ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<returnfilename>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x64,%edi
  call   <malloc@plt>
  mov    %rax,%r14
  mov    $0x0,%ebx
  jmp    L29
L34:
  cmpb   $0x2f,0x0(%rbp,%rbx,1)
  cmove  %r12d,%r13d
  add    $0x1,%rbx
L29:
  mov    %ebx,%r12d
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L34
  lea    0x1(%r13),%ebx
  movslq %ebx,%rbx
  movslq %r13d,%rax
  lea    -0x1(%r14),%r15
  sub    %rax,%r15
  jmp    L23
L6:
  movzbl 0x0(%rbp,%rbx,1),%eax
  mov    %al,(%r15,%rbx,1)
  add    $0x1,%rbx
L23:
  mov    %ebx,%r12d
  sub    %r13d,%r12d
  sub    $0x1,%r12d
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L6
  movslq %r12d,%r12
  movb   $0x0,(%r14,%r12,1)
  mov    %r14,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *returnfilename(char *fullpath) {
  char *filename = (char *)malloc(sizeof(char) * 100);
  int i, j, pos;
  for (i = 0; i < strlen(fullpath); i++) {
    if (fullpath[i] == '/')
      pos = i;
  }
  for (j = 0, i = pos + 1; i < strlen(fullpath); i++, j++) {
    filename[j] = fullpath[i];
  }
  filename[j] = '\0';
  return filename;
}
```
user: Translate the below assembly code into C programming language:

```asm
<returnfilename>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x64,%edi
  push   %rbx
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    %rax,%r13
  jmp    L38
L13:
  cmpb   $0x2f,0x0(%rbp,%rbx,1)
  cmove  %ebx,%r12d
  add    $0x1,%rbx
L38:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L13
  lea    0x1(%r12),%ebx
  lea    -0x1(%r13),%r14
  movslq %r12d,%r12
  movslq %ebx,%rbx
  sub    %r12,%r14
  jmp    L4
  nopl   (%rax)
L0:
  movzbl 0x0(%rbp,%rbx,1),%edx
  add    $0x1,%rbx
  mov    %dl,(%rax)
L4:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r8d
  lea    (%r14,%rbx,1),%rax
  cmp    %ebx,%r8d
  jg     L0
  movb   $0x0,(%rax)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *returnfilename(char *fullpath) {
  char *filename = (char *)malloc(sizeof(char) * 100);
  int i, j, pos;
  for (i = 0; i < strlen(fullpath); i++) {
    if (fullpath[i] == '/')
      pos = i;
  }
  for (j = 0, i = pos + 1; i < strlen(fullpath); i++, j++) {
    filename[j] = fullpath[i];
  }
  filename[j] = '\0';
  return filename;
}
```
user: Translate the below assembly code into C programming language:

```asm
<returnfilename>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x64,%edi
  push   %rbx
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    %rax,%r13
  jmp    L23
L24:
  cmpb   $0x2f,0x0(%rbp,%rbx,1)
  cmove  %ebx,%r12d
  add    $0x1,%rbx
L23:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L24
  lea    0x1(%r12),%ebx
  lea    -0x1(%r13),%r14
  movslq %r12d,%r12
  movslq %ebx,%rbx
  sub    %r12,%r14
  jmp    L8
  nopl   (%rax)
L31:
  movzbl 0x0(%rbp,%rbx,1),%edx
  add    $0x1,%rbx
  mov    %dl,(%rax)
L8:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r8d
  lea    (%r14,%rbx,1),%rax
  cmp    %ebx,%r8d
  jg     L31
  movb   $0x0,(%rax)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *returnfilename(char *fullpath) {
  char *filename = (char *)malloc(sizeof(char) * 100);
  int i, j, pos;
  for (i = 0; i < strlen(fullpath); i++) {
    if (fullpath[i] == '/')
      pos = i;
  }
  for (j = 0, i = pos + 1; i < strlen(fullpath); i++, j++) {
    filename[j] = fullpath[i];
  }
  filename[j] = '\0';
  return filename;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xa,-0x8(%rbp)
  movl   $0x14,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D8
data_type: string

tool: D8: after addition: %d


assistant: ```c
int main(void) {
  int x, y;
  x = 10;
  y = 20;
  x = x + y;
  printf("after addition: %d\n", x);
  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
  mov    $0x1e,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
datatype: string

