ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<Digits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L18:
  addl   $0x1,-0x4(%rbp)
  mov    -0x18(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x18(%rbp)
L7:
  cmpq   $0x0,-0x18(%rbp)
  jne    L18
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int Digits(long long int n) {
  int c = 0;
  while (n != 0) {
    c++;
    n = n / 10;
  }
  return c;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Digits>:
  endbr64
  test   %rdi,%rdi
  je     L26
  mov    $0x0,%ecx
  movabs $0x6666666666666667,%rsi
L27:
  add    $0x1,%ecx
  mov    %rdi,%rax
  imul   %rsi
  sar    $0x2,%rdx
  sar    $0x3f,%rdi
  sub    %rdi,%rdx
  mov    %rdx,%rdi
  jne    L27
L5:
  mov    %ecx,%eax
  ret
L26:
  mov    $0x0,%ecx
  jmp    L5
```

assistant: ```c
int Digits(long long int n) {
  int c = 0;
  while (n != 0) {
    c++;
    n = n / 10;
  }
  return c;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Digits>:
  endbr64
  xor    %r8d,%r8d
  test   %rdi,%rdi
  je     L5
  movabs $0x6666666666666667,%rcx
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    %rdi,%rax
  add    $0x1,%r8d
  imul   %rcx
  mov    %rdi,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  mov    %rdx,%rdi
  sub    %rax,%rdi
  jne    L7
L5:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int Digits(long long int n) {
  int c = 0;
  while (n != 0) {
    c++;
    n = n / 10;
  }
  return c;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Digits>:
  endbr64
  xor    %r8d,%r8d
  test   %rdi,%rdi
  je     L19
  movabs $0x6666666666666667,%rcx
  cs nopw 0x0(%rax,%rax,1)
L5:
  mov    %rdi,%rax
  add    $0x1,%r8d
  imul   %rcx
  mov    %rdi,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  mov    %rdx,%rdi
  sub    %rax,%rdi
  jne    L5
L19:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int Digits(long long int n) {
  int c = 0;
  while (n != 0) {
    c++;
    n = n / 10;
  }
  return c;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_block_relocated_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x5,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    D17(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D36(%rip)
  mov    D36(%rip),%ecx
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  lea    D36(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D36
type: string

tool: D36: %s_%04d

assistant: ```c
char *get_block_relocated_name(char *block_name) {
  static int program_relocator = 0;
  char *relocated_name;
  ;
  {};
  relocated_name = (char *)malloc((5 + strlen(block_name)) * sizeof(char));
  sprintf(relocated_name, "%s_%04d", block_name, (++program_relocator));
  return (relocated_name);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<get_block_relocated_name>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x5(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    D28(%rip),%eax
  lea    0x1(%rax),%ecx
  mov    %ecx,D12(%rip)
  mov    %rbp,%rdx
  lea    D12(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D12
type_infered: string

tool: D12: %s_%04d

assistant: ```c
char *get_block_relocated_name(char *block_name) {
  static int program_relocator = 0;
  char *relocated_name;
  ;
  {};
  relocated_name = (char *)malloc((5 + strlen(block_name)) * sizeof(char));
  sprintf(relocated_name, "%s_%04d", block_name, (++program_relocator));
  return (relocated_name);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x330,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x328(%rsp)
  xor    %eax,%eax
  lea    D22(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%r12
  lea    0x10(%rsp),%rbp
  mov    $0x0,%r14d
  lea    D21(%rip),%r13
L50:
  lea    -0x10(%rbp),%rbx
L15:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L15
  add    $0xa,%r14
  add    $0x28,%rbp
  cmp    $0x28,%r14
  jne    L50
  lea    0x190(%rsp),%rbp
  lea    0xa0(%r12),%rcx
  mov    %rbp,%rax
L45:
  mov    (%r12),%edx
  mov    %edx,(%rax)
  mov    0x4(%r12),%edx
  mov    %edx,0x28(%rax)
  mov    0x8(%r12),%edx
  mov    %edx,0x50(%rax)
