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
<getarg>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  call   <strlen@plt>
  cmp    $0x1,%ebp
  jle    L50
  lea    -0x2(%rbp),%edx
  lea    0x8(%r14),%rbx
  movslq %eax,%r12
  lea    0x10(%r14,%rdx,8),%r14
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L40:
  add    $0x8,%rbx
  cmp    %r14,%rbx
  je     L50
L18:
  mov    (%rbx),%rbp
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L40
  lea    0x0(%rbp,%r12,1),%rdi
  call   <atoi@plt>
  mov    (%rbx),%r13
  mov    $0x6b,%esi
  mov    %eax,%ebp
  add    %r13,%r12
  mov    %r12,%rdi
  call   <rindex@plt>
  test   %rax,%rax
  je     L23
  shl    $0xa,%ebp
L29:
  mov    %r13,%rdi
  mov    $0x5,%edx
  lea    D4(%rip),%rsi
  call   <strncmp@plt>
  pop    %rbx
  mov    %eax,%r8d
  movslq %ebp,%rax
  pop    %rbp
  test   %r8d,%r8d
  cmove  %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L50:
  pop    %rbx
  mov    $0xffffffffffffffff,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  mov    $0x6d,%esi
  mov    %r12,%rdi
  call   <rindex@plt>
  mov    %rax,%r8
  mov    %ebp,%eax
  shl    $0x14,%eax
  test   %r8,%r8
  cmovne %eax,%ebp
  jmp    L29
```

tool_calls: name: parse_data
arguments: data_tag: D4
type_infered: string

tool: D4: label

assistant: ```c
long getarg(s, ac, av)
char *s;
char **av;
{
  register len, i;
  len = strlen(s);
  for (i = 1; i < ac; ++i)
    if (!strncmp(av[i], s, len)) {
      register bs = atoi(&av[i][len]);
      if (rindex(&av[i][len], 'k'))
        bs *= 1024;
      else if (rindex(&av[i][len], 'm'))
        bs *= (1024 * 1024);
      if (!strncmp(av[i], "label", 5)) {
        return (long)(&av[i][len]);
      }
      return (bs);
    }
  return (-1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getarg>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  call   <strlen@plt>
  cmp    $0x1,%ebp
  jle    L6
  lea    -0x2(%rbp),%edx
  lea    0x8(%r14),%rbx
  movslq %eax,%r12
  lea    0x10(%r14,%rdx,8),%r14
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L42:
  add    $0x8,%rbx
  cmp    %r14,%rbx
  je     L6
L26:
  mov    (%rbx),%rbp
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L42
  lea    0x0(%rbp,%r12,1),%rdi
  call   <atoi@plt>
  mov    (%rbx),%r13
  mov    $0x6b,%esi
  mov    %eax,%ebp
  add    %r13,%r12
  mov    %r12,%rdi
  call   <rindex@plt>
  test   %rax,%rax
  je     L24
  shl    $0xa,%ebp
L55:
  mov    %r13,%rdi
  mov    $0x5,%edx
  lea    D0(%rip),%rsi
  call   <strncmp@plt>
  pop    %rbx
  mov    %eax,%r8d
  movslq %ebp,%rax
  pop    %rbp
  test   %r8d,%r8d
  cmove  %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  pop    %rbx
  mov    $0xffffffffffffffff,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L24:
  mov    $0x6d,%esi
  mov    %r12,%rdi
  call   <rindex@plt>
  mov    %rax,%r8
  mov    %ebp,%eax
  shl    $0x14,%eax
  test   %r8,%r8
  cmovne %eax,%ebp
  jmp    L55
```

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

tool: D0: label

assistant: ```c
long getarg(s, ac, av)
char *s;
char **av;
{
  register len, i;
  len = strlen(s);
  for (i = 1; i < ac; ++i)
    if (!strncmp(av[i], s, len)) {
      register bs = atoi(&av[i][len]);
      if (rindex(&av[i][len], 'k'))
        bs *= 1024;
      else if (rindex(&av[i][len], 'm'))
        bs *= (1024 * 1024);
      if (!strncmp(av[i], "label", 5)) {
        return (long)(&av[i][len]);
      }
      return (bs);
    }
  return (-1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fib>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L27
  mov    $0x1,%eax
  jmp    L28
L27:
  cmpl   $0x1,-0x14(%rbp)
  jne    L18
  mov    $0x1,%eax
  jmp    L28
L18:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fib>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <fib>
  add    %ebx,%eax
L28:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int fib(int n) {
  if (n == 0)
    return 1;
  if (n == 1)
    return 1;
  return fib(n - 1) + fib(n - 2);
}
```
user: What is the c source code of the assembly code below:

