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
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x989680,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rax,%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rdx,%rax,1),%r13d
  sar    %r13d
  lea    0x1(%r13),%ebx
  movslq %ebx,%rbx
  mov    %rbx,%rdi
  call   <malloc@plt>
  cmp    $0x1,%rbx
  jle    L22
  mov    %rax,%r12
  mov    %rbx,%r13
  mov    $0x0,%r14d
  mov    $0x1,%ebx
L18:
  movzbl -0x1(%rbp,%rbx,1),%eax
  mov    %al,-0x1(%r12,%rbx,1)
  movb   $0x0,(%r12,%rbx,1)
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %r12,%rsi
  call   <strstr@plt>
  test   %rax,%rax
  je     L4
  sub    %rbp,%rax
  mov    %rax,%r14
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L18
  mov    %r13,%rbx
L4:
  lea    0x1(%r14),%rdx
  lea    -0x1(%rbx),%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L22:
  mov    $0x0,%r14d
  mov    $0x1,%ebx
  jmp    L4
```

tool_calls: name: retrieve_data
arguments: tag: D15
datatype: string
name: retrieve_data
arguments: tag: D3
datatype: string

tool: D15: %s

tool: D3: %ld %ld


assistant: ```c
int main() {
  char *x = malloc(10000000);
  scanf("%s", x);
  long len = strlen(x) / 2 + 1, i, idx = 0;
  char *dup;
  char *s = malloc(len);
  for (i = 1; i < len; i++) {
    s[i - 1] = x[i - 1];
    s[i] = 0;
    dup = strstr(x + i, s);
    if (dup == NULL)
      break;
    else
      idx = dup - x;
  }
  printf("%ld %ld\n", i - 1, idx + 1);
  return 0;
}
```
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
  mov    $0x989680,%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  lea    D5(%rip),%rdi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rdx,%rax,1),%ebx
  sar    %ebx
  add    $0x1,%ebx
  movslq %ebx,%r13
  mov    %r13,%rdi
  call   <malloc@plt>
  cmp    $0x1,%ebx
  jle    L45
  mov    %rax,%r14
  xor    %r12d,%r12d
  mov    $0x1,%ebx
  jmp    L22
  nopw   0x0(%rax,%rax,1)
L13:
  sub    %rbp,%rax
  add    $0x1,%rbx
  mov    %rax,%r12
  cmp    %r13,%rbx
  je     L5
L22:
  movzbl -0x1(%rbp,%rbx,1),%eax
  movb   $0x0,(%r14,%rbx,1)
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %r14,%rsi
  mov    %al,-0x1(%r14,%rbx,1)
  call   <strstr@plt>
  test   %rax,%rax
  jne    L13
L5:
  lea    0x1(%r12),%rdx
  lea    -0x1(%rbx),%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L45:
  xor    %r12d,%r12d
  mov    $0x1,%ebx
  jmp    L5
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
type_guessed: string
name: disassemble_data
arguments: data_tag: D7
type_guessed: string

