ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D5(%rip),%r14
  push   %r13
  lea    D18(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%r12
  xchg   %ax,%ax
L24:
  add    $0x1,%ebx
  mov    %r14,%rdi
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x4,%r12
  call   <scanf@plt>
  cmp    $0xa,%ebx
  jne    L24
  lea    0x30(%rsp),%r15
  xor    %ebx,%ebx
  mov    %r15,%r12
  nopl   0x0(%rax,%rax,1)
L48:
  add    $0x1,%ebx
  mov    %r14,%rdi
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x4,%r12
  call   <scanf@plt>
  cmp    $0xa,%ebx
  jne    L48
  xor    %eax,%eax
  lea    0x60(%rsp),%rbx
  nopl   0x0(%rax,%rax,1)
L2:
  mov    (%r15,%rax,1),%edx
  add    0x0(%rbp,%rax,1),%edx
  mov    %edx,(%rbx,%rax,1)
  add    $0x4,%rax
  cmp    $0x28,%rax
  jne    L2
  lea    0x88(%rsp),%r12
  lea    D28(%rip),%rbp
  nopl   0x0(%rax)
L39:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L39
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x98,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D5
type_infered: string
name: disassemble_data
arguments: data_label: D18
type_infered: string
name: disassemble_data
arguments: data_label: D28
type_infered: string

tool: D5: Digite o %d numero

tool: D18: %d

tool: D28: %d


assistant: ```c
int main() {
  int i, a[10], b[10], c[10];
  for (i = 0; i <= 9; i++) {
    printf("Digite o %d numero", i + 1);
    scanf("%d", &a[i]);
  }
  for (i = 0; i <= 9; i++) {
    printf("Digite o %d numero", i + 1);
    scanf("%d", &b[i]);
  }
  for (i = 0; i <= 9; i++) {
    c[i] = a[i] + b[i];
  }
  for (i = 0; i <= 9; i++) {
    printf("%d\n", c[i]);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r13
  lea    D14(%rip),%r13
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  nopw   0x0(%rax,%rax,1)
L39:
  add    $0x1,%ebx
  mov    %r12,%rdi
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    $0xa,%ebx
  jne    L39
  lea    0x30(%rsp),%rbp
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L37:
  add    $0x1,%ebx
  mov    %r12,%rdi
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    $0xa,%ebx
  jne    L37
  movdqa (%rsp),%xmm0
  paddd  0x30(%rsp),%xmm0
  lea    0x60(%rsp),%rbx
  movq   0x50(%rsp),%xmm1
  lea    0x88(%rsp),%r12
  lea    D19(%rip),%rbp
  movaps %xmm0,0x60(%rsp)
  movdqa 0x10(%rsp),%xmm0
  paddd  0x40(%rsp),%xmm0
  movaps %xmm0,0x70(%rsp)
  movq   0x20(%rsp),%xmm0
  paddd  %xmm1,%xmm0
  movq   %xmm0,0x80(%rsp)
  nop
L18:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L18
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x98,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D14
type_guessed: string
name: disassemble_data
arguments: data_tag: D10
type_guessed: string
name: disassemble_data
arguments: data_tag: D19
type_guessed: string

tool: D14: %d

tool: D10: Digite o %d numero

tool: D19: %d


assistant: ```c
int main() {
  int i, a[10], b[10], c[10];
  for (i = 0; i <= 9; i++) {
    printf("Digite o %d numero", i + 1);
    scanf("%d", &a[i]);
  }
  for (i = 0; i <= 9; i++) {
    printf("Digite o %d numero", i + 1);
    scanf("%d", &b[i]);
  }
  for (i = 0; i <= 9; i++) {
    c[i] = a[i] + b[i];
  }
  for (i = 0; i <= 9; i++) {
    printf("%d\n", c[i]);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x340,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x33c(%rbp)
  lea    -0x338(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L6
L22:
  mov    -0x338(%rbp),%rdx
  mov    -0x33c(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movsd  -0x338(%rbp),%xmm0
  movsd  D19(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x338(%rbp)
  addl   $0x1,-0x33c(%rbp)
L6:
  cmpl   $0x63,-0x33c(%rbp)
  jle    L22
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D24
type_guessed: string
name: parse_data
arguments: data_label: D17
type_guessed: string
name: parse_data
arguments: data_label: D19
type_guessed: byte[8]

