ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<itoa>:
  endbr64
  mov    %esi,%edx
  mov    %esi,%eax
  mov    $0xcccccccd,%r8d
  xor    %esi,%esi
L9:
  mov    %eax,%eax
  add    $0x1,%esi
  mov    %rax,%rcx
  imul   %r8,%rax
  shr    $0x23,%rax
  cmp    $0x9,%ecx
  ja     L9
  add    %rsi,%rdi
  mov    $0xcccccccd,%r9d
  movb   $0x0,(%rdi)
  lea    -0x1(%rdi),%rcx
  nopl   (%rax)
L13:
  mov    %edx,%eax
  mov    %edx,%esi
  mov    %rcx,%r8
  sub    $0x1,%rcx
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edi
  add    %edi,%edi
  sub    %edi,%esi
  add    $0x30,%esi
  mov    %sil,0x1(%rcx)
  mov    %edx,%esi
  mov    %eax,%edx
  cmp    $0x9,%esi
  ja     L13
  mov    %r8,%rax
  ret
```

assistant: ```c
char *itoa(char *a, unsigned int i) {
  unsigned int _i = i, x = 0;
  do {
    x++;
    _i /= 10;
  } while (_i != 0);
  a += x;
  *a-- = '\0';
  do {
    *a-- = (i % 10) + '0';
    i /= 10;
  } while (i != 0);
  return a + 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<itoa>:
  endbr64
  mov    %esi,%edx
  mov    %esi,%eax
  mov    $0xcccccccd,%r8d
  xor    %esi,%esi
L19:
  mov    %eax,%eax
  add    $0x1,%esi
  mov    %rax,%rcx
  imul   %r8,%rax
  shr    $0x23,%rax
  cmp    $0x9,%ecx
  ja     L19
  add    %rsi,%rdi
  mov    $0xcccccccd,%r9d
  movb   $0x0,(%rdi)
  lea    -0x1(%rdi),%rcx
  nopl   (%rax)
L17:
  mov    %edx,%eax
  mov    %edx,%esi
  mov    %rcx,%r8
  sub    $0x1,%rcx
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edi
  add    %edi,%edi
  sub    %edi,%esi
  add    $0x30,%esi
  mov    %sil,0x1(%rcx)
  mov    %edx,%esi
  mov    %eax,%edx
  cmp    $0x9,%esi
  ja     L17
  mov    %r8,%rax
  ret
```

assistant: ```c
char *itoa(char *a, unsigned int i) {
  unsigned int _i = i, x = 0;
  do {
    x++;
    _i /= 10;
  } while (_i != 0);
  a += x;
  *a-- = '\0';
  do {
    *a-- = (i % 10) + '0';
    i /= 10;
  } while (i != 0);
  return a + 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<fib3_helper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x14(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0xffffffff,%eax
  je     L2
  mov    -0x14(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  jmp    L3
L2:
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %ecx,%edi
  call   <fib3_helper>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  lea    -0x2(%rax),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %ecx,%edi
  call   <fib3_helper>
  lea    (%rbx,%rax,1),%edx
  mov    -0x14(%rbp),%eax
  lea    0x0(,%rax,4),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    %edx,(%rax)
  mov    -0x14(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
L3:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D9
data_type: string

tool: D9: fib(%d)

assistant: ```c
signed int fib3_helper(unsigned int n, unsigned int *nchamadas, unsigned int resultados[]) {
  printf("fib(%d)", n);
  ++*nchamadas;
  if (resultados[n] != (unsigned int)-1) {
    return resultados[n];
  }
  resultados[n] = (fib3_helper(n - 1, nchamadas, resultados) + fib3_helper(n - 2, nchamadas, resultados));
  return resultados[n];
}
```
user: What is the c source code of the assembly code below?