  mov    0xc(%r12),%edx
  mov    %edx,0x78(%rax)
  add    $0x28,%r12
  add    $0x4,%rax
  cmp    %rcx,%r12
  jne    L45
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    0xa0(%rbp),%r13
  lea    D1(%rip),%r12
L57:
  mov    $0x0,%ebx
L37:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x4,%rbx
  jne    L37
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x28,%rbp
  cmp    %r13,%rbp
  jne    L57
  mov    0x328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    $0x0,%eax
  add    $0x330,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D22
datatype: string
name: retrieve_data
arguments: data_label: D21
datatype: string
name: retrieve_data
arguments: data_label: D1
datatype: string

tool: D22: Vuvedete z (sled vsqka cifra da ima raztoqnie)

tool: D21: %d

tool: D1: %d	

assistant: ```c
int main() {
  int m = 4, n = 4, c, d, z[10][10], t[10][10];
  printf("Vuvedete z (sled vsqka cifra da ima raztoqnie)\n");
  for (c = 0; c < m; c++)
    for (d = 0; d < n; d++)
      scanf("%d", &z[c][d]);
  for (c = 0; c < m; c++)
    for (d = 0; d < n; d++)
      t[d][c] = z[c][d];
  printf("\n");
  for (c = 0; c < n; c++) {
    for (d = 0; d < m; d++)
      printf("%d\t", t[c][d]);
    printf("\n");
  }
  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
  lea    D27(%rip),%rdi
  xor    %r14d,%r14d
  push   %r13
  lea    D17(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x330,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x328(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x10(%rsp),%rbp
  call   <puts@plt>
L8:
  lea    -0x10(%rbp),%rbx
L5:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbp,%rbx
  jne    L5
  add    $0xa,%r14
  lea    0x28(%rbx),%rbp
  cmp    $0x28,%r14
  jne    L8
  lea    0x190(%rsp),%rbp
  lea    0xa0(%r12),%rcx
  mov    %rbp,%rax
L3:
  mov    (%r12),%edx
  add    $0x28,%r12
  add    $0x4,%rax
  mov    %edx,-0x4(%rax)
  mov    -0x24(%r12),%edx
  mov    %edx,0x24(%rax)
  mov    -0x20(%r12),%edx
  mov    %edx,0x4c(%rax)
  mov    -0x1c(%r12),%edx
  mov    %edx,0x74(%rax)
  cmp    %rcx,%r12
  jne    L3
  mov    $0xa,%edi
  lea    0xa0(%rbp),%r13
  lea    D2(%rip),%r12
  call   <putchar@plt>
L12:
  xor    %ebx,%ebx
L30:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x4,%rbx
  jne    L30
  mov    $0xa,%edi
  add    $0x28,%rbp
  call   <putchar@plt>
  cmp    %rbp,%r13
  jne    L12
  mov    0x328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x330,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D27
data_type: string
name: disassemble_data
arguments: label: D17
data_type: string
name: disassemble_data
arguments: label: D2
data_type: string

tool: D27: Vuvedete z (sled vsqka cifra da ima raztoqnie)

tool: D17: %d

tool: D2: %d	

assistant: ```c
int main() {
  int m = 4, n = 4, c, d, z[10][10], t[10][10];
  printf("Vuvedete z (sled vsqka cifra da ima raztoqnie)\n");
  for (c = 0; c < m; c++)
    for (d = 0; d < n; d++)
      scanf("%d", &z[c][d]);
  for (c = 0; c < m; c++)
    for (d = 0; d < n; d++)
      t[d][c] = z[c][d];
  printf("\n");
  for (c = 0; c < n; c++) {
    for (d = 0; d < m; d++)
      printf("%d\t", t[c][d]);
    printf("\n");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r13
  lea    D10(%rip),%rdi
  xor    %r13d,%r13d
  push   %r12
  lea    D3(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x338,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x328(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  call   <puts@plt>
L18:
  lea    -0x10(%rbp),%rbx
L31:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbx,%rbp
  jne    L31
  add    $0xa,%r13
  add    $0x28,%rbp
  cmp    $0x28,%r13
  jne    L18
  movd   0x78(%rsp),%xmm2
  movd   0x28(%rsp),%xmm3
  mov    $0xa,%edi
  lea    0x190(%rsp),%rbp
  movd   0x50(%rsp),%xmm1
  movd   (%rsp),%xmm0
  lea    0x230(%rsp),%r13
  lea    D2(%rip),%r12
  movd   0x7c(%rsp),%xmm4
  movd   0x2c(%rsp),%xmm5
  punpckldq %xmm2,%xmm1
  punpckldq %xmm3,%xmm0
  movd   0x30(%rsp),%xmm7
  movd   0x80(%rsp),%xmm6
  punpcklqdq %xmm1,%xmm0
  movd   0x54(%rsp),%xmm1
