea    D38(%rip),%rsi
  mov    %r14,%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L105
  mov    %r14,%rsi
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  nopl   0x0(%rax)
L117:
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L59:
  mov    0x808(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L61
  add    $0x818,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  lea    D12(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L59
L61:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D31
type: string
name: disassemble_data
arguments: tag: D48
type: string
name: disassemble_data
arguments: tag: D36
type: string
name: disassemble_data
arguments: tag: D38
type: string
name: disassemble_data
arguments: tag: D10
type: string
name: disassemble_data
arguments: tag: D12
type: string

tool: D31: r

tool: D48: card.raw

tool: D36: %03d.jpg

tool: D38: w

tool: D10: Couldn't open output file %s!


tool: D12: Could not open file!

assistant: ```c
int main(int argc, char *argv[]) {
  int filenum = -1;
  char filename[8];
  FILE *output = NULL;
  FILE *input = fopen("card.raw", "r");
  if (input == NULL) {
    printf("Could not open file!\n");
    return 1;
  }
  uint8_t buffer[512];
  while ((fread(&buffer, sizeof(buffer), 1, input)) != 0) {
    if ((buffer[0] == 0xFF) && (buffer[1] == 0xD8) && (buffer[2] == 0xFF) && ((buffer[3] == 0xE0) || (buffer[3] == 0xE1))) {
      if (output != NULL) {
        fclose(output);
      }
      filenum++;
      sprintf(filename, "%03d.jpg", filenum);
      output = fopen(filename, "w");
      if (output == NULL) {
        printf("Couldn't open output file %s!\n", filename);
        break;
      }
    }
    if (output != NULL) {
      fwrite(&buffer, sizeof(buffer), 1, output);
    }
  }
  fclose(output);
  fclose(input);
  return 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
<is_anagram>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <length@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <length@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  je     L33
  mov    $0x0,%eax
  jmp    L54
L33:
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <quickSort@plt>
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <quickSort@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L1
L46:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L12
  mov    $0x0,%eax
  jmp    L54
L12:
  addl   $0x1,-0xc(%rbp)
L1:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L46
  mov    $0x1,%eax
L54:
  leave
  ret
```

assistant: ```c
short is_anagram(char *s1, char *s2) {
  int n1 = length(s1);
  int n2 = length(s2);
  if (n1 != n2)
    return 0;
  quickSort(s1, 0, n1 - 1);
  quickSort(s2, 0, n2 - 1);
  int i;
  for (i = 0; i < n1; i++)
    if (s1[i] != s2[i])
      return 0;
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<is_anagram>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <length@plt>
  mov    %eax,%r13d
  mov    %rbp,%rdi
  call   <length@plt>
  mov    %eax,%r12d
  mov    $0x0,%eax
  cmp    %r12d,%r13d
  je     L24
L62:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L24:
  lea    -0x1(%r13),%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <quickSort@plt>
  lea    -0x1(%r12),%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <quickSort@plt>
  test   %r13d,%r13d
  jle    L42
  lea    -0x1(%r13),%ecx
  mov    $0x0,%eax
  jmp    L2
L32:
  mov    %rdx,%rax
L2:
  movzbl 0x0(%rbp,%rax,1),%esi
  cmp    %sil,(%rbx,%rax,1)
  jne    L68
  lea    0x1(%rax),%rdx
  cmp    %rcx,%rax
  jne    L32
  mov    $0x1,%eax
  jmp    L62
L42:
  mov    $0x1,%eax
  jmp    L62
L68:
  mov    $0x0,%eax
  jmp    L62
```

