ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  lea    D20(%rip),%rdi
  lea    D54(%rip),%r13
  push   %r12
  lea    D27(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D18(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  call   <puts@plt>
  jmp    L58
  cs nopw 0x0(%rax,%rax,1)
L34:
  mov    %r13,%rdi
  call   <puts@plt>
L58:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  lea    -0x3(%rcx),%eax
  cmp    $0x3e4,%eax
  ja     L34
  mov    $0x1,%ebx
  mov    $0x2,%esi
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  cmove  %edi,%ebx
  add    $0x1,%esi
  cmp    %esi,%ecx
  jg     L3
  lea    D65(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%esi
  test   %ebx,%ebx
  jne    L4
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  lea    D33(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L8
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D20
type_infered: string
name: disassemble_data
arguments: data_label: D54
type_infered: string
name: disassemble_data
arguments: data_label: D27
type_infered: string
name: disassemble_data
arguments: data_label: D18
type_infered: string
name: disassemble_data
arguments: data_label: D65
type_infered: string
name: disassemble_data
arguments: data_label: D0
type_infered: string
name: disassemble_data
arguments: data_label: D33
type_infered: string

tool: D20: Input: 

tool: D54: That's not a number between 2 and 1000, try again. 


tool: D27: Enter a number between 2 and 1000: 

tool: D18: %d

tool: D65: Output: 

tool: D0: The number you have entered, %d, IS NOT prime. 



tool: D33: The number you have entered, %d, IS prime. 



assistant: ```c
int main() {
  int num, i = 0, test = 1, prime = 1;
  printf("Input: \n");
  do {
    printf("Enter a number between 2 and 1000: ");
    scanf("%d", &num);
    if (num < 1000 && num > 2)
      test = 0;
    else
      printf("That's not a number between 2 and 1000, try again. \n\n");
  } while (test);
  for (i = 2; i < num; i++) {
    if (num % i == 0)
      prime = 0;
  }
  printf("Output: \n");
  if (prime == 0)
    printf("The number you have entered, %d, IS NOT prime. \n\n", num);
  else
    printf("The number you have entered, %d, IS prime. \n\n", num);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r13
  lea    D40(%rip),%rdi
  lea    D56(%rip),%r13
  push   %r12
  lea    D9(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D16(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  call   <puts@plt>
  jmp    L51
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    %r13,%rdi
  call   <puts@plt>
L51:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  lea    -0x3(%rcx),%eax
  cmp    $0x3e4,%eax
  ja     L14
  mov    $0x1,%ebx
  mov    $0x2,%esi
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L21:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  cmove  %edi,%ebx
  add    $0x1,%esi
  cmp    %esi,%ecx
  jg     L21
  lea    D20(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%esi
  test   %ebx,%ebx
  jne    L2
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  lea    D66(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L7
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D40
type_infered: string
name: unpack_data
arguments: data_label: D56
type_infered: string
name: unpack_data
arguments: data_label: D9
type_infered: string
name: unpack_data
arguments: data_label: D16
type_infered: string
name: unpack_data
arguments: data_label: D20
type_infered: string
name: unpack_data
arguments: data_label: D3
type_infered: string
name: unpack_data
arguments: data_label: D66
type_infered: string

tool: D40: Input: 

tool: D56: That's not a number between 2 and 1000, try again. 


tool: D9: Enter a number between 2 and 1000: 

tool: D16: %d

tool: D20: Output: 

tool: D3: The number you have entered, %d, IS NOT prime. 



tool: D66: The number you have entered, %d, IS prime. 



assistant: ```c
int main() {
  int num, i = 0, test = 1, prime = 1;
  printf("Input: \n");
  do {
    printf("Enter a number between 2 and 1000: ");
    scanf("%d", &num);
    if (num < 1000 && num > 2)
      test = 0;
    else
      printf("That's not a number between 2 and 1000, try again. \n\n");
  } while (test);
  for (i = 2; i < num; i++) {
    if (num % i == 0)
      prime = 0;
  }
  printf("Output: \n");
  if (prime == 0)
    printf("The number you have entered, %d, IS NOT prime. \n\n", num);
  else
    printf("The number you have entered, %d, IS prime. \n\n", num);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print_hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L7
L27:
  cmpl   $0x0,-0x10(%rbp)
  jle    L15
  mov    -0x10(%rbp),%eax
  and    $0x7,%eax
  test   %eax,%eax
  jne    L15
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L15:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L7:
  mov    -0x10(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L27
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D18
data_type: string
name: disassemble_data
arguments: tag: D8
data_type: string
name: disassemble_data
arguments: tag: D20
data_type: string

