ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<phi>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    $0x2,%ebx
  jmp    L43
  cs nopw 0x0(%rax,%rax,1)
L17:
  add    $0x1,%rbx
L43:
  mov    %r12,%rdi
  call   <sqrt@plt>
  cmp    %rbx,%rax
  jl     L14
  mov    %r12,%rax
  cqto
  idiv   %rbx
  test   %rdx,%rdx
  jne    L17
  xchg   %ax,%ax
L56:
  mov    %r12,%rax
  cqto
  idiv   %rbx
  cqto
  mov    %rax,%r12
  idiv   %rbx
  test   %rdx,%rdx
  je     L56
  mov    %r13,%rax
  cqto
  idiv   %rbx
  sub    %rax,%r13
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L14:
  cmp    $0x1,%r12
  je     L15
  mov    %r13,%rax
  cqto
  idiv   %r12
  sub    %rax,%r13
L15:
  mov    %r13,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
long long int phi(long long int n) {
  long long int res = n, i, j, k;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      while (n % i == 0)
        n = n / i;
      res = res - res / i;
    }
  }
  if (n != 1) {
    res = res - res / n;
  }
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<phi>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    $0x2,%ebx
  jmp    L3
  cs nopw 0x0(%rax,%rax,1)
L17:
  add    $0x1,%rbx
L3:
  mov    %r12,%rdi
  call   <sqrt@plt>
  cmp    %rbx,%rax
  jl     L38
  mov    %r12,%rax
  cqto
  idiv   %rbx
  test   %rdx,%rdx
  jne    L17
  xchg   %ax,%ax
L48:
  mov    %r12,%rax
  cqto
  idiv   %rbx
  cqto
  mov    %rax,%r12
  idiv   %rbx
  test   %rdx,%rdx
  je     L48
  mov    %r13,%rax
  cqto
  idiv   %rbx
  sub    %rax,%r13
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L38:
  cmp    $0x1,%r12
  je     L13
  mov    %r13,%rax
  cqto
  idiv   %r12
  sub    %rax,%r13
L13:
  mov    %r13,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
long long int phi(long long int n) {
  long long int res = n, i, j, k;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      while (n % i == 0)
        n = n / i;
      res = res - res / i;
    }
  }
  if (n != 1) {
    res = res - res / n;
  }
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<modulo_signed_Good2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  shl    $0x2,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  js     L7
  mov    -0x14(%rbp),%eax
  cltq
  movb   $0x7b,-0xd(%rbp,%rax,1)
L7:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
void modulo_signed_Good2(int i) {
  char arr[5];
  int j = i % 5;
  if (j >= 0) {
    arr[j] = 123;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<modulo_signed_Good2>:
  endbr64
  ret
```

assistant: ```c
void modulo_signed_Good2(int i) {
  char arr[5];
  int j = i % 5;
  if (j >= 0) {
    arr[j] = 123;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<modulo_signed_Good2>:
  endbr64
  ret
```

assistant: ```c
void modulo_signed_Good2(int i) {
  char arr[5];
  int j = i % 5;
  if (j >= 0) {
    arr[j] = 123;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<modulo_signed_Good2>:
  endbr64
  ret
```

assistant: ```c
void modulo_signed_Good2(int i) {
  char arr[5];
  int j = i % 5;
  if (j >= 0) {
    arr[j] = 123;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<wczyt_vat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L48
L31:
  movl   $0x1,-0x10(%rbp)
L59:
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  je     L73
  mov    -0x10(%rbp),%eax
  cmp    $0x5,%eax
  je     L73
  mov    -0x10(%rbp),%eax
  cmp    $0x8,%eax
  je     L73
  mov    -0x10(%rbp),%eax
  cmp    $0x17,%eax
  je     L73
  jmp    L59
L73:
  mov    -0x10(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  lea    0x14(%rax),%rdx
  pxor   %xmm0,%xmm0
  cvtsi2ss %ecx,%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  movss  %xmm0,(%rdx,%rax,4)
