ylfeng/ReF-Decompile-dataset huggingface dataset:
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   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  lea    D9(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  nopl   0x0(%rax,%rax,1)
L20:
  mov    %r13,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    %ebp,%eax
  jle    L18
  mov    %ebx,%r14d
  mov    %eax,%ebp
L18:
  cmp    $0x64,%ebx
  jne    L20
  mov    %r14d,%edx
  mov    %ebp,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
data_type: string
name: extract_symbolic_data
arguments: data_tag: D18
data_type: string

tool: D9:  %d

tool: D18: %d
%d


assistant: ```c
int main() {
  int X, i = 0, maior = 0, maior_p = 0;
  while (i < 100) {
    scanf(" %d", &X);
    if (X > maior) {
      maior = X;
      maior_p = i + 1;
    }
    i++;
  }
  printf("%d\n%d\n", maior, maior_p);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<would_overflow>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jle    L1
  cmpl   $0x0,-0xc(%rbp)
  jle    L1
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  test   %eax,%eax
  jns    L1
  mov    $0x1,%eax
  jmp    L14
L1:
  mov    $0x0,%eax
L14:
  pop    %rbp
  ret
```

assistant: ```c
int would_overflow(long long a, int b) {
  if (a > 0 && b > 0 && (int)a + b < 0)
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<would_overflow>:
  endbr64
  test   %rdi,%rdi
  jle    L2
  test   %esi,%esi
  jle    L2
  lea    (%rsi,%rdi,1),%eax
  shr    $0x1f,%eax
  ret
L2:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int would_overflow(long long a, int b) {
  if (a > 0 && b > 0 && (int)a + b < 0)
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<would_overflow>:
  endbr64
  test   %rdi,%rdi
  jle    L1
  test   %esi,%esi
  jle    L1
  lea    (%rsi,%rdi,1),%eax
  shr    $0x1f,%eax
  ret
  nopl   0x0(%rax)
L1:
  xor    %eax,%eax
  ret
```

assistant: ```c
int would_overflow(long long a, int b) {
  if (a > 0 && b > 0 && (int)a + b < 0)
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<would_overflow>:
  endbr64
  test   %rdi,%rdi
  jle    L3
  test   %esi,%esi
  jle    L3
  lea    (%rsi,%rdi,1),%eax
  shr    $0x1f,%eax
  ret
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
  ret
```

assistant: ```c
int would_overflow(long long a, int b) {
  if (a > 0 && b > 0 && (int)a + b < 0)
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mkfifo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  mov    D19(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int mkfifo(const char *pathname, mode_t mode) {
  errno = ENOSYS;
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mkfifo>:
  endbr64
  mov    D15(%rip),%rax
  mov    (%rax),%edx
  mov    D7(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int mkfifo(const char *pathname, mode_t mode) {
  errno = ENOSYS;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mkfifo>:
  endbr64
  mov    D15(%rip),%rax
  mov    (%rax),%edx
  mov    D11(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int mkfifo(const char *pathname, mode_t mode) {
  errno = ENOSYS;
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mkfifo>:
  endbr64
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  mov    D10(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int mkfifo(const char *pathname, mode_t mode) {
  errno = ENOSYS;
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<NESRemoveTitle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L15
  mov    $0x0,%eax
  jmp    L25
L15:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <NESHasTitle@plt>
  test   %eax,%eax
  jne    L53
  mov    $0x1,%eax
  jmp    L25
L53:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <NESGetPrgBankCount@plt>
  shl    $0xe,%eax
  lea    0x10(%rax),%ebx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <NESGetChrBankCount@plt>
  shl    $0xd,%eax
  add    %eax,%ebx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fileno@plt>
  mov    %ebx,%esi
  mov    %eax,%edi
  call   <ftruncate@plt>
  test   %rax,%rax
  je     L49
  mov    $0x0,%eax
  jmp    L25
L49:
  mov    $0x1,%eax
L25:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int NESRemoveTitle(FILE *ofile) {
  if (!ofile)
    return 0;
  if (!NESHasTitle(ofile))
    return 1;
  if (ftruncate(fileno(ofile), 16 + (NESGetPrgBankCount(ofile) * 16384) + (NESGetChrBankCount(ofile) * 8192)) != 0) {
    return 0;
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<NESRemoveTitle>:
  endbr64
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L27
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <NESHasTitle@plt>
  mov    %eax,%edx
  mov    $0x1,%eax
  test   %edx,%edx
  jne    L6
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  mov    %rbx,%rdi
  call   <NESGetPrgBankCount@plt>
  mov    %eax,%ebp
  mov    %rbx,%rdi
  call   <NESGetChrBankCount@plt>
  shl    $0xe,%ebp
  shl    $0xd,%eax
  lea    0x10(%rbp,%rax,1),%ebp
  mov    %rbx,%rdi
  call   <fileno@plt>
  mov    %eax,%edi
  mov    %ebp,%esi
  call   <ftruncate@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L3
L27:
  ret
```

