ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<concatenation>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  lea    0x0(%r13,%rax,1),%edi
  mov    %eax,%ebx
  call   <malloc@plt>
  test   %r13d,%r13d
  jle    L17
  movslq %r13d,%rcx
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L33:
  movzbl (%r12,%rdx,1),%esi
  mov    %sil,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L33
L17:
  test   %ebx,%ebx
  jle    L29
  movslq %r13d,%rcx
  movslq %ebx,%rdi
  xor    %edx,%edx
  add    %rax,%rcx
  nopl   0x0(%rax)
L31:
  movzbl 0x0(%rbp,%rdx,1),%esi
  mov    %sil,(%rcx,%rdx,1)
  add    $0x1,%rdx
  cmp    %rdx,%rdi
  jne    L31
L29:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *concatenation(char *ch1, char *ch2) {
  int size1 = strlen(ch1);
  int size2 = strlen(ch2);
  char *concat = malloc((size1 + size2) * sizeof(char));
  int i = 0;
  for (i = 0; i < size1; i++) {
    concat[i] = ch1[i];
  }
  for (i = 0; i < size2; i++) {
    concat[size1 + i] = ch2[i];
  }
  return concat;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<rotr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int swap1 = 0, swap2 = 0;
  swap1 = *a;
  *a = *c;
  swap2 = *b;
  *b = swap1;
  *c = swap2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rotr>:
  endbr64
  mov    (%rdi),%ecx
  mov    (%rdx),%eax
  mov    %eax,(%rdi)
  mov    (%rsi),%eax
  mov    %ecx,(%rsi)
  mov    %eax,(%rdx)
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int swap1 = 0, swap2 = 0;
  swap1 = *a;
  *a = *c;
  swap2 = *b;
  *b = swap1;
  *c = swap2;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<rotr>:
  endbr64
  mov    (%rdx),%eax
  mov    (%rdi),%ecx
  mov    %eax,(%rdi)
  mov    (%rsi),%eax
  mov    %ecx,(%rsi)
  mov    %eax,(%rdx)
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int swap1 = 0, swap2 = 0;
  swap1 = *a;
  *a = *c;
  swap2 = *b;
  *b = swap1;
  *c = swap2;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rotr>:
  endbr64
  mov    (%rdx),%eax
  mov    (%rdi),%ecx
  mov    %eax,(%rdi)
  mov    (%rsi),%eax
  mov    %ecx,(%rsi)
  mov    %eax,(%rdx)
  ret
```

assistant: ```c
void rotr(int *a, int *b, int *c) {
  int swap1 = 0, swap2 = 0;
  swap1 = *a;
  *a = *c;
  swap2 = *b;
  *b = swap1;
  *c = swap2;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<any_NaN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L33
L17:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  call   <isnan@plt>
  test   %rax,%rax
  je     L15
  mov    $0x1,%eax
  jmp    L29
L15:
  addl   $0x1,-0x4(%rbp)
L33:
  mov    -0x1c(%rbp),%eax
  imul   -0x20(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L17
  mov    $0x0,%eax
L29:
  leave
  ret
```

assistant: ```c
int any_NaN(double *A, int n, int L) {
  int i;
  for (i = 0; i < n * L; i++) {
    if (isnan(A[i]))
      return 1;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<any_NaN>:
  endbr64
  imul   %edx,%esi
  test   %esi,%esi
  jle    L32
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rbp
L11:
  movsd  (%rbx),%xmm0
  call   <isnan@plt>
  test   %rax,%rax
  jne    L5
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L11
  jmp    L1
L32:
  mov    $0x0,%eax
  ret
L5:
  mov    $0x1,%eax
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int any_NaN(double *A, int n, int L) {
  int i;
  for (i = 0; i < n * L; i++) {
    if (isnan(A[i]))
      return 1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<any_NaN>:
  endbr64
  imul   %edx,%esi
  test   %esi,%esi
  jle    L36
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x8(%rdi,%rax,8),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L26
  xchg   %ax,%ax
L34:
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  je     L16
L26:
  movsd  (%rbx),%xmm0
  call   <isnan@plt>
  test   %rax,%rax
  je     L34
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L36:
  xor    %eax,%eax
  ret
```

assistant: ```c
int any_NaN(double *A, int n, int L) {
  int i;
  for (i = 0; i < n * L; i++) {
    if (isnan(A[i]))
      return 1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<any_NaN>:
  endbr64
  imul   %edx,%esi
  test   %esi,%esi
  jle    L11
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x8(%rdi,%rax,8),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L1
  xchg   %ax,%ax
L35:
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  je     L24
L1:
  movsd  (%rbx),%xmm0
  call   <isnan@plt>
  test   %rax,%rax
  je     L35
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L24:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L11:
  xor    %eax,%eax
  ret
```

