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

```asm
<_write_bytes>:
  endbr64
  mov    $0x0,%eax
  cmp    $0xffffffff,%edi
  je     L3
  sub    $0x8,%rsp
  call   <write@plt>
  add    $0x8,%rsp
  ret
L3:
  ret
```

assistant: ```c
int _write_bytes(int fd, char *buf, size_t len) {
  if (-1 == fd)
    return 0;
  return write(fd, buf, len);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_write_bytes>:
  endbr64
  cmp    $0xffffffff,%edi
  je     L5
  jmp    1050 <write@plt>
  xchg   %ax,%ax
L5:
  xor    %eax,%eax
  ret
```

assistant: ```c
int _write_bytes(int fd, char *buf, size_t len) {
  if (-1 == fd)
    return 0;
  return write(fd, buf, len);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<_write_bytes>:
  endbr64
  cmp    $0xffffffff,%edi
  je     L0
  jmp    1050 <write@plt>
  xchg   %ax,%ax
L0:
  xor    %eax,%eax
  ret
```

assistant: ```c
int _write_bytes(int fd, char *buf, size_t len) {
  if (-1 == fd)
    return 0;
  return write(fd, buf, len);
}
```
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    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x14(%rbp),%rsi
  lea    -0x18(%rbp),%rcx
  lea    -0x1c(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %eax,%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  lea    0x3(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x2,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jge    L23
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L29
L23:
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L29:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D20
datatype: string
name: parse_data
arguments: label: D27
datatype: string
name: parse_data
arguments: label: D43
datatype: string
name: parse_data
arguments: label: D2
datatype: string
name: parse_data
arguments: label: D57
datatype: string
name: parse_data
arguments: label: D24
datatype: string
name: parse_data
arguments: label: D0
datatype: string
name: parse_data
arguments: label: D50
datatype: string

tool: D20: Qual foi sua media no ano?

tool: D27: Digite a media de sua escola.


tool: D43: %d

tool: D2: Digite as notas dos 4 bimestres.


tool: D57: %d%d%d%d

tool: D24: Sua media foi %d


tool: D0: Voce repetiu de ano.

tool: D50: Voce passou de ano!

assistant: ```c
int main() {
  int n1, n2, n3, n4, media, mediaesc;
  printf("Qual foi sua media no ano?\n");
  printf("Digite a media de sua escola.\n\n");
  scanf("%d", &mediaesc);
  printf("Digite as notas dos 4 bimestres.\n\n");
  scanf("%d"
        "%d"
        "%d"
        "%d",
        &n1, &n2, &n3, &n4);
  media = (n1 + n2 + n3 + n4) / 4;
  printf("Sua media foi %d\n", media);
  if (media < mediaesc) {
    printf("Voce repetiu de ano.\n");
  } else {
    printf("Voce passou de ano!\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D58(%rip),%rdi
  call   <puts@plt>
  lea    D65(%rip),%rdi
  call   <puts@plt>
  lea    0x14(%rsp),%rsi
  lea    D70(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D19(%rip),%rdi
  call   <puts@plt>
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  lea    0x4(%rsp),%rsi
  lea    0x10(%rsp),%r8
  lea    D59(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  add    0x4(%rsp),%eax
  add    0xc(%rsp),%eax
  add    0x10(%rsp),%eax
  lea    0x3(%rax),%ebx
  cmovns %eax,%ebx
  sar    $0x2,%ebx
  mov    %ebx,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %ebx,0x14(%rsp)
  jle    L7
  lea    D7(%rip),%rdi
  call   <puts@plt>
L19:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L7:
  lea    D74(%rip),%rdi
  call   <puts@plt>
  jmp    L19
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D58
type: string
name: disassemble_data
arguments: data_tag: D65
type: string
name: disassemble_data
arguments: data_tag: D70
type: string
name: disassemble_data
arguments: data_tag: D19
type: string
name: disassemble_data
arguments: data_tag: D59
type: string
name: disassemble_data
arguments: data_tag: D6
type: string
name: disassemble_data
arguments: data_tag: D7
type: string
name: disassemble_data
arguments: data_tag: D74
type: string

tool: D58: Qual foi sua media no ano?

