ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<itoa>:
  endbr64
  movslq %edi,%r9
  push   %rbp
  mov    %rsi,%rbp
  mov    %r9,%rax
  imul   $0x66666667,%r9,%r9
  push   %rbx
  mov    %edx,%ebx
  cltd
  sar    $0x22,%r9
  sub    $0x8,%rsp
  sub    %edx,%r9d
  lea    (%r9,%r9,4),%ecx
  mov    %r9d,%edx
  mov    %eax,%r9d
  add    %ecx,%ecx
  sub    %ecx,%r9d
  mov    %r9d,%ecx
  neg    %ecx
  cmovs  %r9d,%ecx
  add    $0x30,%ecx
  mov    %cl,(%rsi)
  cmp    $0xfffffff7,%eax
  jge    L57
  mov    %edx,%r9d
  neg    %edx
L57:
  mov    $0x2,%ecx
  mov    $0xcccccccd,%r10d
  test   %edx,%edx
  jle    L3
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %edx,%eax
  mov    %edx,%esi
  mov    %ecx,%r8d
  imul   %r10,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edi
  add    %edi,%edi
  sub    %edi,%esi
  add    $0x30,%esi
  mov    %sil,-0x1(%rbp,%rcx,1)
  mov    %edx,%esi
  add    $0x1,%rcx
  mov    %eax,%edx
  cmp    $0x9,%esi
  jg     L26
  movslq %r8d,%rax
L25:
  test   %r9d,%r9d
  jns    L42
  add    $0x1,%r8d
  movb   $0x2d,0x0(%rbp,%rax,1)
  movslq %r8d,%rax
L42:
  cmp    %ebx,%r8d
  jge    L46
  lea    -0x1(%rbx),%edx
  movslq %r8d,%rdi
  mov    $0x7e,%esi
  sub    %r8d,%edx
  add    %rbp,%rdi
  add    $0x1,%rdx
  call   <memset@plt>
  movslq %ebx,%rax
L46:
  movb   $0x0,0x0(%rbp,%rax,1)
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1060 <reverse@plt>
  nopl   0x0(%rax)
L3:
  mov    $0x1,%eax
  mov    $0x1,%r8d
  jmp    L25
```

assistant: ```c
void itoa(int n, char s[], int w) {
  int i = 0;
  int sign = n % 10;
  s[i++] = (sign < 0 ? -sign : sign) + '0';
  n /= 10;
  if (n < 0) {
    sign = n;
    n = -n;
  }
  while (n > 0) {
    s[i++] = n % 10 + '0';
    n /= 10;
  }
  if (sign < 0) {
    s[i++] = '-';
  }
  if (i < w) {
    while (w > i) {
      s[i++] = '~';
    }
  }
  s[i] = '\0';
  reverse(s);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test15>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8d,-0x24(%rbp)
  mov    %r9d,-0x28(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  je     L15
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jne    L10
L15:
  mov    $0x1,%eax
  jmp    L28
L10:
  mov    $0x0,%eax
L28:
  pop    %rbp
  ret
```

assistant: ```c
int test15(int a, int b, int c, int d, int e, int f) {
  int x = a != b;
  int y = c == 0;
  int z = d == e;
  return x == y || x == z;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<test15>:
  endbr64
  mov    %edx,%eax
  cmp    %esi,%edi
  setne  %dl
  movzbl %dl,%edx
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  cmp    %edx,%eax
  sete   %al
  cmp    %r8d,%ecx
  sete   %cl
  movzbl %cl,%ecx
  cmp    %edx,%ecx
  sete   %dl
  or     %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int test15(int a, int b, int c, int d, int e, int f) {
  int x = a != b;
  int y = c == 0;
  int z = d == e;
  return x == y || x == z;
}
```
user: What is the c source code of the assembly code below:

