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
<KEY_256_ASSIST_2>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rdi),%edi
  mov    %rsi,%rbx
  xor    %esi,%esi
  call   <_mm_aeskeygenassist_si128@plt>
  mov    $0xaa,%esi
  mov    %eax,%edi
  call   <_mm_shuffle_epi32@plt>
  mov    (%rbx),%edi
  mov    $0x4,%esi
  mov    %eax,%r12d
  call   <_mm_slli_si128@plt>
  mov    (%rbx),%edi
  mov    %eax,%esi
  mov    %eax,%ebp
  call   <_mm_xor_si128@plt>
  mov    %ebp,%edi
  mov    $0x4,%esi
  mov    %eax,(%rbx)
  call   <_mm_slli_si128@plt>
  mov    (%rbx),%edi
  mov    %eax,%esi
  mov    %eax,%ebp
  call   <_mm_xor_si128@plt>
  mov    %ebp,%edi
  mov    $0x4,%esi
  mov    %eax,(%rbx)
  call   <_mm_slli_si128@plt>
  mov    (%rbx),%edi
  mov    %eax,%esi
  call   <_mm_xor_si128@plt>
  mov    %r12d,%esi
  mov    %eax,(%rbx)
  mov    %eax,%edi
  call   <_mm_xor_si128@plt>
  mov    %eax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void KEY_256_ASSIST_2(__m128i *temp1, __m128i *temp3) {
  __m128i temp2, temp4;
  temp4 = _mm_aeskeygenassist_si128(*temp1, 0x0);
  temp2 = _mm_shuffle_epi32(temp4, 0xaa);
  temp4 = _mm_slli_si128(*temp3, 0x4);
  *temp3 = _mm_xor_si128(*temp3, temp4);
  temp4 = _mm_slli_si128(temp4, 0x4);
  *temp3 = _mm_xor_si128(*temp3, temp4);
  temp4 = _mm_slli_si128(temp4, 0x4);
  *temp3 = _mm_xor_si128(*temp3, temp4);
  *temp3 = _mm_xor_si128(*temp3, temp2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<KEY_256_ASSIST_2>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rdi),%edi
  mov    %rsi,%rbx
  xor    %esi,%esi
  call   <_mm_aeskeygenassist_si128@plt>
  mov    $0xaa,%esi
  mov    %eax,%edi
  call   <_mm_shuffle_epi32@plt>
  mov    (%rbx),%edi
  mov    $0x4,%esi
  mov    %eax,%r12d
  call   <_mm_slli_si128@plt>
  mov    (%rbx),%edi
  mov    %eax,%esi
  mov    %eax,%ebp
  call   <_mm_xor_si128@plt>
  mov    %ebp,%edi
  mov    $0x4,%esi
  mov    %eax,(%rbx)
  call   <_mm_slli_si128@plt>
  mov    (%rbx),%edi
  mov    %eax,%esi
  mov    %eax,%ebp
  call   <_mm_xor_si128@plt>
  mov    %ebp,%edi
  mov    $0x4,%esi
  mov    %eax,(%rbx)
  call   <_mm_slli_si128@plt>
  mov    (%rbx),%edi
  mov    %eax,%esi
  call   <_mm_xor_si128@plt>
  mov    %r12d,%esi
  mov    %eax,(%rbx)
  mov    %eax,%edi
  call   <_mm_xor_si128@plt>
  mov    %eax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void KEY_256_ASSIST_2(__m128i *temp1, __m128i *temp3) {
  __m128i temp2, temp4;
  temp4 = _mm_aeskeygenassist_si128(*temp1, 0x0);
  temp2 = _mm_shuffle_epi32(temp4, 0xaa);
  temp4 = _mm_slli_si128(*temp3, 0x4);
  *temp3 = _mm_xor_si128(*temp3, temp4);
  temp4 = _mm_slli_si128(temp4, 0x4);
  *temp3 = _mm_xor_si128(*temp3, temp4);
  temp4 = _mm_slli_si128(temp4, 0x4);
  *temp3 = _mm_xor_si128(*temp3, temp4);
  *temp3 = _mm_xor_si128(*temp3, temp2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getVMOutputFileName>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x76,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    -0x3(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x6d,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    -0x2(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void getVMOutputFileName(char *string) {
  int length;
  length = strlen(string);
  string[length - 4] = 'v';
  string[length - 3] = 'm';
  string[length - 2] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<getVMOutputFileName>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  movb   $0x76,-0x4(%rbx,%rax,1)
  movb   $0x6d,-0x3(%rbx,%rax,1)
  movb   $0x0,-0x2(%rbx,%rax,1)
  pop    %rbx
  ret
```

assistant: ```c
void getVMOutputFileName(char *string) {
  int length;
  length = strlen(string);
  string[length - 4] = 'v';
  string[length - 3] = 'm';
  string[length - 2] = '\0';
}
```
user: Translate the below assembly code into C programming language:

```asm
<getVMOutputFileName>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  movb   $0x76,-0x4(%rbx,%rax,1)
  movb   $0x6d,-0x3(%rbx,%rax,1)
  movb   $0x0,-0x2(%rbx,%rax,1)
  pop    %rbx
  ret
```

