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

```asm
<get_next_line>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x818,%rsp
  mov    %rsi,0x8(%rsp)
  mov    %fs:0x28,%rdx
  mov    %rdx,0x808(%rsp)
  xor    %edx,%edx
  test   %rsi,%rsi
  je     L14
  mov    %edi,%r14d
  cmp    $0x2a,%edi
  ja     L14
  lea    D7(%rip),%rbx
  movslq %edi,%r12
  mov    (%rbx,%r12,8),%rdi
  test   %rdi,%rdi
  je     L31
L10:
  lea    (%rbx,%r12,8),%r15
  mov    $0x1,%ebp
  lea    0x20(%rsp),%r13
  jmp    L13
  nopl   0x0(%rax)
L17:
  test   %ebp,%ebp
  je     L15
  mov    $0x7e1,%edx
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <read@plt>
  mov    %eax,%ebp
  cmp    $0xffffffff,%eax
  je     L14
  mov    (%rbx,%r12,8),%rdi
  cltq
  movb   $0x0,0x20(%rsp,%rax,1)
  call   <ft_strdup@plt>
  mov    %r15,%rdi
  mov    %rax,0x18(%rsp)
  call   <ft_strdel@plt>
  mov    0x18(%rsp),%rdi
  mov    %r13,%rsi
  call   <ft_strjoin@plt>
  lea    0x18(%rsp),%rdi
  mov    %rax,(%rbx,%r12,8)
  call   <ft_strdel@plt>
  mov    (%rbx,%r12,8),%rdi
L13:
  call   <cs@plt>
  cmp    $0xffffffff,%eax
  je     L17
L15:
  lea    0x8(%rsp),%rdx
  mov    %r15,%rsi
  mov    %ebp,%edi
  call   <processline@plt>
L1:
  mov    0x808(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L67
  add    $0x818,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L14:
  mov    $0xffffffff,%eax
  jmp    L1
  nop
L31:
  mov    $0x7e1,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx,%r12,8)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L14
  mov    $0x7e1,%esi
  call   <ft_bzero@plt>
  mov    (%rbx,%r12,8),%rdi
  jmp    L10
L67:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_next_line(const int fd, char **line) {
  char buffer[2017 + 1];
  int ret;
  static char *str[42];
  char *swap;
  ret = 1;
  if (!line || fd > 42 || fd < 0 || 2017 < 1)
    return (-1);
  if (!str[fd]) {
    if (!(str[fd] = (char *)malloc(sizeof(char) * 2017)))
      return (-1);
    ft_bzero(str[fd], 2017);
  }
  while (cs(str[fd]) == -1 && ret) {
    if ((ret = read(fd, buffer, 2017)) == -1)
      return (-1);
    buffer[ret] = '\0';
    swap = ft_strdup(str[fd]);
    ft_strdel(&str[fd]);
    str[fd] = ft_strjoin(swap, buffer);
    ft_strdel(&swap);
  }
  return (processline(ret, &str[fd], &line));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L18
L3:
  mov    -0x8(%rbp),%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L18:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L3
  mov    -0x4(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type: string

tool: D2: The sum of 1-%d is %d


assistant: ```c
int sum(int m) {
  int i = 0, n = 0;
  for (i = 1; i <= m; i++) {
    n += i;
  }
  printf("The sum of 1-%d is %d\r\n", m, n);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sum>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  test   %edi,%edi
  jle    L20
  lea    0x1(%rdi),%ecx
  mov    $0x0,%edx
  mov    $0x1,%eax
L28:
  add    %eax,%edx
  add    $0x1,%eax
  cmp    %ecx,%eax
  jne    L28
L21:
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L20:
  mov    $0x0,%edx
  jmp    L21
```

