ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__aeabi_uidivmod>:
  endbr64
  mov    $0x0,%edx
  mov    $0x0,%r9d
  mov    $0x1f,%eax
  mov    $0x80000000,%r10d
  jmp    L38
L45:
  add    $0x1,%edx
  cmp    $0x20,%edx
  je     L46
L38:
  mov    %eax,%ecx
  sub    %edx,%ecx
  mov    %esi,%r8d
  shl    %cl,%r8d
  mov    %r8d,%r11d
  shr    %cl,%r11d
  cmp    %esi,%r11d
  jne    L45
  cmp    %edi,%r8d
  ja     L45
  mov    %r10d,%r11d
  mov    %edx,%ecx
  shr    %cl,%r11d
  movslq %r11d,%rcx
  or     %rcx,%r9
  sub    %r8d,%edi
  jne    L45
L46:
  shl    $0x20,%rdi
  mov    %rdi,%rax
  or     %r9,%rax
  ret
```

assistant: ```c
unsigned long long __aeabi_uidivmod(unsigned int value, unsigned int divisor) {
  unsigned long long answer = 0;
  unsigned int i;
  for (i = 0; i < 32; i++) {
    if ((divisor << (31 - i)) >> (31 - i) == divisor) {
      if (value >= divisor << (31 - i)) {
        value -= divisor << (31 - i);
        answer |= (unsigned long long)(1 << (31 - i));
        if (value == 0)
          break;
      }
    }
  }
  answer |= (unsigned long long)value << 32;
  return answer;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<__aeabi_uidivmod>:
  endbr64
  xor    %eax,%eax
  xor    %r9d,%r9d
  mov    $0x1f,%r8d
  mov    $0x80000000,%r10d
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L33:
  add    $0x1,%eax
  cmp    $0x20,%eax
  je     L25
L27:
  mov    %r8d,%ecx
  mov    %esi,%edx
  sub    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%r11d
  shr    %cl,%r11d
  cmp    %esi,%r11d
  jne    L33
  cmp    %edi,%edx
  ja     L33
  mov    %eax,%ecx
  mov    %r10d,%r11d
  shr    %cl,%r11d
  movslq %r11d,%rcx
  or     %rcx,%r9
  sub    %edx,%edi
  jne    L33
  mov    %r9,%rax
  ret
  cs nopw 0x0(%rax,%rax,1)
L25:
  shl    $0x20,%rdi
  or     %rdi,%r9
  mov    %r9,%rax
  ret
```

assistant: ```c
unsigned long long __aeabi_uidivmod(unsigned int value, unsigned int divisor) {
  unsigned long long answer = 0;
  unsigned int i;
  for (i = 0; i < 32; i++) {
    if ((divisor << (31 - i)) >> (31 - i) == divisor) {
      if (value >= divisor << (31 - i)) {
        value -= divisor << (31 - i);
        answer |= (unsigned long long)(1 << (31 - i));
        if (value == 0)
          break;
      }
    }
  }
  answer |= (unsigned long long)value << 32;
  return answer;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<__aeabi_uidivmod>:
  endbr64
  xor    %eax,%eax
  xor    %r9d,%r9d
  mov    $0x1f,%r8d
  mov    $0x80000000,%r10d
  jmp    L45
  nopw   0x0(%rax,%rax,1)
L44:
  add    $0x1,%eax
  cmp    $0x20,%eax
  je     L33
L45:
  mov    %r8d,%ecx
  mov    %esi,%edx
  sub    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%r11d
  shr    %cl,%r11d
  cmp    %esi,%r11d
  jne    L44
  cmp    %edi,%edx
  ja     L44
  mov    %eax,%ecx
  mov    %r10d,%r11d
  shr    %cl,%r11d
  movslq %r11d,%rcx
  or     %rcx,%r9
  sub    %edx,%edi
  jne    L44
  mov    %r9,%rax
  ret
  cs nopw 0x0(%rax,%rax,1)
L33:
  shl    $0x20,%rdi
  or     %rdi,%r9
  mov    %r9,%rax
  ret
```

assistant: ```c
unsigned long long __aeabi_uidivmod(unsigned int value, unsigned int divisor) {
  unsigned long long answer = 0;
  unsigned int i;
  for (i = 0; i < 32; i++) {
    if ((divisor << (31 - i)) >> (31 - i) == divisor) {
      if (value >= divisor << (31 - i)) {
        value -= divisor << (31 - i);
        answer |= (unsigned long long)(1 << (31 - i));
        if (value == 0)
          break;
      }
    }
  }
  answer |= (unsigned long long)value << 32;
  return answer;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  je     L15
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L18
L15:
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L18:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D10
type: string
name: retrieve_data
arguments: label: D11
type: string

tool: D10: Accepts any flags


tool: D11: Hello world!


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2) {
    printf("Accepts any flags\n");
    return 0;
  }
  printf("Hello world!\n");
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L6
  lea    D8(%rip),%rdi
  call   <printf@plt>
L5:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L6:
  lea    D13(%rip),%rdi
  call   <printf@plt>
  jmp    L5
```