  addl   $0x1,-0xc(%rbp)
L48:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L31
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D18
type_guessed: string
name: read_data
arguments: data_label: D12
type_guessed: string

tool: D18: Podaj stawke VAT towaru %d (dopuszczalne wartosci: 0, 5, 8, 23): 

tool: D12: %d

assistant: ```c
void wczyt_vat(float matrix[][5], int max_y) {
  int i;
  int buf = 1;
  for (i = 0; i < max_y; i++) {
    buf = 1;
    do {
      printf("Podaj stawke VAT towaru %d (dopuszczalne wartosci: 0, 5, 8, 23): ", i + 1);
      scanf("%d", &buf);
      if (buf == 0 || buf == 5 || buf == 8 || buf == 23) {
        break;
      }
    } while (1 == 1);
    matrix[1][i] = (float)buf;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<wczyt_vat>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L19
  lea    0x14(%rdi),%r15
  movl   $0x1,0x8(%rsp)
  lea    D2(%rip),%r13
  lea    0x14(%rsp),%r12
  lea    D8(%rip),%rbp
  mov    $0x800121,%r14d
L41:
  movl   $0x1,0x14(%rsp)
  mov    0x8(%rsp),%ebx
L2:
  mov    %ebx,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  cmp    $0x17,%eax
  ja     L2
  bt     %rax,%r14
  jae    L2
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,(%r15)
  addl   $0x1,0x8(%rsp)
  add    $0x4,%r15
  cmp    0xc(%rsp),%ebx
  jl     L41
L19:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type: string
name: retrieve_data
arguments: data_label: D8
type: string

tool: D2: Podaj stawke VAT towaru %d (dopuszczalne wartosci: 0, 5, 8, 23): 

tool: D8: %d

assistant: ```c
void wczyt_vat(float matrix[][5], int max_y) {
  int i;
  int buf = 1;
  for (i = 0; i < max_y; i++) {
    buf = 1;
    do {
      printf("Podaj stawke VAT towaru %d (dopuszczalne wartosci: 0, 5, 8, 23): ", i + 1);
      scanf("%d", &buf);
      if (buf == 0 || buf == 5 || buf == 8 || buf == 23) {
        break;
      }
    } while (1 == 1);
    matrix[1][i] = (float)buf;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<wczyt_vat>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L27
  lea    0x1(%rsi),%eax
  mov    %rdi,%r15
  mov    $0x1,%r14d
  mov    %rax,0x8(%rsp)
  lea    D25(%rip),%r13
  lea    0x14(%rsp),%r12
  lea    D7(%rip),%rbp
  xchg   %ax,%ax
L72:
  movl   $0x1,0x14(%rsp)
  mov    %r14d,%ebx
  jmp    L24
  nopl   (%rax)
L62:
  cmp    $0x17,%eax
  je     L18
L24:
  mov    %ebx,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  je     L76
  cmp    $0x5,%eax
  je     L76
  cmp    $0x8,%eax
  jne    L62
  movss  D24(%rip),%xmm0
L65:
  movss  %xmm0,0x10(%r15,%r14,4)
  add    $0x1,%r14
  cmp    0x8(%rsp),%r14
  jne    L72
L27:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L76:
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  jmp    L65
  nopw   0x0(%rax,%rax,1)
L18:
  movss  D35(%rip),%xmm0
  jmp    L65
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D25
data_type: string
name: disassemble_data
arguments: tag: D7
data_type: string
name: disassemble_data
arguments: tag: D24
data_type: byte[4]
name: disassemble_data
arguments: tag: D35
data_type: byte[4]

tool: D25: Podaj stawke VAT towaru %d (dopuszczalne wartosci: 0, 5, 8, 23): 

tool: D7: %d

tool: {"D24": [0x00,0x00,0x00,0x41]}

tool: {"D35": [0x00,0x00,0xb8,0x41]}

assistant: ```c
void wczyt_vat(float matrix[][5], int max_y) {
  int i;
  int buf = 1;
  for (i = 0; i < max_y; i++) {
    buf = 1;
    do {
      printf("Podaj stawke VAT towaru %d (dopuszczalne wartosci: 0, 5, 8, 23): ", i + 1);
      scanf("%d", &buf);
      if (buf == 0 || buf == 5 || buf == 8 || buf == 23) {
        break;
      }
    } while (1 == 1);
    matrix[1][i] = (float)buf;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<wczyt_vat>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L43
