ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fibo>:
  endbr64
  push   %r15
  mov    $0x1,%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %edi,0xc(%rsp)
  movq   $0x0,0x10(%rsp)
  cmp    $0x2,%edi
  jle    L163
L0:
  movq   $0x0,0x18(%rsp)
  mov    0xc(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x8(%rsp)
  cmp    $0x2,%eax
  je     L58
L136:
  movq   $0x0,0x20(%rsp)
  mov    0x8(%rsp),%eax
  lea    -0x1(%rax),%r9d
  cmp    $0x2,%r9d
  je     L172
L114:
  lea    -0x1(%r9),%esi
  xor    %r8d,%r8d
  cmp    $0x2,%esi
  je     L258
L102:
  lea    -0x1(%rsi),%ebp
  xor    %ecx,%ecx
  cmp    $0x2,%ebp
  je     L200
L10:
  lea    -0x1(%rbp),%ebx
  xor    %r14d,%r14d
  mov    %ebx,%r10d
  cmp    $0x2,%ebx
  je     L116
L164:
  lea    -0x1(%r10),%edx
  xor    %ebx,%ebx
  cmp    $0x2,%edx
  je     L232
L149:
  lea    -0x1(%rdx),%r12d
  xor    %r15d,%r15d
  mov    %rbx,%r11
  cmp    $0x2,%r12d
  je     L255
L132:
  lea    -0x1(%r12),%r13d
  xor    %ebx,%ebx
  cmp    $0x2,%r13d
  je     L229
L20:
  lea    -0x1(%r13),%edi
  mov    %edx,0x4c(%rsp)
  sub    $0x2,%r13d
  mov    %r10d,0x48(%rsp)
  mov    %r11,0x40(%rsp)
  mov    %rcx,0x38(%rsp)
  mov    %r8,0x30(%rsp)
  mov    %esi,0x2c(%rsp)
  mov    %r9d,0x28(%rsp)
  call   <fibo>
  mov    0x28(%rsp),%r9d
  mov    0x2c(%rsp),%esi
  add    %rax,%rbx
  cmp    $0x2,%r13d
  mov    0x30(%rsp),%r8
  mov    0x38(%rsp),%rcx
  mov    0x40(%rsp),%r11
  mov    0x48(%rsp),%r10d
  mov    0x4c(%rsp),%edx
  jg     L20
  add    $0x1,%rbx
L156:
  sub    $0x2,%r12d
  add    %rbx,%r15
  cmp    $0x2,%r12d
  jg     L132
  mov    %r11,%rbx
  add    $0x1,%r15
L214:
  sub    $0x2,%edx
  add    %r15,%rbx
  cmp    $0x2,%edx
  jg     L149
  add    $0x1,%rbx
L78:
  sub    $0x2,%r10d
  add    %rbx,%r14
  cmp    $0x2,%r10d
  jg     L164
  add    $0x1,%r14
L134:
  sub    $0x2,%ebp
  add    %r14,%rcx
  cmp    $0x2,%ebp
  jg     L10
  add    $0x1,%rcx
L252:
  sub    $0x2,%esi
  add    %rcx,%r8
  cmp    $0x2,%esi
  jg     L102
  add    $0x1,%r8
L195:
  sub    $0x2,%r9d
  add    %r8,0x20(%rsp)
  cmp    $0x2,%r9d
  jg     L114
  mov    0x20(%rsp),%rax
  add    $0x1,%rax
L43:
  subl   $0x2,0x8(%rsp)
  mov    0x8(%rsp),%edi
  add    %rax,0x18(%rsp)
  cmp    $0x2,%edi
  jg     L136
  mov    0x18(%rsp),%rax
  add    $0x1,%rax
L126:
  subl   $0x2,0xc(%rsp)
  mov    0xc(%rsp),%edi
  add    %rax,0x10(%rsp)
  cmp    $0x2,%edi
  jg     L0
  mov    0x10(%rsp),%rax
  add    $0x1,%rax
L163:
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L232:
  mov    $0x1,%ebx
  jmp    L78
  nopl   0x0(%rax)
L255:
  mov    $0x1,%r15d
  jmp    L214
  nopl   0x0(%rax,%rax,1)
L229:
  mov    $0x1,%ebx
  jmp    L156
L116:
  mov    $0x1,%r14d
  jmp    L134
L200:
  mov    $0x1,%ecx
  jmp    L252
L258:
  mov    $0x1,%r8d
  jmp    L195
L172:
  mov    $0x1,%eax
  jmp    L43
L58:
  mov    $0x1,%eax
  jmp    L126
```

assistant: ```c
unsigned long long fibo(int n) {
  if (n <= 2)
    return 1;
  return fibo(n - 1) + fibo(n - 2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <foo@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int e = 1;
  foo(e);
  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
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <foo@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int e = 1;
  foo(e);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <foo@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int e = 1;
  foo(e);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x19,%edx
  mov    $0x3,%esi
  mov    $0x7de,%edi
  call   <dayofweek@plt>
  cmp    $0x6,%eax
  ja     L25
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D61(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D61(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L25