tool_calls: name: unpack_data
arguments: tag: D8
type_guessed: string
name: unpack_data
arguments: tag: D13
type_guessed: string

tool: D8: Accepts any flags


tool: D13: Hello world!


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2) {
    printf("Accepts any flags\n");
    return 0;
  }
  printf("Hello world!\n");
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L10
  lea    D17(%rip),%rdi
  call   <printf@plt>
L0:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L10:
  lea    D15(%rip),%rdi
  call   <printf@plt>
  jmp    L0
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D17
datatype: string
name: extract_symbolic_data
arguments: data_label: D15
datatype: string

tool: D17: Accepts any flags


tool: D15: Hello world!


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2) {
    printf("Accepts any flags\n");
    return 0;
  }
  printf("Hello world!\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L15
  lea    D13(%rip),%rdi
  call   <printf@plt>
L14:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L15:
  lea    D12(%rip),%rdi
  call   <printf@plt>
  jmp    L14
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D13
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D12
type_guessed: string

tool: D13: Accepts any flags


tool: D12: Hello world!


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2) {
    printf("Accepts any flags\n");
    return 0;
  }
  printf("Hello world!\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<power>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  cmpl   $0x0,-0x28(%rbp)
  jns    L35
  mov    $0xffffffff,%eax
  jmp    L39
L35:
  cmpl   $0x0,-0x28(%rbp)
  jne    L52
  mov    $0x1,%eax
  jmp    L39
L52:
  mov    -0x28(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L0
  mov    -0x28(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <power>
  imul   -0x24(%rbp),%eax
  mov    %eax,%ebx
  mov    -0x28(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <power>
  imul   %ebx,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L5
L0:
  mov    -0x28(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <power>
  mov    %eax,%ebx
  mov    -0x28(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <power>
  imul   %ebx,%eax
  mov    %eax,-0x14(%rbp)
L5:
  mov    -0x14(%rbp),%eax
L39:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int power(int a, int pow) {
  int product;
  if (pow < 0) {
    return (-1);
  }
  if (pow == 0) {
    return (1);
  }
  if ((pow % 2) == 1) {
    product = a * power(a, pow % 2) * power(a, pow % 2);
  } else {
    product = power(a, pow % 2) * power(a, pow % 2);
  }
  return (product);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<power>:
  endbr64
  test   %esi,%esi
  js     L7
  push   %rbx
  mov    %edi,%ebx
  mov    $0x1,%eax
  je     L20
  mov    %esi,%eax
  shr    $0x1f,%eax
  add    %eax,%esi
  and    $0x1,%esi
  sub    %eax,%esi
  cmp    $0x1,%esi
  je     L19
  mov    $0x0,%esi
  call   <power>
  imul   %eax,%eax
L20:
  pop    %rbx
  ret
L19:
  call   <power>
  mov    %eax,%edx
  mov    %ebx,%eax
  imul   %edx,%eax
  imul   %edx,%eax
  jmp    L20
L7:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int power(int a, int pow) {
  int product;
  if (pow < 0) {
    return (-1);
  }
  if (pow == 0) {
    return (1);
  }
  if ((pow % 2) == 1) {
    product = a * power(a, pow % 2) * power(a, pow % 2);
  } else {
    product = power(a, pow % 2) * power(a, pow % 2);
  }
  return (product);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<power>:
  endbr64
  test   %esi,%esi
  js     L21
  je     L1
  and    $0x1,%esi
  push   %rbx
  mov    $0x1,%eax
  mov    %edi,%ebx
  jne    L23
  pop    %rbx
  ret
L23:
  mov    %ebx,%edi
  mov    $0x1,%esi
  call   <power>
  mov    $0x1,%esi
  mov    %ebx,%edi
  call   <power>
  jmp    L23
  nopl   0x0(%rax,%rax,1)
L21:
  mov    $0xffffffff,%eax
  ret
L1:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int power(int a, int pow) {
  int product;
  if (pow < 0) {
    return (-1);
  }
  if (pow == 0) {
    return (1);
  }
  if ((pow % 2) == 1) {
    product = a * power(a, pow % 2) * power(a, pow % 2);
  } else {
    product = power(a, pow % 2) * power(a, pow % 2);
  }
  return (product);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<power>:
  endbr64
  test   %esi,%esi
  js     L36
  je     L24
  and    $0x1,%esi
  push   %rbx
  mov    $0x1,%eax
  mov    %edi,%ebx
  jne    L6
  pop    %rbx
  ret
L6:
  mov    %ebx,%edi
  mov    $0x1,%esi
  call   <power>
  mov    $0x1,%esi
  mov    %ebx,%edi
  call   <power>
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L36:
  mov    $0xffffffff,%eax
  ret
L24:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int power(int a, int pow) {
  int product;
  if (pow < 0) {
    return (-1);
  }
  if (pow == 0) {
    return (1);
  }
  if ((pow % 2) == 1) {
    product = a * power(a, pow % 2) * power(a, pow % 2);
  } else {
    product = power(a, pow % 2) * power(a, pow % 2);
  }
  return (product);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D3
data_type: string