tool: D4: after addition: %d


assistant: ```c
int main(void) {
  int x, y;
  x = 10;
  y = 20;
  x = x + y;
  printf("after addition: %d\n", x);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1e,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D6: after addition: %d


assistant: ```c
int main(void) {
  int x, y;
  x = 10;
  y = 20;
  x = x + y;
  printf("after addition: %d\n", x);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1e,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_infered: string

tool: D9: after addition: %d


assistant: ```c
int main(void) {
  int x, y;
  x = 10;
  y = 20;
  x = x + y;
  printf("after addition: %d\n", x);
  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
<e>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x12(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rax,%rdi
  call   *%rdx
  lea    -0x12(%rbp),%rax
  mov    %rax,%rdi
  call   <atof@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x20(%rbp)
  mov    $0x0,%eax
  call   <fork@plt>
  test   %rax,%rax
  je     L0
  mov    $0x1,%edi
  call   <exit@plt>
L0:
  movsd  -0x20(%rbp),%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double e(void (*f)(char *)) {
  char tmp[10];
  double x;
  f(tmp);
  x = atof(tmp);
  if (fork() != 0)
    exit(1);
  return x;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<e>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rax
  mov    %fs:0x28,%rdx
  mov    %rdx,0x18(%rsp)
  xor    %edx,%edx
  lea    0xe(%rsp),%rbx
  mov    %rbx,%rdi
  call   *%rax
  mov    %rbx,%rdi
  call   <atof@plt>
  movq   %xmm0,%rbx
  mov    $0x0,%eax
  call   <fork@plt>
  test   %rax,%rax
  jne    L23
L11:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  movq   %rbx,%xmm0
  add    $0x20,%rsp
  pop    %rbx
  ret
L23:
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L11
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double e(void (*f)(char *)) {
  char tmp[10];
  double x;
  f(tmp);
  x = atof(tmp);
  if (fork() != 0)
    exit(1);
  return x;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<e>:
  endbr64
  push   %rbp
  mov    %rdi,%r8
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x1e(%rsp),%rbp
  mov    %rbp,%rdi
  call   *%r8
  mov    %rbp,%rdi
  call   <atof@plt>
  xor    %eax,%eax
  movsd  %xmm0,0x8(%rsp)
  call   <fork@plt>
  movsd  0x8(%rsp),%xmm0
  test   %rax,%rax
  jne    L20
L3:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x30,%rsp
  pop    %rbp
  ret
  nopl   (%rax)
L20:
  mov    $0x1,%edi
  call   <exit@plt>
  movsd  0x8(%rsp),%xmm0
  jmp    L3
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double e(void (*f)(char *)) {
  char tmp[10];
  double x;
  f(tmp);
  x = atof(tmp);
  if (fork() != 0)
    exit(1);
  return x;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<e>:
  endbr64
  push   %rbp
  mov    %rdi,%r8
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x1e(%rsp),%rbp
  mov    %rbp,%rdi
  call   *%r8
  mov    %rbp,%rdi
  call   <atof@plt>
  xor    %eax,%eax
  movsd  %xmm0,0x8(%rsp)
  call   <fork@plt>
  movsd  0x8(%rsp),%xmm0
  test   %rax,%rax
  jne    L0
L18:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x30,%rsp
  pop    %rbp
  ret
  nopl   (%rax)
L0:
  mov    $0x1,%edi
  call   <exit@plt>
  movsd  0x8(%rsp),%xmm0
  jmp    L18
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double e(void (*f)(char *)) {
  char tmp[10];
  double x;
  f(tmp);
  x = atof(tmp);
  if (fork() != 0)
    exit(1);
  return x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<buf_string_compare_advance>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <buf_string_match_head_str@plt>
  test   %rax,%rax
  je     L17
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <buf_advance@plt>
  mov    $0x1,%eax
  jmp    L6
L17:
  mov    $0x0,%eax
L6:
  leave
  ret
```

assistant: ```c
int buf_string_compare_advance(struct buffer *src, const char *match) {
  if (buf_string_match_head_str(src, match)) {
    buf_advance(src, strlen(match));
    return 1;
  } else
    return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<buf_string_compare_advance>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <buf_string_match_head_str@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L0
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <buf_advance@plt>
  mov    $0x1,%eax
  jmp    L9
```