tool: D18: 
[0000] 

tool: D8: 
[%04x] 

tool: D20: %02x 

assistant: ```c
void print_hex(void *in, int start, int end) {
  uint8_t *array = (uint8_t *)in;
  int i;
  printf("\n[0000] ");
  for (i = start; i < end; i++) {
    if (i > 0 && i % 8 == 0) {
      printf("\n[%04x] ", i);
    }
    uint8_t foo = array[i];
    printf("%02x ", foo);
  }
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print_hex>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r13d
  mov    %edx,%r12d
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%r13d
  jge    L22
  movslq %r13d,%rbx
  sub    $0x1,%r12d
  sub    %r13d,%r12d
  lea    0x1(%rbx,%r12,1),%r13
  lea    D19(%rip),%r14
  lea    D16(%rip),%r12
  jmp    L33
L58:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r13,%rbx
  je     L22
L33:
  mov    %ebx,%esi
  test   %ebx,%ebx
  jle    L58
  test   $0x7,%bl
  jne    L58
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L58
L22:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D1
type_guessed: string
name: disassemble_data
arguments: tag: D19
type_guessed: string
name: disassemble_data
arguments: tag: D16
type_guessed: string

tool: D1: 
[0000] 

tool: D19: 
[%04x] 

tool: D16: %02x 

assistant: ```c
void print_hex(void *in, int start, int end) {
  uint8_t *array = (uint8_t *)in;
  int i;
  printf("\n[0000] ");
  for (i = start; i < end; i++) {
    if (i > 0 && i % 8 == 0) {
      printf("\n[%04x] ", i);
    }
    uint8_t foo = array[i];
    printf("%02x ", foo);
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print_hex>:
  endbr64
  push   %r15
  xor    %eax,%eax
  push   %r14
  push   %r13
  mov    %rdi,%r13
  lea    D29(%rip),%rdi
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <printf@plt>
  cmp    %r12d,%ebp
  jge    L33
  movslq %ebp,%rbx
  sub    $0x1,%r12d
  lea    D15(%rip),%r14
  sub    %ebp,%r12d
  lea    0x1(%rbx),%rbp
  lea    D13(%rip),%r15
  add    %rbp,%r12
  jmp    L31
  xchg   %ax,%ax
L20:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %rbp,%rbx
  call   <printf@plt>
  cmp    %r12,%rbp
  je     L33
  add    $0x1,%rbp
L31:
  mov    %ebx,%esi
  test   %ebx,%ebx
  jle    L20
  test   $0x7,%bl
  jne    L20
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L20
  nopw   0x0(%rax,%rax,1)
L33:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <putchar@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D29
datatype: string
name: extract_symbolic_data
arguments: label: D15
datatype: string
name: extract_symbolic_data
arguments: label: D13
datatype: string

tool: D29: 
[0000] 

tool: D15: %02x 

tool: D13: 
[%04x] 

assistant: ```c
void print_hex(void *in, int start, int end) {
  uint8_t *array = (uint8_t *)in;
  int i;
  printf("\n[0000] ");
  for (i = start; i < end; i++) {
    if (i > 0 && i % 8 == 0) {
      printf("\n[%04x] ", i);
    }
    uint8_t foo = array[i];
    printf("%02x ", foo);
  }
  printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<print_hex>:
  endbr64
  push   %r15
  xor    %eax,%eax
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rdi,%r13
  lea    D24(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  cmp    %r14d,%ebx
  jge    L12
  test   %ebx,%ebx
  jle    L14
  lea    D38(%rip),%r12
L27:
  movslq %ebx,%rbx
  lea    D37(%rip),%rbp
  jmp    L56
  nopl   0x0(%rax,%rax,1)
L62:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,%r14d
  jle    L12
L56:
  mov    %ebx,%esi
  test   $0x7,%bl
  jne    L62
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L62
  nopl   0x0(%rax,%rax,1)
L14:
  lea    -0x1(%r14),%r15d
  xor    %eax,%eax
  lea    D38(%rip),%r12
  test   %r15d,%r15d
  cmovg  %eax,%r15d
  movslq %ebx,%rax
  lea    0x0(%r13,%rax,4),%rbp
  nopl   0x0(%rax)
L69:
  mov    0x0(%rbp),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %r15d,%ebx
  jle    L69
  cmp    %ebx,%r14d
  jg     L27
L12:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <putchar@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D24
datatype: string
name: retrieve_data
arguments: data_tag: D38
datatype: string
name: retrieve_data
arguments: data_tag: D37
datatype: string