assistant: ```c
short is_anagram(char *s1, char *s2) {
  int n1 = length(s1);
  int n2 = length(s2);
  if (n1 != n2)
    return 0;
  quickSort(s1, 0, n1 - 1);
  quickSort(s2, 0, n2 - 1);
  int i;
  for (i = 0; i < n1; i++)
    if (s1[i] != s2[i])
      return 0;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_anagram>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <length@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <length@plt>
  cmp    %eax,%ebx
  je     L19
L42:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L19:
  lea    -0x1(%rbx),%r13d
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    %r13d,%edx
  call   <quickSort@plt>
  xor    %esi,%esi
  mov    %r13d,%edx
  mov    %r12,%rdi
  call   <quickSort@plt>
  test   %ebx,%ebx
  jle    L57
  movslq %ebx,%rax
  xor    %edx,%edx
  jmp    L48
  nopl   0x0(%rax)
L6:
  add    $0x1,%rdx
  cmp    %rax,%rdx
  je     L57
L48:
  movzbl (%r12,%rdx,1),%ecx
  cmp    %cl,0x0(%rbp,%rdx,1)
  je     L6
  jmp    L42
  xchg   %ax,%ax
L57:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
short is_anagram(char *s1, char *s2) {
  int n1 = length(s1);
  int n2 = length(s2);
  if (n1 != n2)
    return 0;
  quickSort(s1, 0, n1 - 1);
  quickSort(s2, 0, n2 - 1);
  int i;
  for (i = 0; i < n1; i++)
    if (s1[i] != s2[i])
      return 0;
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<is_anagram>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <length@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <length@plt>
  cmp    %eax,%ebx
  je     L56
L51:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L56:
  lea    -0x1(%rbx),%r13d
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    %r13d,%edx
  call   <quickSort@plt>
  xor    %esi,%esi
  mov    %r13d,%edx
  mov    %r12,%rdi
  call   <quickSort@plt>
  test   %ebx,%ebx
  jle    L16
  movslq %ebx,%rax
  xor    %edx,%edx
  jmp    L43
  nopl   0x0(%rax)
L50:
  add    $0x1,%rdx
  cmp    %rax,%rdx
  je     L16
L43:
  movzbl (%r12,%rdx,1),%ecx
  cmp    %cl,0x0(%rbp,%rdx,1)
  je     L50
  jmp    L51
  xchg   %ax,%ax
L16:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
short is_anagram(char *s1, char *s2) {
  int n1 = length(s1);
  int n2 = length(s2);
  if (n1 != n2)
    return 0;
  quickSort(s1, 0, n1 - 1);
  quickSort(s2, 0, n2 - 1);
  int i;
  for (i = 0; i < n1; i++)
    if (s1[i] != s2[i])
      return 0;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<randomize_buffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L19
L11:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %al,(%rbx)
  addl   $0x1,-0x14(%rbp)
L19:
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,-0x30(%rbp)
  jg     L11
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void randomize_buffer(char *buf, long long size) {
  int i;
  for (i = 0; i < size; i++) {
    buf[i] = rand();
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<randomize_buffer>:
  endbr64
  test   %rsi,%rsi
  jle    L1
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    (%rdi,%rsi,1),%rbp
L6:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  ret
```

assistant: ```c
void randomize_buffer(char *buf, long long size) {
  int i;
  for (i = 0; i < size; i++) {
    buf[i] = rand();
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<randomize_buffer>:
  endbr64
  test   %rsi,%rsi
  jle    L3
  push   %rbp
  lea    (%rdi,%rsi,1),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <rand@plt>
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  ret
```

