ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x20796b7320656874,%rax
  movabs $0x65756c62207369,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <reverse@plt>
  movl   $0x0,-0x28(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L38
L43:
  mov    -0x24(%rbp),%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  cmp    $0x20,%al
  jne    L4
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x28(%rbp),%ecx
  lea    -0x20(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <reverse@plt>
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x28(%rbp)
  jmp    L44
L4:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  cmp    %eax,-0x24(%rbp)
  jne    L44
  mov    -0x24(%rbp),%edx
  mov    -0x28(%rbp),%ecx
  lea    -0x20(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <reverse@plt>
L44:
  addl   $0x1,-0x24(%rbp)
L38:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x24(%rbp)
  jl     L43
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
data_type: string

tool: D5: %s 


assistant: ```c
int main() {
  char s[] = "the sky is blue";
  printf("%s \n", s);
  reverse(s, 0, strlen(s) - 1);
  int last = 0;
  for (int i = 0; i < strlen(s); i++) {
    if (s[i] == ' ') {
      reverse(s, last, i - 1);
      last = i + 1;
      continue;
    }
    if (i == strlen(s) - 1) {
      reverse(s, last, i);
    }
  }
  printf("%s \n", s);
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x20796b7320656874,%rax
  movabs $0x65756c62207369,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <reverse@plt>
  mov    $0x0,%ebx
  mov    $0x0,%r14d
  mov    %rsp,%rbp
  jmp    L3
L11:
  lea    -0x1(%r15),%edx
  mov    %r14d,%esi
  mov    %rbp,%rdi
  call   <reverse@plt>
  lea    0x1(%r15),%r14d
L43:
  add    $0x1,%rbx
L3:
  mov    %ebx,%r15d
  mov    %ebx,%r13d
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %ebx,%r12d
  cmp    %ebx,%eax
  jle    L12
  cmpb   $0x20,(%rbx,%rbp,1)
  je     L11
  mov    %rbp,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  cmp    %r12d,%eax
  jne    L43
  mov    %r13d,%edx
  mov    %r14d,%esi
  mov    %rbp,%rdi
  call   <reverse@plt>
  jmp    L43
L12:
  mov    %rsp,%rsi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  mov    $0x1,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D15
type_guessed: string

tool: D15: %s 


assistant: ```c
int main() {
  char s[] = "the sky is blue";
  printf("%s \n", s);
  reverse(s, 0, strlen(s) - 1);
  int last = 0;
  for (int i = 0; i < strlen(s); i++) {
    if (s[i] == ' ') {
      reverse(s, last, i - 1);
      last = i + 1;
      continue;
    }
    if (i == strlen(s) - 1) {
      reverse(s, last, i);
    }
  }
  printf("%s \n", s);
  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
  push   %r15
  push   %r14
  lea    D7(%rip),%r14
  push   %r13
  mov    %r14,%rdi
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x28,%rsp
  movdqa D10(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  movaps %xmm0,(%rsp)
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  lea    -0x1(%rax),%edx
  call   <reverse@plt>
  jmp    L3
  nop
L51:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  cmp    %r12d,%edx
  je     L17
L57:
  add    $0x1,%rbx
L3:
  mov    %rbp,%rdi
  mov    %ebx,%r12d
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L15
  cmpb   $0x20,0x0(%rbp,%rbx,1)
  lea    0x1(%rbx),%r15d
  jne    L51
  mov    %r13d,%esi
  lea    -0x1(%r12),%edx
  mov    %rbp,%rdi
  mov    %r15d,%r13d
  call   <reverse@plt>
  jmp    L57
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <reverse@plt>
  jmp    L57
  nopl   (%rax)
L15:
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  add    $0x28,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L61:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D7
type: string
name: unpack_data
arguments: data_label: D10
type: byte[16]

tool: D7: %s 


tool: {"D10": [0x74,0x68,0x65,0x20,0x73,0x6b,0x79,0x20,0x69,0x73,0x20,0x62,0x6c,0x75,0x65,0x00]}

assistant: ```c
int main() {
  char s[] = "the sky is blue";
  printf("%s \n", s);
  reverse(s, 0, strlen(s) - 1);
  int last = 0;
  for (int i = 0; i < strlen(s); i++) {
    if (s[i] == ' ') {
      reverse(s, last, i - 1);
      last = i + 1;
      continue;
    }
    if (i == strlen(s) - 1) {
      reverse(s, last, i);
    }
  }
  printf("%s \n", s);
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D17(%rip),%r14
  push   %r13
  mov    %r14,%rdi
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x28,%rsp
  movdqa D5(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  movaps %xmm0,(%rsp)
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  lea    -0x1(%rax),%edx
  call   <reverse@plt>
  jmp    L0
  nop
L27:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  cmp    %r12d,%edx
  je     L10
L47:
  add    $0x1,%rbx
L0:
  mov    %rbp,%rdi
  mov    %ebx,%r12d
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L26
  cmpb   $0x20,0x0(%rbp,%rbx,1)
  lea    0x1(%rbx),%r15d
  jne    L27
  mov    %r13d,%esi
  lea    -0x1(%r12),%edx
  mov    %rbp,%rdi
  mov    %r15d,%r13d
  call   <reverse@plt>
  jmp    L47
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <reverse@plt>
  jmp    L47
  nopl   (%rax)
L26:
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D17
type_infered: string
name: retrieve_data
arguments: label: D5
type_infered: byte[16]