tool: D24:  %lf

tool: D17: N[%d] = %0.4lf


tool: {"D19": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x40]}

assistant: ```c
int main() {
  int i = 0;
  double N[101], X;
  scanf(" %lf", &X);
  while (i < 100) {
    printf("N[%d] = %0.4lf\n", i, X);
    X /= 2;
    i++;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  mov    $0x0,%ebx
  lea    D13(%rip),%rbp
L12:
  movsd  (%rsp),%xmm0
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movsd  D24(%rip),%xmm0
  mulsd  (%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  add    $0x1,%ebx
  cmp    $0x64,%ebx
  jne    L12
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D15
datatype: string
name: read_data
arguments: data_label: D13
datatype: string
name: read_data
arguments: data_label: D24
datatype: byte[8]

tool: D15:  %lf

tool: D13: N[%d] = %0.4lf


tool: {"D24": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

assistant: ```c
int main() {
  int i = 0;
  double N[101], X;
  scanf(" %lf", &X);
  while (i < 100) {
    printf("N[%d] = %0.4lf\n", i, X);
    X /= 2;
    i++;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D17(%rip),%rdi
  lea    D20(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  nopw   0x0(%rax,%rax,1)
L10:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  movsd  D28(%rip),%xmm0
  mulsd  (%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  cmp    $0x64,%ebx
  jne    L10
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D17
datatype: string
name: unpack_data
arguments: data_tag: D20
datatype: string
name: unpack_data
arguments: data_tag: D28
datatype: byte[8]

tool: D17:  %lf

tool: D20: N[%d] = %0.4lf


tool: {"D28": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

assistant: ```c
int main() {
  int i = 0;
  double N[101], X;
  scanf(" %lf", &X);
  while (i < 100) {
    printf("N[%d] = %0.4lf\n", i, X);
    X /= 2;
    i++;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D27(%rip),%rdi
  lea    D10(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  movsd  (%rsp),%xmm0
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  movsd  D7(%rip),%xmm0
  mulsd  (%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  cmp    $0x64,%ebx
  jne    L5
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D27
datatype: string
name: retrieve_data
arguments: label: D10
datatype: string
name: retrieve_data
arguments: label: D7
datatype: byte[8]

tool: D27:  %lf

tool: D10: N[%d] = %0.4lf


tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

assistant: ```c
int main() {
  int i = 0;
  double N[101], X;
  scanf(" %lf", &X);
  while (i < 100) {
    printf("N[%d] = %0.4lf\n", i, X);
    X /= 2;
    i++;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<tmp_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D39(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D0(%rip)
  lea    -0x30(%rbp),%rax
  lea    D0(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    D0(%rip),%eax
  add    $0x30,%eax
  mov    %al,-0x2d(%rbp)
  movb   $0x0,-0x2c(%rbp)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D0
data_type: string