tool: D5: %s

tool: D7: %ld %ld


assistant: ```c
int main() {
  char *x = malloc(10000000);
  scanf("%s", x);
  long len = strlen(x) / 2 + 1, i, idx = 0;
  char *dup;
  char *s = malloc(len);
  for (i = 1; i < len; i++) {
    s[i - 1] = x[i - 1];
    s[i] = 0;
    dup = strstr(x + i, s);
    if (dup == NULL)
      break;
    else
      idx = dup - x;
  }
  printf("%ld %ld\n", i - 1, idx + 1);
  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   %r14
  mov    $0x989680,%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  lea    D7(%rip),%rdi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rdx,%rax,1),%ebx
  sar    %ebx
  add    $0x1,%ebx
  movslq %ebx,%r13
  mov    %r13,%rdi
  call   <malloc@plt>
  cmp    $0x1,%ebx
  jle    L46
  mov    %rax,%r14
  xor    %r12d,%r12d
  mov    $0x1,%ebx
  jmp    L36
  nopw   0x0(%rax,%rax,1)
L32:
  sub    %rbp,%rax
  add    $0x1,%rbx
  mov    %rax,%r12
  cmp    %r13,%rbx
  je     L12
L36:
  movzbl -0x1(%rbp,%rbx,1),%eax
  movb   $0x0,(%r14,%rbx,1)
  lea    0x0(%rbp,%rbx,1),%rdi
  mov    %r14,%rsi
  mov    %al,-0x1(%r14,%rbx,1)
  call   <strstr@plt>
  test   %rax,%rax
  jne    L32
L12:
  lea    0x1(%r12),%rdx
  lea    -0x1(%rbx),%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L46:
  xor    %r12d,%r12d
  mov    $0x1,%ebx
  jmp    L12
```

tool_calls: name: retrieve_data
arguments: data_label: D7
type: string
name: retrieve_data
arguments: data_label: D1
type: string

tool: D7: %s

tool: D1: %ld %ld


assistant: ```c
int main() {
  char *x = malloc(10000000);
  scanf("%s", x);
  long len = strlen(x) / 2 + 1, i, idx = 0;
  char *dup;
  char *s = malloc(len);
  for (i = 1; i < len; i++) {
    s[i - 1] = x[i - 1];
    s[i] = 0;
    dup = strstr(x + i, s);
    if (dup == NULL)
      break;
    else
      idx = dup - x;
  }
  printf("%ld %ld\n", i - 1, idx + 1);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <prime@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L9
L10:
  addl   $0x1,-0xc(%rbp)
L9:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L10
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  leave
  ret
```