assistant: ```c
int NESRemoveTitle(FILE *ofile) {
  if (!ofile)
    return 0;
  if (!NESHasTitle(ofile))
    return 1;
  if (ftruncate(fileno(ofile), 16 + (NESGetPrgBankCount(ofile) * 16384) + (NESGetChrBankCount(ofile) * 8192)) != 0) {
    return 0;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<NESRemoveTitle>:
  endbr64
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L19
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <NESHasTitle@plt>
  mov    %eax,%r8d
  mov    $0x1,%eax
  test   %r8d,%r8d
  jne    L10
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L10:
  mov    %rbp,%rdi
  call   <NESGetPrgBankCount@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <NESGetChrBankCount@plt>
  shl    $0xe,%ebx
  mov    %rbp,%rdi
  shl    $0xd,%eax
  lea    0x10(%rbx,%rax,1),%r12d
  call   <fileno@plt>
  mov    %r12d,%esi
  mov    %eax,%edi
  call   <ftruncate@plt>
  pop    %rbx
  pop    %rbp
  test   %rax,%rax
  pop    %r12
  sete   %al
  movzbl %al,%eax
  ret
  nopl   (%rax)
L19:
  ret
```

assistant: ```c
int NESRemoveTitle(FILE *ofile) {
  if (!ofile)
    return 0;
  if (!NESHasTitle(ofile))
    return 1;
  if (ftruncate(fileno(ofile), 16 + (NESGetPrgBankCount(ofile) * 16384) + (NESGetChrBankCount(ofile) * 8192)) != 0) {
    return 0;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<NESRemoveTitle>:
  endbr64
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <NESHasTitle@plt>
  mov    %eax,%r8d
  mov    $0x1,%eax
  test   %r8d,%r8d
  jne    L11
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L11:
  mov    %rbp,%rdi
  call   <NESGetPrgBankCount@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <NESGetChrBankCount@plt>
  shl    $0xe,%ebx
  mov    %rbp,%rdi
  shl    $0xd,%eax
  lea    0x10(%rbx,%rax,1),%r12d
  call   <fileno@plt>
  mov    %r12d,%esi
  mov    %eax,%edi
  call   <ftruncate@plt>
  pop    %rbx
  pop    %rbp
  test   %rax,%rax
  pop    %r12
  sete   %al
  movzbl %al,%eax
  ret
  nopl   (%rax)
L13:
  ret
```

assistant: ```c
int NESRemoveTitle(FILE *ofile) {
  if (!ofile)
    return 0;
  if (!NESHasTitle(ofile))
    return 1;
  if (ftruncate(fileno(ofile), 16 + (NESGetPrgBankCount(ofile) * 16384) + (NESGetChrBankCount(ofile) * 8192)) != 0) {
    return 0;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<t>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <q1@plt>
  mov    $0x0,%eax
  call   <q@plt>
  pop    %rbp
  ret
```

assistant: ```c
int t(void) {
  q1();
  return q();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<t>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <q1@plt>
  mov    $0x0,%eax
  call   <q@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int t(void) {
  q1();
  return q();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<t>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <q1@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <q@plt>
```

