ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<isThereASum>:
  endbr64
  push   %r13
  lea    0x28(%rdi),%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L24:
  mov    (%rbx),%eax
  mov    %ebp,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  je     L11
  mov    %r12,%rdi
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  jne    L22
L11:
  add    $0x4,%rbx
  cmp    %r13,%rbx
  jne    L24
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L22:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int isThereASum(int *A, int value) {
  for (int i = 0; i < 10; i++) {
    if (value - A[i] != A[i] && binarySearch(A, value - A[i]) != -1) {
      return 1;
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<isThereASum>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  mov    (%rdi),%eax
  sub    %eax,%esi
  cmp    %esi,%eax
  jne    L152
L141:
  mov    0x4(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  jne    L278
L60:
  mov    0x8(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  jne    L186
  mov    0xc(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  jne    L112
L174:
  mov    0x10(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  jne    L153
L222:
  mov    0x14(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  jne    L279
L162:
  mov    0x18(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  jne    L232
L18:
  mov    0x1c(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  jne    L82
L166:
  mov    0x20(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  jne    L95
L158:
  mov    0x24(%rbp),%eax
  sub    %eax,%ebx
  mov    %ebx,%esi
  cmp    %ebx,%eax
  jne    L27
L171:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L278:
  mov    %rbp,%rdi
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  je     L60
L159:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L186:
  mov    %rbp,%rdi
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  jne    L159
  mov    0xc(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  je     L174
L112:
  mov    %rbp,%rdi
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  jne    L159
  mov    0x10(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  je     L222
L153:
  mov    %rbp,%rdi
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  jne    L159
  mov    0x14(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  je     L162
L279:
  mov    %rbp,%rdi
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  jne    L159
  mov    0x18(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  je     L18
L232:
  mov    %rbp,%rdi
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  jne    L159
  mov    0x1c(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  je     L166
L82:
  mov    %rbp,%rdi
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  jne    L159
  mov    0x20(%rbp),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  cmp    %esi,%eax
  je     L158
L95:
  mov    %rbp,%rdi
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  jne    L159
  mov    0x24(%rbp),%eax
  sub    %eax,%ebx
  mov    %ebx,%esi
  cmp    %ebx,%eax
  je     L171
L27:
  mov    %rbp,%rdi
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  je     L171
  jmp    L159
  cs nopw 0x0(%rax,%rax,1)
L152:
  call   <binarySearch@plt>
  cmp    $0xffffffff,%eax
  je     L141
  jmp    L159
```

assistant: ```c
int isThereASum(int *A, int value) {
  for (int i = 0; i < 10; i++) {
    if (value - A[i] != A[i] && binarySearch(A, value - A[i]) != -1) {
      return 1;
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bridge_receiveFromSocket>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <recv@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L16
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L10
L16:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L10:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type_infered: string

tool: D2: Socket receive error


assistant: ```c
int bridge_receiveFromSocket(int sd, char *buffer, int size) {
  ssize_t recvSize = recv(sd, buffer, size - 1, 0);
  if (-1 == recvSize) {
    printf("Socket receive error\n");
  } else {
    buffer[recvSize] = 0;
  }
  return recvSize;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<bridge_receiveFromSocket>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  sub    $0x1,%edx
  mov    $0x0,%ecx
  call   <recv@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L4
  cltq
  movb   $0x0,0x0(%rbp,%rax,1)
L1:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  jmp    L1
```

tool_calls: name: disassemble_data
arguments: label: D9
type: string

tool: D9: Socket receive error


assistant: ```c
int bridge_receiveFromSocket(int sd, char *buffer, int size) {
  ssize_t recvSize = recv(sd, buffer, size - 1, 0);
  if (-1 == recvSize) {
    printf("Socket receive error\n");
  } else {
    buffer[recvSize] = 0;
  }
  return recvSize;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bridge_receiveFromSocket>:
  endbr64
  push   %r12
  sub    $0x1,%edx
  xor    %ecx,%ecx
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <recv@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L0
  cltq
  movb   $0x0,(%rbx,%rax,1)
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L0:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
datatype: string

tool: D4: Socket receive error


assistant: ```c
int bridge_receiveFromSocket(int sd, char *buffer, int size) {
  ssize_t recvSize = recv(sd, buffer, size - 1, 0);
  if (-1 == recvSize) {
    printf("Socket receive error\n");
  } else {
    buffer[recvSize] = 0;
  }
  return recvSize;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<bridge_receiveFromSocket>:
  endbr64
  push   %r12
  sub    $0x1,%edx
  xor    %ecx,%ecx
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <recv@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L8
  cltq
  movb   $0x0,(%rbx,%rax,1)
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L8:
  lea    D0(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D0
type_guessed: string