assistant: ```c
int any_NaN(double *A, int n, int L) {
  int i;
  for (i = 0; i < n * L; i++) {
    if (isnan(A[i]))
      return 1;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,%rax
  mov    %rsi,%rcx
  mov    %rcx,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  cmpq   $0xfffffffffffffff9,-0x30(%rbp)
  sbb    $0xffffffffffffffff,%rax
  jl     L18
  mov    $0x2,%edx
  mov    $0x0,%eax
  cmp    -0x30(%rbp),%rdx
  sbb    -0x28(%rbp),%rax
  jl     L4
  mov    -0x30(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  imul   -0x20(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  imul   -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x20(%rbp),%rax
  mulq   -0x20(%rbp)
  add    %rdx,%rcx
  mov    %rcx,%rdx
  mov    %rax,-0x10(%rbp)
  mov    %rdx,-0x8(%rbp)
  mov    %rax,-0x10(%rbp)
  mov    %rdx,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  xor    $0x29a,%rax
  or     -0x8(%rbp),%rax
  test   %rax,%rax
  jne    L20
  mov    $0x0,%eax
  call   <impossible@plt>
  jmp    L20
L18:
  nop
  jmp    L20
L4:
  nop
L20:
  leave
  ret
```

assistant: ```c
void f(__int128 x) {
  unsigned __int128 y;
  unsigned __int128 z;
  if (x < -7)
    return;
  if (x > 2)
    return;
  y = x;
  z = y * y;
  if (z == 666)
    impossible();
}
```
user: What is the c source code of the assembly code below:

```asm
<f>:
  endbr64
  ret
```

assistant: ```c
void f(__int128 x) {
  unsigned __int128 y;
  unsigned __int128 z;
  if (x < -7)
    return;
  if (x > 2)
    return;
  y = x;
  z = y * y;
  if (z == 666)
    impossible();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<f>:
  endbr64
  ret
```

assistant: ```c
void f(__int128 x) {
  unsigned __int128 y;
  unsigned __int128 z;
  if (x < -7)
    return;
  if (x > 2)
    return;
  y = x;
  z = y * y;
  if (z == 666)
    impossible();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<f>:
  endbr64
  ret
```

assistant: ```c
void f(__int128 x) {
  unsigned __int128 y;
  unsigned __int128 z;
  if (x < -7)
    return;
  if (x > 2)
    return;
  y = x;
  z = y * y;
  if (z == 666)
    impossible();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<l1_norm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L1
L39:
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jbe    L13
  movsd  -0x8(%rbp),%xmm0
  jmp    L35
L13:
  movsd  -0x8(%rbp),%xmm0
  movq   D5(%rip),%xmm1
  xorpd  %xmm1,%xmm0
L35:
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L1:
  mov    -0x2c(%rbp),%eax
  imul   -0x30(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L39
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
datatype: f32[2]

tool: {"D5": [0.0,-0.0]}

assistant: ```c
double l1_norm(double *mat, int rows, int cols) {
  double sum = 0;
  int i;
  for (i = 0; i < rows * cols; i++) {
    double val = *mat++;
    sum += ((val) > 0 ? (val) : -(val));
  }
  return sum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<l1_norm>:
  endbr64
  imul   %edx,%esi
  test   %esi,%esi
  jle    L32
  mov    %esi,%esi
  lea    (%rdi,%rsi,8),%rax
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm2
  movq   D2(%rip),%xmm3
  jmp    L44
L15:
  addsd  %xmm0,%xmm1
  cmp    %rax,%rdi
  je     L22
L44:
  add    $0x8,%rdi
  movsd  -0x8(%rdi),%xmm0
  comisd %xmm2,%xmm0
  ja     L15
  xorpd  %xmm3,%xmm0
  jmp    L15
L32:
  pxor   %xmm1,%xmm1
L22:
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type: f32[2]