assistant: ```c
int buf_string_compare_advance(struct buffer *src, const char *match) {
  if (buf_string_match_head_str(src, match)) {
    buf_advance(src, strlen(match));
    return 1;
  } else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<buf_string_compare_advance>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <buf_string_match_head_str@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L0
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <buf_advance@plt>
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int buf_string_compare_advance(struct buffer *src, const char *match) {
  if (buf_string_match_head_str(src, match)) {
    buf_advance(src, strlen(match));
    return 1;
  } else
    return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<buf_string_compare_advance>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <buf_string_match_head_str@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L4
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <buf_advance@plt>
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int buf_string_compare_advance(struct buffer *src, const char *match) {
  if (buf_string_match_head_str(src, match)) {
    buf_advance(src, strlen(match));
    return 1;
  } else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Vector_Cross_Product>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mulss  %xmm1,%xmm0
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm2
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  mulss  %xmm2,%xmm1
  subss  %xmm1,%xmm0
  mov    -0x8(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm1
  mov    -0x18(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  %xmm1,%xmm0
  mov    -0x10(%rbp),%rax
  movss  (%rax),%xmm2
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm1
  mulss  %xmm2,%xmm1
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x10(%rbp),%rax
  movss  (%rax),%xmm1
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mulss  %xmm1,%xmm0
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm2
  mov    -0x18(%rbp),%rax
  movss  (%rax),%xmm1
  mulss  %xmm2,%xmm1
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void Vector_Cross_Product(float vectorOut[3], float v1[3], float v2[3]) {
  vectorOut[0] = (v1[1] * v2[2]) - (v1[2] * v2[1]);
  vectorOut[1] = (v1[2] * v2[0]) - (v1[0] * v2[2]);
  vectorOut[2] = (v1[0] * v2[1]) - (v1[1] * v2[0]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Vector_Cross_Product>:
  endbr64
  movss  0x4(%rsi),%xmm0
  mulss  0x8(%rdx),%xmm0
  movss  0x8(%rsi),%xmm1
  mulss  0x4(%rdx),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rdi)
  movss  0x8(%rsi),%xmm0
  mulss  (%rdx),%xmm0
  movss  (%rsi),%xmm1
  mulss  0x8(%rdx),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x4(%rdi)
  movss  (%rsi),%xmm0
  mulss  0x4(%rdx),%xmm0
  movss  0x4(%rsi),%xmm1
  mulss  (%rdx),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void Vector_Cross_Product(float vectorOut[3], float v1[3], float v2[3]) {
  vectorOut[0] = (v1[1] * v2[2]) - (v1[2] * v2[1]);
  vectorOut[1] = (v1[2] * v2[0]) - (v1[0] * v2[2]);
  vectorOut[2] = (v1[0] * v2[1]) - (v1[1] * v2[0]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Vector_Cross_Product>:
  endbr64
  movss  0x4(%rsi),%xmm0
  movss  0x8(%rsi),%xmm1
  mulss  0x8(%rdx),%xmm0
  mulss  0x4(%rdx),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rdi)
  movss  0x8(%rsi),%xmm0
  movss  (%rsi),%xmm1
  mulss  (%rdx),%xmm0
  mulss  0x8(%rdx),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x4(%rdi)
  movss  (%rsi),%xmm0
  movss  0x4(%rsi),%xmm1
  mulss  0x4(%rdx),%xmm0
  mulss  (%rdx),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void Vector_Cross_Product(float vectorOut[3], float v1[3], float v2[3]) {
  vectorOut[0] = (v1[1] * v2[2]) - (v1[2] * v2[1]);
  vectorOut[1] = (v1[2] * v2[0]) - (v1[0] * v2[2]);
  vectorOut[2] = (v1[0] * v2[1]) - (v1[1] * v2[0]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Vector_Cross_Product>:
  endbr64
  movss  0x4(%rsi),%xmm0
  movss  0x8(%rsi),%xmm1
  mulss  0x8(%rdx),%xmm0
  mulss  0x4(%rdx),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rdi)
  movss  0x8(%rsi),%xmm0
  movss  (%rsi),%xmm1
  mulss  (%rdx),%xmm0
  mulss  0x8(%rdx),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x4(%rdi)
  movss  (%rsi),%xmm0
  movss  0x4(%rsi),%xmm1
  mulss  0x4(%rdx),%xmm0
  mulss  (%rdx),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void Vector_Cross_Product(float vectorOut[3], float v1[3], float v2[3]) {
  vectorOut[0] = (v1[1] * v2[2]) - (v1[2] * v2[1]);
  vectorOut[1] = (v1[2] * v2[0]) - (v1[0] * v2[2]);
  vectorOut[2] = (v1[0] * v2[1]) - (v1[1] * v2[0]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x14(%rbp)
  jmp    L27
L2:
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x14(%rbp),%xmm1
  movsd  D8(%rip),%xmm0
  divsd  %xmm1,%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L27:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jle    L2
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D5
type: string
name: parse_data
arguments: label: D8
type: f64
name: parse_data
arguments: label: D17
type: string