  lea    0x1(%rsi),%eax
  mov    %rdi,%r15
  mov    $0x1,%r14d
  mov    %rax,0x8(%rsp)
  lea    D18(%rip),%r13
  lea    0x14(%rsp),%r12
  lea    D3(%rip),%rbp
  xchg   %ax,%ax
L80:
  movl   $0x1,0x14(%rsp)
  mov    %r14d,%ebx
  jmp    L78
  nopl   (%rax)
L13:
  cmp    $0x17,%eax
  je     L98
L78:
  mov    %ebx,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  je     L74
  cmp    $0x5,%eax
  je     L74
  cmp    $0x8,%eax
  jne    L13
  movss  D4(%rip),%xmm0
L51:
  movss  %xmm0,0x10(%r15,%r14,4)
  add    $0x1,%r14
  cmp    0x8(%rsp),%r14
  jne    L80
L43:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L74:
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  jmp    L51
  nopw   0x0(%rax,%rax,1)
L98:
  movss  D25(%rip),%xmm0
  jmp    L51
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D18
data_type: string
name: extract_symbolic_data
arguments: data_label: D3
data_type: string
name: extract_symbolic_data
arguments: data_label: D4
data_type: byte[4]
name: extract_symbolic_data
arguments: data_label: D25
data_type: byte[4]

tool: D18: Podaj stawke VAT towaru %d (dopuszczalne wartosci: 0, 5, 8, 23): 

tool: D3: %d

tool: {"D4": [0x00,0x00,0x00,0x41]}

tool: {"D25": [0x00,0x00,0xb8,0x41]}

assistant: ```c
void wczyt_vat(float matrix[][5], int max_y) {
  int i;
  int buf = 1;
  for (i = 0; i < max_y; i++) {
    buf = 1;
    do {
      printf("Podaj stawke VAT towaru %d (dopuszczalne wartosci: 0, 5, 8, 23): ", i + 1);
      scanf("%d", &buf);
      if (buf == 0 || buf == 5 || buf == 8 || buf == 23) {
        break;
      }
    } while (1 == 1);
    matrix[1][i] = (float)buf;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x2c(%rbp)
  jmp    L62
L20:
  lea    -0x20(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x2c(%rbp)
L62:
  cmpl   $0x4,-0x2c(%rbp)
  jle    L20
L49:
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x2c(%rbp)
  jmp    L31
L70:
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%edx
  mov    -0x30(%rbp),%eax
  cmp    %eax,%edx
  jne    L89
  lea    -0x20(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L89:
  addl   $0x1,-0x2c(%rbp)
L31:
  cmpl   $0x4,-0x2c(%rbp)
  jle    L70
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x31(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movzbl -0x31(%rbp),%eax
  cmp    $0x6e,%al
  je     L85
  movzbl -0x31(%rbp),%eax
  cmp    $0x4e,%al
  je     L85
  jmp    L49
L85:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L77
  call   <__stack_chk_fail@plt>
L77:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D54
type_infered: string
name: disassemble_data
arguments: tag: D13
type_infered: string
name: disassemble_data
arguments: tag: D28
type_infered: string
name: disassemble_data
arguments: tag: D26
type_infered: string
name: disassemble_data
arguments: tag: D14
type_infered: string

tool: D54: enter elements

tool: D13: %d

tool: D28: enter element which position you want to find:

tool: D26: will u want to continue(y/n):

tool: D14: %c

assistant: ```c
main() {
  int arr[6], i, *p, k;
  char c;
  printf("enter elements");
  for (i = 0; i < 5; i++)
    scanf("%d", &arr[i]);
  while (1) {
    printf("enter element which position you want to find:");
    scanf("%d", &k);
    for (i = 0; i < 5; i++) {
      if (arr[i] == k) {
        p = &arr[i];
        printf("%d", p);
      }
    }
    printf("will u want to continue(y/n):");
    scanf("%c", &c);
    if (c == 'n' || c == 'N')
      break;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%r13
  lea    0x24(%rsp),%rbp
  mov    %r13,%rbx
  lea    D53(%rip),%r12
L58:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L58
  lea    D2(%rip),%r15