```asm
<fib>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  ja     L4
  ret
L4:
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <fib>
  mov    %eax,%ebp
  lea    -0x2(%rbx),%edi
  call   <fib>
  add    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int fib(int n) {
  if (n == 0)
    return 1;
  if (n == 1)
    return 1;
  return fib(n - 1) + fib(n - 2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<BREAKIF>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L9
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <raise@plt>
L9:
  nop
  leave
  ret
```

assistant: ```c
void BREAKIF(int cond) {
  if (cond) {
    raise(SIGINT);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<BREAKIF>:
  endbr64
  test   %edi,%edi
  jne    L3
  ret
L3:
  sub    $0x8,%rsp
  mov    D0(%rip),%rax
  mov    (%rax),%edi
  call   <raise@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void BREAKIF(int cond) {
  if (cond) {
    raise(SIGINT);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<BREAKIF>:
  endbr64
  test   %edi,%edi
  jne    L9
  ret
  nopl   0x0(%rax)
L9:
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  jmp    1050 <raise@plt>
```

assistant: ```c
void BREAKIF(int cond) {
  if (cond) {
    raise(SIGINT);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<BREAKIF>:
  endbr64
  test   %edi,%edi
  jne    L7
  ret
  nopl   0x0(%rax)
L7:
  mov    D0(%rip),%rax
  mov    (%rax),%edi
  jmp    1050 <raise@plt>
```

assistant: ```c
void BREAKIF(int cond) {
  if (cond) {
    raise(SIGINT);
  }
}
```
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    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  addl   $0x5,-0x8(%rbp)
  addl   $0x5,-0x4(%rbp)
  movl   $0x500,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x10(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  movl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L19
L2:
  mov    -0x10(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  mov    %eax,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L19:
  cmpl   $0x7,-0xc(%rbp)
  jle    L2
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x10(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int res, res1 = 0, res2 = 0;
#pragma acc parallel num_gangs(GANGS) num_workers(32) vector_length(32) reduction(+ : res1) copy(res2, res1)
  {
    res1 += 5;
#pragma acc atomic
    res2 += 5;
  }
  res = 256 * 5;
  assert(res == res1);
  assert(res == res2);
  res = res1 = res2 = 1;
#pragma acc parallel num_gangs(GANGS) num_workers(32) vector_length(32) reduction(* : res1) copy(res1, res2)
  {
    res1 *= 5;
#pragma acc atomic
    res2 *= 5;
  }
  for (int i = 0; i < 8; ++i)
    res *= 5;
  assert(res == res1);
  assert(res == res2);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int res, res1 = 0, res2 = 0;
#pragma acc parallel num_gangs(GANGS) num_workers(32) vector_length(32) reduction(+ : res1) copy(res2, res1)
  {
    res1 += 5;
#pragma acc atomic
    res2 += 5;
  }
  res = 256 * 5;
  assert(res == res1);
  assert(res == res2);
  res = res1 = res2 = 1;
#pragma acc parallel num_gangs(GANGS) num_workers(32) vector_length(32) reduction(* : res1) copy(res1, res2)
  {
    res1 *= 5;
#pragma acc atomic
    res2 *= 5;
  }
  for (int i = 0; i < 8; ++i)
    res *= 5;
  assert(res == res1);
  assert(res == res2);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int res, res1 = 0, res2 = 0;
#pragma acc parallel num_gangs(GANGS) num_workers(32) vector_length(32) reduction(+ : res1) copy(res2, res1)
  {
    res1 += 5;
#pragma acc atomic
    res2 += 5;
  }
  res = 256 * 5;
  assert(res == res1);
  assert(res == res2);
  res = res1 = res2 = 1;
#pragma acc parallel num_gangs(GANGS) num_workers(32) vector_length(32) reduction(* : res1) copy(res1, res2)
  {
    res1 *= 5;
#pragma acc atomic
    res2 *= 5;