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L25
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L25
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L25
  lea    D81(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L25
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L25
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
L25:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D36
datatype: string
name: unpack_data
arguments: tag: D31
datatype: string
name: unpack_data
arguments: tag: D10
datatype: string
name: unpack_data
arguments: tag: D19
datatype: string
name: unpack_data
arguments: tag: D81
datatype: string
name: unpack_data
arguments: tag: D25
datatype: string
name: unpack_data
arguments: tag: D24
datatype: string

tool: D36: 7

tool: D31: 1

tool: D10: 2

tool: D19: 3

tool: D81: 4

tool: D25: 5

tool: D24: 6

assistant: ```c
int main() {
  switch (dayofweek(2014, 3, 25)) {
  case 0:
    printf("7");
    break;
  case 1:
    printf("1");
    break;
  case 2:
    printf("2");
    break;
  case 3:
    printf("3");
    break;
  case 4:
    printf("4");
    break;
  case 5:
    printf("5");
    break;
  case 6:
    printf("6");
    break;
  }
  getchar();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x19,%edx
  mov    $0x3,%esi
  mov    $0x7de,%edi
  call   <dayofweek@plt>
  cmp    $0x6,%eax
  ja     L24
  mov    %eax,%eax
  lea    D19(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D48(%rip),%rdi
  call   <printf@plt>
L24:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
  lea    D18(%rip),%rdi
  call   <printf@plt>
  jmp    L24
  lea    D12(%rip),%rdi
  call   <printf@plt>
  jmp    L24
  lea    D27(%rip),%rdi
  call   <printf@plt>
  jmp    L24
  lea    D76(%rip),%rdi
  call   <printf@plt>
  jmp    L24
  lea    D34(%rip),%rdi
  call   <printf@plt>
  jmp    L24
  lea    D43(%rip),%rdi
  call   <printf@plt>
  jmp    L24
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D48
datatype: string
name: extract_symbolic_data
arguments: data_tag: D18
datatype: string
name: extract_symbolic_data
arguments: data_tag: D12
datatype: string
name: extract_symbolic_data
arguments: data_tag: D27
datatype: string
name: extract_symbolic_data
arguments: data_tag: D76
datatype: string
name: extract_symbolic_data
arguments: data_tag: D34
datatype: string
name: extract_symbolic_data
arguments: data_tag: D43
datatype: string

tool: D48: 7

tool: D18: 1

tool: D12: 2

tool: D27: 3

tool: D76: 4

tool: D34: 5

tool: D43: 6

assistant: ```c
int main() {
  switch (dayofweek(2014, 3, 25)) {
  case 0:
    printf("7");
    break;
  case 1:
    printf("1");
    break;
  case 2:
    printf("2");
    break;
  case 3:
    printf("3");
    break;
  case 4:
    printf("4");
    break;
  case 5:
    printf("5");
    break;
  case 6:
    printf("6");
    break;
  }
  getchar();
  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
  sub    $0x8,%rsp
  mov    $0x19,%edx
  mov    $0x3,%esi
  mov    $0x7de,%edi
  call   <dayofweek@plt>
  cmp    $0x6,%eax
  ja     L37
  lea    D59(%rip),%rdx
  mov    %eax,%eax
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D45(%rip),%rdi
  call   <printf@plt>
L37:
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
  lea    D50(%rip),%rdi
  call   <printf@plt>
  jmp    L37
  lea    D79(%rip),%rdi
  call   <printf@plt>
  jmp    L37
  lea    D25(%rip),%rdi
  call   <printf@plt>
  jmp    L37
  lea    D13(%rip),%rdi
  call   <printf@plt>
  jmp    L37
  lea    D70(%rip),%rdi
  call   <printf@plt>
  jmp    L37
  lea    D52(%rip),%rdi
  call   <printf@plt>
  jmp    L37
```