```asm
<fib3_helper>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    %rdx,%r14
  mov    %edi,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  addl   $0x1,(%r12)
  mov    %ebp,%eax
  lea    (%r14,%rax,4),%r13
  mov    0x0(%r13),%eax
  cmp    $0xffffffff,%eax
  je     L17
L14:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  lea    -0x1(%rbp),%edi
  mov    %r14,%rdx
  mov    %r12,%rsi
  call   <fib3_helper>
  mov    %eax,%ebx
  lea    -0x2(%rbp),%edi
  mov    %r14,%rdx
  mov    %r12,%rsi
  call   <fib3_helper>
  add    %ebx,%eax
  mov    %eax,0x0(%r13)
  jmp    L14
```

tool_calls: name: retrieve_data
arguments: tag: D8
type_infered: string

tool: D8: fib(%d)

assistant: ```c
signed int fib3_helper(unsigned int n, unsigned int *nchamadas, unsigned int resultados[]) {
  printf("fib(%d)", n);
  ++*nchamadas;
  if (resultados[n] != (unsigned int)-1) {
    return resultados[n];
  }
  resultados[n] = (fib3_helper(n - 1, nchamadas, resultados) + fib3_helper(n - 2, nchamadas, resultados));
  return resultados[n];
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fib3_helper>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  mov    %edi,%esi
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%eax
  addl   $0x1,(%r12)
  lea    0x0(%r13,%rax,4),%r14
  mov    (%r14),%eax
  cmp    $0xffffffff,%eax
  je     L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L6:
  lea    -0x1(%rbx),%edi
  mov    %r13,%rdx
  mov    %r12,%rsi
  call   <fib3_helper>
  lea    -0x2(%rbx),%edi
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %eax,%ebp
  call   <fib3_helper>
  add    %ebp,%eax
  mov    %eax,(%r14)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D7
type: string

tool: D7: fib(%d)

assistant: ```c
signed int fib3_helper(unsigned int n, unsigned int *nchamadas, unsigned int resultados[]) {
  printf("fib(%d)", n);
  ++*nchamadas;
  if (resultados[n] != (unsigned int)-1) {
    return resultados[n];
  }
  resultados[n] = (fib3_helper(n - 1, nchamadas, resultados) + fib3_helper(n - 2, nchamadas, resultados));
  return resultados[n];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<return_time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  movsd  -0x20(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  movsd  D3(%rip),%xmm2
  divsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
data_type: f64

tool: D3: 1000000

assistant: ```c
double return_time() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (1.0 * (tv.tv_sec + tv.tv_usec / 1000000.0));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<return_time>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  movsd  0x8(%rsp),%xmm0
  divsd  D6(%rip),%xmm0
  addsd  (%rsp),%xmm0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
data_type: f64

tool: D6: 1000000

assistant: ```c
double return_time() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (1.0 * (tv.tv_sec + tv.tv_usec / 1000000.0));
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<return_time>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  movsd  0x8(%rsp),%xmm0
  divsd  D0(%rip),%xmm0
  addsd  (%rsp),%xmm0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D0
type: f64

tool: D0: 1000000

assistant: ```c
double return_time() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (1.0 * (tv.tv_sec + tv.tv_usec / 1000000.0));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<return_time>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  movsd  0x8(%rsp),%xmm0
  divsd  D9(%rip),%xmm0
  addsd  (%rsp),%xmm0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type_guessed: f64

tool: D9: 1000000

assistant: ```c
double return_time() {
  struct timeval tv;
  gettimeofday(&tv, 0);
  return (1.0 * (tv.tv_sec + tv.tv_usec / 1000000.0));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<display>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L3:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L3
  nop
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
data_type: string

tool: D9: %d	

assistant: ```c
void display(int a[], int n) {
  int i;
  printf("\n");
  for (i = 0; i < n; i++) {
    printf("%d\t", a[i]);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<display>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    $0xa,%edi
  call   <putchar@plt>
  test   %ebp,%ebp
  jle    L6
  mov    %r12,%rbx
  lea    -0x1(%rbp),%eax
  lea    0x4(%r12,%rax,4),%r12
  lea    D3(%rip),%rbp
L1:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L1
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
data_type: string

tool: D3: %d	

assistant: ```c