assistant: ```c
void getVMOutputFileName(char *string) {
  int length;
  length = strlen(string);
  string[length - 4] = 'v';
  string[length - 3] = 'm';
  string[length - 2] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getVMOutputFileName>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  movb   $0x76,-0x4(%rbx,%rax,1)
  movb   $0x6d,-0x3(%rbx,%rax,1)
  movb   $0x0,-0x2(%rbx,%rax,1)
  pop    %rbx
  ret
```

assistant: ```c
void getVMOutputFileName(char *string) {
  int length;
  length = strlen(string);
  string[length - 4] = 'v';
  string[length - 3] = 'm';
  string[length - 2] = '\0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<final_index_val0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rdi
  mov    %rax,%rsi
  call   <hsearch@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L11
  mov    $0x0,%eax
  jmp    L12
L11:
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%rax
L12:
  leave
  ret
```

assistant: ```c
int final_index_val0(const char *s) {
  ENTRY *p;
  ENTRY e;
  e.key = (char *)s;
  if (!(p = hsearch(e, FIND))) {
    return 0;
  }
  return (int)(p->data);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<final_index_val0>:
  endbr64
  sub    $0x8,%rsp
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  mov    $0x0,%esi
  call   <hsearch@plt>
  test   %rax,%rax
  je     L19
  mov    0x8(%rax),%eax
L4:
  add    $0x8,%rsp
  ret
L19:
  mov    $0x0,%eax
  jmp    L4
```

assistant: ```c
int final_index_val0(const char *s) {
  ENTRY *p;
  ENTRY e;
  e.key = (char *)s;
  if (!(p = hsearch(e, FIND))) {
    return 0;
  }
  return (int)(p->data);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<final_index_val0>:
  endbr64
  sub    $0x8,%rsp
  mov    D6(%rip),%rax
  xor    %esi,%esi
  mov    (%rax),%edx
  call   <hsearch@plt>
  test   %rax,%rax
  je     L7
  mov    0x8(%rax),%eax
  add    $0x8,%rsp
  ret
  nopl   (%rax)
L7:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int final_index_val0(const char *s) {
  ENTRY *p;
  ENTRY e;
  e.key = (char *)s;
  if (!(p = hsearch(e, FIND))) {
    return 0;
  }
  return (int)(p->data);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<final_index_val0>:
  endbr64
  sub    $0x8,%rsp
  mov    D4(%rip),%rax
  xor    %esi,%esi
  mov    (%rax),%edx
  call   <hsearch@plt>
  test   %rax,%rax
  je     L5
  mov    0x8(%rax),%eax
  add    $0x8,%rsp
  ret
  nopl   (%rax)
L5:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int final_index_val0(const char *s) {
  ENTRY *p;
  ENTRY e;
  e.key = (char *)s;
  if (!(p = hsearch(e, FIND))) {
    return 0;
  }
  return (int)(p->data);
}
```
user: Translate the below assembly code into C programming language:

```asm
<UARTModemStatusGet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x4000d000,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x4(%rbp),%eax
  add    $0x18,%eax
  cltq
  mov    (%rax),%eax
  and    $0x107,%eax
  leave
  ret
```