tool: D24: 
[0000] 

tool: D38: %02x 

tool: D37: 
[%04x] 

assistant: ```c
void print_hex(void *in, int start, int end) {
  uint8_t *array = (uint8_t *)in;
  int i;
  printf("\n[0000] ");
  for (i = start; i < end; i++) {
    if (i > 0 && i % 8 == 0) {
      printf("\n[%04x] ", i);
    }
    uint8_t foo = array[i];
    printf("%02x ", foo);
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<read_INPUT>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L24
  mov    $0x41004420,%eax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L10
L24:
  cmpl   $0x80,-0x14(%rbp)
  jne    L10
  mov    $0x410044a0,%eax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
L10:
  mov    -0x18(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  and    %edx,%eax
  test   %eax,%eax
  je     L38
  mov    $0x1,%eax
  jmp    L35
L38:
  mov    $0x0,%eax
L35:
  pop    %rbp
  ret
```

assistant: ```c
volatile int read_INPUT(int port, int pin) {
  unsigned volatile int buf;
  if (port == 0x00) {
    buf = *(unsigned int volatile *)(0x41004400 + 0x20);
  } else if (port == 0x80) {
    buf = *(unsigned int volatile *)(0x41004400 + 0xA0);
  }
  if (buf & (1 << pin)) {
    return 1;
  } else {
    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
<read_INPUT>:
  endbr64
  mov    %esi,%ecx
  test   %edi,%edi
  jne    L27
  mov    0x41004420,%eax
  mov    %eax,-0x4(%rsp)
L25:
  mov    -0x4(%rsp),%edx
  mov    $0x1,%eax
  shl    %cl,%eax
  test   %edx,%eax
  setne  %al
  movzbl %al,%eax
  ret
L27:
  cmp    $0x80,%edi
  jne    L25
  mov    0x410044a0,%eax
  mov    %eax,-0x4(%rsp)
  jmp    L25
```

assistant: ```c
volatile int read_INPUT(int port, int pin) {
  unsigned volatile int buf;
  if (port == 0x00) {
    buf = *(unsigned int volatile *)(0x41004400 + 0x20);
  } else if (port == 0x80) {
    buf = *(unsigned int volatile *)(0x41004400 + 0xA0);
  }
  if (buf & (1 << pin)) {
    return 1;
  } else {
    return 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<read_INPUT>:
  endbr64
  mov    %esi,%ecx
  test   %edi,%edi
  jne    L3
  mov    0x41004420,%eax
  mov    %eax,-0x4(%rsp)
L11:
  mov    -0x4(%rsp),%edx
  mov    $0x1,%eax
  shl    %cl,%eax
  test   %edx,%eax
  setne  %al
  movzbl %al,%eax
  ret
  nopl   0x0(%rax)
L3:
  cmp    $0x80,%edi
  jne    L11
  mov    0x410044a0,%eax
  mov    %eax,-0x4(%rsp)
  jmp    L11
```

assistant: ```c
volatile int read_INPUT(int port, int pin) {
  unsigned volatile int buf;
  if (port == 0x00) {
    buf = *(unsigned int volatile *)(0x41004400 + 0x20);
  } else if (port == 0x80) {
    buf = *(unsigned int volatile *)(0x41004400 + 0xA0);
  }
  if (buf & (1 << pin)) {
    return 1;
  } else {
    return 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<read_INPUT>:
  endbr64
  mov    %esi,%ecx
  test   %edi,%edi
  jne    L18
  mov    0x41004420,%eax
  mov    %eax,-0x4(%rsp)
L22:
  mov    -0x4(%rsp),%edx
  mov    $0x1,%eax
  shl    %cl,%eax
  test   %edx,%eax
  setne  %al
  movzbl %al,%eax
  ret
  nopl   0x0(%rax)
L18:
  cmp    $0x80,%edi
  jne    L22
  mov    0x410044a0,%eax
  mov    %eax,-0x4(%rsp)
  jmp    L22
```