tool: D3: Usage:

progname test x86|mmx|mmx2|sse2 count

   Run x86 or mmx routine count times



assistant: ```c
void usage() {
  puts("Usage:\n"
       "\n"
       "progname test x86|mmx|mmx2|sse2 count\n"
       "\n"
       "   Run x86 or mmx routine count times\n"
       "\n");
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type: string

tool: D1: Usage:

progname test x86|mmx|mmx2|sse2 count

   Run x86 or mmx routine count times



assistant: ```c
void usage() {
  puts("Usage:\n"
       "\n"
       "progname test x86|mmx|mmx2|sse2 count\n"
       "\n"
       "   Run x86 or mmx routine count times\n"
       "\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<usage>:
  endbr64
  lea    D9(%rip),%rdi
  jmp    1050 <puts@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
type_guessed: string

tool: D9: Usage:

progname test x86|mmx|mmx2|sse2 count

   Run x86 or mmx routine count times



assistant: ```c
void usage() {
  puts("Usage:\n"
       "\n"
       "progname test x86|mmx|mmx2|sse2 count\n"
       "\n"
       "   Run x86 or mmx routine count times\n"
       "\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<usage>:
  endbr64
  lea    D1(%rip),%rdi
  jmp    1050 <puts@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type_infered: string

tool: D1: Usage:

progname test x86|mmx|mmx2|sse2 count

   Run x86 or mmx routine count times



assistant: ```c
void usage() {
  puts("Usage:\n"
       "\n"
       "progname test x86|mmx|mmx2|sse2 count\n"
       "\n"
       "   Run x86 or mmx routine count times\n"
       "\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test_all>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <test_neurogen_genome@plt>
  mov    $0x0,%eax
  call   <test_neurogen_population@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void test_all() {
  test_neurogen_genome();
  test_neurogen_population();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<test_all>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <test_neurogen_genome@plt>
  mov    $0x0,%eax
  call   <test_neurogen_population@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void test_all() {
  test_neurogen_genome();
  test_neurogen_population();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_all>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_neurogen_genome@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <test_neurogen_population@plt>
```

assistant: ```c
void test_all() {
  test_neurogen_genome();
  test_neurogen_population();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<test_all>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_neurogen_genome@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <test_neurogen_population@plt>
```

assistant: ```c
void test_all() {
  test_neurogen_genome();
  test_neurogen_population();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x64,-0x14(%rbp)
  movl   $0x3e7,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L28
L30:
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  jmp    L35
L22:
  mov    -0x1c(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <isPalindrome@plt>
  test   %eax,%eax
  je     L40
  mov    -0x20(%rbp),%edx
  mov    -0x4(%rbp),%eax
  cmp    %eax,%edx
  cmovge %edx,%eax
  mov    %eax,-0x20(%rbp)
  jmp    L8
L40:
  nop
L8:
  addl   $0x1,-0x18(%rbp)
L35:
  mov    -0x18(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jle    L22
  addl   $0x1,-0x1c(%rbp)
L28:
  mov    -0x1c(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jle    L30
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D7: Largest palindrome is %d