tool: D17: %s 


tool: {"D5": [0x74,0x68,0x65,0x20,0x73,0x6b,0x79,0x20,0x69,0x73,0x20,0x62,0x6c,0x75,0x65,0x00]}

assistant: ```c
int main() {
  char s[] = "the sky is blue";
  printf("%s \n", s);
  reverse(s, 0, strlen(s) - 1);
  int last = 0;
  for (int i = 0; i < strlen(s); i++) {
    if (s[i] == ' ') {
      reverse(s, last, i - 1);
      last = i + 1;
      continue;
    }
    if (i == strlen(s) - 1) {
      reverse(s, last, i);
    }
  }
  printf("%s \n", s);
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ft_strsub>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L4
  mov    $0x0,%eax
  jmp    L6
L4:
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <ft_memcpy@plt>
L6:
  leave
  ret
```

assistant: ```c
char *ft_strsub(char *str, size_t start, size_t len) {
  char *s;
  s = (char *)malloc(len + 1);
  if (!s)
    return (NULL);
  return ((char *)ft_memcpy(s, str + start, len));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ft_strsub>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  lea    0x1(%rdx),%rdi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L9
  lea    (%rbx,%r12,1),%rsi
  mov    %rbp,%rdx
  call   <ft_memcpy@plt>
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ft_strsub(char *str, size_t start, size_t len) {
  char *s;
  s = (char *)malloc(len + 1);
  if (!s)
    return (NULL);
  return ((char *)ft_memcpy(s, str + start, len));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_strsub>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x1(%rdx),%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L2
  lea    (%rbx,%rbp,1),%rsi
  mov    %r12,%rdx
  pop    %rbx
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <ft_memcpy@plt>
  nopw   0x0(%rax,%rax,1)
L2:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ft_strsub(char *str, size_t start, size_t len) {
  char *s;
  s = (char *)malloc(len + 1);
  if (!s)
    return (NULL);
  return ((char *)ft_memcpy(s, str + start, len));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_strsub>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x1(%rdx),%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L2
  lea    (%rbx,%rbp,1),%rsi
  mov    %r12,%rdx
  pop    %rbx
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <ft_memcpy@plt>
  nopw   0x0(%rax,%rax,1)
L2:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *ft_strsub(char *str, size_t start, size_t len) {
  char *s;
  s = (char *)malloc(len + 1);
  if (!s)
    return (NULL);
  return ((char *)ft_memcpy(s, str + start, len));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  sub    $0x30,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  cmpl   $0x4,-0x3c(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <__builtin_assume@plt>
  addl   $0x1,-0x3c(%rbp)
  cmpl   $0x2,-0x3c(%rbp)
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <__builtin_assume@plt>
  cmpl   $0x5,-0x3c(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <__builtin_assume@plt>
  mov    $0x7,%eax
  movslq %eax,%rdx
  mov    %rdx,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  mov    %eax,-0x2c(%rbp)
  mov    -0x3c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    $0x30,%rsp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
int foo(int *a, int i) {
  __builtin_assume(i != 4);
  __builtin_assume(++i > 2);
  int test = sizeof(struct { char qq[(__builtin_assume(i != 5), 7)]; });
  return a[i];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<foo>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  cmp    $0x4,%esi
  setne  %dil
  movzbl %dil,%edi
  mov    $0x0,%eax
  call   <__builtin_assume@plt>
  add    $0x1,%ebx
  cmp    $0x2,%ebx
  setg   %dil
  movzbl %dil,%edi
  mov    $0x0,%eax
  call   <__builtin_assume@plt>
  cmp    $0x5,%ebx
  setne  %dil
  movzbl %dil,%edi
  mov    $0x0,%eax
  call   <__builtin_assume@plt>
  movslq %ebx,%rbx
  mov    0x0(%rbp,%rbx,4),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int foo(int *a, int i) {
  __builtin_assume(i != 4);
  __builtin_assume(++i > 2);
  int test = sizeof(struct { char qq[(__builtin_assume(i != 5), 7)]; });
  return a[i];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  cmp    $0x4,%esi
  setne  %dil
  add    $0x1,%ebx
  xor    %eax,%eax
  call   <__builtin_assume@plt>
  xor    %edi,%edi
  cmp    $0x2,%ebx
  setg   %dil
  xor    %eax,%eax
  call   <__builtin_assume@plt>