  }
  for (int i = 0; i < 8; ++i)
    res *= 5;
  assert(res == res1);
  assert(res == res2);
  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
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int res, res1 = 0, res2 = 0;
#pragma acc parallel num_gangs(GANGS) num_workers(32) vector_length(32) reduction(+ : res1) copy(res2, res1)
  {
    res1 += 5;
#pragma acc atomic
    res2 += 5;
  }
  res = 256 * 5;
  assert(res == res1);
  assert(res == res2);
  res = res1 = res2 = 1;
#pragma acc parallel num_gangs(GANGS) num_workers(32) vector_length(32) reduction(* : res1) copy(res1, res2)
  {
    res1 *= 5;
#pragma acc atomic
    res2 *= 5;
  }
  for (int i = 0; i < 8; ++i)
    res *= 5;
  assert(res == res1);
  assert(res == res2);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<do_rounding>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <__builtin_clzll@plt>
  mov    %eax,%edx
  mov    $0x3f,%eax
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x34,-0x4(%rbp)
  jle    L6
  mov    -0x4(%rbp),%eax
  sub    $0x35,%eax
  mov    %eax,%ecx
  sarl   %cl,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  add    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x34,%eax
  mov    %eax,%ecx
  shll   %cl,-0x14(%rbp)
L6:
  mov    -0x14(%rbp),%eax
  leave
  ret
```

assistant: ```c
uint64_t do_rounding(uint64_t x) {
  int idx = 63 - __builtin_clzll(x);
  if (idx > 52) {
    x >>= idx - 52 - 1;
    x = (x >> 1) + (x & 1);
    x <<= idx - 52;
  }
  return x;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<do_rounding>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <__builtin_clzll@plt>
  mov    $0x3f,%edx
  sub    %eax,%edx
  cmp    $0x34,%edx
  jle    L4
  mov    $0xa,%ecx
  sub    %eax,%ecx
  sar    %cl,%ebx
  mov    %ebx,%edx
  sar    %edx
  and    $0x1,%ebx
  add    %ebx,%edx
  mov    $0xb,%ecx
  sub    %eax,%ecx
  mov    %edx,%ebx
  shl    %cl,%ebx
L4:
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
uint64_t do_rounding(uint64_t x) {
  int idx = 63 - __builtin_clzll(x);
  if (idx > 52) {
    x >>= idx - 52 - 1;
    x = (x >> 1) + (x & 1);
    x <<= idx - 52;
  }
  return x;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<do_rounding>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  call   <__builtin_clzll@plt>
  mov    $0x3f,%edx
  sub    %eax,%edx
  cmp    $0x34,%edx
  jle    L3
  mov    $0xa,%ecx
  sub    %eax,%ecx
  sar    %cl,%r12d
  mov    $0xb,%ecx
  mov    %r12d,%edx
  and    $0x1,%r12d
  sub    %eax,%ecx
  sar    %edx
  add    %edx,%r12d
  shl    %cl,%r12d
L3:
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint64_t do_rounding(uint64_t x) {
  int idx = 63 - __builtin_clzll(x);
  if (idx > 52) {
    x >>= idx - 52 - 1;
    x = (x >> 1) + (x & 1);
    x <<= idx - 52;
  }
  return x;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<do_rounding>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  call   <__builtin_clzll@plt>
  mov    $0x3f,%edx
  sub    %eax,%edx
  cmp    $0x34,%edx
  jle    L7
  mov    $0xa,%ecx
  sub    %eax,%ecx
  sar    %cl,%r12d
  mov    $0xb,%ecx
  mov    %r12d,%edx
  and    $0x1,%r12d
  sub    %eax,%ecx
  sar    %edx
  add    %edx,%r12d
  shl    %cl,%r12d
L7:
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint64_t do_rounding(uint64_t x) {
  int idx = 63 - __builtin_clzll(x);
  if (idx > 52) {
    x >>= idx - 52 - 1;
    x = (x >> 1) + (x & 1);
    x <<= idx - 52;
  }
  return x;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ft_isalpha>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x60,-0x4(%rbp)
  jle    L35
  cmpl   $0x7a,-0x4(%rbp)
  jg     L35
  mov    $0x1,%eax
  jmp    L41
L35:
  cmpl   $0x40,-0x4(%rbp)
  jle    L47
  cmpl   $0x5a,-0x4(%rbp)
  jg     L47
  mov    $0x1,%eax
  jmp    L41
L47:
  mov    $0x0,%eax
L41:
  pop    %rbp
  ret
```