tool: D5: %d

tool: D8: 1

tool: D17: %.3lf


assistant: ```c
int main() {
  int i, n;
  double sum = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    sum += 1.0 / i;
  printf("%.3lf\n", sum);
  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    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L24
  add    $0x1,%edx
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  movsd  D23(%rip),%xmm3
L39:
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  movapd %xmm3,%xmm1
  divsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  add    $0x1,%eax
  cmp    %edx,%eax
  jne    L39
L16:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L24:
  pxor   %xmm0,%xmm0
  jmp    L16
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D22
type: string
name: parse_data
arguments: data_tag: D23
type: f64
name: parse_data
arguments: data_tag: D6
type: string

tool: D22: %d

tool: D23: 1

tool: D6: %.3lf


assistant: ```c
int main() {
  int i, n;
  double sum = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    sum += 1.0 / i;
  printf("%.3lf\n", sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D13(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L18
  movsd  D20(%rip),%xmm3
  add    $0x1,%edx
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  nopl   (%rax)
L20:
  pxor   %xmm2,%xmm2
  movapd %xmm3,%xmm1
  cvtsi2sd %eax,%xmm2
  add    $0x1,%eax
  divsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %eax,%edx
  jne    L20
L11:
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L18:
  pxor   %xmm0,%xmm0
  jmp    L11
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D13
type_infered: string
name: disassemble_data
arguments: tag: D20
type_infered: f64
name: disassemble_data
arguments: tag: D19
type_infered: string

tool: D13: %d

tool: D20: 1

tool: D19: %.3lf


assistant: ```c
int main() {
  int i, n;
  double sum = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    sum += 1.0 / i;
  printf("%.3lf\n", sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D28(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L7
  movsd  D27(%rip),%xmm3
  add    $0x1,%edx
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  nopl   (%rax)
L24:
  pxor   %xmm2,%xmm2
  movapd %xmm3,%xmm1
  cvtsi2sd %eax,%xmm2
  add    $0x1,%eax
  divsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %eax,%edx
  jne    L24
L22:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  pxor   %xmm0,%xmm0
  jmp    L22
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D28
data_type: string
name: retrieve_data
arguments: data_label: D27
data_type: f64
name: retrieve_data
arguments: data_label: D1
data_type: string

tool: D28: %d

tool: D27: 1

tool: D1: %.3lf


assistant: ```c
int main() {
  int i, n;
  double sum = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    sum += 1.0 / i;
  printf("%.3lf\n", sum);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ns>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  mov    -0x20(%rbp),%rax
  imul   $0x3b9aca00,%rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
uint64_t ns() {
  struct timespec ts;
  clock_gettime(CLOCK_MONOTONIC, &ts);
  return ts.tv_sec * 1000000000ULL + ts.tv_nsec;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ns>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    D5(%rip),%rax
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,(%rsp),%rax
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t ns() {
  struct timespec ts;
  clock_gettime(CLOCK_MONOTONIC, &ts);
  return ts.tv_sec * 1000000000ULL + ts.tv_nsec;
}
```
user: What is the c source code of the assembly code below?

```asm
<ns>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D2(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,(%rsp),%rax
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t ns() {
  struct timespec ts;
  clock_gettime(CLOCK_MONOTONIC, &ts);
  return ts.tv_sec * 1000000000ULL + ts.tv_nsec;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ns>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D5(%rip),%rax
  mov    %rsp,%rsi
  mov    (%rax),%edi
  call   <clock_gettime@plt>
  imul   $0x3b9aca00,(%rsp),%rax
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t ns() {
  struct timespec ts;
  clock_gettime(CLOCK_MONOTONIC, &ts);
  return ts.tv_sec * 1000000000ULL + ts.tv_nsec;
}
```