```asm
<test15>:
  endbr64
  mov    %edx,%r9d
  xor    %edx,%edx
  cmp    %esi,%edi
  setne  %dl
  xor    %eax,%eax
  test   %r9d,%r9d
  sete   %al
  cmp    %edx,%eax
  sete   %al
  cmp    %r8d,%ecx
  sete   %cl
  movzbl %cl,%ecx
  cmp    %edx,%ecx
  sete   %dl
  or     %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int test15(int a, int b, int c, int d, int e, int f) {
  int x = a != b;
  int y = c == 0;
  int z = d == e;
  return x == y || x == z;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test15>:
  endbr64
  mov    %edx,%r9d
  xor    %edx,%edx
  cmp    %esi,%edi
  setne  %dl
  xor    %eax,%eax
  test   %r9d,%r9d
  sete   %al
  cmp    %edx,%eax
  sete   %al
  cmp    %r8d,%ecx
  sete   %cl
  movzbl %cl,%ecx
  cmp    %edx,%ecx
  sete   %dl
  or     %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int test15(int a, int b, int c, int d, int e, int f) {
  int x = a != b;
  int y = c == 0;
  int z = d == e;
  return x == y || x == z;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<WriteUnsigned2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  shl    $0x20,%rax
  or     %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  pop    %rbp
  ret
```

assistant: ```c
__attribute__((always_)) unsigned *WriteUnsigned2(unsigned *ptr, unsigned out0, unsigned out1) {
  *(unsigned long long *)(ptr) = out0 | ((unsigned long long)(out1) << 32);
  return ptr + 2;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<WriteUnsigned2>:
  endbr64
  shl    $0x20,%rdx
  mov    %esi,%esi
  or     %rsi,%rdx
  mov    %rdx,(%rdi)
  lea    0x8(%rdi),%rax
  ret
```

assistant: ```c
__attribute__((always_)) unsigned *WriteUnsigned2(unsigned *ptr, unsigned out0, unsigned out1) {
  *(unsigned long long *)(ptr) = out0 | ((unsigned long long)(out1) << 32);
  return ptr + 2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<WriteUnsigned2>:
  endbr64
  shl    $0x20,%rdx
  mov    %esi,%esi
  lea    0x8(%rdi),%rax
  or     %rsi,%rdx
  mov    %rdx,(%rdi)
  ret
```

assistant: ```c
__attribute__((always_)) unsigned *WriteUnsigned2(unsigned *ptr, unsigned out0, unsigned out1) {
  *(unsigned long long *)(ptr) = out0 | ((unsigned long long)(out1) << 32);
  return ptr + 2;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<WriteUnsigned2>:
  endbr64
  shl    $0x20,%rdx
  mov    %esi,%esi
  lea    0x8(%rdi),%rax
  or     %rsi,%rdx
  mov    %rdx,(%rdi)
  ret
```

assistant: ```c
__attribute__((always_)) unsigned *WriteUnsigned2(unsigned *ptr, unsigned out0, unsigned out1) {
  *(unsigned long long *)(ptr) = out0 | ((unsigned long long)(out1) << 32);
  return ptr + 2;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<generate_request>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L46
  cmpl   $0x3,-0x14(%rbp)
  jle    L18
L46:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x14(%rbp)
  cmpl   $0x5,-0x14(%rbp)
  jg     L123
  movl   $0x1,-0x14(%rbp)
  jmp    L18
L123:
  cmpl   $0x9,-0x14(%rbp)
  jne    L110
  movl   $0x3,-0x14(%rbp)
  jmp    L18
L110:
  movl   $0x2,-0x14(%rbp)
L18:
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <generate_word@plt>
  mov    -0x28(%rbp),%rax
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  cmpl   $0x3,-0x14(%rbp)
  je     L116
  cmpl   $0x3,-0x14(%rbp)
  jg     L48
  cmpl   $0x1,-0x14(%rbp)
  je     L39
  cmpl   $0x2,-0x14(%rbp)
  je     L92
  jmp    L48
L39:
  mov    -0x8(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  lea    D22(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x20(%rbp),%rdx
  mov    %rax,(%rdx)
  jmp    L48
L92:
  mov    -0x8(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  lea    D38(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x20(%rbp),%rdx
  mov    %rax,(%rdx)
  jmp    L48
L116:
  mov    -0x8(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  lea    D23(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x20(%rbp),%rdx
  mov    %rax,(%rdx)
  nop
L48:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D22
datatype: string
name: unpack_data
arguments: data_tag: D38
datatype: string
name: unpack_data
arguments: data_tag: D23
datatype: string
name: unpack_data
arguments: data_tag: D24
datatype: string