tool: D0: tmp

assistant: ```c
char *tmp_name(void) {
  char name[30];
  static int sequence = 0;
  ++sequence;
  strcpy(name, "tmp");
  name[3] = sequence + '0';
  name[4] = '\0';
  return (name);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<twoscomplement>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  jmp    L2
L7:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
  subl   $0x1,-0x14(%rbp)
L2:
  cmpl   $0x0,-0x14(%rbp)
  jne    L7
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  neg    %eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void twoscomplement(unsigned char *cs, unsigned char *buf, unsigned int size) {
  *cs = 0;
  while (size > 0) {
    *cs += *buf;
    buf++;
    size--;
  }
  *cs = (~*cs) + 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<twoscomplement>:
  endbr64
  movb   $0x0,(%rdi)
  test   %edx,%edx
  je     L6
  mov    %edx,%edx
  add    %rsi,%rdx
L12:
  movzbl (%rsi),%eax
  add    %al,(%rdi)
  add    $0x1,%rsi
  cmp    %rdx,%rsi
  jne    L12
L6:
  negb   (%rdi)
  ret
```

assistant: ```c
void twoscomplement(unsigned char *cs, unsigned char *buf, unsigned int size) {
  *cs = 0;
  while (size > 0) {
    *cs += *buf;
    buf++;
    size--;
  }
  *cs = (~*cs) + 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<twoscomplement>:
  endbr64
  movb   $0x0,(%rdi)
  test   %edx,%edx
  je     L5
  mov    %edx,%edx
  xor    %eax,%eax
  add    %rsi,%rdx
  nopw   0x0(%rax,%rax,1)
L12:
  add    (%rsi),%al
  add    $0x1,%rsi
  mov    %al,(%rdi)
  cmp    %rsi,%rdx
  jne    L12
  neg    %eax
  mov    %al,(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void twoscomplement(unsigned char *cs, unsigned char *buf, unsigned int size) {
  *cs = 0;
  while (size > 0) {
    *cs += *buf;
    buf++;
    size--;
  }
  *cs = (~*cs) + 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<twoscomplement>:
  endbr64
  movb   $0x0,(%rdi)
  test   %edx,%edx
  je     L18
  mov    %edx,%edx
  xor    %eax,%eax
  add    %rsi,%rdx
  nopw   0x0(%rax,%rax,1)
L7:
  add    (%rsi),%al
  add    $0x1,%rsi
  mov    %al,(%rdi)
  cmp    %rsi,%rdx
  jne    L7
  neg    %eax
  mov    %al,(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  xor    %eax,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void twoscomplement(unsigned char *cs, unsigned char *buf, unsigned int size) {
  *cs = 0;
  while (size > 0) {
    *cs += *buf;
    buf++;
    size--;
  }
  *cs = (~*cs) + 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x7a1000(%rsp),%rax
L45:
  cmp    %rax,%rsp
  je     L65
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L45
L65:
  sub    $0x200,%rsp
  orq    $0x0,0x1f8(%rsp)
  lea    0x3(%rsp),%rdx
  mov    %rdx,%rsi
  shr    $0x2,%rsi
  and    $0xfffffffffffffffc,%rdx
  mov    %rdx,%rax
  lea    0x7a1200(%rdx),%rcx
L97:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L97
  movl   $0x0,0x0(,%rsi,4)
  movl   $0x0,0x4(,%rsi,4)
  mov    $0x4,%edi
  mov    $0x2,%ecx
  jmp    L63
L72:
  add    $0x1,%eax
  add    $0x1,%rcx
  add    $0x2,%edi
  imul   %eax,%eax
  cmp    $0x1e8480,%eax
  jg     L18
L63:
  mov    %ecx,%eax
  cmpl   $0x1,(%rdx,%rcx,4)
  jne    L72
  cmp    $0x1e8480,%edi
  jg     L72
  movslq %edi,%rsi
L137:
  movl   $0x0,(%rdx,%rsi,4)
  add    %rcx,%rsi
  cmp    $0x1e8480,%esi
  jle    L137
  jmp    L72
L18:
  mov    $0x0,%eax
  mov    $0x0,%esi
  jmp    L120
L121:
  add    $0x1,%rax
  cmp    $0x1e8480,%rax
  je     L7
L120:
  cmpl   $0x1,(%rdx,%rax,4)
  jne    L121
  add    %rax,%rsi
  jmp    L121
L7:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  mov    $0x0,%eax
  leave
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D7
data_type: string

tool: D7: result: %lu

assistant: ```c
int main(int argc, char *argv[]) {
  int limit = 2000000;
  long result = 0;
  int primes[limit];
  for (int i = 0; i < limit; i++) {
    primes[i] = 1;
  }
  primes[0] = 0;
  primes[1] = 0;
  for (int i = 2; i * i <= limit; i++) {