assistant: ```c
int t(void) {
  q1();
  return q();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<t>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <q1@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <q@plt>
```

assistant: ```c
int t(void) {
  q1();
  return q();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<getBEdword>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x18,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x10,%eax
  or     %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  or     %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int getBEdword(char buf[4]) {
  unsigned char *b = (unsigned char *)buf;
  return (b[0] << 24) | (b[1] << 16) | (b[2] << 8) | (b[3] << 0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getBEdword>:
  endbr64
  movzbl (%rdi),%eax
  shl    $0x18,%eax
  movzbl 0x1(%rdi),%edx
  shl    $0x10,%edx
  or     %edx,%eax
  movzbl 0x3(%rdi),%edx
  or     %edx,%eax
  movzbl 0x2(%rdi),%edx
  shl    $0x8,%edx
  or     %edx,%eax
  ret
```

assistant: ```c
int getBEdword(char buf[4]) {
  unsigned char *b = (unsigned char *)buf;
  return (b[0] << 24) | (b[1] << 16) | (b[2] << 8) | (b[3] << 0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getBEdword>:
  endbr64
  mov    (%rdi),%eax
  bswap  %eax
  ret
```

assistant: ```c
int getBEdword(char buf[4]) {
  unsigned char *b = (unsigned char *)buf;
  return (b[0] << 24) | (b[1] << 16) | (b[2] << 8) | (b[3] << 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<c_get_cmsg_overhead>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x38(%rbp)
  lea    -0x9(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  movl   $0x100,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <CMSG_FIRSTHDR@plt>
  mov    %rax,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <CMSG_DATA@plt>
  cltq
  sub    -0x38(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
uint64_t c_get_cmsg_overhead() {
  char buf[1];
  struct msghdr mhdr;
  struct cmsghdr *p = 0;
  mhdr.msg_control = buf;
  mhdr.msg_controllen = 256;
  mhdr.msg_flags = 0;
  p = CMSG_FIRSTHDR(&mhdr);
  return ((char *)(CMSG_DATA(p)) - ((char *)p));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<c_get_cmsg_overhead>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x27(%rsp),%rax
  mov    %rax,(%rsp)
  movl   $0x100,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    %rsp,%rdi
  call   <CMSG_FIRSTHDR@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <CMSG_DATA@plt>
  cltq
  sub    %rbx,%rax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x30,%rsp
  pop    %rbx
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t c_get_cmsg_overhead() {
  char buf[1];
  struct msghdr mhdr;
  struct cmsghdr *p = 0;
  mhdr.msg_control = buf;
  mhdr.msg_controllen = 256;
  mhdr.msg_flags = 0;
  p = CMSG_FIRSTHDR(&mhdr);
  return ((char *)(CMSG_DATA(p)) - ((char *)p));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<c_get_cmsg_overhead>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    0x27(%rsp),%rax
  movl   $0x100,0x8(%rsp)
  mov    %rax,(%rsp)
  movq   $0x0,0x10(%rsp)
  call   <CMSG_FIRSTHDR@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  xor    %eax,%eax
  call   <CMSG_DATA@plt>
  cltq
  sub    %rbx,%rax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x30,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t c_get_cmsg_overhead() {
  char buf[1];
  struct msghdr mhdr;
  struct cmsghdr *p = 0;
  mhdr.msg_control = buf;
  mhdr.msg_controllen = 256;
  mhdr.msg_flags = 0;
  p = CMSG_FIRSTHDR(&mhdr);
  return ((char *)(CMSG_DATA(p)) - ((char *)p));
}
```
user: What is the c source code of the assembly code below?

```asm
<c_get_cmsg_overhead>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    0x27(%rsp),%rax
  movl   $0x100,0x8(%rsp)
  mov    %rax,(%rsp)
  movq   $0x0,0x10(%rsp)
  call   <CMSG_FIRSTHDR@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  xor    %eax,%eax
  call   <CMSG_DATA@plt>
  cltq
  sub    %rbx,%rax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x30,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t c_get_cmsg_overhead() {
  char buf[1];
  struct msghdr mhdr;
  struct cmsghdr *p = 0;
  mhdr.msg_control = buf;
  mhdr.msg_controllen = 256;
  mhdr.msg_flags = 0;
  p = CMSG_FIRSTHDR(&mhdr);
  return ((char *)(CMSG_DATA(p)) - ((char *)p));
}
```
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
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D48
data_type: string
name: parse_data
arguments: tag: D15
data_type: string
name: parse_data
arguments: tag: D5
data_type: string
name: parse_data
arguments: tag: D44
data_type: string
name: parse_data
arguments: tag: D49
data_type: string

