ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  lea    -0x2(%rdi),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%r13,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r13d,%r13d
  jle    L10
  lea    -0x3(%rbx),%r12d
  xor    %ebx,%ebx
L15:
  mov    0x10(%r14,%rbx,8),%rdi
  call   <atoi@plt>
  mov    %eax,0x0(%rbp,%rbx,4)
  mov    %rbx,%rax
  add    $0x1,%rbx
  cmp    %r12,%rax
  jne    L15
L10:
  mov    0x8(%r14),%rdi
  call   <atoi@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  mov    %eax,%edx
  call   <searchInsert@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: tag: D4
type_infered: string

tool: D4: %d


assistant: ```c
int main(int argc, char **argv) {
  int i;
  int *nums = malloc((argc - 2) * sizeof(int));
  for (i = 0; i < argc - 2; i++) {
    nums[i] = atoi(argv[i + 2]);
  }
  printf("%d\n", searchInsert(nums, argc - 2, atoi(argv[1])));
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x30011,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <foo@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
int main() {
  int i = 0x30011;
  foo(&i);
  exit(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
  movl   $0x30011,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  call   <foo@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i = 0x30011;
  foo(&i);
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  movl   $0x30011,0x4(%rsp)
  call   <foo@plt>
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i = 0x30011;
  foo(&i);
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  movl   $0x30011,0x4(%rsp)
  call   <foo@plt>
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int i = 0x30011;
  foo(&i);
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<numbtrop_quaddobl_retrieve_tropism>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rcx,%rbx
  mov    %r8,%r13
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  lea    0x4(,%rdi,4),%r14d
  movslq %r14d,%rax
  shl    $0x3,%rax
  mov    %rax,%r8
  and    $0xfffffffffffff000,%r8
  mov    %rsp,%rcx
  sub    %r8,%rcx
L50:
  cmp    %rcx,%rsp
  je     L27
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L50
L27:
  mov    %rax,%rcx
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  je     L55
  orq    $0x0,-0x8(%rsp,%rcx,1)
L55:
  mov    %rsp,%r12
  mov    %edi,-0x30(%rbp)
  mov    %esi,-0x2c(%rbp)
  lea    -0x30(%rbp),%rsi
  mov    $0x0,%r8d
  mov    %r12,%rcx
  mov    $0x2d5,%edi
  call   <_ada_use_c2phc4c@plt>
  lea    -0x4(%r14),%esi
  test   %esi,%esi
  jle    L4
  mov    %esi,%ecx
  mov    $0x0,%edx
L10:
  movsd  (%r12,%rdx,8),%xmm0
  movsd  %xmm0,(%rbx,%rdx,8)
  add    $0x1,%rdx
  cmp    %rdx,%rcx
  jne    L10
L4:
  movslq %esi,%rdx
  lea    (%r12,%rdx,8),%rdx
  movsd  (%rdx),%xmm0
  movsd  %xmm0,0x0(%r13)
  movsd  0x8(%rdx),%xmm0
  movsd  %xmm0,0x8(%r13)
  movsd  0x10(%rdx),%xmm0
  movsd  %xmm0,0x10(%r13)
  movsd  0x18(%rdx),%xmm0
  movsd  %xmm0,0x18(%r13)
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L43
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L43:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int numbtrop_quaddobl_retrieve_tropism(int dim, int idx, int *wnd, double *dir, double *err) {
  int fail, k;
  int nd[2];
  double cff[4 * (dim + 1)];
  nd[0] = dim;
  nd[1] = idx;
  fail = _ada_use_c2phc4c(725, nd, wnd, cff, 0);
  for (k = 0; k < 4 * dim; k++)
    dir[k] = cff[k];
  err[0] = cff[4 * dim];
  err[1] = cff[4 * dim + 1];
  err[2] = cff[4 * dim + 2];
  err[3] = cff[4 * dim + 3];
  return fail;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<numbtrop_quaddobl_retrieve_tropism>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  lea    0x4(,%rdi,4),%r12d
  push   %rbx
  mov    %r8,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %r12d,%rax
  mov    %rsp,%rcx
  shl    $0x3,%rax
  mov    %rax,%r8
  and    $0xfffffffffffff000,%r8
  sub    %r8,%rcx
  cmp    %rcx,%rsp
  je     L27
L30:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L30
L27:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L0
L55:
  mov    %rsp,%r14
  mov    %edi,-0x40(%rbp)
  xor    %r8d,%r8d
  sub    $0x4,%r12d
  mov    %esi,-0x3c(%rbp)
  mov    %r14,%rcx
  lea    -0x40(%rbp),%rsi
  mov    $0x2d5,%edi
  call   <_ada_use_c2phc4c@plt>
  test   %r12d,%r12d
  movslq %r12d,%r12
  mov    %eax,%r15d
  jle    L22
  lea    0x0(,%r12,8),%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
L22:
  lea    (%r14,%r12,8),%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,(%rbx)
  movsd  0x8(%rax),%xmm0
  movsd  %xmm0,0x8(%rbx)
  movsd  0x10(%rax),%xmm0
  movsd  %xmm0,0x10(%rbx)
  movsd  0x18(%rax),%xmm0
  movsd  %xmm0,0x18(%rbx)
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  lea    -0x28(%rbp),%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  xchg   %ax,%ax
L0:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L55
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int numbtrop_quaddobl_retrieve_tropism(int dim, int idx, int *wnd, double *dir, double *err) {
  int fail, k;
  int nd[2];
  double cff[4 * (dim + 1)];
  nd[0] = dim;
  nd[1] = idx;
  fail = _ada_use_c2phc4c(725, nd, wnd, cff, 0);
  for (k = 0; k < 4 * dim; k++)
    dir[k] = cff[k];
  err[0] = cff[4 * dim];
  err[1] = cff[4 * dim + 1];
  err[2] = cff[4 * dim + 2];
  err[3] = cff[4 * dim + 3];
  return fail;
}
```
user: What is the c source code of the assembly code below?