  lea    D53(%rip),%r12
  lea    D45(%rip),%r14
  jmp    L5
L52:
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  je     L22
L37:
  mov    0xc(%rsp),%eax
  cmp    %eax,(%rbx)
  jne    L52
  mov    %rbx,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L52
L22:
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xb(%rsp),%rsi
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movzbl 0xb(%rsp),%eax
  and    $0xffffffdf,%eax
  cmp    $0x4e,%al
  je     L57
L5:
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %r13,%rbx
  jmp    L37
L57:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D0
data_type: string
name: unpack_data
arguments: tag: D53
data_type: string
name: unpack_data
arguments: tag: D2
data_type: string
name: unpack_data
arguments: tag: D45
data_type: string
name: unpack_data
arguments: tag: D27
data_type: string

tool: D0: enter elements

tool: D53: %d

tool: D2: enter element which position you want to find:

tool: D45: will u want to continue(y/n):

tool: D27: %c

assistant: ```c
main() {
  int arr[6], i, *p, k;
  char c;
  printf("enter elements");
  for (i = 0; i < 5; i++)
    scanf("%d", &arr[i]);
  while (1) {
    printf("enter element which position you want to find:");
    scanf("%d", &k);
    for (i = 0; i < 5; i++) {
      if (arr[i] == k) {
        p = &arr[i];
        printf("%d", p);
      }
    }
    printf("will u want to continue(y/n):");
    scanf("%c", &c);
    if (c == 'n' || c == 'N')
      break;
  }
  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   %r15
  lea    D41(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x34(%rsp),%rbx
  call   <printf@plt>
  lea    0x20(%rsp),%rax
  mov    %rax,(%rsp)
  mov    %rax,%r12
L25:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%r12
  call   <scanf@plt>
  cmp    %rbx,%r12
  jne    L25
  lea    0x1b(%rsp),%rax
  lea    D26(%rip),%r15
  mov    %rax,0x8(%rsp)
  lea    0x1c(%rsp),%r14
  lea    D10(%rip),%r13
  nopl   0x0(%rax)
L2:
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    (%rsp),%r12
L22:
  mov    0x1c(%rsp),%eax
  cmp    %eax,(%r12)
  je     L15
L56:
  add    $0x4,%r12
  cmp    %rbx,%r12
  jne    L22
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rsi
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movzbl 0x1b(%rsp),%eax
  and    $0xffffffdf,%eax
  cmp    $0x4e,%al
  jne    L2
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L56
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D41
type: string
name: retrieve_data
arguments: data_label: D8
type: string
name: retrieve_data
arguments: data_label: D26
type: string
name: retrieve_data
arguments: data_label: D10
type: string
name: retrieve_data
arguments: data_label: D29
type: string

tool: D41: enter elements

tool: D8: %d

tool: D26: enter element which position you want to find:

tool: D10: will u want to continue(y/n):

tool: D29: %c

assistant: ```c
main() {
  int arr[6], i, *p, k;
  char c;
  printf("enter elements");
  for (i = 0; i < 5; i++)
    scanf("%d", &arr[i]);
  while (1) {
    printf("enter element which position you want to find:");
    scanf("%d", &k);
    for (i = 0; i < 5; i++) {
      if (arr[i] == k) {
        p = &arr[i];
        printf("%d", p);
      }
    }
    printf("will u want to continue(y/n):");
    scanf("%c", &c);
    if (c == 'n' || c == 'N')
      break;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<converse_26_to_10>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L17
L16:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  lea    -0x40(%rax),%ebx
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  sub    %rdx,%rax
  sub    $0x1,%rax
  mov    %rax,%rsi
  mov    $0x1a,%edi
  call   <pow@plt>