tool: D48: Input the first integer: 

tool: D15: %d

tool: D5: Input the second integer: 

tool: D44: The values that were entered are: %d and %d

tool: D49: 
The total of the two numbers is: %d

assistant: ```c
void main() {
  int a, b;
  int total;
  printf("Input the first integer: ");
  scanf("%d", &a);
  printf("Input the second integer: ");
  scanf("%d", &b);
  total = a + b;
  printf("The values that were entered are: %d and %d", a, b);
  printf("\nThe total of the two numbers is: %d", total);
}
```
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   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D0(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    (%rsp),%ebx
  mov    0x4(%rsp),%ebp
  mov    %ebp,%edx
  mov    %ebx,%esi
  lea    D48(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    (%rbx,%rbp,1),%esi
  lea    D22(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D7
datatype: string
name: disassemble_data
arguments: label: D0
datatype: string
name: disassemble_data
arguments: label: D33
datatype: string
name: disassemble_data
arguments: label: D48
datatype: string
name: disassemble_data
arguments: label: D22
datatype: string

tool: D7: Input the first integer: 

tool: D0: %d

tool: D33: Input the second integer: 

tool: D48: The values that were entered are: %d and %d

tool: D22: 
The total of the two numbers is: %d

assistant: ```c
void main() {
  int a, b;
  int total;
  printf("Input the first integer: ");
  scanf("%d", &a);
  printf("Input the second integer: ");
  scanf("%d", &b);
  total = a + b;
  printf("The values that were entered are: %d and %d", a, b);
  printf("\nThe total of the two numbers is: %d", total);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D33(%rip),%rdi
  lea    D17(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    (%rsp),%ebx
  mov    0x4(%rsp),%ebp
  xor    %eax,%eax
  lea    D31(%rip),%rdi
  mov    %ebx,%esi
  mov    %ebp,%edx
  call   <printf@plt>
  xor    %eax,%eax
  lea    (%rbx,%rbp,1),%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D33
datatype: string
name: read_data
arguments: tag: D17
datatype: string
name: read_data
arguments: tag: D13
datatype: string
name: read_data
arguments: tag: D31
datatype: string
name: read_data
arguments: tag: D7
datatype: string

tool: D33: Input the first integer: 

tool: D17: %d

tool: D13: Input the second integer: 

tool: D31: The values that were entered are: %d and %d

tool: D7: 
The total of the two numbers is: %d

assistant: ```c
void main() {
  int a, b;
  int total;
  printf("Input the first integer: ");
  scanf("%d", &a);
  printf("Input the second integer: ");
  scanf("%d", &b);
  total = a + b;
  printf("The values that were entered are: %d and %d", a, b);
  printf("\nThe total of the two numbers is: %d", total);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D11(%rip),%rdi
  lea    D1(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    (%rsp),%ebx
  mov    0x4(%rsp),%ebp
  xor    %eax,%eax
  lea    D31(%rip),%rdi
  mov    %ebx,%esi
  mov    %ebp,%edx
  call   <printf@plt>
  xor    %eax,%eax
  lea    (%rbx,%rbp,1),%esi
  lea    D36(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D11
type: string
name: unpack_data
arguments: label: D1
type: string
name: unpack_data
arguments: label: D5
type: string
name: unpack_data
arguments: label: D31
type: string
name: unpack_data
arguments: label: D36
type: string

tool: D11: Input the first integer: 

tool: D1: %d

tool: D5: Input the second integer: 

tool: D31: The values that were entered are: %d and %d

tool: D36: 
The total of the two numbers is: %d

assistant: ```c
void main() {
  int a, b;
  int total;
  printf("Input the first integer: ");
  scanf("%d", &a);
  printf("Input the second integer: ");
  scanf("%d", &b);
  total = a + b;
  printf("The values that were entered are: %d and %d", a, b);
  printf("\nThe total of the two numbers is: %d", total);
}
```