```asm
<numbtrop_quaddobl_retrieve_tropism>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %r8,%r12
  push   %rbx
  lea    0x4(,%rdi,4),%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %ebx,%rax
  mov    %rsp,%rcx
  shl    $0x3,%rax
  mov    %rax,%r8
  and    $0xfffffffffffff000,%r8
  sub    %r8,%rcx
  cmp    %rcx,%rsp
  je     L3
L9:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L9
L3:
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L50
L45:
  movd   %esi,%xmm2
  movd   %edi,%xmm0
  mov    %rsp,%r14
  xor    %r8d,%r8d
  punpckldq %xmm2,%xmm0
  sub    $0x4,%ebx
  lea    -0x40(%rbp),%rsi
  mov    %r14,%rcx
  mov    $0x2d5,%edi
  movq   %xmm0,-0x40(%rbp)
  call   <_ada_use_c2phc4c@plt>
  test   %ebx,%ebx
  movslq %ebx,%rbx
  mov    %eax,%r15d
  jle    L51
  lea    0x0(,%rbx,8),%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
L51:
  lea    (%r14,%rbx,8),%rax
  movupd 0x10(%rax),%xmm0
  movupd (%rax),%xmm1
  movups %xmm0,0x10(%r12)
  movups %xmm1,(%r12)
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  lea    -0x28(%rbp),%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopl   0x0(%rax)
L50:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L45
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int numbtrop_quaddobl_retrieve_tropism(int dim, int idx, int *wnd, double *dir, double *err) {
  int fail, k;
  int nd[2];
  double cff[4 * (dim + 1)];
  nd[0] = dim;
  nd[1] = idx;
  fail = _ada_use_c2phc4c(725, nd, wnd, cff, 0);
  for (k = 0; k < 4 * dim; k++)
    dir[k] = cff[k];
  err[0] = cff[4 * dim];
  err[1] = cff[4 * dim + 1];
  err[2] = cff[4 * dim + 2];
  err[3] = cff[4 * dim + 3];
  return fail;
}
```
user: What is the c source code of the assembly code below?

```asm
<destroyFIFO>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <unlink@plt>
  test   %rax,%rax
  jns    L9
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L9:
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
datatype: string

tool: D4: 	#ERROR# when destroying client FIFO '%s'


assistant: ```c
void destroyFIFO(char *name) {
  if (unlink(name) < 0)
    printf("\t#ERROR# when destroying client FIFO '%s'\n", name);
  else if (0)
    printf("\t#FIFO# Client '%s' has been destroyed\n", name);
}
```
user: Translate the below assembly code into C programming language:

```asm
<destroyFIFO>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <unlink@plt>
  test   %rax,%rax
  js     L2
L4:
  pop    %rbx
  ret
L2:
  mov    %rbx,%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L4
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type: string

tool: D3: 	#ERROR# when destroying client FIFO '%s'


assistant: ```c
void destroyFIFO(char *name) {
  if (unlink(name) < 0)
    printf("\t#ERROR# when destroying client FIFO '%s'\n", name);
  else if (0)
    printf("\t#FIFO# Client '%s' has been destroyed\n", name);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<destroyFIFO>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <unlink@plt>
  test   %rax,%rax
  js     L2
  pop    %rbp
  ret
  nopl   0x0(%rax)
L2:
  mov    %rbp,%rsi
  lea    D0(%rip),%rdi
  pop    %rbp
  jmp    1070 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D0
type_infered: string

tool: D0: 	#ERROR# when destroying client FIFO '%s'


assistant: ```c
void destroyFIFO(char *name) {
  if (unlink(name) < 0)
    printf("\t#ERROR# when destroying client FIFO '%s'\n", name);
  else if (0)
    printf("\t#FIFO# Client '%s' has been destroyed\n", name);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<destroyFIFO>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <unlink@plt>
  test   %rax,%rax
  js     L1
  pop    %rbp
  ret
  nopl   0x0(%rax)
L1:
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  pop    %rbp
  jmp    1070 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
data_type: string

tool: D4: 	#ERROR# when destroying client FIFO '%s'


assistant: ```c
void destroyFIFO(char *name) {
  if (unlink(name) < 0)
    printf("\t#ERROR# when destroying client FIFO '%s'\n", name);
  else if (0)
    printf("\t#FIFO# Client '%s' has been destroyed\n", name);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hash_message>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %rdx,-0x68(%rbp)
  mov    %rcx,-0x70(%rbp)
  mov    %r8,-0x78(%rbp)
  mov    %r9,-0x80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x40(%rbp),%rax
  mov    %rax,-0x48(%rbp)
  mov    -0x80(%rbp),%rax
  lea    -0x30(%rax),%rcx
  mov    -0x70(%rbp),%rax
  mov    $0x10,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x80(%rbp),%rax
  lea    -0x20(%rax),%rcx
  mov    -0x78(%rbp),%rax
  mov    $0x20,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    0x10(%rbp),%rax
  lea    0x30(%rax),%rcx
  mov    -0x80(%rbp),%rax
  lea    -0x30(%rax),%rdx
  lea    -0x40(%rbp),%rax
  mov    $0x2b,%esi
  mov    %rax,%rdi
  call   <shake256@plt>
  mov    -0x48(%rbp),%rcx
  mov    -0x58(%rbp),%rax
  mov    $0x22,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  addq   $0x22,-0x48(%rbp)
  mov    -0x48(%rbp),%rax
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <bytes_to_ull@plt>
  mov    -0x60(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x60(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x60(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x8,-0x48(%rbp)
