ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<program_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L12
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L8
L12:
  mov    -0x18(%rbp),%rax
  jmp    L29
L8:
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
L29:
  leave
  ret
```

assistant: ```c
char *program_name(char *command) {
  char *last_slash = strrchr(command, '/');
  if (!last_slash || !*(last_slash + 1))
    return command;
  return last_slash + 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<program_name>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x2f,%esi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L4
  mov    %rax,%rdx
  mov    %rbx,%rax
  cmpb   $0x0,0x1(%rdx)
  je     L15
  lea    0x1(%rdx),%rax
L15:
  pop    %rbx
  ret
L4:
  mov    %rbx,%rax
  jmp    L15
```

assistant: ```c
char *program_name(char *command) {
  char *last_slash = strrchr(command, '/');
  if (!last_slash || !*(last_slash + 1))
    return command;
  return last_slash + 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<program_name>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  test   %rax,%rax
  je     L16
  mov    %rax,%rdx
  mov    %rbx,%rax
  cmpb   $0x0,0x1(%rdx)
  je     L11
  lea    0x1(%rdx),%rax
L11:
  pop    %rbx
  ret
  nopl   0x0(%rax)
L16:
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *program_name(char *command) {
  char *last_slash = strrchr(command, '/');
  if (!last_slash || !*(last_slash + 1))
    return command;
  return last_slash + 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<program_name>:
  endbr64
  push   %rbx
  mov    $0x2f,%esi
  mov    %rdi,%rbx
  call   <strrchr@plt>
  test   %rax,%rax
  je     L10
  mov    %rax,%rdx
  mov    %rbx,%rax
  cmpb   $0x0,0x1(%rdx)
  je     L3
  lea    0x1(%rdx),%rax
L3:
  pop    %rbx
  ret
  nopl   0x0(%rax)
L10:
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *program_name(char *command) {
  char *last_slash = strrchr(command, '/');
  if (!last_slash || !*(last_slash + 1))
    return command;
  return last_slash + 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0xa,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  sub    $0x1,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  mov    %rax,%r8
  mov    $0x0,%r9d
  mov    -0x2c(%rbp),%eax
  cltq
  mov    %rax,%rsi
  mov    $0x0,%edi
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  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
L24:
  cmp    %rdx,%rsp
  je     L5
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L24
L5:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L18
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L18:
  mov    %rsp,%rax
  add    $0x7,%rax
  shr    $0x3,%rax
  shl    $0x3,%rax
  mov    %rax,-0x20(%rbp)
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x30(%rbp)
  jmp    L77
L62:
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x30(%rbp)
L77:
  mov    -0x30(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L62
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x30(%rbp)
  jmp    L55
L28:
  mov    -0x20(%rbp),%rax
  mov    -0x30(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,8),%rax
  movq   %rax,%xmm0
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  addl   $0x1,-0x30(%rbp)
L55:
  mov    -0x30(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L28
  mov    $0x0,%eax
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D26
type: string
name: parse_data
arguments: label: D17
type: string
name: parse_data
arguments: label: D6
type: string
name: parse_data
arguments: label: D31
type: string

tool: D26: Enter 10 float numbers:

tool: D17: %lf

tool: D6: You entered numbers:

tool: D31: %6.2lf

assistant: ```c
int main(void) {
  const int TEN = 10;
  double arr[TEN];
  int index;
  printf("Enter 10 float numbers:\n");
  for (index = 0; index < TEN; index++)
    scanf("%lf", &arr[index]);
  printf("You entered numbers:");
  for (index = 0; index < TEN; index++)
    printf("%6.2lf", arr[index]);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    D1(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  lea    0x50(%rsp),%r12
  mov    %rbx,%rbp
  lea    D30(%rip),%r13
L2:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x8,%rbp
  cmp    %r12,%rbp
  jne    L2
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D29(%rip),%rbp
L4:
  movsd  (%rbx),%xmm0
  mov    %rbp,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L4
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D1
type: string
name: disassemble_data
arguments: label: D30
type: string
name: disassemble_data
arguments: label: D21
type: string
name: disassemble_data
arguments: label: D29
type: string

tool: D1: Enter 10 float numbers:

tool: D30: %lf

tool: D21: You entered numbers:

tool: D29: %6.2lf

assistant: ```c
int main(void) {
  const int TEN = 10;
  double arr[TEN];
  int index;
  printf("Enter 10 float numbers:\n");
  for (index = 0; index < TEN; index++)
    scanf("%lf", &arr[index]);
  printf("You entered numbers:");
  for (index = 0; index < TEN; index++)
    printf("%6.2lf", arr[index]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  lea    D8(%rip),%rdi
  lea    D6(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x50(%rsp),%r12
  call   <puts@plt>
  mov    %rbx,%rbp
  nopl   0x0(%rax)
L12:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x8,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L12
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  lea    D23(%rip),%rbp
  call   <printf@plt>
  nopl   0x0(%rax)
L28:
  movsd  (%rbx),%xmm0
  mov    %rbp,%rdi
  mov    $0x1,%eax
  add    $0x8,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L28
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
type: string
name: retrieve_data
arguments: data_tag: D6
type: string
name: retrieve_data
arguments: data_tag: D1
type: string
name: retrieve_data
arguments: data_tag: D23
type: string

tool: D8: Enter 10 float numbers:

tool: D6: %lf

tool: D1: You entered numbers:

tool: D23: %6.2lf

assistant: ```c
int main(void) {
  const int TEN = 10;
  double arr[TEN];
  int index;
  printf("Enter 10 float numbers:\n");
  for (index = 0; index < TEN; index++)
    scanf("%lf", &arr[index]);
  printf("You entered numbers:");
  for (index = 0; index < TEN; index++)
    printf("%6.2lf", arr[index]);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D39(%rip),%rdi
  lea    D19(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x50(%rsp),%r12
  call   <puts@plt>
  mov    %rbx,%rbp
  nopl   0x0(%rax)
L29:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x8,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L29
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  lea    D12(%rip),%rbp
  call   <printf@plt>
  nopl   0x0(%rax)
L23:
  movsd  (%rbx),%xmm0
  mov    %rbp,%rdi
  mov    $0x1,%eax
  add    $0x8,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L23
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D39
type_infered: string
name: extract_symbolic_data
arguments: label: D19
type_infered: string
name: extract_symbolic_data
arguments: label: D11
type_infered: string
name: extract_symbolic_data
arguments: label: D12
type_infered: string

tool: D39: Enter 10 float numbers:

tool: D19: %lf

tool: D11: You entered numbers:

tool: D12: %6.2lf

assistant: ```c
int main(void) {
  const int TEN = 10;
  double arr[TEN];
  int index;
  printf("Enter 10 float numbers:\n");
  for (index = 0; index < TEN; index++)
    scanf("%lf", &arr[index]);
  printf("You entered numbers:");
  for (index = 0; index < TEN; index++)
    printf("%6.2lf", arr[index]);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_little_endian>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  movl   $0x1,-0x18(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x1,%al
  jne    L16
  movl   $0x1,-0x14(%rbp)
L16:
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
bool is_little_endian(void) {
  bool ret = false;
  unsigned int test = 1;
  unsigned char *ptr = (unsigned char *)&test;
  if (*ptr == 1) {
    ret = true;
  }
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_little_endian>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool is_little_endian(void) {
  bool ret = false;
  unsigned int test = 1;
  unsigned char *ptr = (unsigned char *)&test;
  if (*ptr == 1) {
    ret = true;
  }
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<is_little_endian>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool is_little_endian(void) {
  bool ret = false;
  unsigned int test = 1;
  unsigned char *ptr = (unsigned char *)&test;
  if (*ptr == 1) {
    ret = true;
  }
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<is_little_endian>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool is_little_endian(void) {
  bool ret = false;
  unsigned int test = 1;
  unsigned char *ptr = (unsigned char *)&test;
  if (*ptr == 1) {
    ret = true;
  }
  return ret;
}
```
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    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L12
L10:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L26
L23:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D25(%rip),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L26:
  cmpl   $0x2,-0x14(%rbp)
  jle    L23
  addl   $0x1,-0x18(%rbp)
L12:
  cmpl   $0x6,-0x18(%rbp)
  jle    L10
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D25
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D29
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D28
type_guessed: string

tool: D25: dnannsum

tool: D29: # c::%s:len=%d


tool: D28: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dnannsum", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D8(%rip),%r15
  jmp    L0
L38:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L36
L0:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L4:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L4
  jmp    L38
L36:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D8
data_type: string
name: parse_data
arguments: label: D6
data_type: string
name: parse_data
arguments: label: D18
data_type: string

tool: D8: dnannsum

tool: D6: # c::%s:len=%d


tool: D18: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dnannsum", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D29(%rip),%r15
  push   %r14
  lea    D23(%rip),%r14
  push   %r13
  lea    D3(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L16:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L1:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L1
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L16
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D29
type_infered: string
name: unpack_data
arguments: data_tag: D23
type_infered: string
name: unpack_data
arguments: data_tag: D3
type_infered: string

tool: D29: dnannsum

tool: D23: # c::%s:len=%d


tool: D3: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dnannsum", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D18(%rip),%r15
  push   %r14
  lea    D5(%rip),%r14
  push   %r13
  lea    D16(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L14:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L9:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L9
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L14
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D18
datatype: string
name: disassemble_data
arguments: data_label: D5
datatype: string
name: disassemble_data
arguments: data_label: D16
datatype: string

tool: D18: dnannsum

tool: D5: # c::%s:len=%d


tool: D16: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dnannsum", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Translate the below assembly code into C programming language:

```asm
<valid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
bool valid(int letter, int new_pos) {
  bool ret = true;
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<valid>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool valid(int letter, int new_pos) {
  bool ret = true;
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<valid>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool valid(int letter, int new_pos) {
  bool ret = true;
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fnGenRandInput>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L2:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rdx,%rcx
  movslq %eax,%rdx
  imul   $0x68db8bad,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0xc,%edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  imul   $0x2710,%edx,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,(%rcx)
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L2
  nop
  nop
  leave
  ret
```

assistant: ```c
void fnGenRandInput(int X[], int n) {
  int i;
  srand(time(NULL));
  for (i = 0; i < n; i++) {
    X[i] = rand() % 10000;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<fnGenRandInput>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %ebp,%ebp
  jle    L12
  mov    %r12,%rbx
  lea    -0x1(%rbp),%eax
  lea    0x4(%r12,%rax,4),%rbp
L17:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x68db8bad,%rdx,%rdx
  sar    $0x2c,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x2710,%edx,%edx
  sub    %edx,%eax
  mov    %eax,(%rbx)
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L17
L12:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void fnGenRandInput(int X[], int n) {
  int i;
  srand(time(NULL));
  for (i = 0; i < n; i++) {
    X[i] = rand() % 10000;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fnGenRandInput>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %ebx,%ebx
  jle    L14
  lea    -0x1(%rbx),%eax
  lea    0x4(%rbp,%rax,4),%rbx
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x68db8bad,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x2c,%rdx
  sub    %ecx,%edx
  imul   $0x2710,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmp    %rbx,%rbp
  jne    L3
L14:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void fnGenRandInput(int X[], int n) {
  int i;
  srand(time(NULL));
  for (i = 0; i < n; i++) {
    X[i] = rand() % 10000;
  }
}
```