assistant: ```c
int ft_isalpha(int c) {
  if (c >= 'a' && c <= 'z')
    return (1);
  if (c >= 'A' && c <= 'Z')
    return (1);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_isalpha>:
  endbr64
  and    $0xffffffdf,%edi
  sub    $0x41,%edi
  cmp    $0x19,%edi
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int ft_isalpha(int c) {
  if (c >= 'a' && c <= 'z')
    return (1);
  if (c >= 'A' && c <= 'Z')
    return (1);
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_isalpha>:
  endbr64
  and    $0xffffffdf,%edi
  xor    %eax,%eax
  sub    $0x41,%edi
  cmp    $0x19,%edi
  setbe  %al
  ret
```

assistant: ```c
int ft_isalpha(int c) {
  if (c >= 'a' && c <= 'z')
    return (1);
  if (c >= 'A' && c <= 'Z')
    return (1);
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %edi,-0x84(%rbp)
  mov    %rsi,-0x90(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <getString@plt>
  mov    %rax,-0x80(%rbp)
  mov    -0x80(%rbp),%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x80(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  call   <getStringNoMalloc@plt>
  mov    %rax,-0x78(%rbp)
  mov    -0x78(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <getStringCallByRef@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D28
data_type: string
name: read_data
arguments: tag: D6
data_type: string
name: read_data
arguments: tag: D29
data_type: string

tool: D28: MALLOC ### a = %s 


tool: D6: NO MALLOC ### b = %s 


tool: D29: CALLBYREF ### c = %s 


assistant: ```c
int main(int argc, char *argv[]) {
  char *a = getString();
  printf("MALLOC ### a = %s \n", a);
  free(a);
  char *b = getStringNoMalloc();
  printf("NO MALLOC ### b = %s \n", b);
  char c[100];
  getStringCallByRef(c);
  printf("CALLBYREF ### c = %s \n", c);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  call   <getString@plt>
  mov    %rax,%rbx
  mov    %rax,%rsi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  call   <getStringNoMalloc@plt>
  mov    %rax,%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <getStringCallByRef@plt>
  mov    %rbx,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D13
type_guessed: string
name: disassemble_data
arguments: tag: D7
type_guessed: string
name: disassemble_data
arguments: tag: D8
type_guessed: string

tool: D13: MALLOC ### a = %s 


tool: D7: NO MALLOC ### b = %s 


tool: D8: CALLBYREF ### c = %s 


assistant: ```c
int main(int argc, char *argv[]) {
  char *a = getString();
  printf("MALLOC ### a = %s \n", a);
  free(a);
  char *b = getStringNoMalloc();
  printf("NO MALLOC ### b = %s \n", b);
  char c[100];
  getStringCallByRef(c);
  printf("CALLBYREF ### c = %s \n", c);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  call   <getString@plt>
  lea    D29(%rip),%rdi
  mov    %rax,%rsi
  mov    %rax,%rbp
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    %rsp,%rbp
  call   <free@plt>
  xor    %eax,%eax
  call   <getStringNoMalloc@plt>
  lea    D23(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <getStringCallByRef@plt>
  mov    %rbp,%rsi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D29
type_guessed: string
name: read_data
arguments: tag: D23
type_guessed: string
name: read_data
arguments: tag: D16
type_guessed: string

tool: D29: MALLOC ### a = %s 


tool: D23: NO MALLOC ### b = %s 


tool: D16: CALLBYREF ### c = %s 


assistant: ```c
int main(int argc, char *argv[]) {
  char *a = getString();
  printf("MALLOC ### a = %s \n", a);
  free(a);
  char *b = getStringNoMalloc();
  printf("NO MALLOC ### b = %s \n", b);
  char c[100];
  getStringCallByRef(c);
  printf("CALLBYREF ### c = %s \n", c);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  call   <getString@plt>
  lea    D15(%rip),%rdi
  mov    %rax,%rsi
  mov    %rax,%rbp
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    %rsp,%rbp
  call   <free@plt>
  xor    %eax,%eax
  call   <getStringNoMalloc@plt>
  lea    D28(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <getStringCallByRef@plt>
  mov    %rbp,%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D15
datatype: string
name: read_data
arguments: tag: D28
datatype: string
name: read_data
arguments: tag: D7
datatype: string

tool: D15: MALLOC ### a = %s 


tool: D28: NO MALLOC ### b = %s 


tool: D7: CALLBYREF ### c = %s 


assistant: ```c
int main(int argc, char *argv[]) {
  char *a = getString();
  printf("MALLOC ### a = %s \n", a);
  free(a);
  char *b = getStringNoMalloc();
  printf("NO MALLOC ### b = %s \n", b);
  char c[100];
  getStringCallByRef(c);
  printf("CALLBYREF ### c = %s \n", c);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x30(%rbp),%rax
  mov    0x8(%rax),%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L35
L15:
  mov    -0x20(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  addl   $0x1,-0x20(%rbp)
L35:
  cmpl   $0xf,-0x20(%rbp)
  jle    L15
  mov    -0x18(%rbp),%rbx
  jmp    L17
L30:
  movzbl (%rbx),%eax
  movsbl %al,%eax
  add    %eax,-0x1c(%rbp)
  add    $0x1,%rbx
L17:
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L30
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D3
datatype: string

tool: D3: Sum is: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i, j;
  int sum = 0;
  const char *str = argv[1];
  register const char *s, *t;
  for (i = 0; i < 16; i++) {
    sum += i;
  }
  for (t = str; *t; ++t) {
    sum += *t;
  }
  printf("Sum is: %d\n", sum);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdx
  movzbl (%rdx),%eax
  test   %al,%al
  je     L12
  mov    $0x78,%esi
L4:
  movsbl %al,%eax
  add    %eax,%esi
  add    $0x1,%rdx
  movzbl (%rdx),%eax
  test   %al,%al
  jne    L4
L9:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L12:
  mov    $0x78,%esi
  jmp    L9
```