void display(int a[], int n) {
  int i;
  printf("\n");
  for (i = 0; i < n; i++) {
    printf("%d\t", a[i]);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<display>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0xa,%edi
  call   <putchar@plt>
  test   %ebp,%ebp
  jle    L18
  lea    -0x1(%rbp),%eax
  lea    D9(%rip),%rbp
  lea    0x4(%rbx,%rax,4),%r12
  nopw   0x0(%rax,%rax,1)
L19:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L19
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

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

tool: D9: %d	

assistant: ```c
void display(int a[], int n) {
  int i;
  printf("\n");
  for (i = 0; i < n; i++) {
    printf("%d\t", a[i]);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<display>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0xa,%edi
  call   <putchar@plt>
  test   %ebp,%ebp
  jle    L6
  lea    -0x1(%rbp),%eax
  lea    D4(%rip),%rbp
  lea    0x4(%rbx,%rax,4),%r12
  nopw   0x0(%rax,%rax,1)
L12:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L12
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
datatype: string

tool: D4: %d	

assistant: ```c
void display(int a[], int n) {
  int i;
  printf("\n");
  for (i = 0; i < n; i++) {
    printf("%d\t", a[i]);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L2
L64:
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <isSubsequence@plt>
  test   %rax,%rax
  jne    L39
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <isSubsequence@plt>
  test   %rax,%rax
  je     L68
L39:
  mov    $0x1,%eax
  jmp    L4
L68:
  mov    $0x0,%eax
L4:
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L62
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L2
L62:
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L2:
  mov    -0x20(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  test   %eax,%eax
  jne    L64
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L41
  call   <__stack_chk_fail@plt>
L41:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D10
datatype: string
name: disassemble_data
arguments: data_label: D23
datatype: string
name: disassemble_data
arguments: data_label: D6
datatype: string
name: disassemble_data
arguments: data_label: D3
datatype: string

tool: D10: %d

tool: D23: %s %s

tool: D6: YES


tool: D3: NO


assistant: ```c
int main(void) {
  int T, isOk;
  char *s1, *s2;
  scanf("%d", &T);
  s1 = (char *)malloc(sizeof(char));
  s2 = (char *)malloc(sizeof(char));
  while (T--) {
    scanf("%s %s", s1, s2);
    isOk = (isSubsequence(s1, s2) || isSubsequence(s2, s1));
    if (isOk) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  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    D8(%rip),%rdi
  call   <scanf@plt>
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L67
  lea    D38(%rip),%r12
  lea    D4(%rip),%r13
  lea    D23(%rip),%r14
  jmp    L10
L41:
  mov    %r13,%rdi
  call   <printf@plt>
L64:
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L67
L10:
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <isSubsequence@plt>
  test   %rax,%rax
  jne    L41
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <isSubsequence@plt>
  test   %rax,%rax
  jne    L41
  mov    %r14,%rdi
  call   <printf@plt>
  jmp    L64
L67:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type_infered: string
name: retrieve_data
arguments: data_label: D38
type_infered: string
name: retrieve_data
arguments: data_label: D4
type_infered: string
name: retrieve_data
arguments: data_label: D23
type_infered: string

tool: D8: %d

tool: D38: %s %s

tool: D4: YES


tool: D23: NO


assistant: ```c
int main(void) {
  int T, isOk;
  char *s1, *s2;
  scanf("%d", &T);
  s1 = (char *)malloc(sizeof(char));
  s2 = (char *)malloc(sizeof(char));
  while (T--) {
    scanf("%s %s", s1, s2);
    isOk = (isSubsequence(s1, s2) || isSubsequence(s2, s1));
    if (isOk) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  lea    D37(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    $0x1,%edi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L39
  lea    D38(%rip),%r12
  lea    D14(%rip),%r14
  lea    D34(%rip),%r13
  jmp    L29
  nopl   (%rax)
L56:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L39
L29:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    %rbp,%rdx
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <isSubsequence@plt>
  test   %rax,%rax
  jne    L56
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <isSubsequence@plt>
  test   %rax,%rax
  jne    L56
  mov    %r14,%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jne    L29
