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

```asm
<arr_median>:
  endbr64
  push   %r13
  lea    0x0(,%rsi,8),%r13
  push   %r12
  mov    %rdi,%r12
  mov    %r13,%rdi
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <memcpy@plt>
  lea    -0x1(%rbx),%rdx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <quicksort@plt>
  mov    %rbx,%rax
  shr    %rax
  and    $0x1,%ebx
  jne    L16
  lea    0x0(,%rax,8),%rdx
  movsd  -0x8(%rbp,%rdx,1),%xmm0
  addsd  0x0(%rbp,%rax,8),%xmm0
  mulsd  D0(%rip),%xmm0
L5:
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <free@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L16:
  movsd  0x0(%rbp,%rax,8),%xmm0
  jmp    L5
```

tool_calls: name: retrieve_data
arguments: data_label: D0
data_type: byte[8]

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

assistant: ```c
double arr_median(double x[], size_t N) {
  double median;
  double *y = malloc(N * sizeof(double));
  memcpy(y, x, N * sizeof(double));
  quicksort(y, 0, N - 1);
  if (N % 2 == 0) {
    median = (y[N / 2] + y[N / 2 - 1]) / 2.0;
  } else {
    median = y[N / 2];
  }
  free(y);
  return median;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<arr_median>:
  endbr64
  push   %r13
  lea    0x0(,%rsi,8),%r13
  push   %r12
  mov    %rdi,%r12
  mov    %r13,%rdi
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <memcpy@plt>
  lea    -0x1(%rbx),%rdx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <quicksort@plt>
  mov    %rbx,%rax
  shr    %rax
  and    $0x1,%ebx
  jne    L2
  lea    0x0(,%rax,8),%rdx
  movsd  -0x8(%rbp,%rdx,1),%xmm0
  addsd  0x0(%rbp,%rax,8),%xmm0
  mulsd  D5(%rip),%xmm0
L11:
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <free@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L2:
  movsd  0x0(%rbp,%rax,8),%xmm0
  jmp    L11
```

tool_calls: name: read_data
arguments: data_label: D5
type_guessed: byte[8]

tool: {"D5": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

assistant: ```c
double arr_median(double x[], size_t N) {
  double median;
  double *y = malloc(N * sizeof(double));
  memcpy(y, x, N * sizeof(double));
  quicksort(y, 0, N - 1);
  if (N % 2 == 0) {
    median = (y[N / 2] + y[N / 2 - 1]) / 2.0;
  } else {
    median = y[N / 2];
  }
  free(y);
  return median;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<utf8_strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  cmpl   $0x0,-0x2c(%rbp)
  jns    L23
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x2c(%rbp)
  jmp    L23
L15:
  lea    -0x14(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <utf8_tounicode@plt>
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  add    %rax,-0x28(%rbp)
  mov    -0xc(%rbp),%eax
  sub    %eax,-0x2c(%rbp)
L23:
  cmpl   $0x0,-0x2c(%rbp)
  jne    L15
  mov    -0x10(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int utf8_strlen(const char *str, int bytelen) {
  int charlen = 0;
  if (bytelen < 0) {
    bytelen = strlen(str);
  }
  while (bytelen) {
    int c;
    int l = utf8_tounicode(str, &c);
    charlen++;
    str += l;
    bytelen -= l;
  }
  return charlen;
}
```
user: Translate the below assembly code into C programming language:

```asm
<utf8_strlen>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  js     L27
L38:
  test   %ebx,%ebx
  je     L12
  mov    $0x0,%r12d
  lea    0x4(%rsp),%r13
L0:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <utf8_tounicode@plt>
  add    $0x1,%r12d
  movslq %eax,%rdx
  add    %rdx,%rbp
  sub    %eax,%ebx
  jne    L0
L26:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  mov    %r12d,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L27:
  call   <strlen@plt>
  mov    %eax,%ebx
  jmp    L38
L12:
  mov    %ebx,%r12d
  jmp    L26
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int utf8_strlen(const char *str, int bytelen) {
  int charlen = 0;
  if (bytelen < 0) {
    bytelen = strlen(str);
  }
  while (bytelen) {
    int c;
    int l = utf8_tounicode(str, &c);
    charlen++;
    str += l;
    bytelen -= l;
  }
  return charlen;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<utf8_strlen>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  js     L35
L39:
  xor    %r12d,%r12d
  test   %ebx,%ebx
  je     L7
  lea    0x4(%rsp),%r13
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %rbp,%rdi