assistant: ```c
volatile int read_INPUT(int port, int pin) {
  unsigned volatile int buf;
  if (port == 0x00) {
    buf = *(unsigned int volatile *)(0x41004400 + 0x20);
  } else if (port == 0x80) {
    buf = *(unsigned int volatile *)(0x41004400 + 0xA0);
  }
  if (buf & (1 << pin)) {
    return 1;
  } else {
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<center_of_mass>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %ecx,-0x34(%rbp)
  mov    %r8,-0x40(%rbp)
  mov    %r9,-0x48(%rbp)
  mov    -0x34(%rbp),%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  mov    (%rax),%eax
  add    %eax,-0x20(%rbp)
  mov    -0x48(%rbp),%rax
  mov    (%rax),%eax
  add    %eax,-0x1c(%rbp)
  addl   $0x1,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L26
L54:
  mov    -0x10(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L74
L36:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  cmp    %eax,-0xc(%rbp)
  jg     L72
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  imul   -0x14(%rbp),%eax
  add    %eax,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  imul   -0x10(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %eax,-0x18(%rbp)
  jmp    L64
L72:
  nop
L64:
  addl   $0x1,-0x14(%rbp)
L74:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L36
  addl   $0x1,-0x10(%rbp)
L26:
  mov    -0x10(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L54
  cmpl   $0x0,-0x18(%rbp)
  jne    L66
  mov    $0x1,%eax
  jmp    L22
L66:
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %eax,%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x1c(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %eax,%edx
  mov    -0x48(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
L22:
  pop    %rbp
  ret
```

assistant: ```c
int center_of_mass(unsigned char *rep, int w, int h, int iter, int *markx, int *marky) {
  int sumx, sumy;
  int sumw;
  unsigned char *repo;
  int x, y;
  int iter2;
  iter2 = iter / 4;
  sumx = 0;
  sumy = 0;
  sumw = 0;
  sumx += markx[0];
  sumy += marky[0];
  sumw += 1;
  for (y = 0; y < h; y++) {
    repo = rep + y * w;
    for (x = 0; x < w; x++) {
      if (repo[x] < iter2)
        continue;
      sumx += x * ((int)repo[x]);
      sumy += y * ((int)repo[x]);
      sumw += repo[x];
    }
  }
  if (!sumw)
    return 1;
  (*markx) = sumx / sumw;
  (*marky) = sumy / sumw;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<center_of_mass>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r11
  mov    %r8,%rbp
  lea    0x3(%rcx),%r8d
  test   %ecx,%ecx
  cmovns %ecx,%r8d
  sar    $0x2,%r8d
  mov    0x0(%rbp),%eax
  mov    (%r9),%edi
  test   %edx,%edx
  jle    L55
  mov    %esi,%ebx
  mov    %edx,%r12d
  movslq %esi,%r15
  mov    $0x0,%r13d
  mov    $0x1,%esi
  mov    %ebx,%r14d
  jmp    L19
L22:
  mov    %ecx,%r10d
  imul   %edx,%r10d
  add    %r10d,%eax
  mov    %ecx,%r10d
  imul   %r13d,%r10d
  add    %r10d,%edi
  add    %ecx,%esi
L49:
  add    $0x1,%rdx
  cmp    %r14,%rdx
  je     L45
L41:
  movzbl (%r11,%rdx,1),%ecx
  cmp    %r8d,%ecx
  jge    L22
  jmp    L49
L45:
  add    $0x1,%r13d
  add    %r15,%r11
  cmp    %r13d,%r12d
  je     L57
L19:
  test   %ebx,%ebx
  jle    L45
  mov    $0x0,%edx
  jmp    L41
L57:
  mov    $0x1,%edx
  test   %esi,%esi
  je     L48
L88:
  cltd
  idiv   %esi
  mov    %eax,0x0(%rbp)
  mov    %edi,%eax
  cltd
  idiv   %esi
  mov    %eax,(%r9)
  mov    $0x0,%edx
L48:
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L55:
  mov    $0x1,%esi
  jmp    L88
```