  cs nopw 0x0(%rax,%rax,1)
L39:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D37
type: string
name: extract_symbolic_data
arguments: data_tag: D38
type: string
name: extract_symbolic_data
arguments: data_tag: D14
type: string
name: extract_symbolic_data
arguments: data_tag: D34
type: string

tool: D37: %d

tool: D38: %s %s

tool: D14: NO


tool: D34: YES


assistant: ```c
int main(void) {
  int T, isOk;
  char *s1, *s2;
  scanf("%d", &T);
  s1 = (char *)malloc(sizeof(char));
  s2 = (char *)malloc(sizeof(char));
  while (T--) {
    scanf("%s %s", s1, s2);
    isOk = (isSubsequence(s1, s2) || isSubsequence(s2, s1));
    if (isOk) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D30(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    $0x1,%edi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L66
  lea    D25(%rip),%r12
  lea    D1(%rip),%r14
  lea    D31(%rip),%r13
  jmp    L47
  nopl   (%rax)
L49:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  je     L66
L47:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    %rbp,%rdx
  call   <scanf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <isSubsequence@plt>
  test   %rax,%rax
  jne    L49
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <isSubsequence@plt>
  test   %rax,%rax
  jne    L49
  mov    %r14,%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jne    L47
  cs nopw 0x0(%rax,%rax,1)
L66:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L61:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D30
datatype: string
name: parse_data
arguments: data_tag: D25
datatype: string
name: parse_data
arguments: data_tag: D1
datatype: string
name: parse_data
arguments: data_tag: D31
datatype: string

tool: D30: %d

tool: D25: %s %s

tool: D1: NO


tool: D31: YES


assistant: ```c
int main(void) {
  int T, isOk;
  char *s1, *s2;
  scanf("%d", &T);
  s1 = (char *)malloc(sizeof(char));
  s2 = (char *)malloc(sizeof(char));
  while (T--) {
    scanf("%s %s", s1, s2);
    isOk = (isSubsequence(s1, s2) || isSubsequence(s2, s1));
    if (isOk) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%eax
  call   <unknown1@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L21
L16:
  addl   $0x1,-0x14(%rbp)
  addl   $0x1,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  add    %eax,-0x18(%rbp)
  mov    -0x10(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L21
  mov    -0x14(%rbp),%eax
  add    %eax,-0x1c(%rbp)
L21:
  mov    $0x0,%eax
  call   <unknown1@plt>
  test   %eax,%eax
  jne    L16
  movl   $0x1,-0xc(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L60
  mov    -0x18(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x1c(%rbp),%eax
  sub    %edx,%eax
  add    $0x2,%eax
  mov    %eax,-0xc(%rbp)
L60:
  movl   $0x0,-0x8(%rbp)
  jmp    L45
L44:
  mov    $0x0,%eax
  call   <unknown2@plt>
  test   %rax,%rax
  je     L35
  addl   $0x1,-0x8(%rbp)
  jmp    L45
L35:
  addl   $0x2,-0x8(%rbp)
L45:
  mov    -0x8(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L44
  cmpl   $0x4,-0x8(%rbp)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int flag = unknown1();
  int t = 0;
  int s = 0;
  int a = 0;
  int b = 0;
  int x, y;
  while (unknown1()) {
    a++;
    b++;
    s += a;
    t += b;
    if (flag) {
      t += a;
    }
    t = t;
  }
  x = 1;
  if (flag) {
    x = t - 2 * s + 2;
  }
  y = 0;
  while (y <= x) {
    if (unknown2())
      y++;
    else
      y += 2;
    y = y;
  }
  if (y >= 5) {
    goto ERROR;
  ERROR:;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <unknown1@plt>
  mov    %eax,%r14d
  mov    $0x0,%r12d
  mov    $0x0,%r13d
  mov    $0x0,%ebp
  jmp    L32
L41:
  add    $0x1,%r12d
  add    %r12d,%r13d
  add    %r12d,%ebp
  lea    0x0(%rbp,%r12,1),%eax
  test   %r14d,%r14d
  cmovne %eax,%ebp
L32:
  mov    $0x0,%eax
  call   <unknown1@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L41
  test   %r14d,%r14d
  je     L5
  add    %r13d,%r13d
  sub    %r13d,%ebp
  add    $0x2,%ebp
  jns    L11
  jmp    L18
L5:
  mov    $0x1,%ebp
L11:
  mov    $0x0,%eax
  call   <unknown2@plt>
  cmp    $0x1,%rax
  adc    $0x1,%ebx
  cmp    %ebp,%ebx
  jle    L11
L18:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int main() {
  int flag = unknown1();
  int t = 0;
  int s = 0;
  int a = 0;
  int b = 0;
  int x, y;
  while (unknown1()) {
    a++;
    b++;
    s += a;
    t += b;
    if (flag) {
      t += a;
    }
    t = t;
  }
  x = 1;
  if (flag) {
    x = t - 2 * s + 2;
  }
  y = 0;
  while (y <= x) {
    if (unknown2())
      y++;
    else
      y += 2;
    y = y;
  }
  if (y >= 5) {
    goto ERROR;
  ERROR:;
  }
}
```