tool_calls: name: unpack_data
arguments: tag: D2
type: string

tool: D2: The sum of 1-%d is %d


assistant: ```c
int sum(int m) {
  int i = 0, n = 0;
  for (i = 1; i <= m; i++) {
    n += i;
  }
  printf("The sum of 1-%d is %d\r\n", m, n);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sum>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  test   %edi,%edi
  jle    L25
  lea    0x1(%rdi),%ecx
  xor    %edx,%edx
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L6:
  add    %eax,%edx
  add    $0x1,%eax
  cmp    %ecx,%eax
  jne    L6
L15:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L25:
  xor    %edx,%edx
  jmp    L15
```

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

tool: D9: The sum of 1-%d is %d


assistant: ```c
int sum(int m) {
  int i = 0, n = 0;
  for (i = 1; i <= m; i++) {
    n += i;
  }
  printf("The sum of 1-%d is %d\r\n", m, n);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sum>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  test   %edi,%edi
  jle    L55
  lea    -0x1(%rdi),%eax
  cmp    $0x9,%eax
  jbe    L99
  mov    %edi,%edx
  movdqa D21(%rip),%xmm1
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movdqa D3(%rip),%xmm3
  shr    $0x2,%edx
  nopl   0x0(%rax)
L58:
  movdqa %xmm1,%xmm2
  add    $0x1,%eax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %eax,%edx
  jne    L58
  movdqa %xmm0,%xmm1
  mov    %esi,%ecx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%ecx
  paddd  %xmm1,%xmm0
  lea    0x1(%rcx),%eax
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%edx
  cmp    %ecx,%esi
  je     L14
L104:
  lea    0x1(%rax),%ecx
  add    %eax,%edx
  cmp    %ecx,%esi
  jl     L14
  add    %ecx,%edx
  lea    0x2(%rax),%ecx
  cmp    %ecx,%esi
  jl     L14
  add    %ecx,%edx
  lea    0x3(%rax),%ecx
  cmp    %ecx,%esi
  jl     L14
  add    %ecx,%edx
  lea    0x4(%rax),%ecx
  cmp    %ecx,%esi
  jl     L14
  add    %ecx,%edx
  lea    0x5(%rax),%ecx
  cmp    %ecx,%esi
  jl     L14
  add    %ecx,%edx
  lea    0x6(%rax),%ecx
  cmp    %ecx,%esi
  jl     L14
  add    %ecx,%edx
  lea    0x7(%rax),%ecx
  cmp    %ecx,%esi
  jl     L14
  add    %ecx,%edx
  lea    0x8(%rax),%ecx
  cmp    %ecx,%esi
  jl     L14
  add    %ecx,%edx
  add    $0x9,%eax
  lea    (%rdx,%rax,1),%ecx
  cmp    %eax,%esi
  cmovge %ecx,%edx
L14:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
  nopl   (%rax)
L55:
  xor    %edx,%edx
  jmp    L14
L99:
  xor    %edx,%edx
  mov    $0x1,%eax
  jmp    L104
```

tool_calls: name: disassemble_data
arguments: data_tag: D21
datatype: byte[16]
name: disassemble_data
arguments: data_tag: D3
datatype: byte[16]
name: disassemble_data
arguments: data_tag: D7
datatype: string

tool: {"D21": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D3": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D7: The sum of 1-%d is %d


assistant: ```c
int sum(int m) {
  int i = 0, n = 0;
  for (i = 1; i <= m; i++) {
    n += i;
  }
  printf("The sum of 1-%d is %d\r\n", m, n);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<psvec2i_divide>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <vec2i_divide@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
__attribute__((always_)) struct vec2i *psvec2i_divide(struct vec2i *result, struct vec2i *a, struct vec2i *b) {
  vec2i_divide((int32_t *)result, (int32_t *)a, (int32_t *)b);
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<psvec2i_divide>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <vec2i_divide@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
__attribute__((always_)) struct vec2i *psvec2i_divide(struct vec2i *result, struct vec2i *a, struct vec2i *b) {
  vec2i_divide((int32_t *)result, (int32_t *)a, (int32_t *)b);
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<psvec2i_divide>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <vec2i_divide@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
__attribute__((always_)) struct vec2i *psvec2i_divide(struct vec2i *result, struct vec2i *a, struct vec2i *b) {
  vec2i_divide((int32_t *)result, (int32_t *)a, (int32_t *)b);
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<psvec2i_divide>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <vec2i_divide@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
__attribute__((always_)) struct vec2i *psvec2i_divide(struct vec2i *result, struct vec2i *a, struct vec2i *b) {
  vec2i_divide((int32_t *)result, (int32_t *)a, (int32_t *)b);
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rcx
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0xa,-0xc(%rbp)
  jmp    L56
L31:
  mov    -0x18(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  add    %ecx,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L72
  mov    -0x14(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x2,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L72
  mov    -0x10(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  sub    %ecx,%eax
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x3,%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L72
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L22
L72:
  addl   $0x1,-0xc(%rbp)
L56:
  cmpl   $0x64,-0xc(%rbp)
  jle    L31
L22:
  cmpl   $0x64,-0xc(%rbp)
  jle    L57
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L57:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D11
type: string
name: read_data
arguments: tag: D21
type: string
name: read_data
arguments: tag: D24
type: string