tool_calls: name: unpack_data
arguments: data_label: D45
data_type: string
name: unpack_data
arguments: data_label: D50
data_type: string
name: unpack_data
arguments: data_label: D79
data_type: string
name: unpack_data
arguments: data_label: D25
data_type: string
name: unpack_data
arguments: data_label: D13
data_type: string
name: unpack_data
arguments: data_label: D70
data_type: string
name: unpack_data
arguments: data_label: D52
data_type: string

tool: D45: 6

tool: D50: 5

tool: D79: 7

tool: D25: 1

tool: D13: 2

tool: D70: 3

tool: D52: 4

assistant: ```c
int main() {
  switch (dayofweek(2014, 3, 25)) {
  case 0:
    printf("7");
    break;
  case 1:
    printf("1");
    break;
  case 2:
    printf("2");
    break;
  case 3:
    printf("3");
    break;
  case 4:
    printf("4");
    break;
  case 5:
    printf("5");
    break;
  case 6:
    printf("6");
    break;
  }
  getchar();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x19,%edx
  mov    $0x3,%esi
  mov    $0x7de,%edi
  call   <dayofweek@plt>
  cmp    $0x6,%eax
  ja     L28
  lea    D16(%rip),%rdx
  mov    %eax,%eax
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D79(%rip),%rdi
  call   <printf@plt>
L28:
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
  lea    D44(%rip),%rdi
  call   <printf@plt>
  jmp    L28
  lea    D35(%rip),%rdi
  call   <printf@plt>
  jmp    L28
  lea    D33(%rip),%rdi
  call   <printf@plt>
  jmp    L28
  lea    D0(%rip),%rdi
  call   <printf@plt>
  jmp    L28
  lea    D26(%rip),%rdi
  call   <printf@plt>
  jmp    L28
  lea    D43(%rip),%rdi
  call   <printf@plt>
  jmp    L28
```

tool_calls: name: extract_symbolic_data
arguments: tag: D79
data_type: string
name: extract_symbolic_data
arguments: tag: D44
data_type: string
name: extract_symbolic_data
arguments: tag: D35
data_type: string
name: extract_symbolic_data
arguments: tag: D33
data_type: string
name: extract_symbolic_data
arguments: tag: D0
data_type: string
name: extract_symbolic_data
arguments: tag: D26
data_type: string
name: extract_symbolic_data
arguments: tag: D43
data_type: string

tool: D79: 6

tool: D44: 5

tool: D35: 7

tool: D33: 1

tool: D0: 2

tool: D26: 3

tool: D43: 4

assistant: ```c
int main() {
  switch (dayofweek(2014, 3, 25)) {
  case 0:
    printf("7");
    break;
  case 1:
    printf("1");
    break;
  case 2:
    printf("2");
    break;
  case 3:
    printf("3");
    break;
  case 4:
    printf("4");
    break;
  case 5:
    printf("5");
    break;
  case 6:
    printf("6");
    break;
  }
  getchar();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<av_fast_realloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jae    L4
  mov    -0x8(%rbp),%rax
  jmp    L19
L4:
  mov    -0x14(%rbp),%eax
  lea    0x2800(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <av_realloc@plt>
L19:
  leave
  ret
```

assistant: ```c
void *av_fast_realloc(void *ptr, unsigned int *size, unsigned int min_size) {
  if (min_size < *size)
    return ptr;
  *size = min_size + 10 * 1024;
  return av_realloc(ptr, *size);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<av_fast_realloc>:
  endbr64
  mov    %rdi,%rax
  cmp    %edx,(%rsi)
  jbe    L7
  ret
L7:
  sub    $0x8,%rsp
  mov    %rsi,%rcx
  lea    0x2800(%rdx),%esi
  mov    %esi,(%rcx)
  call   <av_realloc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *av_fast_realloc(void *ptr, unsigned int *size, unsigned int min_size) {
  if (min_size < *size)
    return ptr;
  *size = min_size + 10 * 1024;
  return av_realloc(ptr, *size);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<av_fast_realloc>:
  endbr64
  mov    %rsi,%rcx
  cmp    %edx,(%rsi)
  jbe    L2
  mov    %rdi,%rax
  ret
  nop
L2:
  lea    0x2800(%rdx),%esi
  mov    %esi,(%rcx)
  jmp    1050 <av_realloc@plt>
```

assistant: ```c
void *av_fast_realloc(void *ptr, unsigned int *size, unsigned int min_size) {
  if (min_size < *size)
    return ptr;
  *size = min_size + 10 * 1024;
  return av_realloc(ptr, *size);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<av_fast_realloc>:
  endbr64
  mov    %rsi,%rcx
  cmp    %edx,(%rsi)
  jbe    L6
  mov    %rdi,%rax
  ret
  nop
L6:
  lea    0x2800(%rdx),%esi
  mov    %esi,(%rcx)
  jmp    1050 <av_realloc@plt>
```