tool: {"D2": [0.0,-0.0]}

assistant: ```c
double l1_norm(double *mat, int rows, int cols) {
  double sum = 0;
  int i;
  for (i = 0; i < rows * cols; i++) {
    double val = *mat++;
    sum += ((val) > 0 ? (val) : -(val));
  }
  return sum;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<l1_norm>:
  endbr64
  imul   %edx,%esi
  test   %esi,%esi
  jle    L27
  movslq %esi,%rsi
  pxor   %xmm2,%xmm2
  movq   D3(%rip),%xmm3
  lea    (%rdi,%rsi,8),%rax
  movapd %xmm2,%xmm1
  nopw   0x0(%rax,%rax,1)
L0:
  movsd  (%rdi),%xmm0
  add    $0x8,%rdi
  comisd %xmm2,%xmm0
  ja     L18
  xorpd  %xmm3,%xmm0
L18:
  addsd  %xmm0,%xmm1
  cmp    %rax,%rdi
  jne    L0
  movapd %xmm1,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L27:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: label: D3
type: f32[2]

tool: {"D3": [0.0,-0.0]}

assistant: ```c
double l1_norm(double *mat, int rows, int cols) {
  double sum = 0;
  int i;
  for (i = 0; i < rows * cols; i++) {
    double val = *mat++;
    sum += ((val) > 0 ? (val) : -(val));
  }
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<l1_norm>:
  endbr64
  imul   %edx,%esi
  test   %esi,%esi
  jle    L15
  cmp    $0x1,%esi
  je     L13
  mov    %esi,%edx
  pxor   %xmm4,%xmm4
  pxor   %xmm5,%xmm5
  mov    %rdi,%rax
  shr    %edx
  movapd D6(%rip),%xmm6
  movapd %xmm4,%xmm0
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
  nop
L8:
  movupd (%rax),%xmm2
  movapd %xmm5,%xmm1
  add    $0x10,%rax
  cmpltpd %xmm2,%xmm1
  movapd %xmm2,%xmm3
  xorpd  %xmm6,%xmm3
  andpd  %xmm1,%xmm2
  andnpd %xmm3,%xmm1
  orpd   %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  unpckhpd %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rax,%rdx
  jne    L8
  mov    %esi,%eax
  and    $0xfffffffe,%eax
  mov    %eax,%edx
  lea    (%rdi,%rdx,8),%rdi
  cmp    %eax,%esi
  je     L6
L54:
  movsd  (%rdi),%xmm1
  comisd %xmm4,%xmm1
  ja     L58
  xorpd  D1(%rip),%xmm1
L58:
  addsd  %xmm1,%xmm0
  ret
  nopl   0x0(%rax)
L6:
  ret
  nopl   0x0(%rax)
L15:
  pxor   %xmm0,%xmm0
  ret
L13:
  pxor   %xmm4,%xmm4
  movapd %xmm4,%xmm0
  jmp    L54
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
datatype: f32[4]
name: extract_symbolic_data
arguments: data_tag: D1
datatype: f32[4]

tool: {"D6": [0.0,-0.0,0.0,-0.0]}

tool: {"D1": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double l1_norm(double *mat, int rows, int cols) {
  double sum = 0;
  int i;
  for (i = 0; i < rows * cols; i++) {
    double val = *mat++;
    sum += ((val) > 0 ? (val) : -(val));
  }
  return sum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<print_sys_msg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D19
type: string
name: parse_data
arguments: data_label: D15
type: string

tool: D19: Welcome to the Daily Report System! :-)


tool: D15: Please Log in.


assistant: ```c
void print_sys_msg() {
  printf("Welcome to the Daily Report System! :-)\n");
  printf("Please Log in.\n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<print_sys_msg>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    D6(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D3
datatype: string
name: unpack_data
arguments: label: D6
datatype: string

tool: D3: Welcome to the Daily Report System! :-)


tool: D6: Please Log in.


assistant: ```c
void print_sys_msg() {
  printf("Welcome to the Daily Report System! :-)\n");
  printf("Please Log in.\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<print_sys_msg>:
  endbr64
  sub    $0x8,%rsp
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    D5(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D11
datatype: string
name: parse_data
arguments: data_tag: D5
datatype: string

tool: D11: Welcome to the Daily Report System! :-)


tool: D5: Please Log in.