tool: D11: %d%d%d

tool: D21: %d


tool: D24: No answer

assistant: ```c
int main() {
  int a, b, c, k;
  scanf("%d%d%d", &a, &b, &c);
  for (k = 10; k <= 100; k++) {
    if ((k - a) % 3 == 0 && (k - b) % 5 == 0 && (k - c) % 7 == 0) {
      printf("%d\n", k);
      break;
    }
  }
  if (k > 100) {
    printf("No answer\n");
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%ecx
  mov    $0xa,%edx
  sub    %ecx,%edx
  mov    $0x65,%esi
  sub    %ecx,%esi
  mov    %ecx,%r8d
  sub    0x10(%rsp),%r8d
  mov    %ecx,%edi
  sub    0x14(%rsp),%edi
  jmp    L20
L46:
  add    $0x1,%edx
  cmp    %esi,%edx
  je     L43
L20:
  lea    (%rdx,%rcx,1),%ebx
  lea    (%r8,%rdx,1),%r9d
  movslq %r9d,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %r9d,%r10d
  sar    $0x1f,%r10d
  sub    %r10d,%eax
  lea    (%rax,%rax,4),%eax
  mov    %r9d,%r10d
  sub    %eax,%r10d
  lea    (%rdi,%rdx,1),%r9d
  movslq %r9d,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %r9d,%eax
  sar    $0x2,%eax
  mov    %r9d,%r11d
  sar    $0x1f,%r11d
  sub    %r11d,%eax
  lea    0x0(,%rax,8),%r11d
  sub    %eax,%r11d
  sub    %r11d,%r9d
  or     %r10d,%r9d
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%r10d
  sar    $0x1f,%r10d
  sub    %r10d,%eax
  lea    (%rax,%rax,2),%r10d
  mov    %edx,%eax
  sub    %r10d,%eax
  or     %eax,%r9d
  jne    L46
  mov    %ebx,%esi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x64,%ebx
  jg     L43
L49:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L43:
  lea    D12(%rip),%rdi
  call   <puts@plt>
  jmp    L49
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D15
datatype: string
name: read_data
arguments: data_tag: D26
datatype: string
name: read_data
arguments: data_tag: D12
datatype: string

tool: D15: %d%d%d

tool: D26: %d


tool: D12: No answer

assistant: ```c
int main() {
  int a, b, c, k;
  scanf("%d%d%d", &a, &b, &c);
  for (k = 10; k <= 100; k++) {
    if ((k - a) % 3 == 0 && (k - b) % 5 == 0 && (k - c) % 7 == 0) {
      printf("%d\n", k);
      break;
    }
  }
  if (k > 100) {
    printf("No answer\n");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbx
  lea    D27(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%ecx
  mov    $0xa,%edx
  mov    $0x65,%esi
  mov    %ecx,%r8d
  mov    %ecx,%edi
  sub    %ecx,%edx
  sub    %ecx,%esi
  sub    0x10(%rsp),%r8d
  sub    0x14(%rsp),%edi
  jmp    L14
  cs nopw 0x0(%rax,%rax,1)
L48:
  add    $0x1,%edx
  cmp    %esi,%edx
  je     L12
L14:
  lea    (%r8,%rdx,1),%r10d
  lea    (%rdx,%rcx,1),%r9d
  movslq %r10d,%rax
  mov    %r10d,%r11d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%r11d
  sar    $0x21,%rax
  sub    %r11d,%eax
  mov    %r10d,%r11d
  lea    (%rdi,%rdx,1),%r10d
  lea    (%rax,%rax,4),%eax
  mov    %r10d,%ebx
  sub    %eax,%r11d
  movslq %r10d,%rax
  sar    $0x1f,%ebx
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %r10d,%eax
  sar    $0x2,%eax
  sub    %ebx,%eax
  lea    0x0(,%rax,8),%ebx
  sub    %eax,%ebx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  sub    %ebx,%r10d
  or     %r11d,%r10d
  mov    %edx,%r11d
  sar    $0x1f,%r11d
  shr    $0x20,%rax
  sub    %r11d,%eax
  lea    (%rax,%rax,2),%r11d
  mov    %edx,%eax
  sub    %r11d,%eax
  or     %eax,%r10d
  jne    L48
  mov    %r9d,%esi
  lea    D19(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L22:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L12:
  lea    D6(%rip),%rdi
  call   <puts@plt>
  jmp    L22
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D27
datatype: string
name: extract_symbolic_data
arguments: data_tag: D19
datatype: string
name: extract_symbolic_data
arguments: data_tag: D6
datatype: string