  mov    -0x48(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <bytes_to_ull@plt>
  mov    -0x68(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x68(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x68(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
void hash_message(unsigned char *digest, uint64_t *tree, uint32_t *leaf_idx, const unsigned char *R, const unsigned char *pk, unsigned char *m, unsigned long long mlen) {
  unsigned char buf[(((9 * 30 + 7) / 8) + ((((60 / 20) * (20 - 1)) + 7) / 8) + (((60 / 20) + 7) / 8))];
  unsigned char *bufp = buf;
  memcpy(m - 16 - (2 * 16), R, 16);
  memcpy(m - (2 * 16), pk, (2 * 16));
  shake256(buf, (((9 * 30 + 7) / 8) + ((((60 / 20) * (20 - 1)) + 7) / 8) + (((60 / 20) + 7) / 8)), m - 16 - (2 * 16), mlen + 16 + (2 * 16));
  memcpy(digest, bufp, ((9 * 30 + 7) / 8));
  bufp += ((9 * 30 + 7) / 8);
  *tree = bytes_to_ull(bufp, ((((60 / 20) * (20 - 1)) + 7) / 8));
  *tree &= (~(uint64_t)0) >> (64 - ((60 / 20) * (20 - 1)));
  bufp += ((((60 / 20) * (20 - 1)) + 7) / 8);
  *leaf_idx = bytes_to_ull(bufp, (((60 / 20) + 7) / 8));
  *leaf_idx &= (~(uint32_t)0) >> (32 - (60 / 20));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<hash_message>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    %rcx,%rsi
  mov    %r8,%r14
  mov    %r9,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    -0x30(%r9),%r15
  mov    $0x10,%edx
  mov    %r15,%rdi
  call   <memcpy@plt>
  lea    -0x20(%rbx),%rdi
  mov    $0x20,%edx
  mov    %r14,%rsi
  call   <memcpy@plt>
  mov    0x80(%rsp),%rax
  lea    0x30(%rax),%rcx
  mov    %rsp,%rbx
  mov    %r15,%rdx
  mov    $0x2b,%esi
  mov    %rbx,%rdi
  call   <shake256@plt>
  mov    $0x22,%edx
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  lea    0x22(%rsp),%rdi
  mov    $0x8,%esi
  call   <bytes_to_ull@plt>
  mov    %eax,(%r12)
  lea    0x2a(%rsp),%rdi
  mov    $0x1,%esi
  call   <bytes_to_ull@plt>
  mov    %eax,0x0(%rbp)
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void hash_message(unsigned char *digest, uint64_t *tree, uint32_t *leaf_idx, const unsigned char *R, const unsigned char *pk, unsigned char *m, unsigned long long mlen) {
  unsigned char buf[(((9 * 30 + 7) / 8) + ((((60 / 20) * (20 - 1)) + 7) / 8) + (((60 / 20) + 7) / 8))];
  unsigned char *bufp = buf;
  memcpy(m - 16 - (2 * 16), R, 16);
  memcpy(m - (2 * 16), pk, (2 * 16));
  shake256(buf, (((9 * 30 + 7) / 8) + ((((60 / 20) * (20 - 1)) + 7) / 8) + (((60 / 20) + 7) / 8)), m - 16 - (2 * 16), mlen + 16 + (2 * 16));
  memcpy(digest, bufp, ((9 * 30 + 7) / 8));
  bufp += ((9 * 30 + 7) / 8);
  *tree = bytes_to_ull(bufp, ((((60 / 20) * (20 - 1)) + 7) / 8));
  *tree &= (~(uint64_t)0) >> (64 - ((60 / 20) * (20 - 1)));
  bufp += ((((60 / 20) * (20 - 1)) + 7) / 8);
  *leaf_idx = bytes_to_ull(bufp, (((60 / 20) + 7) / 8));
  *leaf_idx &= (~(uint32_t)0) >> (32 - (60 / 20));
}
```
user: Translate the below assembly code into C programming language:

```asm
<hash_message>:
  endbr64
  push   %r15
  lea    -0x30(%r9),%r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %rdi,%r13
  mov    %r15,%rdi
  push   %r12
  mov    %rsi,%r12
  mov    %rcx,%rsi
  push   %rbp
  mov    %rdx,%rbp
  mov    $0x10,%edx
  push   %rbx
  mov    %r9,%rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <memcpy@plt>
  mov    %r14,%rsi
  lea    -0x20(%rbx),%rdi
  mov    %rsp,%r14
  mov    $0x20,%edx
  call   <memcpy@plt>
  mov    %r15,%rdx
  mov    $0x2b,%esi
  mov    %r14,%rdi
  mov    0x80(%rsp),%rax
  lea    0x30(%rax),%rcx
  call   <shake256@plt>
  mov    $0x22,%edx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  lea    0x22(%rsp),%rdi
  mov    $0x8,%esi
  call   <bytes_to_ull@plt>
  lea    0x2a(%rsp),%rdi
  mov    $0x1,%esi
  mov    %eax,(%r12)
  call   <bytes_to_ull@plt>
  mov    %eax,0x0(%rbp)
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void hash_message(unsigned char *digest, uint64_t *tree, uint32_t *leaf_idx, const unsigned char *R, const unsigned char *pk, unsigned char *m, unsigned long long mlen) {
  unsigned char buf[(((9 * 30 + 7) / 8) + ((((60 / 20) * (20 - 1)) + 7) / 8) + (((60 / 20) + 7) / 8))];
  unsigned char *bufp = buf;
  memcpy(m - 16 - (2 * 16), R, 16);
  memcpy(m - (2 * 16), pk, (2 * 16));
  shake256(buf, (((9 * 30 + 7) / 8) + ((((60 / 20) * (20 - 1)) + 7) / 8) + (((60 / 20) + 7) / 8)), m - 16 - (2 * 16), mlen + 16 + (2 * 16));
  memcpy(digest, bufp, ((9 * 30 + 7) / 8));
  bufp += ((9 * 30 + 7) / 8);
  *tree = bytes_to_ull(bufp, ((((60 / 20) * (20 - 1)) + 7) / 8));
  *tree &= (~(uint64_t)0) >> (64 - ((60 / 20) * (20 - 1)));
  bufp += ((((60 / 20) * (20 - 1)) + 7) / 8);
  *leaf_idx = bytes_to_ull(bufp, (((60 / 20) + 7) / 8));
  *leaf_idx &= (~(uint32_t)0) >> (32 - (60 / 20));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<hash_message>:
  endbr64
  push   %r15
  lea    -0x30(%r9),%r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %rdi,%r13
  mov    %r15,%rdi
  push   %r12
  mov    %rsi,%r12
  mov    %rcx,%rsi
  push   %rbp
  mov    %rdx,%rbp
  mov    $0x10,%edx
  push   %rbx
  mov    %r9,%rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <memcpy@plt>
  mov    %r14,%rsi
  lea    -0x20(%rbx),%rdi
  mov    %rsp,%r14
  mov    $0x20,%edx
  call   <memcpy@plt>
  mov    %r15,%rdx
  mov    $0x2b,%esi
  mov    %r14,%rdi
  mov    0x80(%rsp),%rax
  lea    0x30(%rax),%rcx
  call   <shake256@plt>
  mov    $0x22,%edx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  lea    0x22(%rsp),%rdi
  mov    $0x8,%esi
  call   <bytes_to_ull@plt>
  lea    0x2a(%rsp),%rdi
  mov    $0x1,%esi
  mov    %eax,(%r12)
  call   <bytes_to_ull@plt>
  mov    %eax,0x0(%rbp)
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void hash_message(unsigned char *digest, uint64_t *tree, uint32_t *leaf_idx, const unsigned char *R, const unsigned char *pk, unsigned char *m, unsigned long long mlen) {
  unsigned char buf[(((9 * 30 + 7) / 8) + ((((60 / 20) * (20 - 1)) + 7) / 8) + (((60 / 20) + 7) / 8))];
  unsigned char *bufp = buf;
  memcpy(m - 16 - (2 * 16), R, 16);
  memcpy(m - (2 * 16), pk, (2 * 16));
  shake256(buf, (((9 * 30 + 7) / 8) + ((((60 / 20) * (20 - 1)) + 7) / 8) + (((60 / 20) + 7) / 8)), m - 16 - (2 * 16), mlen + 16 + (2 * 16));
  memcpy(digest, bufp, ((9 * 30 + 7) / 8));
  bufp += ((9 * 30 + 7) / 8);
  *tree = bytes_to_ull(bufp, ((((60 / 20) * (20 - 1)) + 7) / 8));
  *tree &= (~(uint64_t)0) >> (64 - ((60 / 20) * (20 - 1)));
  bufp += ((((60 / 20) * (20 - 1)) + 7) / 8);
  *leaf_idx = bytes_to_ull(bufp, (((60 / 20) + 7) / 8));
  *leaf_idx &= (~(uint32_t)0) >> (32 - (60 / 20));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<show_dobldobl_solution>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    -0x54(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <padcon_get_dobldobl_solution@plt>
  mov    %eax,-0x4c(%rbp)
  lea    -0x50(%rbp),%rdx
  mov    -0x54(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <solcon_length_dobldobl_solution_string@plt>
  mov    %eax,-0x4c(%rbp)
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x50(%rbp),%eax
  add    $0x1,%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x48(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L36:
  cmp    %rdx,%rsp
  je     L7
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L36
L7:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L21
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L21:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x40(%rbp)
  mov    -0x50(%rbp),%ecx
  mov    -0x40(%rbp),%rdx
  mov    -0x54(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <solcon_write_dobldobl_solution_string@plt>
  mov    %eax,-0x4c(%rbp)
  mov    -0x40(%rbp),%rdx
  mov    -0x54(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    %rbx,%rsp
  nop
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D0
data_type: string

tool: D0: Solution %d :
%s


assistant: ```c
void show_dobldobl_solution(int index) {
  int fail, strlensol;
  fail = padcon_get_dobldobl_solution(index, 0);
  fail = solcon_length_dobldobl_solution_string(index, &strlensol);
  {
    char idxsol[strlensol + 1];
    fail = solcon_write_dobldobl_solution_string(index, strlensol, idxsol);
    printf("Solution %d :\n%s\n", index, idxsol);
  }
}
```