assistant: ```c
uint32_t UARTModemStatusGet(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x4000D000);
  return ((*((volatile uint32_t *)(ui32Base + 0x00000018))) & (0x00000100 | 0x00000004 | 0x00000001 | 0x00000002));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<UARTModemStatusGet>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0x4000d000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  add    $0x18,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  and    $0x107,%eax
  pop    %rbx
  ret
```

assistant: ```c
uint32_t UARTModemStatusGet(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x4000D000);
  return ((*((volatile uint32_t *)(ui32Base + 0x00000018))) & (0x00000100 | 0x00000004 | 0x00000001 | 0x00000002));
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<UARTModemStatusGet>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  cmp    $0x4000d000,%ebx
  sete   %dil
  add    $0x18,%ebx
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  pop    %rbx
  and    $0x107,%eax
  ret
```

assistant: ```c
uint32_t UARTModemStatusGet(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x4000D000);
  return ((*((volatile uint32_t *)(ui32Base + 0x00000018))) & (0x00000100 | 0x00000004 | 0x00000001 | 0x00000002));
}
```
user: What is the c source code of the assembly code below?

```asm
<UARTModemStatusGet>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  cmp    $0x4000d000,%ebx
  sete   %dil
  add    $0x18,%ebx
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  pop    %rbx
  and    $0x107,%eax
  ret
```

assistant: ```c
uint32_t UARTModemStatusGet(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x4000D000);
  return ((*((volatile uint32_t *)(ui32Base + 0x00000018))) & (0x00000100 | 0x00000004 | 0x00000001 | 0x00000002));
}
```
user: Translate the below assembly code into C programming language:

```asm
<hello>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D2(%rip),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L14
L18:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  addl   $0x1,-0xc(%rbp)
L14:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L18
  nop
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
type_infered: string

tool: D2: Hello \"!


assistant: ```c
void hello() {
  char *hello = "Hello \\\"!\r\n";
  int i;
  for (i = 0; hello[i]; i++)
    write(1, hello + i, 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<hello>:
  endbr64
  push   %rbx
  lea    D2(%rip),%rbx
L4:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  jne    L4
  pop    %rbx
  ret
```

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

tool: D2: Hello \"!


assistant: ```c
void hello() {
  char *hello = "Hello \\\"!\r\n";
  int i;
  for (i = 0; hello[i]; i++)
    write(1, hello + i, 1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<hello>:
  endbr64
  push   %rbx
  lea    D1(%rip),%rbx
  nopl   0x0(%rax)
L3:
  mov    %rbx,%rsi
  mov    $0x1,%edx
  mov    $0x1,%edi
  add    $0x1,%rbx
  call   <write@plt>
  cmpb   $0x0,(%rbx)
  jne    L3
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: label: D1
data_type: string

tool: D1: Hello \"!


assistant: ```c
void hello() {
  char *hello = "Hello \\\"!\r\n";
  int i;
  for (i = 0; hello[i]; i++)
    write(1, hello + i, 1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<hello>:
  endbr64
  push   %rbx
  lea    D0(%rip),%rbx
  nopl   0x0(%rax)
L6:
  mov    %rbx,%rsi
  mov    $0x1,%edx
  mov    $0x1,%edi
  add    $0x1,%rbx
  call   <write@plt>
  cmpb   $0x0,(%rbx)
  jne    L6
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: label: D0
type: string

tool: D0: Hello \"!