tool: D27: %d%d%d

tool: D19: %d


tool: D6: No answer

assistant: ```c
int main() {
  int a, b, c, k;
  scanf("%d%d%d", &a, &b, &c);
  for (k = 10; k <= 100; k++) {
    if ((k - a) % 3 == 0 && (k - b) % 5 == 0 && (k - c) % 7 == 0) {
      printf("%d\n", k);
      break;
    }
  }
  if (k > 100) {
    printf("No answer\n");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbx
  lea    D23(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%ecx
  mov    $0xa,%edx
  mov    $0x65,%esi
  mov    %ecx,%r8d
  mov    %ecx,%edi
  sub    %ecx,%edx
  sub    %ecx,%esi
  sub    0x10(%rsp),%r8d
  sub    0x14(%rsp),%edi
  jmp    L17
  cs nopw 0x0(%rax,%rax,1)
L14:
  add    $0x1,%edx
  cmp    %esi,%edx
  je     L7
L17:
  lea    (%r8,%rdx,1),%r10d
  lea    (%rdx,%rcx,1),%r9d
  movslq %r10d,%rax
  mov    %r10d,%r11d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%r11d
  sar    $0x21,%rax
  sub    %r11d,%eax
  mov    %r10d,%r11d
  lea    (%rdi,%rdx,1),%r10d
  lea    (%rax,%rax,4),%eax
  mov    %r10d,%ebx
  sub    %eax,%r11d
  movslq %r10d,%rax
  sar    $0x1f,%ebx
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %r10d,%eax
  sar    $0x2,%eax
  sub    %ebx,%eax
  lea    0x0(,%rax,8),%ebx
  sub    %eax,%ebx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  sub    %ebx,%r10d
  or     %r11d,%r10d
  mov    %edx,%r11d
  sar    $0x1f,%r11d
  shr    $0x20,%rax
  sub    %r11d,%eax
  lea    (%rax,%rax,2),%r11d
  mov    %edx,%eax
  sub    %r11d,%eax
  or     %eax,%r10d
  jne    L14
  mov    %r9d,%esi
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L40:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L7:
  lea    D5(%rip),%rdi
  call   <puts@plt>
  jmp    L40
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D23
type_infered: string
name: unpack_data
arguments: label: D18
type_infered: string
name: unpack_data
arguments: label: D5
type_infered: string