tool: D0: Socket receive error


assistant: ```c
int bridge_receiveFromSocket(int sd, char *buffer, int size) {
  ssize_t recvSize = recv(sd, buffer, size - 1, 0);
  if (-1 == recvSize) {
    printf("Socket receive error\n");
  } else {
    buffer[recvSize] = 0;
  }
  return recvSize;
}
```
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    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <init@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x5,%edi
  call   <foo@plt>
  mov    %rax,-0x38(%rbp)
  lea    -0x30(%rbp),%rax
  cmp    %rax,-0x38(%rbp)
  je     L23
  mov    $0x0,%eax
  call   <abort@plt>
L23:
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x5,%edi
  call   <foo2@plt>
  mov    %rax,-0x38(%rbp)
  lea    -0x30(%rbp),%rax
  add    $0x20,%rax
  cmp    %rax,-0x38(%rbp)
  je     L12
  mov    $0x0,%eax
  call   <abort@plt>
L12:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
int main(void) {
  char *list[5];
  char *res;
  init(list);
  res = foo(5, list);
  if (res != (char *)&list[0])
    abort();
  res = foo2(5, list);
  if (res != (char *)&list[5 - 1])
    abort();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <init@plt>
  mov    %rbx,%rsi
  mov    $0x5,%edi
  call   <foo@plt>
  cmp    %rax,%rbx
  je     L0
  mov    $0x0,%eax
  call   <abort@plt>
L0:
  mov    %rsp,%rsi
  mov    $0x5,%edi
  call   <foo2@plt>
  lea    0x20(%rsp),%rdx
  cmp    %rax,%rdx
  je     L5
  mov    $0x0,%eax
  call   <abort@plt>
L5:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  char *list[5];
  char *res;
  init(list);
  res = foo(5, list);
  if (res != (char *)&list[0])
    abort();
  res = foo2(5, list);
  if (res != (char *)&list[5 - 1])
    abort();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <init@plt>
  mov    %rbp,%rsi
  mov    $0x5,%edi
  call   <foo@plt>
  cmp    %rax,%rbp
  je     L21
  xor    %eax,%eax
  call   <abort@plt>
L21:
  mov    %rbp,%rsi
  mov    $0x5,%edi
  call   <foo2@plt>
  lea    0x20(%rsp),%rdx
  cmp    %rax,%rdx
  je     L3
  xor    %eax,%eax
  call   <abort@plt>
L3:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  char *list[5];
  char *res;
  init(list);
  res = foo(5, list);
  if (res != (char *)&list[0])
    abort();
  res = foo2(5, list);
  if (res != (char *)&list[5 - 1])
    abort();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <init@plt>
  mov    %rbp,%rsi
  mov    $0x5,%edi
  call   <foo@plt>
  cmp    %rax,%rbp
  je     L2
  xor    %eax,%eax
  call   <abort@plt>
L2:
  mov    %rbp,%rsi
  mov    $0x5,%edi
  call   <foo2@plt>
  lea    0x20(%rsp),%rdx
  cmp    %rax,%rdx
  je     L22
  xor    %eax,%eax
  call   <abort@plt>
L22:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  char *list[5];
  char *res;
  init(list);
  res = foo(5, list);
  if (res != (char *)&list[0])
    abort();
  res = foo2(5, list);
  if (res != (char *)&list[5 - 1])
    abort();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <DoTests@plt>
  mov    $0x0,%eax
  call   <getSumDiv@plt>
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%ecx
  mov    $0x3d0900,%edx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <fibSumEven@plt>
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xe589eac7,%edi
  call   <largestPrimeFactor@plt>
  mov    %eax,%esi
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <maxPalProd@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x64,%edi
  call   <sumSqrtDiff@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x2711,%edi
  call   <nPrimeNumber@plt>
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xd,%edi
  call   <maxProd1000dig@plt>
  mov    %eax,%esi
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x3e8,%edi
  call   <pythTriplet@plt>
  mov    %eax,%esi
  lea    D71(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1e8480,%edi
  call   <sumPrimes@plt>
  mov    %eax,%esi
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D28
data_type: string
name: disassemble_data
arguments: data_label: D22
data_type: string
name: disassemble_data
arguments: data_label: D89
data_type: string
name: disassemble_data
arguments: data_label: D7
data_type: string
name: disassemble_data
arguments: data_label: D3
data_type: string
name: disassemble_data
arguments: data_label: D6
data_type: string