tool: D65: Digite a media de sua escola.


tool: D70: %d

tool: D19: Digite as notas dos 4 bimestres.


tool: D59: %d%d%d%d

tool: D6: Sua media foi %d


tool: D7: Voce repetiu de ano.

tool: D74: Voce passou de ano!

assistant: ```c
int main() {
  int n1, n2, n3, n4, media, mediaesc;
  printf("Qual foi sua media no ano?\n");
  printf("Digite a media de sua escola.\n\n");
  scanf("%d", &mediaesc);
  printf("Digite as notas dos 4 bimestres.\n\n");
  scanf("%d"
        "%d"
        "%d"
        "%d",
        &n1, &n2, &n3, &n4);
  media = (n1 + n2 + n3 + n4) / 4;
  printf("Sua media foi %d\n", media);
  if (media < mediaesc) {
    printf("Voce repetiu de ano.\n");
  } else {
    printf("Voce passou de ano!\n");
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D46(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D72(%rip),%rdi
  call   <puts@plt>
  lea    0x14(%rsp),%rsi
  lea    D49(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D31(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  xor    %eax,%eax
  lea    D19(%rip),%rdi
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%r8
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  add    0x4(%rsp),%eax
  lea    D57(%rip),%rdi
  add    0xc(%rsp),%eax
  add    0x10(%rsp),%eax
  lea    0x3(%rax),%ebx
  cmovns %eax,%ebx
  xor    %eax,%eax
  sar    $0x2,%ebx
  mov    %ebx,%esi
  call   <printf@plt>
  cmp    %ebx,0x14(%rsp)
  jle    L9
  lea    D12(%rip),%rdi
  call   <puts@plt>
L1:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L9:
  lea    D38(%rip),%rdi
  call   <puts@plt>
  jmp    L1
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D46
type_infered: string
name: extract_symbolic_data
arguments: label: D72
type_infered: string
name: extract_symbolic_data
arguments: label: D49
type_infered: string
name: extract_symbolic_data
arguments: label: D31
type_infered: string
name: extract_symbolic_data
arguments: label: D19
type_infered: string
name: extract_symbolic_data
arguments: label: D57
type_infered: string
name: extract_symbolic_data
arguments: label: D12
type_infered: string
name: extract_symbolic_data
arguments: label: D38
type_infered: string

tool: D46: Qual foi sua media no ano?

tool: D72: Digite a media de sua escola.


tool: D49: %d

tool: D31: Digite as notas dos 4 bimestres.


tool: D19: %d%d%d%d

tool: D57: Sua media foi %d


tool: D12: Voce repetiu de ano.

tool: D38: Voce passou de ano!

assistant: ```c
int main() {
  int n1, n2, n3, n4, media, mediaesc;
  printf("Qual foi sua media no ano?\n");
  printf("Digite a media de sua escola.\n\n");
  scanf("%d", &mediaesc);
  printf("Digite as notas dos 4 bimestres.\n\n");
  scanf("%d"
        "%d"
        "%d"
        "%d",
        &n1, &n2, &n3, &n4);
  media = (n1 + n2 + n3 + n4) / 4;
  printf("Sua media foi %d\n", media);
  if (media < mediaesc) {
    printf("Voce repetiu de ano.\n");
  } else {
    printf("Voce passou de ano!\n");
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D21(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D42(%rip),%rdi
  call   <puts@plt>
  lea    0x14(%rsp),%rsi
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D66(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  xor    %eax,%eax
  lea    D24(%rip),%rdi
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%r8
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  add    0x4(%rsp),%eax
  lea    D57(%rip),%rdi
  add    0xc(%rsp),%eax
  add    0x10(%rsp),%eax
  lea    0x3(%rax),%ebx
  cmovns %eax,%ebx
  xor    %eax,%eax
  sar    $0x2,%ebx
  mov    %ebx,%esi
  call   <printf@plt>
  cmp    %ebx,0x14(%rsp)
  jle    L21
  lea    D38(%rip),%rdi
  call   <puts@plt>
L14:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L21:
  lea    D4(%rip),%rdi
  call   <puts@plt>
  jmp    L14
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D21
datatype: string
name: parse_data
arguments: label: D42
datatype: string
name: parse_data
arguments: label: D8
datatype: string
name: parse_data
arguments: label: D66
datatype: string
name: parse_data
arguments: label: D24
datatype: string
name: parse_data
arguments: label: D57
datatype: string
name: parse_data
arguments: label: D38
datatype: string
name: parse_data
arguments: label: D4
datatype: string

tool: D21: Qual foi sua media no ano?