tool_calls: name: parse_data
arguments: label: D8
type_guessed: string

tool: D8: Sum is: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i, j;
  int sum = 0;
  const char *str = argv[1];
  register const char *s, *t;
  for (i = 0; i < 16; i++) {
    sum += i;
  }
  for (t = str; *t; ++t) {
    sum += *t;
  }
  printf("Sum is: %d\n", sum);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdx
  mov    $0x78,%esi
  movsbl (%rdx),%eax
  test   %al,%al
  je     L12
  nopl   0x0(%rax,%rax,1)
L18:
  add    $0x1,%rdx
  add    %eax,%esi
  movsbl (%rdx),%eax
  test   %al,%al
  jne    L18
L12:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
type_infered: string

tool: D8: Sum is: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i, j;
  int sum = 0;
  const char *str = argv[1];
  register const char *s, *t;
  for (i = 0; i < 16; i++) {
    sum += i;
  }
  for (t = str; *t; ++t) {
    sum += *t;
  }
  printf("Sum is: %d\n", sum);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdx
  mov    $0x78,%esi
  movsbl (%rdx),%eax
  test   %al,%al
  je     L10
  nopl   0x0(%rax,%rax,1)
L11:
  add    $0x1,%rdx
  add    %eax,%esi
  movsbl (%rdx),%eax
  test   %al,%al
  jne    L11
L10:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_infered: string

tool: D1: Sum is: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i, j;
  int sum = 0;
  const char *str = argv[1];
  register const char *s, *t;
  for (i = 0; i < 16; i++) {
    sum += i;
  }
  for (t = str; *t; ++t) {
    sum += *t;
  }
  printf("Sum is: %d\n", sum);
  return 0;
}
```