assistant: ```c
void hello() {
  char *hello = "Hello \\\"!\r\n";
  int i;
  for (i = 0; hello[i]; i++)
    write(1, hello + i, 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<drawTriangle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %ecx,-0x10(%rbp)
  mov    %r8d,-0x14(%rbp)
  mov    %r9d,-0x18(%rbp)
  mov    -0x10(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%esi
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <drawLine@plt>
  mov    -0x18(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%esi
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <drawLine@plt>
  mov    -0x8(%rbp),%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <drawLine@plt>
  nop
  leave
  ret
```

assistant: ```c
void drawTriangle(int x1, int y1, int x2, int y2, int x3, int y3) {
  drawLine(x1, y1, x2, y2);
  drawLine(x2, y2, x3, y3);
  drawLine(x3, y3, x1, y1);
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<drawTriangle>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %esi,%r13d
  mov    %edx,%r14d
  mov    %ecx,%r15d
  mov    %r8d,%ebx
  mov    %r9d,%ebp
  call   <drawLine@plt>
  mov    %ebp,%ecx
  mov    %ebx,%edx
  mov    %r15d,%esi
  mov    %r14d,%edi
  call   <drawLine@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <drawLine@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void drawTriangle(int x1, int y1, int x2, int y2, int x3, int y3) {
  drawLine(x1, y1, x2, y2);
  drawLine(x2, y2, x3, y3);
  drawLine(x3, y3, x1, y1);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<drawTriangle>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %r9d,%r12d
  push   %rbp
  mov    %r8d,%ebp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x8,%rsp
  call   <drawLine@plt>
  mov    %r12d,%ecx
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r15d,%edi
  call   <drawLine@plt>
  add    $0x8,%rsp
  mov    %r14d,%ecx
  mov    %r13d,%edx
  pop    %rbx
  mov    %r12d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <drawLine@plt>
```

assistant: ```c
void drawTriangle(int x1, int y1, int x2, int y2, int x3, int y3) {
  drawLine(x1, y1, x2, y2);
  drawLine(x2, y2, x3, y3);
  drawLine(x3, y3, x1, y1);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L8
L4:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff88888889,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff88888889,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  imul   $0x3c,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    -0xc(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    %eax,%esi
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L8:
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  jg     L4
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D47
type_guessed: string
name: disassemble_data
arguments: data_tag: D58
type_guessed: string
name: disassemble_data
arguments: data_tag: D26
type_guessed: string
name: disassemble_data
arguments: data_tag: D59
type_guessed: string
name: disassemble_data
arguments: data_tag: D41
type_guessed: string
name: disassemble_data
arguments: data_tag: D25
type_guessed: string

tool: D47: Convert seconds to minutes and seconds!

tool: D58: Enter the number of seconds (<=0 to quit):

tool: D26: %d

tool: D59: %d seconds is %dminutes%dseconds.


tool: D41: Enter next value (<=0 to quit) :

tool: D25: Bingo!

assistant: ```c
int main() {
  int sec, min, left;
  printf("Convert seconds to minutes and seconds!\n");
  printf("Enter the number of seconds (<=0 to quit):\n");
  scanf("%d", &sec);
  while (sec > 0) {
    min = sec / 60;
    left = sec % 60;
    printf("%d seconds is %dminutes%dseconds.\n", sec, min, left);
    printf("Enter next value (<=0 to quit) :\n");
    scanf("%d", &sec);
  }
  printf("Bingo!\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D67(%rip),%rdi
  call   <puts@plt>
  lea    D44(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L8
  lea    D61(%rip),%r13
  lea    D22(%rip),%r12
  lea    0x4(%rsp),%rbp
  lea    D18(%rip),%rbx
L3:
  movslq %esi,%rdx
  imul   $0xffffffff88888889,%rdx,%rdx
  shr    $0x20,%rdx
  add    %esi,%edx
  sar    $0x5,%edx
  mov    %esi,%eax
  sar    $0x1f,%eax
  sub    %eax,%edx
  imul   $0x3c,%edx,%eax
  mov    %esi,%ecx
  sub    %eax,%ecx
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jg     L3
L8:
  lea    D65(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D67
type_guessed: string
name: parse_data
arguments: data_label: D44
type_guessed: string
name: parse_data
arguments: data_label: D18
type_guessed: string
name: parse_data
arguments: data_label: D61
type_guessed: string
name: parse_data
arguments: data_label: D22
type_guessed: string
name: parse_data
arguments: data_label: D65
type_guessed: string

tool: D67: Convert seconds to minutes and seconds!