  movd   0x84(%rsp),%xmm2
  movaps %xmm0,0x190(%rsp)
  movd   0x4(%rsp),%xmm0
  movd   0x34(%rsp),%xmm3
  punpckldq %xmm4,%xmm1
  punpckldq %xmm5,%xmm0
  punpcklqdq %xmm1,%xmm0
  movd   0x58(%rsp),%xmm1
  movups %xmm0,0x1b8(%rsp)
  movd   0x8(%rsp),%xmm0
  punpckldq %xmm6,%xmm1
  punpckldq %xmm7,%xmm0
  punpcklqdq %xmm1,%xmm0
  movd   0x5c(%rsp),%xmm1
  movaps %xmm0,0x1e0(%rsp)
  movd   0xc(%rsp),%xmm0
  punpckldq %xmm2,%xmm1
  punpckldq %xmm3,%xmm0
  punpcklqdq %xmm1,%xmm0
  movups %xmm0,0x208(%rsp)
  call   <putchar@plt>
L42:
  xor    %ebx,%ebx
L12:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x4,%rbx
  jne    L12
  mov    $0xa,%edi
  add    $0x28,%rbp
  call   <putchar@plt>
  cmp    %rbp,%r13
  jne    L42
  mov    0x328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x338,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D10
type: string
name: retrieve_data
arguments: label: D3
type: string
name: retrieve_data
arguments: label: D2
type: string

tool: D10: Vuvedete z (sled vsqka cifra da ima raztoqnie)

tool: D3: %d

tool: D2: %d	

assistant: ```c
int main() {
  int m = 4, n = 4, c, d, z[10][10], t[10][10];
  printf("Vuvedete z (sled vsqka cifra da ima raztoqnie)\n");
  for (c = 0; c < m; c++)
    for (d = 0; d < n; d++)
      scanf("%d", &z[c][d]);
  for (c = 0; c < m; c++)
    for (d = 0; d < n; d++)
      t[d][c] = z[c][d];
  printf("\n");
  for (c = 0; c < n; c++) {
    for (d = 0; d < m; d++)
      printf("%d\t", t[c][d]);
    printf("\n");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<remote_add>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <remote_service_client_init@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  mov    -0x10(%rbp),%rcx
  mov    (%rcx),%ecx
  mov    %eax,%esi
  mov    %ecx,%edi
  mov    $0x0,%eax
  call   <remote_service_add@plt>
  mov    -0x8(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
type: string

tool: D6: failure? %i


assistant: ```c
void *remote_add(void *arguments) {
  int failure;
  int *result;
  int *args;
  args = (int *)arguments;
  failure = remote_service_client_init((unsigned int)args[0]);
  printf("failure? %i\n", failure);
  result = malloc(sizeof(int));
  *result = remote_service_add((unsigned int)args[0], args[1], args[2]);
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<remote_add>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    (%rdi),%edi
  mov    $0x0,%eax
  call   <remote_service_client_init@plt>
  mov    %eax,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    0x8(%rbx),%edx
  mov    0x4(%rbx),%esi
  mov    (%rbx),%edi
  mov    $0x0,%eax
  call   <remote_service_add@plt>
  mov    %eax,0x0(%rbp)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
type: string

tool: D6: failure? %i


assistant: ```c
void *remote_add(void *arguments) {
  int failure;
  int *result;
  int *args;
  args = (int *)arguments;
  failure = remote_service_client_init((unsigned int)args[0]);
  printf("failure? %i\n", failure);
  result = malloc(sizeof(int));
  *result = remote_service_add((unsigned int)args[0], args[1], args[2]);
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<remote_add>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%edi
  call   <remote_service_client_init@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    0x8(%rbx),%edx
  mov    0x4(%rbx),%esi
  mov    (%rbx),%edi
  mov    %rax,%r12
  xor    %eax,%eax
  call   <remote_service_add@plt>
  mov    %eax,(%r12)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D5
type_guessed: string

tool: D5: failure? %i


assistant: ```c
void *remote_add(void *arguments) {
  int failure;
  int *result;
  int *args;
  args = (int *)arguments;
  failure = remote_service_client_init((unsigned int)args[0]);
  printf("failure? %i\n", failure);
  result = malloc(sizeof(int));
  *result = remote_service_add((unsigned int)args[0], args[1], args[2]);
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<remote_add>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%edi