assistant: ```c
void randomize_buffer(char *buf, long long size) {
  int i;
  for (i = 0; i < size; i++) {
    buf[i] = rand();
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<randomize_buffer>:
  endbr64
  test   %rsi,%rsi
  jle    L6
  push   %rbp
  lea    (%rdi,%rsi,1),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L2:
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <rand@plt>
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L2
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  ret
```

assistant: ```c
void randomize_buffer(char *buf, long long size) {
  int i;
  for (i = 0; i < size; i++) {
    buf[i] = rand();
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<LOG__>:
  endbr64
  cmp    $0x7,%edi
  ja     L9
  mov    %edi,%edi
  lea    D54(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D96(%rip),%rax
  ret
  lea    D80(%rip),%rax
  ret
  lea    D41(%rip),%rax
  ret
  lea    D49(%rip),%rax
  ret
  lea    D35(%rip),%rax
  ret
  lea    D89(%rip),%rax
  ret
  lea    D81(%rip),%rax
  ret
L9:
  lea    D31(%rip),%rax
  ret
  lea    D39(%rip),%rax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D96
datatype: string
name: extract_symbolic_data
arguments: label: D80
datatype: string
name: extract_symbolic_data
arguments: label: D41
datatype: string
name: extract_symbolic_data
arguments: label: D49
datatype: string
name: extract_symbolic_data
arguments: label: D35
datatype: string
name: extract_symbolic_data
arguments: label: D89
datatype: string
name: extract_symbolic_data
arguments: label: D81
datatype: string
name: extract_symbolic_data
arguments: label: D31
datatype: string
name: extract_symbolic_data
arguments: label: D39
datatype: string

tool: D96: ALERT

tool: D80: CRIT

tool: D41: ERROR

tool: D49: WARNING

tool: D35: NOTICE

tool: D89: INFO

tool: D81: DEBUG

tool: D31: PANIC(unknown)

tool: D39: PANIC

assistant: ```c
const char *LOG__(int p) {
  switch (p) {
  case 0:
    return "PANIC";
  case 1:
    return "ALERT";
  case 2:
    return "CRIT";
  case 3:
    return "ERROR";
  case 4:
    return "WARNING";
  case 5:
    return "NOTICE";
  case 6:
    return "INFO";
  case 7:
    return "DEBUG";
  }
  return "PANIC(unknown)";
}
```
user: 请从给定的汇编指令推导出C源代码：

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

tool_calls: name: retrieve_data
arguments: tag: D2
type_infered: string

tool: D2: -- %s


assistant: ```c
void print_log(const char *message) {
  if (1) {
    printf("-- %s\n", message);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<print_log>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D1
datatype: string

tool: D1: -- %s


assistant: ```c
void print_log(const char *message) {
  if (1) {
    printf("-- %s\n", message);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<print_log>:
  endbr64
  mov    %rdi,%rsi
  lea    D8(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: tag: D8
type_guessed: string

tool: D8: -- %s


assistant: ```c
void print_log(const char *message) {
  if (1) {
    printf("-- %s\n", message);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<print_log>:
  endbr64
  mov    %rdi,%rsi
  lea    D0(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: tag: D0
data_type: string

tool: D0: -- %s


assistant: ```c
void print_log(const char *message) {
  if (1) {
    printf("-- %s\n", message);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<gmp_mul_ui>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <mpz_mul_ui@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <mpz_mod@plt>
  nop
  leave
  ret
```

assistant: ```c
void gmp_mul_ui(mpz_ptr out, int in, mpz_ptr order) {
  mpz_mul_ui(out, out, in);
  mpz_mod(out, out, order);
}
```
user: Translate the below assembly code into C programming language:

```asm
<gmp_mul_ui>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %edx,%ebp
  mov    %esi,%edx
  mov    %edi,%esi
  call   <mpz_mul_ui@plt>
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %ebx,%edi
  call   <mpz_mod@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void gmp_mul_ui(mpz_ptr out, int in, mpz_ptr order) {
  mpz_mul_ui(out, out, in);
  mpz_mod(out, out, order);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<gmp_mul_ui>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  mov    %esi,%edx
  mov    %edi,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <mpz_mul_ui@plt>
  add    $0x8,%rsp
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1070 <mpz_mod@plt>
```

assistant: ```c
void gmp_mul_ui(mpz_ptr out, int in, mpz_ptr order) {
  mpz_mul_ui(out, out, in);
  mpz_mod(out, out, order);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<gmp_mul_ui>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  mov    %esi,%edx
  mov    %edi,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x8,%rsp
  call   <mpz_mul_ui@plt>
  add    $0x8,%rsp
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1070 <mpz_mod@plt>
```

assistant: ```c
void gmp_mul_ui(mpz_ptr out, int in, mpz_ptr order) {
  mpz_mul_ui(out, out, in);
  mpz_mod(out, out, order);
}
```