tool: D23: %d%d%d

tool: D18: %d


tool: D5: No answer

assistant: ```c
int main() {
  int a, b, c, k;
  scanf("%d%d%d", &a, &b, &c);
  for (k = 10; k <= 100; k++) {
    if ((k - a) % 3 == 0 && (k - b) % 5 == 0 && (k - c) % 7 == 0) {
      printf("%d\n", k);
      break;
    }
  }
  if (k > 100) {
    printf("No answer\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<jam_ltoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x38(%rbp)
  movl   $0x0,-0x34(%rbp)
  cmpq   $0x0,-0x50(%rbp)
  jns    L62
  mov    -0x38(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
  negq   -0x50(%rbp)
  jmp    L10
L62:
  cmpq   $0x0,-0x50(%rbp)
  jne    L10
  mov    -0x38(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  jmp    L10
L60:
  mov    -0x50(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rsi
  sar    $0x3f,%rsi
  sub    %rsi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %edx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x34(%rbp)
  mov    %ecx,%edx
  cltq
  mov    %dl,-0x30(%rbp,%rax,1)
  mov    -0x50(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x50(%rbp)
L10:
  cmpq   $0x0,-0x50(%rbp)
  jne    L60
  jmp    L43
L25:
  subl   $0x1,-0x34(%rbp)
  mov    -0x38(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rax,%rdx
  mov    -0x34(%rbp),%eax
  cltq
  movzbl -0x30(%rbp,%rax,1),%eax
  mov    %al,(%rdx)
L43:
  cmpl   $0x0,-0x34(%rbp)
  jg     L25
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void jam_ltoa(char *buffer, long number) {
  int index = 0;
  int rev_index = 0;
  char reverse[32];
  if (number < 0L) {
    buffer[index++] = '-';
    number = 0 - number;
  } else if (number == 0) {
    buffer[index++] = '0';
  }
  while (number != 0) {
    reverse[rev_index++] = (char)((number % 10) + '0');
    number /= 10;
  }
  while (rev_index > 0) {
    buffer[index++] = reverse[--rev_index];
  }
  buffer[index] = '\0';
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<jam_ltoa>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,%r9
  mov    %rsi,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  js     L29
  jne    L0
  movb   $0x30,(%rdi)
  mov    $0x1,%r10d
  jmp    L4
L29:
  movb   $0x2d,(%rdi)
  neg    %rcx
  mov    $0x1,%r10d
L22:
  mov    $0x1,%esi
  lea    -0x1(%rsp),%r11
  movabs $0x6666666666666667,%r8
L62:
  mov    %esi,%edi
  mov    %rcx,%rax
  imul   %r8
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  add    $0x30,%ecx
  mov    %cl,(%r11,%rsi,1)
  mov    %rdx,%rcx
  add    $0x1,%rsi
  test   %rdx,%rdx
  jne    L62
  test   %edi,%edi
  jle    L4
  movslq %edi,%rax
  lea    (%rsp,%rax,1),%rdx
  movslq %r10d,%rcx
  lea    (%r9,%rcx,1),%rax
  lea    0x1(%r9,%rcx,1),%rsi
  lea    -0x1(%rdi),%ecx
  add    %rcx,%rsi
L78:
  movzbl -0x1(%rdx),%ecx
  mov    %cl,(%rax)
  sub    $0x1,%rdx
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L78
  lea    (%rdi,%r10,1),%r10d
L4:
  movslq %r10d,%r10
  movb   $0x0,(%r9,%r10,1)
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x30,%rsp
  pop    %rbx
  ret
L0:
  mov    $0x0,%r10d
  jmp    L22
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void jam_ltoa(char *buffer, long number) {
  int index = 0;
  int rev_index = 0;
  char reverse[32];
  if (number < 0L) {
    buffer[index++] = '-';
    number = 0 - number;
  } else if (number == 0) {
    buffer[index++] = '0';
  }
  while (number != 0) {
    reverse[rev_index++] = (char)((number % 10) + '0');
    number /= 10;
  }
  while (rev_index > 0) {
    buffer[index++] = reverse[--rev_index];
  }
  buffer[index] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<jam_ltoa>:
  endbr64
  push   %rbx
  mov    %rsi,%rcx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  js     L13
  je     L54
  xor    %r8d,%r8d
L68:
  mov    $0x1,%esi
  lea    -0x1(%rsp),%r11
  movabs $0xcccccccccccccccd,%r10
  nopw   0x0(%rax,%rax,1)
L56:
  mov    %rcx,%rax
  mov    %esi,%r9d
  mul    %r10
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  lea    0x30(%rcx),%ebx
  mov    %rdx,%rcx
  mov    %bl,(%r11,%rsi,1)
  add    $0x1,%rsi
  test   %rdx,%rdx
  jne    L56
  movslq %r8d,%rcx
  movslq %r9d,%rdx
  sub    $0x1,%r9d
  lea    0x1(%rdi,%rcx,1),%rsi
  lea    (%rdi,%rcx,1),%rax
  add    %rsp,%rdx
  mov    %r9,%rcx
  add    %r9,%rsi
  jmp    L29
L59:
  movzbl -0x1(%rdx),%ebx
L29:
  mov    %bl,(%rax)
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %rsi,%rax
  jne    L59
  lea    0x1(%r8,%rcx,1),%eax
  cltq
L3:
  movb   $0x0,(%rdi,%rax,1)
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x30,%rsp
  pop    %rbx
  ret
  nopl   0x0(%rax)
L54:
  movb   $0x30,(%rdi)
  mov    $0x1,%eax
  jmp    L3
  nopw   0x0(%rax,%rax,1)
L13:
  movb   $0x2d,(%rdi)
  neg    %rcx
  mov    $0x1,%r8d
  jmp    L68
L38:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void jam_ltoa(char *buffer, long number) {
  int index = 0;
  int rev_index = 0;
  char reverse[32];
  if (number < 0L) {
    buffer[index++] = '-';
    number = 0 - number;
  } else if (number == 0) {
    buffer[index++] = '0';
  }
  while (number != 0) {
    reverse[rev_index++] = (char)((number % 10) + '0');
    number /= 10;
  }
  while (rev_index > 0) {
    buffer[index++] = reverse[--rev_index];
  }
  buffer[index] = '\0';
}
```