  call   <remote_service_client_init@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    0x8(%rbx),%edx
  mov    0x4(%rbx),%esi
  mov    (%rbx),%edi
  mov    %rax,%r12
  xor    %eax,%eax
  call   <remote_service_add@plt>
  mov    %eax,(%r12)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: tag: D7
data_type: string

tool: D7: failure? %i


assistant: ```c
void *remote_add(void *arguments) {
  int failure;
  int *result;
  int *args;
  args = (int *)arguments;
  failure = remote_service_client_init((unsigned int)args[0]);
  printf("failure? %i\n", failure);
  result = malloc(sizeof(int));
  *result = remote_service_add((unsigned int)args[0], args[1], args[2]);
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<inv2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movsd  (%rax),%xmm1
  mov    -0x18(%rbp),%rax
  add    $0x18,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm2
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  movsd  (%rax),%xmm1
  mulsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <fabs@plt>
  test   %eax,%eax
  jg     L10
  mov    $0xffffffff,%eax
  jmp    L6
L10:
  mov    -0x18(%rbp),%rax
  add    $0x18,%rax
  movsd  (%rax),%xmm0
  divsd  -0x8(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  movq   D13(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  movsd  (%rax),%xmm0
  movq   D13(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  movsd  (%rax),%xmm0
  mov    -0x20(%rbp),%rax
  add    $0x18,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    $0x0,%eax
L6:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
type_infered: f32[2]

tool: {"D13": [0.0,-0.0]}

assistant: ```c
int inv2(const double *a, double *b) {
  double det = a[0] * a[3] - a[1] * a[2];
  if (fabs(det) < (1e-60))
    return -1;
  b[0] = a[3] / det;
  b[1] = -a[1] / det;
  b[2] = -a[2] / det;
  b[3] = a[0] / det;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<inv2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  movsd  (%rdi),%xmm0
  movapd %xmm0,%xmm1
  mulsd  0x18(%rdi),%xmm1
  movsd  0x8(%rdi),%xmm0
  mulsd  0x10(%rdi),%xmm0
  subsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rsp)
  movapd %xmm1,%xmm0
  call   <fabs@plt>
  test   %eax,%eax
  jle    L6
  movsd  0x18(%rbx),%xmm0
  movsd  0x8(%rsp),%xmm2
  divsd  %xmm2,%xmm0
  movsd  %xmm0,0x0(%rbp)
  movsd  0x8(%rbx),%xmm0
  movq   D7(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  movsd  %xmm0,0x8(%rbp)
  movsd  0x10(%rbx),%xmm0
  xorpd  %xmm1,%xmm0
  divsd  %xmm2,%xmm0
  movsd  %xmm0,0x10(%rbp)
  movsd  (%rbx),%xmm0
  divsd  %xmm2,%xmm0
  movsd  %xmm0,0x18(%rbp)
  mov    $0x0,%eax
L18:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  mov    $0xffffffff,%eax
  jmp    L18
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
datatype: f32[2]

tool: {"D7": [0.0,-0.0]}

assistant: ```c
int inv2(const double *a, double *b) {
  double det = a[0] * a[3] - a[1] * a[2];
  if (fabs(det) < (1e-60))
    return -1;
  b[0] = a[3] / det;
  b[1] = -a[1] / det;
  b[2] = -a[2] / det;
  b[3] = a[0] / det;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<inv2>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  movsd  (%rdi),%xmm0
  movsd  0x8(%rdi),%xmm1
  mulsd  0x18(%rdi),%xmm0
  mulsd  0x10(%rdi),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%rsp)
  call   <fabs@plt>
  test   %eax,%eax
  jle    L18
  movsd  0x18(%rbx),%xmm1
  movsd  0x8(%rsp),%xmm0
  xor    %eax,%eax
  movq   D4(%rip),%xmm2
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x0(%rbp)
  movsd  0x8(%rbx),%xmm1
  xorpd  %xmm2,%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rbp)
  movsd  0x10(%rbx),%xmm1
  xorpd  %xmm2,%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x10(%rbp)
  movsd  (%rbx),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x18(%rbp)
L10:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  mov    $0xffffffff,%eax
  jmp    L10
```

tool_calls: name: unpack_data
arguments: data_tag: D4
data_type: f32[2]

tool: {"D4": [0.0,-0.0]}

assistant: ```c
int inv2(const double *a, double *b) {
  double det = a[0] * a[3] - a[1] * a[2];