assistant: ```c
void print_sys_msg() {
  printf("Welcome to the Daily Report System! :-)\n");
  printf("Please Log in.\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_sys_msg>:
  endbr64
  sub    $0x8,%rsp
  lea    D14(%rip),%rdi
  call   <printf@plt>
  lea    D12(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D14
type_guessed: string
name: disassemble_data
arguments: tag: D12
type_guessed: string

tool: D14: Welcome to the Daily Report System! :-)


tool: D12: Please Log in.


assistant: ```c
void print_sys_msg() {
  printf("Welcome to the Daily Report System! :-)\n");
  printf("Please Log in.\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<solve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  movl   $0x0,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L3
L47:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L43
L1:
  mov    -0x14(%rbp),%eax
  imul   -0x10(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L13
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <is_palindrome@plt>
  test   %rax,%rax
  je     L13
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L13:
  addl   $0x1,-0x10(%rbp)
L43:
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L1
  addl   $0x1,-0x14(%rbp)
L3:
  mov    -0x14(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L47
  mov    -0x18(%rbp),%eax
  leave
  ret
```

assistant: ```c
int solve(int p) {
  int max = 0;
  int lo = pow(10, p - 1);
  int hi = pow(10, p);
  int i, j, product;
  for (i = lo; i < hi; i++) {
    for (j = lo; j < hi; j++) {
      product = i * j;
      if (product > max && is_palindrome(product)) {
        max = product;
      }
    }
  }
  return max;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<solve>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  lea    -0x1(%rdi),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%r15d
  mov    %eax,0x8(%rsp)
  mov    %ebx,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  cmp    %eax,%r15d
  jge    L24
  mov    %eax,%r14d
  mov    %r15d,%eax
  imul   %r15d,%eax
  mov    %eax,0xc(%rsp)
  mov    $0x0,%r12d
  jmp    L67
L11:
  lea    0x1(%rbp),%eax
  add    0x4(%rsp),%ebx
  cmp    %eax,%r14d
  je     L54
  mov    %eax,%ebp
L15:
  cmp    %r12d,%ebx
  jle    L11
  mov    %ebx,%edi
  call   <is_palindrome@plt>
  test   %rax,%rax
  cmovne %ebx,%r12d
  jmp    L11
L54:
  lea    0x1(%r15),%eax
  mov    0x8(%rsp),%ecx
  add    %ecx,0xc(%rsp)
  cmp    %ebp,%r15d
  je     L18
  mov    %eax,%r15d
L67:
  mov    %r15d,0x4(%rsp)
  mov    0xc(%rsp),%ebx
  mov    0x8(%rsp),%ebp
  jmp    L15
L24:
  mov    $0x0,%r12d
L18:
  mov    %r12d,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int solve(int p) {
  int max = 0;
  int lo = pow(10, p - 1);
  int hi = pow(10, p);
  int i, j, product;
  for (i = lo; i < hi; i++) {
    for (j = lo; j < hi; j++) {
      product = i * j;
      if (product > max && is_palindrome(product)) {
        max = product;
      }
    }
  }
  return max;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<solve>:
  endbr64
  push   %r15
  lea    -0x1(%rdi),%esi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  mov    $0xa,%edi
  push   %rbx
  sub    $0x18,%rsp
  call   <pow@plt>
  mov    %ebp,%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  mov    %eax,0xc(%rsp)
  call   <pow@plt>
  cmp    %eax,%ebx
  jge    L48
  mov    %ebx,%r13d
  mov    %eax,%r14d
  mov    %ebx,%r15d
  xor    %r12d,%r12d
  imul   %ebx,%r13d
  nopw   0x0(%rax,%rax,1)
L30:
  mov    0xc(%rsp),%ebp
  mov    %r13d,%ebx
  jmp    L46
  nopl   0x0(%rax)
L9:
  lea    0x1(%rbp),%eax
  add    %r15d,%ebx
  cmp    %eax,%r14d
  je     L54
L51:
  mov    %eax,%ebp
L46:
  cmp    %r12d,%ebx
  jle    L9
  mov    %ebx,%edi
  call   <is_palindrome@plt>
  test   %rax,%rax
  lea    0x1(%rbp),%eax
  cmovne %ebx,%r12d
  add    %r15d,%ebx