assistant: ```c
int main() {
  const int kMin = 100;
  const int kMax = 999;
  int a, b, max;
  a = b = kMin;
  max = 0;
  for (int a = kMin; a <= kMax; ++a) {
    for (int b = a; b <= kMax; ++b) {
      int n = a * b;
      if (!isPalindrome(n))
        continue;
      max = max < n ? n : max;
    }
  }
  printf("Largest palindrome is %d\n", max);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x64,%r14d
  mov    $0x0,%ebp
  jmp    L15
L22:
  mov    %r14d,%r13d
  mov    %r14d,%ebx
  imul   %r14d,%ebx
  mov    %r14d,%r12d
L20:
  mov    %ebx,%edi
  call   <isPalindrome@plt>
  mov    %eax,%edx
  cmp    %ebx,%ebp
  mov    %ebx,%eax
  cmovge %ebp,%eax
  test   %edx,%edx
  cmovne %eax,%ebp
  add    $0x1,%r12d
  add    %r13d,%ebx
  cmp    $0x3e8,%r12d
  jne    L20
L37:
  add    $0x1,%r14d
  cmp    $0x3e8,%r14d
  je     L2
L15:
  cmp    $0x3e7,%r14d
  jle    L22
  jmp    L37
L2:
  mov    %ebp,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: tag: D3
type: string

tool: D3: Largest palindrome is %d


assistant: ```c
int main() {
  const int kMin = 100;
  const int kMax = 999;
  int a, b, max;
  a = b = kMin;
  max = 0;
  for (int a = kMin; a <= kMax; ++a) {
    for (int b = a; b <= kMax; ++b) {
      int n = a * b;
      if (!isPalindrome(n))
        continue;
      max = max < n ? n : max;
    }
  }
  printf("Largest palindrome is %d\n", max);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x64,%r13d
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L13:
  mov    %r13d,%ebx
  mov    %r13d,%r12d
  imul   %r13d,%ebx
  nopw   0x0(%rax,%rax,1)
L19:
  mov    %ebx,%edi
  call   <isPalindrome@plt>
  cmp    %ebx,%ebp
  mov    %eax,%r8d
  mov    %ebx,%eax
  cmovge %ebp,%eax
  test   %r8d,%r8d
  cmovne %eax,%ebp
  add    $0x1,%r12d
  add    %r13d,%ebx
  cmp    $0x3e8,%r12d
  jne    L19
  add    $0x1,%r13d
  cmp    $0x3e8,%r13d
  jne    L13
  mov    %ebp,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: tag: D3
type_guessed: string

tool: D3: Largest palindrome is %d


assistant: ```c
int main() {
  const int kMin = 100;
  const int kMax = 999;
  int a, b, max;
  a = b = kMin;
  max = 0;
  for (int a = kMin; a <= kMax; ++a) {
    for (int b = a; b <= kMax; ++b) {
      int n = a * b;
      if (!isPalindrome(n))
        continue;
      max = max < n ? n : max;
    }
  }
  printf("Largest palindrome is %d\n", max);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x64,%r13d
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    %r13d,%ebx
  mov    %r13d,%r12d
  imul   %r13d,%ebx
  nopw   0x0(%rax,%rax,1)
L12:
  mov    %ebx,%edi
  call   <isPalindrome@plt>
  cmp    %ebx,%ebp
  mov    %eax,%r8d
  mov    %ebx,%eax
  cmovge %ebp,%eax
  test   %r8d,%r8d
  cmovne %eax,%ebp
  add    $0x1,%r12d
  add    %r13d,%ebx
  cmp    $0x3e8,%r12d
  jne    L12
  add    $0x1,%r13d
  cmp    $0x3e8,%r13d
  jne    L1
  mov    %ebp,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