  if (fabs(det) < (1e-60))
    return -1;
  b[0] = a[3] / det;
  b[1] = -a[1] / det;
  b[2] = -a[2] / det;
  b[3] = a[0] / det;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<inv2>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  movsd  (%rdi),%xmm0
  movsd  0x8(%rdi),%xmm1
  mulsd  0x18(%rdi),%xmm0
  mulsd  0x10(%rdi),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%rsp)
  call   <fabs@plt>
  test   %eax,%eax
  jle    L2
  movsd  0x18(%rbx),%xmm1
  movsd  0x8(%rsp),%xmm0
  xor    %eax,%eax
  movq   D8(%rip),%xmm2
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x0(%rbp)
  movsd  0x8(%rbx),%xmm1
  xorpd  %xmm2,%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x8(%rbp)
  movsd  0x10(%rbx),%xmm1
  xorpd  %xmm2,%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x10(%rbp)
  movsd  (%rbx),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,0x18(%rbp)
L3:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  mov    $0xffffffff,%eax
  jmp    L3
```

tool_calls: name: read_data
arguments: label: D8
type_guessed: f32[2]

tool: {"D8": [0.0,-0.0]}

assistant: ```c
int inv2(const double *a, double *b) {
  double det = a[0] * a[3] - a[1] * a[2];
  if (fabs(det) < (1e-60))
    return -1;
  b[0] = a[3] / det;
  b[1] = -a[1] / det;
  b[2] = -a[2] / det;
  b[3] = a[0] / det;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<getDirectory>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x120,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,-0x118(%rbp)
  lea    -0x110(%rbp),%rax
  mov    $0x100,%esi
  mov    %rax,%rdi
  call   <getcwd@plt>
  movl   $0x0,-0x120(%rbp)
  movl   $0x0,-0x11c(%rbp)
  jmp    L11
L5:
  mov    -0x11c(%rbp),%eax
  cltq
  movzbl -0x110(%rbp,%rax,1),%eax
  cmp    $0x2f,%al
  jne    L8
  mov    -0x11c(%rbp),%eax
  mov    %eax,-0x120(%rbp)
L8:
  addl   $0x1,-0x11c(%rbp)
L11:
  cmpl   $0xff,-0x11c(%rbp)
  jle    L5
  mov    -0x120(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  lea    -0x110(%rbp),%rax
  add    %rax,%rdx
  mov    -0x118(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x118(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
char *getDirectory() {
  char cwd[256];
  char *result = malloc(256);
  getcwd(cwd, 256);
  int lastSlash = 0;
  for (int i = 0; i < 256; i++) {
    if (cwd[i] == '/')
      lastSlash = i;
  }
  strcpy(result, cwd + lastSlash + 1);
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getDirectory>:
  endbr64
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rsp,%rdi
  mov    $0x100,%esi
  call   <getcwd@plt>
  mov    $0x0,%eax
  mov    $0x0,%edx
  mov    %rsp,%rcx
L19:
  cmpb   $0x2f,(%rax,%rcx,1)
  cmove  %eax,%edx
  add    $0x1,%rax
  cmp    $0x100,%rax
  jne    L19
  movslq %edx,%rdx
  lea    0x1(%rsp,%rdx,1),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  mov    %rbx,%rax
  add    $0x110,%rsp
  pop    %rbx
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *getDirectory() {
  char cwd[256];
  char *result = malloc(256);
  getcwd(cwd, 256);
  int lastSlash = 0;
  for (int i = 0; i < 256; i++) {
    if (cwd[i] == '/')
      lastSlash = i;
  }
  strcpy(result, cwd + lastSlash + 1);
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<getDirectory>:
  endbr64
  push   %r12
  mov    $0x100,%edi
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <malloc@plt>
  mov    $0x100,%esi
  mov    %rbx,%rdi
  mov    %rax,%r12
  call   <getcwd@plt>
  xor    %eax,%eax
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L13:
  cmpb   $0x2f,(%rbx,%rax,1)
  cmove  %eax,%edx
  add    $0x1,%rax
  cmp    $0x100,%rax
  jne    L13
  lea    0x1(%rdx),%eax
  mov    %r12,%rdi
  cltq
  lea    (%rbx,%rax,1),%rsi
  call   <strcpy@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x118,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *getDirectory() {
  char cwd[256];
  char *result = malloc(256);
  getcwd(cwd, 256);
  int lastSlash = 0;
  for (int i = 0; i < 256; i++) {
    if (cwd[i] == '/')
      lastSlash = i;
  }
  strcpy(result, cwd + lastSlash + 1);
  return result;
}
```