assistant: ```c
int center_of_mass(unsigned char *rep, int w, int h, int iter, int *markx, int *marky) {
  int sumx, sumy;
  int sumw;
  unsigned char *repo;
  int x, y;
  int iter2;
  iter2 = iter / 4;
  sumx = 0;
  sumy = 0;
  sumw = 0;
  sumx += markx[0];
  sumy += marky[0];
  sumw += 1;
  for (y = 0; y < h; y++) {
    repo = rep + y * w;
    for (x = 0; x < w; x++) {
      if (repo[x] < iter2)
        continue;
      sumx += x * ((int)repo[x]);
      sumy += y * ((int)repo[x]);
      sumw += repo[x];
    }
  }
  if (!sumw)
    return 1;
  (*markx) = sumx / sumw;
  (*marky) = sumy / sumw;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<center_of_mass>:
  endbr64
  push   %r14
  test   %ecx,%ecx
  push   %r13
  lea    0x3(%rcx),%r13d
  cmovns %ecx,%r13d
  push   %r12
  mov    (%r9),%ecx
  push   %rbp
  mov    %esi,%ebp
  mov    (%r8),%esi
  push   %rbx
  sar    $0x2,%r13d
  test   %edx,%edx
  jle    L27
  mov    %rdi,%r11
  mov    %edx,%ebx
  movslq %ebp,%r12
  xor    %r14d,%r14d
  mov    $0x1,%edi
  cs nopw 0x0(%rax,%rax,1)
L35:
  test   %ebp,%ebp
  jle    L47
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L7:
  movzbl (%r11,%rax,1),%edx
  cmp    %r13d,%edx
  jl     L40
  mov    %edx,%r10d
  add    %edx,%edi
  imul   %eax,%r10d
  add    %r10d,%esi
  mov    %edx,%r10d
  imul   %r14d,%r10d
  add    %r10d,%ecx
L40:
  add    $0x1,%rax
  cmp    %rax,%r12
  jne    L7
L47:
  add    $0x1,%r14d
  add    %r12,%r11
  cmp    %r14d,%ebx
  jne    L35
  mov    %esi,%eax
  cltd
  idiv   %edi
  mov    %eax,%esi
  mov    %ecx,%eax
  cltd
  idiv   %edi
  mov    %eax,%ecx
L27:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  mov    %esi,(%r8)
  pop    %r12
  pop    %r13
  mov    %ecx,(%r9)
  pop    %r14
  ret
```

assistant: ```c
int center_of_mass(unsigned char *rep, int w, int h, int iter, int *markx, int *marky) {
  int sumx, sumy;
  int sumw;
  unsigned char *repo;
  int x, y;
  int iter2;
  iter2 = iter / 4;
  sumx = 0;
  sumy = 0;
  sumw = 0;
  sumx += markx[0];
  sumy += marky[0];
  sumw += 1;
  for (y = 0; y < h; y++) {
    repo = rep + y * w;
    for (x = 0; x < w; x++) {
      if (repo[x] < iter2)
        continue;
      sumx += x * ((int)repo[x]);
      sumy += y * ((int)repo[x]);
      sumw += repo[x];
    }
  }
  if (!sumw)
    return 1;
  (*markx) = sumx / sumw;
  (*marky) = sumy / sumw;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<set_usb_mass_storage_ifsubtype>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  lea    D19(%rip),%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x18(%rbp),%rcx
  mov    -0x30(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  cmp    %rax,-0x30(%rbp)
  je     L78
  cmpl   $0x6,-0x1c(%rbp)
  ja     L64
  mov    -0x1c(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D53(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D53(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D29(%rip),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L78
  lea    D21(%rip),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L78
  lea    D72(%rip),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L78
  lea    D18(%rip),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L78
  lea    D59(%rip),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L78
L64:
  nop
L78:
  mov    -0x10(%rbp),%rdx
  mov    -0x38(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <util_strscpy@plt>
  mov    -0x1c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L60
  call   <__stack_chk_fail@plt>
L60:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D19
type: string
name: retrieve_data
arguments: tag: D29
type: string
name: retrieve_data
arguments: tag: D21
type: string
name: retrieve_data
arguments: tag: D72
type: string
name: retrieve_data
arguments: tag: D18
type: string
name: retrieve_data
arguments: tag: D59
type: string

tool: D19: generic

tool: D29: atapi

tool: D21: tape

tool: D72: floppy

tool: D18: rbc

tool: D59: scsi

assistant: ```c
int set_usb_mass_storage_ifsubtype(char *to, const char *from, size_t len) {
  int type_num = 0;
  char *eptr;
  char *type = "generic";
  type_num = strtoul(from, &eptr, 0);
  if (eptr != from) {
    switch (type_num) {
    case 2:
      type = "atapi";
      break;
    case 3:
      type = "tape";
      break;
    case 4:
    case 5:
      type = "floppy";
      break;
    case 1:
      type = "rbc";
      break;
    case 6:
      type = "scsi";
      break;
    default:
      break;
    }
  }
  util_strscpy(to, len, type);
  return type_num;
}
```