  cmp    %eax,%r14d
  jne    L51
L54:
  lea    0x1(%r15),%eax
  add    0xc(%rsp),%r13d
  cmp    %ebp,%r15d
  je     L19
  mov    %eax,%r15d
  jmp    L30
L48:
  xor    %r12d,%r12d
L19:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int solve(int p) {
  int max = 0;
  int lo = pow(10, p - 1);
  int hi = pow(10, p);
  int i, j, product;
  for (i = lo; i < hi; i++) {
    for (j = lo; j < hi; j++) {
      product = i * j;
      if (product > max && is_palindrome(product)) {
        max = product;
      }
    }
  }
  return max;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<solve>:
  endbr64
  push   %r15
  lea    -0x1(%rdi),%esi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  mov    $0xa,%edi
  push   %rbx
  sub    $0x18,%rsp
  call   <pow@plt>
  mov    %ebp,%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  mov    %eax,0xc(%rsp)
  call   <pow@plt>
  cmp    %eax,%ebx
  jge    L9
  mov    %ebx,%r13d
  mov    %eax,%r14d
  mov    %ebx,%r15d
  xor    %r12d,%r12d
  imul   %ebx,%r13d
  nopw   0x0(%rax,%rax,1)
L36:
  mov    0xc(%rsp),%ebp
  mov    %r13d,%ebx
  jmp    L8
  nopl   0x0(%rax)
L20:
  lea    0x1(%rbp),%eax
  add    %r15d,%ebx
  cmp    %eax,%r14d
  je     L40
L61:
  mov    %eax,%ebp
L8:
  cmp    %r12d,%ebx
  jle    L20
  mov    %ebx,%edi
  call   <is_palindrome@plt>
  test   %rax,%rax
  lea    0x1(%rbp),%eax
  cmovne %ebx,%r12d
  add    %r15d,%ebx
  cmp    %eax,%r14d
  jne    L61
L40:
  lea    0x1(%r15),%eax
  add    0xc(%rsp),%r13d
  cmp    %ebp,%r15d
  je     L58
  mov    %eax,%r15d
  jmp    L36
L9:
  xor    %r12d,%r12d
L58:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int solve(int p) {
  int max = 0;
  int lo = pow(10, p - 1);
  int hi = pow(10, p);
  int i, j, product;
  for (i = lo; i < hi; i++) {
    for (j = lo; j < hi; j++) {
      product = i * j;
      if (product > max && is_palindrome(product)) {
        max = product;
      }
    }
  }
  return max;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_divides>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x1,%esi
  mov    $0x3,%edi
  call   <__CrestCall@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x2,%edi
  call   <__CrestStore@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <__CrestStore@plt>
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x8,%edi
  call   <__CrestLoad@plt>
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x7,%edi
  call   <__CrestLoad@plt>
  mov    -0x18(%rbp),%eax
  mov    -0x14(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  cltq
  mov    %rax,%rdx
  mov    $0x4,%esi
  mov    $0x6,%edi
  call   <__CrestApply2@plt>
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x5,%edi
  call   <__CrestLoad@plt>
  mov    -0x18(%rbp),%eax
  mov    -0x14(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %rax,%rdx
  mov    $0xc,%esi
  mov    $0x4,%edi
  call   <__CrestApply2@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x9,%edi
  call   <__CrestStore@plt>
  mov    -0x18(%rbp),%eax
  mov    -0x14(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xa,%edi
  call   <__CrestLoad@plt>
  mov    $0xb,%edi
  call   <__CrestReturn@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int _divides(int n, int m) {
  int __retres3;
  {
    __CrestCall(3, 1);
    __CrestStore(2, (unsigned long)(&m));
    __CrestStore(1, (unsigned long)(&n));
    __CrestLoad(8, (unsigned long)(&m), (long long)m);
    __CrestLoad(7, (unsigned long)(&n), (long long)n);
    __CrestApply2(6, 4, (long long)(m % n));
    __CrestLoad(5, (unsigned long)0, (long long)0);
    __CrestApply2(4, 12, (long long)(m % n == 0));
    __CrestStore(9, (unsigned long)(&__retres3));
    __retres3 = m % n == 0;
    {
      __CrestLoad(10, (unsigned long)(&__retres3), (long long)__retres3);
      __CrestReturn(11);
      return (__retres3);
    }
  }
}
```