tool: D42: Digite a media de sua escola.


tool: D8: %d

tool: D66: Digite as notas dos 4 bimestres.


tool: D24: %d%d%d%d

tool: D57: Sua media foi %d


tool: D38: Voce repetiu de ano.

tool: D4: Voce passou de ano!

assistant: ```c
int main() {
  int n1, n2, n3, n4, media, mediaesc;
  printf("Qual foi sua media no ano?\n");
  printf("Digite a media de sua escola.\n\n");
  scanf("%d", &mediaesc);
  printf("Digite as notas dos 4 bimestres.\n\n");
  scanf("%d"
        "%d"
        "%d"
        "%d",
        &n1, &n2, &n3, &n4);
  media = (n1 + n2 + n3 + n4) / 4;
  printf("Sua media foi %d\n", media);
  if (media < mediaesc) {
    printf("Voce repetiu de ano.\n");
  } else {
    printf("Voce passou de ano!\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x5,-0x4(%rbp)
  mov    $0x0,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D5
type: string
name: retrieve_data
arguments: data_label: D1
type: string

tool: D5: XOR (n^n) = %d


tool: D1: XOR (n^0) = %d


assistant: ```c
int main() {
  int x = 5;
  printf("XOR (n^n) = %d\n", x ^ x);
  printf("XOR (n^0) = %d\n", x ^ 0);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D15
data_type: string
name: parse_data
arguments: tag: D0
data_type: string

tool: D15: XOR (n^n) = %d


tool: D0: XOR (n^0) = %d


assistant: ```c
int main() {
  int x = 5;
  printf("XOR (n^n) = %d\n", x ^ x);
  printf("XOR (n^0) = %d\n", x ^ 0);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D18
type: string
name: retrieve_data
arguments: data_tag: D11
type: string

tool: D18: XOR (n^n) = %d


tool: D11: XOR (n^0) = %d


assistant: ```c
int main() {
  int x = 5;
  printf("XOR (n^n) = %d\n", x ^ x);
  printf("XOR (n^0) = %d\n", x ^ 0);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D9
type_guessed: string
name: disassemble_data
arguments: data_label: D7
type_guessed: string

tool: D9: XOR (n^n) = %d


tool: D7: XOR (n^0) = %d


assistant: ```c
int main() {
  int x = 5;
  printf("XOR (n^n) = %d\n", x ^ x);
  printf("XOR (n^0) = %d\n", x ^ 0);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<powerOfTen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movsd  D1(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  js     L17
  movl   $0x0,-0xc(%rbp)
  jmp    L56
L53:
  movsd  -0x8(%rbp),%xmm1
  movsd  D14(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L56:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L53
  jmp    L2
L17:
  movl   $0x0,-0xc(%rbp)
  jmp    L38
L42:
  movsd  -0x8(%rbp),%xmm1
  movsd  D3(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L38:
  mov    -0x14(%rbp),%eax
  neg    %eax
  cmp    %eax,-0xc(%rbp)
  jl     L42
L2:
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
type: f64
name: extract_symbolic_data
arguments: data_label: D14
type: f64
name: extract_symbolic_data
arguments: data_label: D3
type: f64

tool: D1: 1

tool: D14: 10

tool: D3: 0.1

assistant: ```c
double powerOfTen(int num) {
  double rst = 1.0;
  int i;
  if (num >= 0) {
    for (i = 0; i < num; i++) {
      rst *= 10.0;
    }
  } else {
    for (i = 0; i < (0 - num); i++) {
      rst *= 0.1;
    }
  }
  return rst;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<powerOfTen>:
  endbr64
  test   %edi,%edi
  js     L6
  jle    L16