assistant: ```c
int is_prime(int n) {
  int i, *p = prime(n);
  for (i = 0; p[i] != 0; i++)
    ;
  return p[i - 1];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<is_prime>:
  endbr64
  sub    $0x8,%rsp
  call   <prime@plt>
  cmpl   $0x0,(%rax)
  je     L17
  mov    $0x4,%edx
L11:
  mov    %rdx,%rcx
  add    $0x4,%rdx
  cmpl   $0x0,-0x4(%rax,%rdx,1)
  jne    L11
L24:
  mov    -0x4(%rax,%rcx,1),%eax
  add    $0x8,%rsp
  ret
L17:
  mov    $0x0,%ecx
  jmp    L24
```

assistant: ```c
int is_prime(int n) {
  int i, *p = prime(n);
  for (i = 0; p[i] != 0; i++)
    ;
  return p[i - 1];
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<is_prime>:
  endbr64
  sub    $0x8,%rsp
  call   <prime@plt>
  mov    (%rax),%edi
  test   %edi,%edi
  je     L14
  mov    $0x4,%edx
  nopl   0x0(%rax,%rax,1)
L18:
  mov    %rdx,%rcx
  add    $0x4,%rdx
  mov    -0x4(%rax,%rdx,1),%esi
  test   %esi,%esi
  jne    L18
  sub    $0x4,%rcx
  mov    (%rax,%rcx,1),%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  mov    $0xfffffffffffffffc,%rcx
  mov    (%rax,%rcx,1),%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int is_prime(int n) {
  int i, *p = prime(n);
  for (i = 0; p[i] != 0; i++)
    ;
  return p[i - 1];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<is_prime>:
  endbr64
  sub    $0x8,%rsp
  call   <prime@plt>
  mov    (%rax),%edi
  test   %edi,%edi
  je     L19
  mov    $0x4,%edx
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %rdx,%rcx
  add    $0x4,%rdx
  mov    -0x4(%rax,%rdx,1),%esi
  test   %esi,%esi
  jne    L11
  sub    $0x4,%rcx
  mov    (%rax,%rcx,1),%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  mov    $0xfffffffffffffffc,%rcx
  mov    (%rax,%rcx,1),%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int is_prime(int n) {
  int i, *p = prime(n);
  for (i = 0; p[i] != 0; i++)
    ;
  return p[i - 1];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<longestPalindrome>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  lea    0x2(%rax),%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  lea    0x1(%rbx),%rdi
  mov    %r12,%rsi
  call   <strcpy@plt>
  movb   $0xff,(%rbx)
  mov    $0x1,%r8d
  mov    $0x1,%r11d
  cmpb   $0x0,0x1(%rbx)
  je     L75
  mov    $0x1,%r8d
  mov    $0x1,%r11d
  mov    $0x1,%r9d
  mov    $0x1,%edi
  lea    0x2(%rbx),%r10
  jmp    L61
L48:
  mov    %edx,%r12d
  jmp    L37
L110:
  lea    0x1(%rdx),%edi
  movslq %edi,%rax
  cmpb   $0x0,(%rbx,%rax,1)
  je     L28
L61:
  movslq %edi,%r12
  lea    (%r10,%r12,1),%rax
  mov    %edi,%edx
  movzbl (%rbx,%r12,1),%ecx
  cmp    %cl,0x1(%rbx,%r12,1)
  jne    L30
L84:
  add    $0x1,%edx
  movzbl (%rax),%esi
  movzbl -0x1(%rax),%ecx
  add    $0x1,%rax
  cmp    %cl,%sil
  je     L84
L30:
  movslq %edx,%rcx
  movzbl 0x1(%rbx,%rcx,1),%eax
  cmp    %al,-0x1(%rbx,%r12,1)
  jne    L48
  lea    0x1(%rdx),%eax
  cltq
  lea    (%rdi,%rdx,1),%r14d
  add    %r12,%rcx
  lea    -0x1(%rbx,%rcx,1),%r13
L26:
  mov    %r14d,%edi
  sub    %eax,%edi
  mov    %eax,%r12d
  mov    %r13,%rsi
  sub    %rax,%rsi
  movzbl 0x1(%rax,%rbx,1),%ecx
  add    $0x1,%rax
  cmp    %cl,(%rsi)
  je     L26
L37:
  mov    %r12d,%eax
  sub    %edi,%eax
  cmp    %r9d,%eax
  jl     L110
  lea    0x1(%rax),%r9d
  mov    %r12d,%r8d
  mov    %edi,%r11d
  jmp    L110
L28:
  cmp    %r8d,%r11d
  jg     L1
L75:
  movslq %r11d,%rdx
  mov    %rdx,%rax
  mov    %rbp,%rcx
  sub    %rdx,%rcx
L22:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%r8d
  jge    L22
L1:
  sub    %r11d,%r8d
  movslq %r8d,%rax
  movb   $0x0,0x1(%rbp,%rax,1)
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *longestPalindrome(char *s) {
  int len = strlen(s);
  char *x = (char *)(malloc(sizeof(char) * (len + 2)));
  char *ans = (char *)(malloc(sizeof(char) * (len + 2)));
  strcpy(x + 1, s);
  int i;
  x[0] = -1;
  int ma = 1;
  int l, r;
  l = r = 1;
  for (i = 1; x[i]; ++i) {
    int s = i, e = i;
    while (x[e + 1] == x[e])
      ++e;
    i = e;
    while (x[s - 1] == x[e + 1])
      --s, ++e;
    if (e - s + 1 > ma) {
      ma = e - s + 1;
      l = s;
      r = e;
    }
  }
  for (i = l; i <= r; i++) {
    ans[i - l] = x[i];
  }
  ans[r - l + 1] = '\0';
  free(x);
  return ans;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<longestPalindrome>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x2(%rax),%r12d
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %r12d,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %r13,%rsi
  lea    0x1(%rbp),%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  movzbl 0x1(%rbp),%esi
  movb   $0xff,0x0(%rbp)
  test   %sil,%sil
  je     L20
  mov    $0xffffffff,%r8d
  mov    $0x1,%r11d
  mov    $0x1,%ebx
  mov    $0x1,%r10d
  mov    $0x1,%r14d
  xchg   %ax,%ax
L47:
  movslq %r14d,%rdi
  movzbl 0x1(%rbp,%rdi,1),%eax
  lea    0x1(%rdi),%rdx
  cmp    %sil,%al
  jne    L76
  nopl   (%rax)
L26:
  mov    %rdx,%rax
  movslq %edx,%rcx
  add    $0x1,%rdx
  movzbl 0x1(%rax,%rbp,1),%esi
  cmp    0x0(%rbp,%rax,1),%sil
  je     L26
L36:
  lea    0x1(%rcx),%r13d
  cmp    %sil,%r8b
  jne    L58
  add    %rbp,%rdi
  movslq %r13d,%rax
  lea    (%rcx,%r14,1),%r9d
  lea    -0x1(%rcx,%rdi,1),%r15
  cs nopw 0x0(%rax,%rax,1)
L99:
  mov    %r15,%r8
  mov    %r9d,%r14d
  movzbl 0x1(%rbp,%rax,1),%edi
  mov    %eax,%ecx
  sub    %rax,%r8
  sub    %eax,%r14d
  add    $0x1,%rax
  cmp    %dil,(%r8)
  je     L99
L58:
  mov    %ecx,%eax
  sub    %r14d,%eax
  cmp    %ebx,%eax
  jl     L7
  lea    0x1(%rax),%ebx
  mov    %ecx,%r11d
  mov    %r14d,%r10d
L7:
  mov    %r13d,%r14d
  test   %sil,%sil
  je     L32
  movzbl -0x1(%rbp,%rdx,1),%r8d
  jmp    L47
  nopl   0x0(%rax)
L76:
  mov    %eax,%esi
  movslq %r14d,%rcx
  jmp    L36
  nop
L32:
  mov    %r11d,%ecx
  sub    %r10d,%ecx
  movslq %ecx,%rcx
  add    $0x1,%rcx
  cmp    %r11d,%r10d
  jg     L35
  movslq %r10d,%rdi
  movzbl 0x0(%rbp,%rdi,1),%esi
L89:
  lea    0x1(%r10),%eax
  lea    -0x1(%r12),%rdx
  cltq
  sub    %rdi,%rdx
  jmp    L60
  nopw   0x0(%rax,%rax,1)
L53:
  movzbl 0x0(%rbp,%rax,1),%esi
  add    $0x1,%rax
L60:
  mov    %sil,(%rdx,%rax,1)
  cmp    %eax,%r11d
  jge    L53
L35:
  movb   $0x0,(%r12,%rcx,1)
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  mov    $0x1,%ecx
  mov    $0x1,%r11d
  mov    $0x1,%r10d
  mov    $0x1,%edi
  jmp    L89
```