tool: D22: FIND %s

tool: D38: INSERT %s

tool: D23: REMOVE %s

tool: D24: %s


assistant: ```c
void generate_request(int method, size_t *size, char *request) {
  if (method <= 0 || method > 3) {
    method = rand() % 10;
    if (method < 6)
      method = 1;
    else if (method == 9)
      method = 3;
    else
      method = 2;
  }
  char *word = malloc(10);
  generate_word(word);
  memset(request, 0, sizeof(request));
  switch (method) {
  case 1:
    *size = sprintf(request, "FIND %s", word);
    break;
  case 2:
    *size = sprintf(request, "INSERT %s", word);
    break;
  case 3:
    *size = sprintf(request, "REMOVE %s", word);
    break;
  }
  printf("%s\n", request);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<generate_request>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%r13
  mov    %rdx,%rbp
  lea    -0x1(%rdi),%eax
  cmp    $0x2,%eax
  ja     L81
  mov    %edi,%ebx
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <generate_word@plt>
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  cmp    $0x2,%ebx
  je     L86
  cmp    $0x3,%ebx
  je     L28
  cmp    $0x1,%ebx
  je     L12
L74:
  mov    %rbp,%rsi
  lea    D20(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L81:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  cmp    $0x5,%eax
  jle    L45
  cmp    $0x9,%eax
  je     L7
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <generate_word@plt>
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L86:
  mov    %r12,%rdx
  lea    D7(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rax,0x0(%r13)
  jmp    L74
L45:
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <generate_word@plt>
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L12:
  mov    %r12,%rdx
  lea    D37(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rax,0x0(%r13)
  jmp    L74
L7:
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <generate_word@plt>
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L28:
  mov    %r12,%rdx
  lea    D32(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rax,0x0(%r13)
  jmp    L74
```

tool_calls: name: read_data
arguments: data_label: D20
datatype: string
name: read_data
arguments: data_label: D7
datatype: string
name: read_data
arguments: data_label: D37
datatype: string
name: read_data
arguments: data_label: D32
datatype: string

tool: D20: %s


tool: D7: INSERT %s

tool: D37: FIND %s

tool: D32: REMOVE %s

assistant: ```c
void generate_request(int method, size_t *size, char *request) {
  if (method <= 0 || method > 3) {
    method = rand() % 10;
    if (method < 6)
      method = 1;
    else if (method == 9)
      method = 3;
    else
      method = 2;
  }
  char *word = malloc(10);
  generate_word(word);
  memset(request, 0, sizeof(request));
  switch (method) {
  case 1:
    *size = sprintf(request, "FIND %s", word);
    break;
  case 2:
    *size = sprintf(request, "INSERT %s", word);
    break;
  case 3:
    *size = sprintf(request, "REMOVE %s", word);
    break;
  }
  printf("%s\n", request);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<generate_request>:
  endbr64
  push   %r13
  lea    -0x1(%rdi),%eax
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x2,%eax
  ja     L56
  mov    %edi,%ebx
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <generate_word@plt>
  xor    %esi,%esi
  mov    $0x8,%edx
  mov    %rbp,%rdi
  call   <memset@plt>
  cmp    $0x2,%ebx
  je     L30
  cmp    $0x3,%ebx
  jne    L67
L7:
  mov    %r12,%rdx
  lea    D29(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
L79:
  mov    %rax,0x0(%r13)
  add    $0x8,%rsp
  mov    %rbp,%rsi
  lea    D38(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10b0 <printf@plt>
  nop
L56:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  cmp    $0x5,%eax
  jg     L9
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <generate_word@plt>
  mov    $0x8,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L67:
  mov    %r12,%rdx
  lea    D0(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L79
L9:
  cmp    $0x9,%eax
  je     L65
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <generate_word@plt>
  mov    $0x8,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L30:
  mov    %r12,%rdx
  lea    D20(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L79
L65:
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <generate_word@plt>
  mov    $0x8,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  jmp    L7
```

tool_calls: name: retrieve_data
arguments: label: D29
data_type: string
name: retrieve_data
arguments: label: D38
data_type: string
name: retrieve_data
arguments: label: D0
data_type: string
name: retrieve_data
arguments: label: D20
data_type: string