name: disassemble_data
arguments: data_label: D68
data_type: string
name: disassemble_data
arguments: data_label: D71
data_type: string
name: disassemble_data
arguments: data_label: D52
data_type: string

tool: D28: Problem 1 result: %d


tool: D22: Problem 2 result: %d


tool: D89: Problem 3 result: %lld


tool: D7: Problem 4 result: %d


tool: D3: Problem 6 result: %d


tool: D6: Problem 7 result: %lld


tool: D68: Problem 8 result: %lld


tool: D71: Problem 9 result: %d


tool: D52: Problem 10 result: %lld


assistant: ```c
int main() {
  DoTests();
  printf("Problem 1 result: %d\n", getSumDiv());
  printf("Problem 2 result: %d\n", fibSumEven(1, 1, 4000000, 0));
  printf("Problem 3 result: %lld\n", largestPrimeFactor(600851475143));
  printf("Problem 4 result: %d\n", maxPalProd(3));
  printf("Problem 6 result: %d\n", sumSqrtDiff(100));
  printf("Problem 7 result: %lld\n", nPrimeNumber(10001));
  printf("Problem 8 result: %lld\n", maxProd1000dig(13));
  printf("Problem 9 result: %d\n", pythTriplet(1000));
  printf("Problem 10 result: %lld\n", sumPrimes(2000000));
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <DoTests@plt>
  mov    $0x0,%eax
  call   <getSumDiv@plt>
  mov    %eax,%esi
  lea    D69(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%ecx
  mov    $0x3d0900,%edx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <fibSumEven@plt>
  mov    %eax,%esi
  lea    D39(%rip),%rdi
  call   <printf@plt>
  mov    $0xe589eac7,%edi
  call   <largestPrimeFactor@plt>
  mov    %eax,%esi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <maxPalProd@plt>
  mov    %eax,%esi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0x64,%edi
  call   <sumSqrtDiff@plt>
  mov    %eax,%esi
  lea    D89(%rip),%rdi
  call   <printf@plt>
  mov    $0x2711,%edi
  call   <nPrimeNumber@plt>
  mov    %eax,%esi
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    $0xd,%edi
  call   <maxProd1000dig@plt>
  mov    %eax,%esi
  lea    D78(%rip),%rdi
  call   <printf@plt>
  mov    $0x3e8,%edi
  call   <pythTriplet@plt>
  mov    %eax,%esi
  lea    D86(%rip),%rdi
  call   <printf@plt>
  mov    $0x1e8480,%edi
  call   <sumPrimes@plt>
  mov    %eax,%esi
  lea    D71(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D69
type_infered: string
name: read_data
arguments: label: D39
type_infered: string
name: read_data
arguments: label: D19
type_infered: string
name: read_data
arguments: label: D16
type_infered: string
name: read_data
arguments: label: D89
type_infered: string
name: read_data
arguments: label: D38
type_infered: string
name: read_data
arguments: label: D78
type_infered: string
name: read_data
arguments: label: D86
type_infered: string
name: read_data
arguments: label: D71
type_infered: string

tool: D69: Problem 1 result: %d


tool: D39: Problem 2 result: %d


tool: D19: Problem 3 result: %lld


tool: D16: Problem 4 result: %d


tool: D89: Problem 6 result: %d


tool: D38: Problem 7 result: %lld


tool: D78: Problem 8 result: %lld


tool: D86: Problem 9 result: %d


tool: D71: Problem 10 result: %lld


assistant: ```c
int main() {
  DoTests();
  printf("Problem 1 result: %d\n", getSumDiv());
  printf("Problem 2 result: %d\n", fibSumEven(1, 1, 4000000, 0));
  printf("Problem 3 result: %lld\n", largestPrimeFactor(600851475143));
  printf("Problem 4 result: %d\n", maxPalProd(3));
  printf("Problem 6 result: %d\n", sumSqrtDiff(100));
  printf("Problem 7 result: %lld\n", nPrimeNumber(10001));
  printf("Problem 8 result: %lld\n", maxProd1000dig(13));
  printf("Problem 9 result: %d\n", pythTriplet(1000));
  printf("Problem 10 result: %lld\n", sumPrimes(2000000));
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <DoTests@plt>
  xor    %eax,%eax
  call   <getSumDiv@plt>
  lea    D50(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %ecx,%ecx
  mov    $0x3d0900,%edx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <fibSumEven@plt>
  lea    D24(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xe589eac7,%edi
  call   <largestPrimeFactor@plt>
  lea    D77(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <maxPalProd@plt>
  lea    D40(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x64,%edi
  call   <sumSqrtDiff@plt>
  lea    D58(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x2711,%edi