  xor    %edi,%edi
  cmp    $0x5,%ebx
  movslq %ebx,%rbx
  setne  %dil
  xor    %eax,%eax
  call   <__builtin_assume@plt>
  mov    0x0(%rbp,%rbx,4),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int foo(int *a, int i) {
  __builtin_assume(i != 4);
  __builtin_assume(++i > 2);
  int test = sizeof(struct { char qq[(__builtin_assume(i != 5), 7)]; });
  return a[i];
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  cmp    $0x4,%esi
  setne  %dil
  add    $0x1,%ebx
  xor    %eax,%eax
  call   <__builtin_assume@plt>
  xor    %edi,%edi
  cmp    $0x2,%ebx
  setg   %dil
  xor    %eax,%eax
  call   <__builtin_assume@plt>
  xor    %edi,%edi
  cmp    $0x5,%ebx
  movslq %ebx,%rbx
  setne  %dil
  xor    %eax,%eax
  call   <__builtin_assume@plt>
  mov    0x0(%rbp,%rbx,4),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int foo(int *a, int i) {
  __builtin_assume(i != 4);
  __builtin_assume(++i > 2);
  int test = sizeof(struct { char qq[(__builtin_assume(i != 5), 7)]; });
  return a[i];
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x2625a00,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0x2625a00,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    0xc(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    $0x0,%r13d
  lea    D13(%rip),%r12
  cmpl   $0x0,0xc(%rsp)
  jle    L66
L11:
  mov    $0x0,%ebx
L87:
  mov    $0x0,%eax
  call   <getchar@plt>
  movzbl (%r12,%rax,1),%eax
  movslq %ebx,%rdx
  mov    %al,0x11(%rsp,%rdx,1)
  cmp    $0x20,%al
  setne  %al
  movzbl %al,%eax
  add    %eax,%ebx
  cmp    $0x6,%ebx
  jle    L87
  lea    0x11(%rsp),%rdi
  call   <atoi@plt>
  addl   $0x1,0x0(%rbp,%rax,4)
  add    $0x1,%r13d
  cmp    %r13d,0xc(%rsp)
  jg     L11
L66:
  mov    $0x0,%ebx
  mov    $0x1,%eax
  lea    D30(%rip),%r12
  jmp    L38
L21:
  add    $0x1,%rbx
  cmp    $0x989680,%rbx
  je     L85
L38:
  mov    0x0(%rbp,%rbx,4),%ecx
  cmp    $0x1,%ecx
  jle    L21
  movslq %ebx,%rsi
  imul   $0x68db8bad,%rsi,%rsi
  sar    $0x2c,%rsi
  mov    %ebx,%eax
  sar    $0x1f,%eax
  sub    %eax,%esi
  imul   $0x2710,%esi,%eax
  mov    %ebx,%edx
  sub    %eax,%edx
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L21
L85:
  test   %eax,%eax
  jne    L61
L9:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L61:
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L9
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D24
datatype: string
name: disassemble_data
arguments: data_tag: D13
datatype: string
name: disassemble_data
arguments: data_tag: D30
datatype: string
name: disassemble_data
arguments: data_tag: D0
datatype: string

tool: D24: %d

tool: D13:                                                 0123456789       2223334445556667 77888999                                       

tool: D30: %03d-%04d %d


tool: D0: No duplicates.

assistant: ```c
int main() {
  const char *trans = "                                                0123456789       2223334445556667 77888999                                       ";
  int *dict, n, i, l, no_duplicates = 1;
  char num_str[7];
  dict = malloc(sizeof(int[10000000]));
  memset(dict, 0, sizeof(int[10000000]));
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    l = 0;
    while (l < 7) {
      num_str[l] = trans[getchar()];
      if (num_str[l] != ' ')
        l++;
    }
    dict[atoi(num_str)]++;
  }
  for (i = 0; i < 10000000; i++)
    if (dict[i] > 1) {
      printf("%03d-%04d %d\n", i / 10000, i % 10000, dict[i]);
      no_duplicates = 0;
    }
  if (no_duplicates)
    printf("No duplicates.");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x2625a00,%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    $0x2625a00,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <memset@plt>
  lea    0xc(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L16
  xor    %r13d,%r13d
  lea    D18(%rip),%r12
  lea    0x11(%rsp),%r14
  nopl   0x0(%rax)
L23:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L54:
  xor    %eax,%eax
  call   <getchar@plt>
  movslq %ebx,%rdx
  movzbl (%r12,%rax,1),%eax
  cmp    $0x20,%al
  mov    %al,0x11(%rsp,%rdx,1)
  setne  %al
  movzbl %al,%eax
  add    %eax,%ebx
  cmp    $0x7,%ebx
  jne    L54
  mov    %r14,%rdi
  add    $0x1,%r13d
  call   <atoi@plt>
  addl   $0x1,0x0(%rbp,%rax,4)
  cmp    %r13d,0xc(%rsp)
  jg     L23
L16:
  xor    %ebx,%ebx
  mov    $0x1,%eax
  mov    $0xd1b71759,%r13d
  lea    D21(%rip),%r12