assistant: ```c
char *longestPalindrome(char *s) {
  int len = strlen(s);
  char *x = (char *)(malloc(sizeof(char) * (len + 2)));
  char *ans = (char *)(malloc(sizeof(char) * (len + 2)));
  strcpy(x + 1, s);
  int i;
  x[0] = -1;
  int ma = 1;
  int l, r;
  l = r = 1;
  for (i = 1; x[i]; ++i) {
    int s = i, e = i;
    while (x[e + 1] == x[e])
      ++e;
    i = e;
    while (x[s - 1] == x[e + 1])
      --s, ++e;
    if (e - s + 1 > ma) {
      ma = e - s + 1;
      l = s;
      r = e;
    }
  }
  for (i = l; i <= r; i++) {
    ans[i - l] = x[i];
  }
  ans[r - l + 1] = '\0';
  free(x);
  return ans;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x80(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x80(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D12
type: string
name: parse_data
arguments: data_label: D4
type: string

tool: D12: %[^
]

tool: D4: Hello, World.

assistant: ```c
int main() {
  char input_string[105];
  scanf("%[^\n]", input_string);
  printf("Hello, World.\n");
  printf("%s\n", input_string);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  lea    D12(%rip),%rdi
  call   <puts@plt>
  mov    %rbx,%rdi
  call   <puts@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D9
datatype: string
name: read_data
arguments: data_label: D12
datatype: string

tool: D9: %[^
]

tool: D12: Hello, World.

assistant: ```c
int main() {
  char input_string[105];
  scanf("%[^\n]", input_string);
  printf("Hello, World.\n");
  printf("%s\n", input_string);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rdi
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <scanf@plt>
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  sub    $0xffffffffffffff80,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
data_type: string
name: extract_symbolic_data
arguments: data_label: D0
data_type: string

tool: D4: %[^
]

tool: D0: Hello, World.