  movsbl %al,%eax
  imul   %ebx,%eax
  add    %eax,-0x18(%rbp)
  addl   $0x1,-0x14(%rbp)
L17:
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,-0x30(%rbp)
  ja     L16
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
type: string

tool: D3: %d

assistant: ```c
void converse_26_to_10(char *temp, size_t length) {
  int sum = 0;
  for (int i = 0; i < length; i++) {
    sum += (temp[i] - 'A' + 1) * pow(26, length - i - 1);
  }
  printf("%d", sum);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<converse_26_to_10>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L23
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    $0x1,%ebx
  mov    $0x0,%r12d
L17:
  movsbl -0x1(%r13,%rbx,1),%r14d
  sub    $0x40,%r14d
  mov    %rbp,%rsi
  sub    %rbx,%rsi
  mov    $0x1a,%edi
  call   <pow@plt>
  movsbl %al,%eax
  imul   %r14d,%eax
  add    %eax,%r12d
  mov    %rbx,%rax
  add    $0x1,%rbx
  cmp    %rbp,%rax
  jne    L17
L0:
  mov    %r12d,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  mov    $0x0,%r12d
  jmp    L0
```

tool_calls: name: read_data
arguments: data_tag: D0
datatype: string

tool: D0: %d

assistant: ```c
void converse_26_to_10(char *temp, size_t length) {
  int sum = 0;
  for (int i = 0; i < length; i++) {
    sum += (temp[i] - 'A' + 1) * pow(26, length - i - 1);
  }
  printf("%d", sum);
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<converse_26_to_10>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L10
  lea    -0x1(%rsi),%rbx
  mov    %rdi,%r13
  xor    %r12d,%r12d
  nopl   (%rax)
L16:
  movsbl 0x0(%r13),%ebp
  mov    %rbx,%rsi
  mov    $0x1a,%edi
  add    $0x1,%r13
  call   <pow@plt>
  sub    $0x40,%ebp
  movsbl %al,%eax
  imul   %ebp,%eax
  add    %eax,%r12d
  sub    $0x1,%rbx
  jae    L16
L15:
  add    $0x8,%rsp
  mov    %r12d,%esi
  lea    D3(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <printf@plt>
  nopl   0x0(%rax)
L10:
  xor    %r12d,%r12d
  jmp    L15
```

tool_calls: name: parse_data
arguments: tag: D3
data_type: string

tool: D3: %d

assistant: ```c
void converse_26_to_10(char *temp, size_t length) {
  int sum = 0;
  for (int i = 0; i < length; i++) {
    sum += (temp[i] - 'A' + 1) * pow(26, length - i - 1);
  }
  printf("%d", sum);
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<converse_26_to_10>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L13
  lea    -0x1(%rsi),%rbx
  mov    %rdi,%r13
  xor    %r12d,%r12d
  nopl   (%rax)
L1:
  movsbl 0x0(%r13),%ebp
  mov    %rbx,%rsi
  mov    $0x1a,%edi
  add    $0x1,%r13
  call   <pow@plt>
  sub    $0x40,%ebp
  movsbl %al,%eax
  imul   %ebp,%eax
  add    %eax,%r12d
  sub    $0x1,%rbx
  jae    L1
L25:
  add    $0x8,%rsp
  mov    %r12d,%esi
  lea    D0(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <printf@plt>
  nopl   0x0(%rax)
L13:
  xor    %r12d,%r12d
  jmp    L25
```

tool_calls: name: parse_data
arguments: tag: D0
type: string

tool: D0: %d

assistant: ```c
void converse_26_to_10(char *temp, size_t length) {
  int sum = 0;
  for (int i = 0; i < length; i++) {
    sum += (temp[i] - 'A' + 1) * pow(26, length - i - 1);
  }
  printf("%d", sum);
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<imprimir>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L15:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <scrn_print@plt>
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  cltd
  shr    $0x1e,%edx
  add    %edx,%eax
  and    $0x3,%eax
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
L2:
  cmpl   $0x3,-0x4(%rbp)
  jle    L15
  nop
  nop
  leave
  ret
```

assistant: ```c
void imprimir(char *buf[], int row) {
  int i;
  for (i = 0; i < 4;) {
    scrn_print(row, 0, buf[i]);
    i = (i + 1) % 4;
  }
}
```