  call   <nPrimeNumber@plt>
  lea    D68(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xd,%edi
  call   <maxProd1000dig@plt>
  lea    D12(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3e8,%edi
  call   <pythTriplet@plt>
  lea    D26(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x1e8480,%edi
  call   <sumPrimes@plt>
  lea    D32(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D50
type_guessed: string
name: unpack_data
arguments: tag: D24
type_guessed: string
name: unpack_data
arguments: tag: D77
type_guessed: string
name: unpack_data
arguments: tag: D40
type_guessed: string
name: unpack_data
arguments: tag: D58
type_guessed: string
name: unpack_data
arguments: tag: D68
type_guessed: string
name: unpack_data
arguments: tag: D12
type_guessed: string
name: unpack_data
arguments: tag: D26
type_guessed: string
name: unpack_data
arguments: tag: D32
type_guessed: string

tool: D50: Problem 1 result: %d


tool: D24: Problem 2 result: %d


tool: D77: Problem 3 result: %lld


tool: D40: Problem 4 result: %d


tool: D58: Problem 6 result: %d


tool: D68: Problem 7 result: %lld


tool: D12: Problem 8 result: %lld


tool: D26: Problem 9 result: %d


tool: D32: Problem 10 result: %lld


assistant: ```c
int main() {
  DoTests();
  printf("Problem 1 result: %d\n", getSumDiv());
  printf("Problem 2 result: %d\n", fibSumEven(1, 1, 4000000, 0));
  printf("Problem 3 result: %lld\n", largestPrimeFactor(600851475143));
  printf("Problem 4 result: %d\n", maxPalProd(3));
  printf("Problem 6 result: %d\n", sumSqrtDiff(100));
  printf("Problem 7 result: %lld\n", nPrimeNumber(10001));
  printf("Problem 8 result: %lld\n", maxProd1000dig(13));
  printf("Problem 9 result: %d\n", pythTriplet(1000));
  printf("Problem 10 result: %lld\n", sumPrimes(2000000));
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <DoTests@plt>
  xor    %eax,%eax
  call   <getSumDiv@plt>
  lea    D73(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %ecx,%ecx
  mov    $0x3d0900,%edx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <fibSumEven@plt>
  lea    D65(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xe589eac7,%edi
  call   <largestPrimeFactor@plt>
  lea    D14(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3,%edi
  call   <maxPalProd@plt>
  lea    D25(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x64,%edi
  call   <sumSqrtDiff@plt>
  lea    D38(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x2711,%edi
  call   <nPrimeNumber@plt>
  lea    D87(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xd,%edi
  call   <maxProd1000dig@plt>
  lea    D76(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3e8,%edi
  call   <pythTriplet@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x1e8480,%edi
  call   <sumPrimes@plt>
  lea    D55(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D73
data_type: string
name: parse_data
arguments: label: D65
data_type: string
name: parse_data
arguments: label: D14
data_type: string
name: parse_data
arguments: label: D25
data_type: string
name: parse_data
arguments: label: D38
data_type: string
name: parse_data
arguments: label: D87
data_type: string
name: parse_data
arguments: label: D76
data_type: string
name: parse_data
arguments: label: D7
data_type: string
name: parse_data
arguments: label: D55
data_type: string

tool: D73: Problem 1 result: %d


tool: D65: Problem 2 result: %d


tool: D14: Problem 3 result: %lld


tool: D25: Problem 4 result: %d


tool: D38: Problem 6 result: %d


tool: D87: Problem 7 result: %lld


tool: D76: Problem 8 result: %lld


tool: D7: Problem 9 result: %d


tool: D55: Problem 10 result: %lld


assistant: ```c
int main() {
  DoTests();
  printf("Problem 1 result: %d\n", getSumDiv());
  printf("Problem 2 result: %d\n", fibSumEven(1, 1, 4000000, 0));
  printf("Problem 3 result: %lld\n", largestPrimeFactor(600851475143));
  printf("Problem 4 result: %d\n", maxPalProd(3));
  printf("Problem 6 result: %d\n", sumSqrtDiff(100));
  printf("Problem 7 result: %lld\n", nPrimeNumber(10001));
  printf("Problem 8 result: %lld\n", maxProd1000dig(13));
  printf("Problem 9 result: %d\n", pythTriplet(1000));
  printf("Problem 10 result: %lld\n", sumPrimes(2000000));
  return 1;
}
```