assistant: ```c
int main() {
  char input_string[105];
  scanf("%[^\n]", input_string);
  printf("Hello, World.\n");
  printf("%s\n", input_string);
  return 0;
}
```
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
  lea    D18(%rip),%rdi
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <scanf@plt>
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  sub    $0xffffffffffffff80,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D18
data_type: string
name: extract_symbolic_data
arguments: data_label: D11
data_type: string

tool: D18: %[^
]

tool: D11: Hello, World.

assistant: ```c
int main() {
  char input_string[105];
  scanf("%[^\n]", input_string);
  printf("Hello, World.\n");
  printf("%s\n", input_string);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<none>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L5:
  jmp    L5
```

assistant: ```c
int none() {
  for (;;) {
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<none>:
  endbr64
L3:
  jmp    L3
```

assistant: ```c
int none() {
  for (;;) {
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<none>:
  endbr64
  nopl   0x0(%rax)
L1:
  jmp    L1
```

assistant: ```c
int none() {
  for (;;) {
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<none>:
  endbr64
  nopl   0x0(%rax)
L3:
  jmp    L3
```

assistant: ```c
int none() {
  for (;;) {
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  movl   $0x0,-0x40(%rbp)
  jmp    L48
L83:
  mov    -0x40(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x40(%rbp)
L48:
  cmpl   $0x9,-0x40(%rbp)
  jle    L83
  movl   $0x0,-0x3c(%rbp)
  jmp    L74
L43:
  mov    -0x3c(%rbp),%eax
  cltq
  movl   $0x1,-0x30(%rbp,%rax,4)
  addl   $0x1,-0x3c(%rbp)
L74:
  cmpl   $0x9,-0x3c(%rbp)
  jle    L43
  movl   $0x0,-0x38(%rbp)
  jmp    L55
L58:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x38(%rbp)
L55:
  cmpl   $0x9,-0x38(%rbp)
  jle    L58
  lea    -0x30(%rbp),%rax
  mov    $0x28,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L71
L47:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L71:
  cmpl   $0x9,-0x34(%rbp)
  jle    L47
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D25
datatype: string

tool: D25: %d

assistant: ```c
int main(void) {
  int vetor[10] = {0};
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
  for (int i = 0; i < 10; i++)
    vetor[i] = 1;
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
  memset(vetor, 0, sizeof(vetor));
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  mov    %rsp,%rbx
  lea    0x28(%rsp),%rbp
  mov    %rbx,%r12
  lea    D11(%rip),%r13
L8:
  mov    (%r12),%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x4,%r12
  cmp    %rbp,%r12
  jne    L8
  mov    %rbx,%rax
L9:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rbp,%rax
  jne    L9
  mov    %rbx,%r12
  lea    D11(%rip),%r13
L43:
  mov    (%r12),%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x4,%r12
  cmp    %rbp,%r12
  jne    L43
  mov    %rsp,%rdi
  mov    $0x28,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  lea    D11(%rip),%r12
L29:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L29
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D11
type_guessed: string

tool: D11: %d

assistant: ```c
int main(void) {
  int vetor[10] = {0};
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
  for (int i = 0; i < 10; i++)
    vetor[i] = 1;
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
  memset(vetor, 0, sizeof(vetor));
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  pxor   %xmm0,%xmm0
  xor    %esi,%esi
  push   %r13
  lea    D5(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  movaps %xmm0,(%rsp)
  lea    0x4(%rsp),%r12
  movaps %xmm0,0x10(%rsp)
  mov    %r14,%rbx
  lea    0x28(%rsp),%rbp
  movq   $0x0,0x20(%rsp)
  jmp    L59
  nopl   0x0(%rax)
L0:
  mov    (%r12),%esi
  add    $0x4,%r12
L59:
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    %rbp,%r12
  jne    L0
  mov    %r14,%rax
L2:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rbp,%rax
  jne    L2
  mov    %r14,%r12
  nopw   0x0(%rax,%rax,1)
L31:
  mov    (%r12),%esi
  mov    %r13,%rdi
  add    $0x4,%r12
  call   <printf@plt>
  cmp    %rbp,%r12
  jne    L31
  mov    $0x28,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <memset@plt>
  nopl   0x0(%rax)
L37:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L37
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D5
type_infered: string

tool: D5: %d

assistant: ```c
int main(void) {
  int vetor[10] = {0};
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
  for (int i = 0; i < 10; i++)
    vetor[i] = 1;
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
  memset(vetor, 0, sizeof(vetor));
  for (int i = 0; i < 10; i++)
    printf("%d", vetor[i]);
}
```