tool: D29: REMOVE %s

tool: D38: %s


tool: D0: FIND %s

tool: D20: INSERT %s

assistant: ```c
void generate_request(int method, size_t *size, char *request) {
  if (method <= 0 || method > 3) {
    method = rand() % 10;
    if (method < 6)
      method = 1;
    else if (method == 9)
      method = 3;
    else
      method = 2;
  }
  char *word = malloc(10);
  generate_word(word);
  memset(request, 0, sizeof(request));
  switch (method) {
  case 1:
    *size = sprintf(request, "FIND %s", word);
    break;
  case 2:
    *size = sprintf(request, "INSERT %s", word);
    break;
  case 3:
    *size = sprintf(request, "REMOVE %s", word);
    break;
  }
  printf("%s\n", request);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<generate_request>:
  endbr64
  push   %r13
  lea    -0x1(%rdi),%eax
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x2,%eax
  ja     L60
  mov    %edi,%ebx
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <generate_word@plt>
  xor    %esi,%esi
  mov    $0x8,%edx
  mov    %rbp,%rdi
  call   <memset@plt>
  cmp    $0x2,%ebx
  je     L0
  cmp    $0x3,%ebx
  jne    L39
L7:
  mov    %r12,%rdx
  lea    D2(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
L29:
  mov    %rax,0x0(%r13)
  add    $0x8,%rsp
  mov    %rbp,%rsi
  lea    D8(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10b0 <printf@plt>
  nop
L60:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  cmp    $0x5,%eax
  jg     L24
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <generate_word@plt>
  mov    $0x8,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L39:
  mov    %r12,%rdx
  lea    D11(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L29
L24:
  cmp    $0x9,%eax
  je     L50
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <generate_word@plt>
  mov    $0x8,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L0:
  mov    %r12,%rdx
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  jmp    L29
L50:
  mov    $0xa,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <generate_word@plt>
  mov    $0x8,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  jmp    L7
```

tool_calls: name: retrieve_data
arguments: label: D2
type_guessed: string
name: retrieve_data
arguments: label: D8
type_guessed: string
name: retrieve_data
arguments: label: D11
type_guessed: string
name: retrieve_data
arguments: label: D9
type_guessed: string

tool: D2: REMOVE %s

tool: D8: %s


tool: D11: FIND %s

tool: D9: INSERT %s

assistant: ```c
void generate_request(int method, size_t *size, char *request) {
  if (method <= 0 || method > 3) {
    method = rand() % 10;
    if (method < 6)
      method = 1;
    else if (method == 9)
      method = 3;
    else
      method = 2;
  }
  char *word = malloc(10);
  generate_word(word);
  memset(request, 0, sizeof(request));
  switch (method) {
  case 1:
    *size = sprintf(request, "FIND %s", word);
    break;
  case 2:
    *size = sprintf(request, "INSERT %s", word);
    break;
  case 3:
    *size = sprintf(request, "REMOVE %s", word);
    break;
  }
  printf("%s\n", request);
}
```
user: Translate the below assembly code into C programming language:

```asm
<seq_utils_clock_pqq_to_divisor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x8,-0x4(%rbp)
  ja     L31
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D1(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D1(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x60,%eax
  jmp    L28
  mov    $0x30,%eax
  jmp    L28
  mov    $0x20,%eax
  jmp    L28
  mov    $0x18,%eax
  jmp    L28
  mov    $0x10,%eax
  jmp    L28
  mov    $0xc,%eax
  jmp    L28
  mov    $0x8,%eax
  jmp    L28
  mov    $0x4,%eax
  jmp    L28
L31:
  mov    $0x0,%eax
L28:
  pop    %rbp
  ret
```