tool: D44: Enter the number of seconds (<=0 to quit):

tool: D18: %d

tool: D61: %d seconds is %dminutes%dseconds.


tool: D22: Enter next value (<=0 to quit) :

tool: D65: Bingo!

assistant: ```c
int main() {
  int sec, min, left;
  printf("Convert seconds to minutes and seconds!\n");
  printf("Enter the number of seconds (<=0 to quit):\n");
  scanf("%d", &sec);
  while (sec > 0) {
    min = sec / 60;
    left = sec % 60;
    printf("%d seconds is %dminutes%dseconds.\n", sec, min, left);
    printf("Enter next value (<=0 to quit) :\n");
    scanf("%d", &sec);
  }
  printf("Bingo!\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r14
  lea    D6(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D57(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  call   <puts@plt>
  lea    D12(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L14
  lea    D59(%rip),%r14
  lea    D58(%rip),%r13
  mov    $0x88888889,%r12d
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %esi,%edx
  mov    %esi,%ecx
  mov    %r14,%rdi
  imul   %r12,%rdx
  shr    $0x25,%rdx
  imul   $0x3c,%edx,%eax
  sub    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jg     L15
L14:
  lea    D52(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D6
datatype: string
name: read_data
arguments: tag: D57
datatype: string
name: read_data
arguments: tag: D12
datatype: string
name: read_data
arguments: tag: D59
datatype: string
name: read_data
arguments: tag: D58
datatype: string
name: read_data
arguments: tag: D52
datatype: string

tool: D6: Convert seconds to minutes and seconds!

tool: D57: %d

tool: D12: Enter the number of seconds (<=0 to quit):

tool: D59: %d seconds is %dminutes%dseconds.


tool: D58: Enter next value (<=0 to quit) :

tool: D52: Bingo!

assistant: ```c
int main() {
  int sec, min, left;
  printf("Convert seconds to minutes and seconds!\n");
  printf("Enter the number of seconds (<=0 to quit):\n");
  scanf("%d", &sec);
  while (sec > 0) {
    min = sec / 60;
    left = sec % 60;
    printf("%d seconds is %dminutes%dseconds.\n", sec, min, left);
    printf("Enter next value (<=0 to quit) :\n");
    scanf("%d", &sec);
  }
  printf("Bingo!\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D23(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D15(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  call   <puts@plt>
  lea    D47(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L2
  lea    D55(%rip),%r14
  lea    D43(%rip),%r13
  mov    $0x88888889,%r12d
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %esi,%edx
  mov    %esi,%ecx
  mov    %r14,%rdi
  imul   %r12,%rdx
  shr    $0x25,%rdx
  imul   $0x3c,%edx,%eax
  sub    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jg     L1
L2:
  lea    D30(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D23
type_infered: string
name: read_data
arguments: data_label: D15
type_infered: string
name: read_data
arguments: data_label: D47
type_infered: string
name: read_data
arguments: data_label: D55
type_infered: string
name: read_data
arguments: data_label: D43
type_infered: string
name: read_data
arguments: data_label: D30
type_infered: string

tool: D23: Convert seconds to minutes and seconds!

tool: D15: %d

tool: D47: Enter the number of seconds (<=0 to quit):

tool: D55: %d seconds is %dminutes%dseconds.


tool: D43: Enter next value (<=0 to quit) :

tool: D30: Bingo!

assistant: ```c
int main() {
  int sec, min, left;
  printf("Convert seconds to minutes and seconds!\n");
  printf("Enter the number of seconds (<=0 to quit):\n");
  scanf("%d", &sec);
  while (sec > 0) {
    min = sec / 60;
    left = sec % 60;
    printf("%d seconds is %dminutes%dseconds.\n", sec, min, left);
    printf("Enter next value (<=0 to quit) :\n");
    scanf("%d", &sec);
  }
  printf("Bingo!\n");
  return 0;
}
```