assistant: ```c
void *av_fast_realloc(void *ptr, unsigned int *size, unsigned int min_size) {
  if (min_size < *size)
    return ptr;
  *size = min_size + 10 * 1024;
  return av_realloc(ptr, *size);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %edx,%edi
  sub    %eax,%edi
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    -0xc(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%r8d
  sar    $0x1f,%r8d
  sub    %r8d,%eax
  mov    %eax,%esi
  mov    %esi,%eax
  shl    $0x2,%eax
  add    %esi,%eax
  add    %eax,%eax
  mov    %ecx,%esi
  sub    %eax,%esi
  mov    %edi,%ecx
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_infered: string
name: unpack_data
arguments: data_label: D15
type_infered: string

tool: D1: %d

tool: D15: %d%d%d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  printf("%d%d%d\n", n % 10, n / 10 % 10, n / 100);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  movslq %esi,%rax
  imul   $0x51eb851f,%rax,%rcx
  sar    $0x25,%rcx
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sub    %edx,%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%edi
  sar    $0x1f,%edi
  sub    %edi,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  mov    %eax,%edi
  sub    %edx,%edi
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %edi,%edx
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D11
type_guessed: string
name: read_data
arguments: tag: D19
type_guessed: string

tool: D11: %d

tool: D19: %d%d%d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  printf("%d%d%d\n", n % 10, n / 10 % 10, n / 100);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D8(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rax
  imul   $0x51eb851f,%rax,%rcx
  mov    %rax,%rsi
  mov    %eax,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x25,%rcx
  sar    $0x22,%rax
  sub    %edx,%ecx
  sub    %edx,%eax
  movslq %eax,%rdx
  mov    %eax,%edi
  mov    %eax,%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%edi
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  sar    $0x22,%rdx
  sub    %edi,%edx
  lea    D12(%rip),%rdi
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%r8d
  mov    %r8d,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type_infered: string
name: extract_symbolic_data
arguments: data_label: D12
type_infered: string

tool: D8: %d

tool: D12: %d%d%d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  printf("%d%d%d\n", n % 10, n / 10 % 10, n / 100);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rax
  imul   $0x51eb851f,%rax,%rcx
  mov    %rax,%rsi
  mov    %eax,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x25,%rcx
  sar    $0x22,%rax
  sub    %edx,%ecx
  sub    %edx,%eax
  movslq %eax,%rdx
  mov    %eax,%edi
  mov    %eax,%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%edi
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  sar    $0x22,%rdx
  sub    %edi,%edx
  lea    D5(%rip),%rdi
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%r8d
  mov    %r8d,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D16
type: string
name: parse_data
arguments: data_tag: D5
type: string

tool: D16: %d

tool: D5: %d%d%d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  printf("%d%d%d\n", n % 10, n / 10 % 10, n / 100);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<isPrime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  jne    L36
  mov    $0x0,%eax
  jmp    L70
L36:
  mov    -0x28(%rbp),%rcx
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    %rax
  mov    %rax,%rdx
  add    %rdx,%rdx
  add    %rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  test   %rax,%rax
  jne    L17
  mov    $0x0,%eax
  jmp    L70
L17:
  movl   $0x4,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <sqrtl@plt>
  add    $0x1,%eax
  cltq
  mov    %rax,-0x8(%rbp)
  movq   $0x5,-0x10(%rbp)
  jmp    L39
L67:
  mov    -0x28(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x10(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L16
  mov    $0x0,%eax
  jmp    L70
L16:
  mov    $0x6,%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  add    %rax,-0x10(%rbp)
L39:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L67
  mov    $0x1,%eax
L70:
  leave
  ret
```

assistant: ```c
bool isPrime(unsigned long n) {
  if (!(n & 1)) {
    return false;
  } else if (n % 3 == 0) {
    return false;
  }
  int step = 4;
  unsigned long root = sqrtl(n) + 1;
  for (unsigned long i = 5; i < root; step = 6 - step, i += step) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<isPrime>:
  endbr64
  mov    $0x0,%eax
  test   $0x1,%dil
  je     L25
  push   %rbx
  mov    %rdi,%rbx
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mov    %rdi,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    %rax
  and    $0xfffffffffffffffe,%rdx
  add    %rax,%rdx
  mov    $0x0,%eax
  cmp    %rdx,%rdi
  jne    L63
L59:
  pop    %rbx
  ret
L63:
  call   <sqrtl@plt>
  lea    0x1(%rax),%edi
  movslq %edi,%rdi
  cmp    $0x5,%rdi
  jbe    L39
  movabs $0xcccccccccccccccd,%rdx
  mov    %rbx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x2,%rax
  and    $0xfffffffffffffffc,%rdx
  add    %rax,%rdx
  cmp    %rdx,%rbx
  je     L71
  mov    $0x5,%ecx
  mov    $0x4,%esi
  mov    $0x6,%r8d
L24:
  mov    %r8d,%eax
  sub    %esi,%eax
  mov    %eax,%esi
  cltq
  add    %rax,%rcx
  cmp    %rcx,%rdi
  jbe    L2
  mov    %rbx,%rax
  mov    $0x0,%edx
  div    %rcx
  test   %rdx,%rdx
  jne    L24
  mov    $0x0,%eax
  jmp    L59
L2:
  mov    $0x1,%eax
  jmp    L59
L39:
  mov    $0x1,%eax
  jmp    L59
L71:
  mov    $0x0,%eax
  jmp    L59
L25:
  ret
```