  mov    $0x0,%eax
  movsd  D13(%rip),%xmm0
  movsd  D45(%rip),%xmm1
L17:
  mulsd  %xmm1,%xmm0
  add    $0x1,%eax
  cmp    %eax,%edi
  jne    L17
  ret
L6:
  neg    %edi
  mov    $0x0,%eax
  movsd  D13(%rip),%xmm0
  movsd  D43(%rip),%xmm1
L19:
  mulsd  %xmm1,%xmm0
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L19
  ret
L16:
  movsd  D13(%rip),%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D13
type_infered: f64
name: disassemble_data
arguments: tag: D45
type_infered: f64
name: disassemble_data
arguments: tag: D43
type_infered: f64

tool: D13: 1

tool: D45: 10

tool: D43: 0.1

assistant: ```c
double powerOfTen(int num) {
  double rst = 1.0;
  int i;
  if (num >= 0) {
    for (i = 0; i < num; i++) {
      rst *= 10.0;
    }
  } else {
    for (i = 0; i < (0 - num); i++) {
      rst *= 0.1;
    }
  }
  return rst;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<powerOfTen>:
  endbr64
  test   %edi,%edi
  js     L1
  je     L38
  movsd  D49(%rip),%xmm0
  movsd  D37(%rip),%xmm1
  xor    %eax,%eax
  nopl   0x0(%rax)
L12:
  mulsd  %xmm1,%xmm0
  add    $0x1,%eax
  cmp    %eax,%edi
  jne    L12
  ret
  nopl   0x0(%rax)
L1:
  movsd  D49(%rip),%xmm0
  movsd  D31(%rip),%xmm1
  neg    %edi
  xor    %eax,%eax
  nopl   0x0(%rax)
L27:
  mulsd  %xmm1,%xmm0
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L27
  ret
  nopl   0x0(%rax)
L38:
  movsd  D49(%rip),%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D49
type_infered: f64
name: disassemble_data
arguments: data_tag: D37
type_infered: f64
name: disassemble_data
arguments: data_tag: D31
type_infered: f64

tool: D49: 1

tool: D37: 10

tool: D31: 0.1

assistant: ```c
double powerOfTen(int num) {
  double rst = 1.0;
  int i;
  if (num >= 0) {
    for (i = 0; i < num; i++) {
      rst *= 10.0;
    }
  } else {
    for (i = 0; i < (0 - num); i++) {
      rst *= 0.1;
    }
  }
  return rst;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<powerOfTen>:
  endbr64
  test   %edi,%edi
  js     L27
  je     L25
  movsd  D47(%rip),%xmm0
  movsd  D26(%rip),%xmm1
  xor    %eax,%eax
  nopl   0x0(%rax)
L33:
  mulsd  %xmm1,%xmm0
  add    $0x1,%eax
  cmp    %eax,%edi
  jne    L33
  ret
  nopl   0x0(%rax)
L27:
  movsd  D47(%rip),%xmm0
  movsd  D29(%rip),%xmm1
  neg    %edi
  xor    %eax,%eax
  nopl   0x0(%rax)
L13:
  mulsd  %xmm1,%xmm0
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L13
  ret
  nopl   0x0(%rax)
L25:
  movsd  D47(%rip),%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D47
data_type: f64
name: extract_symbolic_data
arguments: data_label: D26
data_type: f64
name: extract_symbolic_data
arguments: data_label: D29
data_type: f64

tool: D47: 1

tool: D26: 10

tool: D29: 0.1

assistant: ```c
double powerOfTen(int num) {
  double rst = 1.0;
  int i;
  if (num >= 0) {
    for (i = 0; i < num; i++) {
      rst *= 10.0;
    }
  } else {
    for (i = 0; i < (0 - num); i++) {
      rst *= 0.1;
    }
  }
  return rst;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  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   $0x0,-0x10(%rbp)
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x18(%rbp)
  jmp    L3
L30:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  add    %eax,-0x14(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    $0x40,%eax