    if (primes[i] == 1) {
      for (int j = 2; i * j <= limit; j++) {
        primes[j * i] = 0;
      }
    }
  }
  for (int i = 0; i < limit; i++) {
    if (primes[i] == 1) {
      result += i;
    }
  }
  printf("result: %lu", result);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x7a1000(%rsp),%rax
  cmp    %rax,%rsp
  je     L90
L32:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rax,%rsp
  jne    L32
L90:
  sub    $0x200,%rsp
  orq    $0x0,0x1f8(%rsp)
  lea    0x3(%rsp),%rdx
  mov    %rdx,%rsi
  and    $0xfffffffffffffffc,%rdx
  shr    $0x2,%rsi
  mov    %rdx,%rax
  lea    0x7a1200(%rdx),%rcx
  nopw   0x0(%rax,%rax,1)
L111:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L111
  mov    $0x4,%edi
  mov    $0x2,%ecx
  movq   $0x0,0x0(,%rsi,4)
  jmp    L13
  nopw   0x0(%rax,%rax,1)
L92:
  add    $0x1,%eax
  add    $0x1,%rcx
  add    $0x2,%edi
  imul   %eax,%eax
  cmp    $0x1e8480,%eax
  jg     L23
L13:
  cmpl   $0x1,(%rdx,%rcx,4)
  mov    %ecx,%eax
  jne    L92
  cmp    $0x1e8480,%edi
  jg     L92
  movslq %edi,%rsi
  nopw   0x0(%rax,%rax,1)
L60:
  movl   $0x0,(%rdx,%rsi,4)
  add    %rcx,%rsi
  cmp    $0x1e8480,%esi
  jle    L60
  jmp    L92
L23:
  mov    $0x1,%eax
  xor    %esi,%esi
  nopl   0x0(%rax,%rax,1)
L73:
  cmpl   $0x1,(%rdx,%rax,4)
  jne    L12
  add    %rax,%rsi
L12:
  add    $0x1,%rax
  cmp    $0x1e8480,%rax
  jne    L73
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  leave
  xor    %eax,%eax
  ret
L64:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D0
data_type: string

tool: D0: result: %lu

assistant: ```c
int main(int argc, char *argv[]) {
  int limit = 2000000;
  long result = 0;
  int primes[limit];
  for (int i = 0; i < limit; i++) {
    primes[i] = 1;
  }
  primes[0] = 0;
  primes[1] = 0;
  for (int i = 2; i * i <= limit; i++) {
    if (primes[i] == 1) {
      for (int j = 2; i * j <= limit; j++) {
        primes[j * i] = 0;
      }
    }
  }
  for (int i = 0; i < limit; i++) {
    if (primes[i] == 1) {
      result += i;
    }
  }
  printf("result: %lu", result);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x7a1000(%rsp),%rax
  cmp    %rax,%rsp
  je     L58
L47:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rax,%rsp
  jne    L47
L58:
  sub    $0x200,%rsp
  orq    $0x0,0x1f8(%rsp)
  movdqa D36(%rip),%xmm4
  lea    0x3(%rsp),%rdx
  mov    %rdx,%r8
  and    $0xfffffffffffffffc,%rdx
  shr    $0x2,%r8
  lea    0x7a1200(%rdx),%rcx
  mov    %rdx,%rax
  nop
L60:
  movups %xmm4,(%rax)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L60
  mov    $0x4,%edi
  mov    $0x2,%ecx
  movq   $0x0,0x0(,%r8,4)
  jmp    L129
  nopl   0x0(%rax)
L37:
  add    $0x1,%eax
  add    $0x1,%rcx
  add    $0x2,%edi
  imul   %eax,%eax
  cmp    $0x1e8480,%eax
  jg     L81
L129:
  cmpl   $0x1,(%rdx,%rcx,4)
  mov    %ecx,%eax
  jne    L37
  cmp    $0x1e8480,%edi
  jg     L37
  movslq %edi,%rsi
  nop
L44:
  movl   $0x0,(%rdx,%rsi,4)
  add    %rcx,%rsi
  cmp    $0x1e8480,%esi
  jle    L44
  jmp    L37
L81:
  movdqa D0(%rip),%xmm2
  lea    0x4(%rdx),%rax
  pxor   %xmm1,%xmm1
  pxor   %xmm6,%xmm6
  movdqa D28(%rip),%xmm7
  add    $0x7a11f4,%rdx
  nopw   0x0(%rax,%rax,1)
L26:
  movdqu (%rax),%xmm0
  movdqa %xmm2,%xmm3
  add    $0x10,%rax
  paddd  %xmm7,%xmm2
  pcmpeqd %xmm4,%xmm0
  pand   %xmm3,%xmm0
  movdqa %xmm6,%xmm3
  pcmpgtd %xmm0,%xmm3
  movdqa %xmm0,%xmm5
  punpckldq %xmm3,%xmm5
  punpckhdq %xmm3,%xmm0
  paddq  %xmm5,%xmm1
  paddq  %xmm0,%xmm1
  cmp    %rax,%rdx
  jne    L26
  cmpl   $0x1,0x7a11f4(,%r8,4)
  movdqa %xmm1,%xmm0
  psrldq $0x8,%xmm0
  paddq  %xmm0,%xmm1
  movq   %xmm1,%rsi
  jne    L110
  add    $0x1e847d,%rsi
L110:
  cmpl   $0x1,0x7a11f8(,%r8,4)
  jne    L48
  add    $0x1e847e,%rsi
L48:
  cmpl   $0x1,0x7a11fc(,%r8,4)
  jne    L67
  add    $0x1e847f,%rsi
L67:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  leave
  xor    %eax,%eax
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D36