assistant: ```c
int seq_utils_clock_pqq_to_divisor(int ppq) {
  switch (ppq) {
  case 1:
    return (96 / 1);
  case 2:
    return (96 / 2);
  case 3:
    return (96 / 3);
  case 4:
    return (96 / 4);
  case 5:
    return (96 / 6);
  case 6:
    return (96 / 8);
  case 7:
    return (96 / 12);
  case 8:
    return (96 / 24);
  default:
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<seq_utils_clock_pqq_to_divisor>:
  endbr64
  cmp    $0x8,%edi
  ja     L3
  mov    %edi,%edi
  lea    D7(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x30,%eax
  ret
  mov    $0x20,%eax
  ret
  mov    $0x18,%eax
  ret
  mov    $0x10,%eax
  ret
  mov    $0xc,%eax
  ret
  mov    $0x8,%eax
  ret
  mov    $0x4,%eax
  ret
L3:
  mov    $0x0,%eax
  ret
  mov    $0x60,%eax
  ret
```

assistant: ```c
int seq_utils_clock_pqq_to_divisor(int ppq) {
  switch (ppq) {
  case 1:
    return (96 / 1);
  case 2:
    return (96 / 2);
  case 3:
    return (96 / 3);
  case 4:
    return (96 / 4);
  case 5:
    return (96 / 6);
  case 6:
    return (96 / 8);
  case 7:
    return (96 / 12);
  case 8:
    return (96 / 24);
  default:
    return 0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<seq_utils_clock_pqq_to_divisor>:
  endbr64
  sub    $0x1,%edi
  xor    %eax,%eax
  cmp    $0x7,%edi
  ja     L5
  lea    D9(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L5:
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D9
datatype: string

tool: D9: `

assistant: ```c
int seq_utils_clock_pqq_to_divisor(int ppq) {
  switch (ppq) {
  case 1:
    return (96 / 1);
  case 2:
    return (96 / 2);
  case 3:
    return (96 / 3);
  case 4:
    return (96 / 4);
  case 5:
    return (96 / 6);
  case 6:
    return (96 / 8);
  case 7:
    return (96 / 12);
  case 8:
    return (96 / 24);
  default:
    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
<seq_utils_clock_pqq_to_divisor>:
  endbr64
  sub    $0x1,%edi
  xor    %eax,%eax
  cmp    $0x7,%edi
  ja     L4
  lea    D3(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L4:
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type: string

tool: D3: `

assistant: ```c
int seq_utils_clock_pqq_to_divisor(int ppq) {
  switch (ppq) {
  case 1:
    return (96 / 1);
  case 2:
    return (96 / 2);
  case 3:
    return (96 / 3);
  case 4:
    return (96 / 4);
  case 5:
    return (96 / 6);
  case 6:
    return (96 / 8);
  case 7:
    return (96 / 12);
  case 8:
    return (96 / 24);
  default:
    return 0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_version>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D7(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type_infered: string
name: extract_symbolic_data
arguments: tag: D12
type_infered: string

tool: D7: 2.0

tool: D12: LWFQZip %s


assistant: ```c
void print_version() {
  char *VERSION = "2.0";
  printf("LWFQZip %s\n", VERSION);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<print_version>:
  endbr64
  sub    $0x8,%rsp
  lea    D12(%rip),%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D12
type_guessed: string
name: read_data
arguments: tag: D5
type_guessed: string

tool: D12: 2.0

tool: D5: LWFQZip %s


assistant: ```c
void print_version() {
  char *VERSION = "2.0";
  printf("LWFQZip %s\n", VERSION);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<print_version>:
  endbr64
  lea    D10(%rip),%rsi
  lea    D8(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D10
type: string
name: unpack_data
arguments: data_label: D8
type: string

tool: D10: 2.0

tool: D8: LWFQZip %s


assistant: ```c
void print_version() {
  char *VERSION = "2.0";
  printf("LWFQZip %s\n", VERSION);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<print_version>:
  endbr64
  lea    D8(%rip),%rsi
  lea    D19(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: tag: D8
data_type: string
name: read_data
arguments: tag: D19
data_type: string

tool: D8: 2.0

tool: D19: LWFQZip %s


assistant: ```c
void print_version() {
  char *VERSION = "2.0";
  printf("LWFQZip %s\n", VERSION);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<notmain>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0x0,%eax
  call   <uart_init@plt>
  mov    $0x0,%eax
  call   <i2c1_init@plt>
  mov    $0x0,%eax
  call   <ar1010_main@plt>
L2:
  jmp    L2
```

assistant: ```c
int notmain(unsigned int earlypc) {
  uart_init();
  i2c1_init();
  ar1010_main();
  while (1)
    ;
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<notmain>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <uart_init@plt>
  mov    $0x0,%eax
  call   <i2c1_init@plt>
  mov    $0x0,%eax
  call   <ar1010_main@plt>
L5:
  jmp    L5
```

assistant: ```c
int notmain(unsigned int earlypc) {
  uart_init();
  i2c1_init();
  ar1010_main();
  while (1)
    ;
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<notmain>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <uart_init@plt>
  xor    %eax,%eax
  call   <i2c1_init@plt>
  xor    %eax,%eax
  call   <ar1010_main@plt>
L5:
  jmp    L5
```

assistant: ```c
int notmain(unsigned int earlypc) {
  uart_init();
  i2c1_init();
  ar1010_main();
  while (1)
    ;
  return (0);
}
```