  mov    %r13,%rsi
  add    $0x1,%r12d
  call   <utf8_tounicode@plt>
  movslq %eax,%rdx
  add    %rdx,%rbp
  sub    %eax,%ebx
  jne    L11
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L35:
  call   <strlen@plt>
  mov    %eax,%ebx
  jmp    L39
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int utf8_strlen(const char *str, int bytelen) {
  int charlen = 0;
  if (bytelen < 0) {
    bytelen = strlen(str);
  }
  while (bytelen) {
    int c;
    int l = utf8_tounicode(str, &c);
    charlen++;
    str += l;
    bytelen -= l;
  }
  return charlen;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<utf8_strlen>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  js     L30
L12:
  xor    %r12d,%r12d
  test   %ebx,%ebx
  je     L38
  lea    0x4(%rsp),%r13
  nopl   0x0(%rax,%rax,1)
L19:
  mov    %rbp,%rdi
  mov    %r13,%rsi
  add    $0x1,%r12d
  call   <utf8_tounicode@plt>
  movslq %eax,%rdx
  add    %rdx,%rbp
  sub    %eax,%ebx
  jne    L19
L38:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L30:
  call   <strlen@plt>
  mov    %eax,%ebx
  jmp    L12
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int utf8_strlen(const char *str, int bytelen) {
  int charlen = 0;
  if (bytelen < 0) {
    bytelen = strlen(str);
  }
  while (bytelen) {
    int c;
    int l = utf8_tounicode(str, &c);
    charlen++;
    str += l;
    bytelen -= l;
  }
  return charlen;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x18(%rbp)
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L57
L25:
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x1c(%rbp)
  movl   $0x1,-0x20(%rbp)
  jmp    L40
L8:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  jle    L38
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x1c(%rbp)
L38:
  addl   $0x1,-0x20(%rbp)
L40:
  mov    -0x24(%rbp),%eax
  cmp    %eax,-0x20(%rbp)
  jl     L8
  mov    -0x24(%rbp),%eax
  cmp    $0x1,%eax
  jne    L29
  movq   $0x0,-0x18(%rbp)
  jmp    L85
L29:
  movl   $0x0,-0x20(%rbp)
  jmp    L4
L18:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%eax
  sub    %edx,%eax
  mov    %eax,%edi
  call   <abs@plt>
  add    %rax,-0x18(%rbp)
  addl   $0x1,-0x20(%rbp)
L4:
  mov    -0x24(%rbp),%eax
  cmp    %eax,-0x20(%rbp)
  jl     L18
L85:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movq   $0x0,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L57:
  mov    -0x28(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x28(%rbp)
  test   %eax,%eax
  jne    L25
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D35
type_guessed: string
name: retrieve_data
arguments: data_label: D23
type_guessed: string

tool: D35: %d

tool: D23: %lld


assistant: ```c
int main(void) {
  int t, n, *w, i, min_no;
  unsigned long long no_of_op = 0;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    w = malloc(n * sizeof(int));
    scanf("%d", &w[0]);
    min_no = w[0];
    for (i = 1; i < n; i++) {
      scanf("%d", &w[i]);
      if (min_no > w[i]) {
        min_no = w[i];
      }
    }
    if (n == 1) {
      no_of_op = 0;
    } else {
      for (i = 0; i < n; i++) {
        no_of_op += abs(min_no - w[i]);
      }
    }
    printf("%lld\n", no_of_op);
    no_of_op = 0;
    free(w);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L17
  lea    0x4(%rsp),%r15
  lea    D3(%rip),%r14
  jmp    L37
L70:
  mov    $0x0,%r12d
L31:
  mov    %r12,%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L17
L37:
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    %rax,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    0x0(%r13),%ebp
  mov    0x4(%rsp),%eax
  cmp    $0x1,%eax
  jle    L21
  lea    0x4(%r13),%rbx
  mov    $0x1,%r12d
L6:
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    (%rbx),%eax
  cmp    %eax,%ebp
  cmovg  %eax,%ebp
  add    $0x1,%r12d
  mov    0x4(%rsp),%eax
  add    $0x4,%rbx
  cmp    %r12d,%eax
  jg     L6
L21:
  mov    $0x0,%r12d
  cmp    $0x1,%eax
  je     L31
  cmpl   $0x0,0x4(%rsp)
  jle    L70
  mov    $0x0,%ebx
L88:
  mov    %ebp,%edi
  sub    0x0(%r13,%rbx,4),%edi
  call   <abs@plt>
  add    %rax,%r12
  add    $0x1,%rbx
  cmp    %ebx,0x4(%rsp)
  jg     L88
  jmp    L31
L17:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type_guessed: string
name: disassemble_data
arguments: data_label: D1