  jg     L0
  addl   $0x1,-0x10(%rbp)
L0:
  addl   $0x1,-0x18(%rbp)
L3:
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jle    L30
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x14(%rbp),%xmm0
  mov    -0x20(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0xc(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D47
type: string
name: unpack_data
arguments: data_tag: D1
type: string
name: unpack_data
arguments: data_tag: D7
type: string
name: unpack_data
arguments: data_tag: D49
type: string
name: unpack_data
arguments: data_tag: D50
type: string

tool: D47: How many grades will you be entering? 

tool: D1: %i

tool: D7: Enter a grade #%i: 

tool: D49: 
Grade average = %.2f


tool: D50: Number of failures = %i


assistant: ```c
int main(void) {
  int numberOfGrades, i, grade;
  int gradeTotal = 0;
  int failureCount = 0;
  float average;
  printf("How many grades will you be entering? ");
  scanf("%i", &numberOfGrades);
  for (i = 1; i <= numberOfGrades; ++i) {
    printf("Enter a grade #%i: ", i);
    scanf("%i", &grade);
    gradeTotal += grade;
    if (grade < 65) {
      ++failureCount;
    }
  }
  average = (float)gradeTotal / numberOfGrades;
  printf("\nGrade average = %.2f\n", average);
  printf("Number of failures = %i\n", failureCount);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```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
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  test   %eax,%eax
  jle    L23
  mov    $0x0,%ebp
  mov    $0x0,%r12d
  mov    $0x1,%ebx
  lea    D32(%rip),%r15
  lea    0x4(%rsp),%r14
  lea    D1(%rip),%r13
L35:
  mov    %ebx,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  add    %eax,%r12d
  cmp    $0x41,%eax
  setl   %al
  movzbl %al,%eax
  add    %eax,%ebp
  add    $0x1,%ebx
  mov    (%rsp),%eax
  cmp    %ebx,%eax
  jge    L35
L37:
  pxor   %xmm0,%xmm0
  cvtsi2ss %r12d,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  divss  %xmm1,%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D33(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  mov    $0x0,%ebp
  mov    $0x0,%r12d
  jmp    L37
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D14
type_guessed: string
name: disassemble_data
arguments: label: D1
type_guessed: string
name: disassemble_data
arguments: label: D32
type_guessed: string
name: disassemble_data
arguments: label: D33
type_guessed: string
name: disassemble_data
arguments: label: D26
type_guessed: string

tool: D14: How many grades will you be entering? 

tool: D1: %i

tool: D32: Enter a grade #%i: 

tool: D33: 
Grade average = %.2f


tool: D26: Number of failures = %i


assistant: ```c
int main(void) {
  int numberOfGrades, i, grade;
  int gradeTotal = 0;
  int failureCount = 0;
  float average;
  printf("How many grades will you be entering? ");
  scanf("%i", &numberOfGrades);
  for (i = 1; i <= numberOfGrades; ++i) {
    printf("Enter a grade #%i: ", i);
    scanf("%i", &grade);
    gradeTotal += grade;
    if (grade < 65) {
      ++failureCount;
    }
  }
  average = (float)gradeTotal / numberOfGrades;
  printf("\nGrade average = %.2f\n", average);
  printf("Number of failures = %i\n", failureCount);
  return 0;
}
```