type_infered: byte[16]
name: disassemble_data
arguments: data_tag: D0
type_infered: byte[16]
name: disassemble_data
arguments: data_tag: D28
type_infered: byte[16]
name: disassemble_data
arguments: data_tag: D7
type_infered: string

tool: {"D36": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: {"D0": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D28": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D7: result: %lu

assistant: ```c
int main(int argc, char *argv[]) {
  int limit = 2000000;
  long result = 0;
  int primes[limit];
  for (int i = 0; i < limit; i++) {
    primes[i] = 1;
  }
  primes[0] = 0;
  primes[1] = 0;
  for (int i = 2; i * i <= limit; i++) {
    if (primes[i] == 1) {
      for (int j = 2; i * j <= limit; j++) {
        primes[j * i] = 0;
      }
    }
  }
  for (int i = 0; i < limit; i++) {
    if (primes[i] == 1) {
      result += i;
    }
  }
  printf("result: %lu", result);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<bpf_verify_program>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x18(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    %r9,-0x38(%rbp)
  mov    0x10(%rbp),%rax
  mov    %rax,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,-0x10(%rbp)
  mov    0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    0x20(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x40(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  je     L15
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  jmp    L12
L15:
  mov    $0x0,%eax
L12:
  mov    %eax,-0x10(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  lea    -0x10(%rbp),%rcx
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <sys_bpf@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
int bpf_verify_program(enum bpf_prog_type type, const struct bpf_insn *insns, size_t insns_cnt, int strict_alignment, const char *license, __u32 kern_version, char *log_buf, size_t log_buf_sz, int log_level) {
  union bpf_attr attr;
  bzero(&attr, sizeof(attr));
  attr.prog_type = type;
  attr.insn_cnt = (__u32)insns_cnt;
  attr.insns = ptr_to_u64(insns);
  attr.license = ptr_to_u64(license);
  attr.log_buf = ptr_to_u64(log_buf);
  attr.log_size = log_buf_sz;
  attr.log_level = log_level;
  log_buf[0] = 0;
  attr.kern_version = kern_version;
  attr.prog_flags = strict_alignment ? BPF_F_STRICT_ALIGNMENT : 0;
  return sys_bpf(BPF_PROG_LOAD, &attr, sizeof(attr));
}
```
user: Translate the below assembly code into C programming language:

```asm
<bpf_verify_program>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%r13
  mov    %rdx,%r14
  mov    %ecx,%ebx
  mov    %r8,%r12
  mov    %r9,%rbp
  mov    0x50(%rsp),%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x8,%esi
  call   <bzero@plt>
  mov    %r14,(%rsp)
  mov    %r13,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,(%rsp)
  mov    %r12,%rdi
  call   <ptr_to_u64@plt>
  mov    %rax,(%rsp)
  mov    %r15,%rdi
  call   <ptr_to_u64@plt>
  movb   $0x0,(%r15)
  mov    %rbp,(%rsp)
  test   %ebx,%ebx
  je     L15
  mov    D11(%rip),%rax
  mov    (%rax),%ebx
L15:
  mov    %ebx,(%rsp)
  mov    %rsp,%rsi
  mov    $0x8,%edx
  mov    D14(%rip),%rax
  mov    (%rax),%edi
  call   <sys_bpf@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bpf_verify_program(enum bpf_prog_type type, const struct bpf_insn *insns, size_t insns_cnt, int strict_alignment, const char *license, __u32 kern_version, char *log_buf, size_t log_buf_sz, int log_level) {
  union bpf_attr attr;
  bzero(&attr, sizeof(attr));
  attr.prog_type = type;
  attr.insn_cnt = (__u32)insns_cnt;
  attr.insns = ptr_to_u64(insns);
  attr.license = ptr_to_u64(license);
  attr.log_buf = ptr_to_u64(log_buf);
  attr.log_size = log_buf_sz;
  attr.log_level = log_level;
  log_buf[0] = 0;
  attr.kern_version = kern_version;
  attr.prog_flags = strict_alignment ? BPF_F_STRICT_ALIGNMENT : 0;
  return sys_bpf(BPF_PROG_LOAD, &attr, sizeof(attr));
}
```