type_guessed: string

tool: D3: %d

tool: D1: %lld


assistant: ```c
int main(void) {
  int t, n, *w, i, min_no;
  unsigned long long no_of_op = 0;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    w = malloc(n * sizeof(int));
    scanf("%d", &w[0]);
    min_no = w[0];
    for (i = 1; i < n; i++) {
      scanf("%d", &w[i]);
      if (min_no > w[i]) {
        min_no = w[i];
      }
    }
    if (n == 1) {
      no_of_op = 0;
    } else {
      for (i = 0; i < n; i++) {
        no_of_op += abs(min_no - w[i]);
      }
    }
    printf("%lld\n", no_of_op);
    no_of_op = 0;
    free(w);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    D7(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L1
  lea    0x4(%rsp),%r15
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %r12,%rdi
  mov    %rax,%rsi
  mov    %rax,%r13
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    0x0(%r13),%ebp
  cmp    $0x1,%eax
  jle    L42
  lea    0x4(%r13),%rbx
  mov    $0x1,%r14d
L25:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    (%rbx),%eax
  cmp    %eax,%ebp
  cmovg  %eax,%ebp
  mov    0x4(%rsp),%eax
  add    $0x1,%r14d
  add    $0x4,%rbx
  cmp    %r14d,%eax
  jg     L25
L42:
  xor    %r14d,%r14d
  cmp    $0x1,%eax
  jne    L10
L34:
  mov    %r14,%rsi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jne    L9
L1:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L56
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L10:
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L34
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %ebp,%edi
  sub    0x0(%r13,%rbx,4),%edi
  add    $0x1,%rbx
  call   <abs@plt>
  add    %rax,%r14
  cmp    %ebx,0x4(%rsp)
  jg     L17
  jmp    L34
L56:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type_infered: string
name: extract_symbolic_data
arguments: label: D13
type_infered: string

tool: D7: %d

tool: D13: %lld


assistant: ```c
int main(void) {
  int t, n, *w, i, min_no;
  unsigned long long no_of_op = 0;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    w = malloc(n * sizeof(int));
    scanf("%d", &w[0]);
    min_no = w[0];
    for (i = 1; i < n; i++) {
      scanf("%d", &w[i]);
      if (min_no > w[i]) {
        min_no = w[i];
      }
    }
    if (n == 1) {
      no_of_op = 0;
    } else {
      for (i = 0; i < n; i++) {
        no_of_op += abs(min_no - w[i]);
      }
    }
    printf("%lld\n", no_of_op);
    no_of_op = 0;
    free(w);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  lea    D16(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    0x4(%rsp),%r15
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L12
  nopw   0x0(%rax,%rax,1)
L61:
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %r12,%rdi
  mov    %rax,%rsi
  mov    %rax,%r13
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    0x0(%r13),%ebp
  cmp    $0x1,%eax
  jle    L24
  lea    0x4(%r13),%rbx
  mov    $0x1,%r14d
L9:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    (%rbx),%eax
  cmp    %eax,%ebp
  cmovg  %eax,%ebp
  mov    0x4(%rsp),%eax
  add    $0x1,%r14d
  add    $0x4,%rbx
  cmp    %r14d,%eax
  jg     L9
L24:
  xor    %r14d,%r14d
  cmp    $0x1,%eax
  jne    L2
L78:
  mov    %r14,%rsi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jne    L61
L12:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L2:
  test   %eax,%eax
  jle    L78
  xor    %ebx,%ebx
  xchg   %ax,%ax
L14:
  mov    %ebp,%edi
  sub    0x0(%r13,%rbx,4),%edi
  add    $0x1,%rbx
  call   <abs@plt>
  add    %rax,%r14
  cmp    %ebx,0x4(%rsp)
  jg     L14
  jmp    L78
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D16
type_infered: string
name: parse_data
arguments: data_tag: D10
type_infered: string

tool: D16: %d

tool: D10: %lld


assistant: ```c
int main(void) {
  int t, n, *w, i, min_no;
  unsigned long long no_of_op = 0;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    w = malloc(n * sizeof(int));
    scanf("%d", &w[0]);
    min_no = w[0];
    for (i = 1; i < n; i++) {
      scanf("%d", &w[i]);
      if (min_no > w[i]) {
        min_no = w[i];
      }
    }
    if (n == 1) {
      no_of_op = 0;
    } else {
      for (i = 0; i < n; i++) {
        no_of_op += abs(min_no - w[i]);
      }
    }
    printf("%lld\n", no_of_op);
    no_of_op = 0;