  xchg   %ax,%ax
L59:
  mov    0x0(%rbp,%rbx,4),%ecx
  cmp    $0x1,%ecx
  jle    L72
  mov    %ebx,%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  imul   %r13,%rsi
  shr    $0x2d,%rsi
  imul   $0x2710,%esi,%eax
  sub    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
L72:
  add    $0x1,%rbx
  cmp    $0x989680,%rbx
  jne    L59
  test   %eax,%eax
  jne    L34
L4:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L77
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L34:
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L4
L77:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D1
datatype: string
name: parse_data
arguments: tag: D18
datatype: string
name: parse_data
arguments: tag: D21
datatype: string
name: parse_data
arguments: tag: D29
datatype: string

tool: D1: %d

tool: D18:                                                 0123456789       2223334445556667 77888999                                       

tool: D21: %03d-%04d %d


tool: D29: No duplicates.

assistant: ```c
int main() {
  const char *trans = "                                                0123456789       2223334445556667 77888999                                       ";
  int *dict, n, i, l, no_duplicates = 1;
  char num_str[7];
  dict = malloc(sizeof(int[10000000]));
  memset(dict, 0, sizeof(int[10000000]));
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    l = 0;
    while (l < 7) {
      num_str[l] = trans[getchar()];
      if (num_str[l] != ' ')
        l++;
    }
    dict[atoi(num_str)]++;
  }
  for (i = 0; i < 10000000; i++)
    if (dict[i] > 1) {
      printf("%03d-%04d %d\n", i / 10000, i % 10000, dict[i]);
      no_duplicates = 0;
    }
  if (no_duplicates)
    printf("No duplicates.");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x2625a00,%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    $0x2625a00,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <memset@plt>
  lea    0xc(%rsp),%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L5
  xor    %r13d,%r13d
  lea    D17(%rip),%r12
  lea    0x11(%rsp),%r14
  nopl   0x0(%rax)
L70:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L65:
  xor    %eax,%eax
  call   <getchar@plt>
  movslq %ebx,%rdx
  movzbl (%r12,%rax,1),%eax
  cmp    $0x20,%al
  mov    %al,0x11(%rsp,%rdx,1)
  setne  %al
  movzbl %al,%eax
  add    %eax,%ebx
  cmp    $0x7,%ebx
  jne    L65
  mov    %r14,%rdi
  add    $0x1,%r13d
  call   <atoi@plt>
  addl   $0x1,0x0(%rbp,%rax,4)
  cmp    %r13d,0xc(%rsp)
  jg     L70
L5:
  xor    %ebx,%ebx
  mov    $0x1,%eax
  mov    $0xd1b71759,%r13d
  lea    D33(%rip),%r12
  xchg   %ax,%ax
L36:
  mov    0x0(%rbp,%rbx,4),%ecx
  cmp    $0x1,%ecx
  jle    L25
  mov    %ebx,%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  imul   %r13,%rsi
  shr    $0x2d,%rsi
  imul   $0x2710,%esi,%eax
  sub    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
L25:
  add    $0x1,%rbx
  cmp    $0x989680,%rbx
  jne    L36
  test   %eax,%eax
  jne    L46
L11:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L72
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L46:
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L11
L72:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D20
type_infered: string
name: disassemble_data
arguments: tag: D17
type_infered: string
name: disassemble_data
arguments: tag: D33
type_infered: string
name: disassemble_data
arguments: tag: D24
type_infered: string

tool: D20: %d

tool: D17:                                                 0123456789       2223334445556667 77888999                                       

tool: D33: %03d-%04d %d


tool: D24: No duplicates.

assistant: ```c
int main() {
  const char *trans = "                                                0123456789       2223334445556667 77888999                                       ";
  int *dict, n, i, l, no_duplicates = 1;
  char num_str[7];
  dict = malloc(sizeof(int[10000000]));
  memset(dict, 0, sizeof(int[10000000]));
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    l = 0;
    while (l < 7) {
      num_str[l] = trans[getchar()];
      if (num_str[l] != ' ')
        l++;
    }
    dict[atoi(num_str)]++;
  }
  for (i = 0; i < 10000000; i++)
    if (dict[i] > 1) {
      printf("%03d-%04d %d\n", i / 10000, i % 10000, dict[i]);
      no_duplicates = 0;
    }
  if (no_duplicates)
    printf("No duplicates.");
  return 0;
}
```