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

tool: D4: Largest palindrome is %d


assistant: ```c
int main() {
  const int kMin = 100;
  const int kMax = 999;
  int a, b, max;
  a = b = kMin;
  max = 0;
  for (int a = kMin; a <= kMax; ++a) {
    for (int b = a; b <= kMax; ++b) {
      int n = a * b;
      if (!isPalindrome(n))
        continue;
      max = max < n ? n : max;
    }
  }
  printf("Largest palindrome is %d\n", max);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L79
  cmpl   $0x0,-0x18(%rbp)
  jle    L79
  movl   $0x1,-0x8(%rbp)
  jmp    L108
L10:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    $0x1,%edi
  call   <hash@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
  addl   $0x1,-0x8(%rbp)
L108:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L10
L79:
  cmpl   $0x0,-0x14(%rbp)
  jle    L114
  movl   $0x2,-0xc(%rbp)
  jmp    L94
L29:
  movl   $0x1,-0x8(%rbp)
  jmp    L65
L26:
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x2,%edi
  call   <hash@plt>
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <hash@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
  addl   $0x1,-0x8(%rbp)
L65:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L26
  movl   $0x1,-0x8(%rbp)
  jmp    L21
L88:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
  addl   $0x1,-0x8(%rbp)
L21:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L88
  addl   $0x1,-0xc(%rbp)
L94:
  mov    -0xc(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L29
L114:
  cmpl   $0x0,-0x14(%rbp)
  jle    L6
  cmpl   $0x0,-0x18(%rbp)
  jle    L6
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L128
L82:
  mov    $0x2,%edi
  call   <hash@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <hash@plt>
  addl   $0x1,-0x8(%rbp)
L128:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L82
L6:
  nop
  leave
  ret
```

assistant: ```c
void test(int M, int N) {
  int c2, c4;
  int i, j;
  if ((M >= 1) && (N >= 1)) {
    for (c4 = 1; c4 <= M; c4++) {
      {
        hash(1);
        hash(1);
        hash(c4);
      };
    }
  }
  if (M >= 1) {
    for (c2 = 2; c2 <= N; c2++) {
      for (c4 = 1; c4 <= M; c4++) {
        i = c2 - 1;
        {
          hash(2);
          hash(c2 - 1);
          hash(c4);
        };
      }
      for (c4 = 1; c4 <= M; c4++) {
        {
          hash(1);
          hash(c2);
          hash(c4);
        };
      }
    }
  }
  if ((M >= 1) && (N >= 1)) {
    c2 = N + 1;
    for (c4 = 1; c4 <= M; c4++) {
      {
        hash(2);
        hash(N);
        hash(c4);
      };
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebp
  mov    %esi,%r15d
  test   %edi,%edi
  setg   %dl
  test   %esi,%esi
  setg   %al
  and    %al,%dl
  mov    %dl,0xf(%rsp)
  je     L69
  mov    $0x1,%ebx
L43:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %ebx,%edi
  call   <hash@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jge    L43
  cmp    $0x1,%r15d
  jle    L108
L105:
  mov    $0x1,%r14d
L9:
  mov    %r14d,%r12d
  add    $0x1,%r14d
  mov    %r14d,%r13d
  mov    $0x1,%ebx
L75:
  mov    $0x2,%edi
  call   <hash@plt>
  mov    %r12d,%edi
  call   <hash@plt>
  mov    %ebx,%edi
  call   <hash@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jge    L75
  mov    $0x1,%ebx
L59:
  mov    $0x1,%edi
  call   <hash@plt>
  mov    %r13d,%edi
  call   <hash@plt>
  mov    %ebx,%edi
  call   <hash@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jge    L59
  add    $0x2,%r12d
  cmp    %r12d,%r15d
  jge    L9
  cmpb   $0x0,0xf(%rsp)
  je     L58
L108:
  mov    $0x1,%ebx
L24:
  mov    $0x2,%edi
  call   <hash@plt>
  mov    %r15d,%edi
  call   <hash@plt>
  mov    %ebx,%edi
  call   <hash@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jge    L24
L58:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L69:
  test   %edi,%edi
  jle    L58
  cmp    $0x1,%esi
  jg     L105
  jmp    L58
```

assistant: ```c
void test(int M, int N) {
  int c2, c4;
  int i, j;
  if ((M >= 1) && (N >= 1)) {
    for (c4 = 1; c4 <= M; c4++) {
      {
        hash(1);
        hash(1);
        hash(c4);
      };
    }
  }
  if (M >= 1) {
    for (c2 = 2; c2 <= N; c2++) {
      for (c4 = 1; c4 <= M; c4++) {
        i = c2 - 1;
        {
          hash(2);
          hash(c2 - 1);
          hash(c4);
        };
      }
      for (c4 = 1; c4 <= M; c4++) {
        {
          hash(1);
          hash(c2);
          hash(c4);
        };
      }
    }
  }
  if ((M >= 1) && (N >= 1)) {
    c2 = N + 1;
    for (c4 = 1; c4 <= M; c4++) {
      {
        hash(2);
        hash(N);
        hash(c4);
      };
    }
  }
}
```