    free(w);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<config_setting_get_int64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <__config_setting_get_int64@plt>
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
long long config_setting_get_int64(const struct config_setting_t *setting) {
  long long value = 0;
  __config_setting_get_int64(setting, &value);
  return (value);
}
```
user: What is the c source code of the assembly code below:

```asm
<config_setting_get_int64>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rsi
  call   <__config_setting_get_int64@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long config_setting_get_int64(const struct config_setting_t *setting) {
  long long value = 0;
  __config_setting_get_int64(setting, &value);
  return (value);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<config_setting_get_int64>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <__config_setting_get_int64@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long config_setting_get_int64(const struct config_setting_t *setting) {
  long long value = 0;
  __config_setting_get_int64(setting, &value);
  return (value);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<config_setting_get_int64>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <__config_setting_get_int64@plt>
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long config_setting_get_int64(const struct config_setting_t *setting) {
  long long value = 0;
  __config_setting_get_int64(setting, &value);
  return (value);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<slaFk54z>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  movsd  %xmm0,-0x48(%rbp)
  movsd  %xmm1,-0x50(%rbp)
  movsd  %xmm2,-0x58(%rbp)
  mov    %rdi,-0x60(%rbp)
  mov    %rsi,-0x68(%rbp)
  mov    %rdx,-0x70(%rbp)
  mov    %rcx,-0x78(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  -0x48(%rbp),%xmm0
  movsd  %xmm0,-0x40(%rbp)
  movsd  -0x50(%rbp),%xmm0
  movsd  %xmm0,-0x38(%rbp)
  movsd  -0x58(%rbp),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  lea    -0x18(%rbp),%r9
  lea    -0x20(%rbp),%r8
  lea    -0x28(%rbp),%rcx
  lea    -0x30(%rbp),%rdx
  lea    -0x38(%rbp),%rsi
  lea    -0x40(%rbp),%rax
  sub    $0x8,%rsp
  lea    -0x10(%rbp),%rdi
  push   %rdi
  mov    %rax,%rdi
  call   <sla_fk54z_@plt>
  add    $0x10,%rsp
  movsd  -0x28(%rbp),%xmm0
  mov    -0x60(%rbp),%rax
  movsd  %xmm0,(%rax)
  movsd  -0x20(%rbp),%xmm0
  mov    -0x68(%rbp),%rax
  movsd  %xmm0,(%rax)
  movsd  -0x18(%rbp),%xmm0
  mov    -0x70(%rbp),%rax
  movsd  %xmm0,(%rax)
  movsd  -0x10(%rbp),%xmm0
  mov    -0x78(%rbp),%rax
  movsd  %xmm0,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void slaFk54z(double r2000, double d2000, double bepoch, double *r1950, double *d1950, double *dr1950, double *dd1950) {
  double R2000;
  double D2000;
  double BEPOCH;
  double R1950;
  double D1950;
  double DR1950;
  double DD1950;
  R2000 = r2000;
  D2000 = d2000;
  BEPOCH = bepoch;
  sla_fk54z_(&R2000, &D2000, &BEPOCH, &R1950, &D1950, &DR1950, &DD1950);
  *r1950 = R1950;
  *d1950 = D1950;
  *dr1950 = DR1950;
  *dd1950 = DD1950;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<slaFk54z>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    %rcx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  movsd  %xmm1,0x8(%rsp)
  movsd  %xmm2,0x10(%rsp)
  lea    0x18(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  mov    %rsp,%rdi
  sub    $0x8,%rsp
  lea    0x38(%rsp),%rax
  push   %rax
  lea    0x38(%rsp),%r9
  lea    0x30(%rsp),%r8
  call   <sla_fk54z_@plt>
  movsd  0x28(%rsp),%xmm0
  movsd  %xmm0,0x0(%r13)
  movsd  0x30(%rsp),%xmm0
  movsd  %xmm0,(%r12)
  movsd  0x38(%rsp),%xmm0
  movsd  %xmm0,0x0(%rbp)
  movsd  0x40(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  add    $0x10,%rsp
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void slaFk54z(double r2000, double d2000, double bepoch, double *r1950, double *d1950, double *dr1950, double *dd1950) {
  double R2000;
  double D2000;
  double BEPOCH;
  double R1950;
  double D1950;
  double DR1950;
  double DD1950;
  R2000 = r2000;
  D2000 = d2000;
  BEPOCH = bepoch;
  sla_fk54z_(&R2000, &D2000, &BEPOCH, &R1950, &D1950, &DR1950, &DD1950);
  *r1950 = R1950;
  *d1950 = D1950;
  *dr1950 = DR1950;
  *dd1950 = DD1950;
}
```