assistant: ```c
bool isPrime(unsigned long n) {
  if (!(n & 1)) {
    return false;
  } else if (n % 3 == 0) {
    return false;
  }
  int step = 4;
  unsigned long root = sqrtl(n) + 1;
  for (unsigned long i = 5; i < root; step = 6 - step, i += step) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<isPrime>:
  endbr64
  test   $0x1,%dil
  je     L12
  movabs $0xaaaaaaaaaaaaaaab,%rax
  push   %r12
  mov    %rdi,%r12
  movabs $0x5555555555555555,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%rax
  ja     L19
L56:
  xor    %eax,%eax
  pop    %r12
  ret
  nopl   (%rax)
L19:
  call   <sqrtl@plt>
  lea    0x1(%rax),%edi
  movslq %edi,%rdi
  cmp    $0x5,%rdi
  jbe    L30
  movabs $0xcccccccccccccccd,%rax
  movabs $0x3333333333333333,%rdx
  imul   %r12,%rax
  cmp    %rdx,%rax
  jbe    L56
  mov    $0x5,%ecx
  mov    $0x4,%esi
  mov    $0x6,%r8d
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L18:
  mov    %r12,%rax
  xor    %edx,%edx
  div    %rcx
  test   %rdx,%rdx
  je     L56
L4:
  mov    %r8d,%eax
  sub    %esi,%eax
  mov    %eax,%esi
  cltq
  add    %rax,%rcx
  cmp    %rcx,%rdi
  ja     L18
L30:
  mov    $0x1,%eax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L12:
  xor    %eax,%eax
  ret
```

assistant: ```c
bool isPrime(unsigned long n) {
  if (!(n & 1)) {
    return false;
  } else if (n % 3 == 0) {
    return false;
  }
  int step = 4;
  unsigned long root = sqrtl(n) + 1;
  for (unsigned long i = 5; i < root; step = 6 - step, i += step) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
```
user: Translate the below assembly code into C programming language:

```asm
<isPrime>:
  endbr64
  test   $0x1,%dil
  je     L67
  movabs $0xaaaaaaaaaaaaaaab,%rax
  push   %r12
  mov    %rdi,%r12
  movabs $0x5555555555555555,%rdx
  imul   %rdi,%rax
  cmp    %rdx,%rax
  ja     L17
L53:
  xor    %eax,%eax
  pop    %r12
  ret
  nopl   (%rax)
L17:
  call   <sqrtl@plt>
  lea    0x1(%rax),%edi
  movslq %edi,%rdi
  cmp    $0x5,%rdi
  jbe    L42
  movabs $0xcccccccccccccccd,%rax
  movabs $0x3333333333333333,%rdx
  imul   %r12,%rax
  cmp    %rdx,%rax
  jbe    L53
  mov    $0x5,%ecx
  mov    $0x4,%esi
  mov    $0x6,%r8d
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %r12,%rax
  xor    %edx,%edx
  div    %rcx
  test   %rdx,%rdx
  je     L53
L1:
  mov    %r8d,%eax
  sub    %esi,%eax
  mov    %eax,%esi
  cltq
  add    %rax,%rcx
  cmp    %rcx,%rdi
  ja     L0
L42:
  mov    $0x1,%eax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L67:
  xor    %eax,%eax
  ret
```

assistant: ```c
bool isPrime(unsigned long n) {
  if (!(n & 1)) {
    return false;
  } else if (n % 3 == 0) {
    return false;
  }
  int step = 4;
  unsigned long root = sqrtl(n) + 1;
  for (unsigned long i = 5; i < root; step = 6 - step, i += step) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
```