ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D16(%rip),%rdi
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  call   <puts@plt>
  nopl   (%rax)
L39:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x23,%al
  je     L48
L7:
  cmp    $0x20,%al
  je     L41
  cmp    $0xa,%al
  je     L3
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <getchar@plt>
  cmp    $0x23,%al
  jne    L7
L48:
  mov    %r12d,%ecx
  mov    %r13d,%edx
  mov    %ebp,%esi
  xor    %eax,%eax
  lea    D2(%rip),%rdi
  call   <printf@plt>
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
L41:
  add    $0x1,%ebp
  jmp    L39
  nopl   (%rax)
L3:
  add    $0x1,%r13d
  jmp    L39
```

tool_calls: name: parse_data
arguments: label: D16
type_infered: string
name: parse_data
arguments: label: D2
type_infered: string

tool: D16: Please enter something (ends with #)

tool: D2: Space: %d
Line:%d
Other:%d


assistant: ```c
int main(void) {
  char c;
  int spaces, lines, other;
  spaces = lines = other = 0;
  printf("Please enter something (ends with #)\n");
  while ((c = getchar()) != '#') {
    if (' ' == c)
      spaces++;
    else if ('\n' == c)
      lines++;
    else
      other++;
  }
  printf("Space: %d\nLine:%d\nOther:%d\n", spaces, lines, other);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r13
  lea    D10(%rip),%rdi
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  call   <puts@plt>
  nopl   (%rax)
L18:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x23,%al
  je     L8
L33:
  cmp    $0x20,%al
  je     L37
  cmp    $0xa,%al
  je     L14
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <getchar@plt>
  cmp    $0x23,%al
  jne    L33
L8:
  mov    %r12d,%ecx
  mov    %r13d,%edx
  mov    %ebp,%esi
  xor    %eax,%eax
  lea    D0(%rip),%rdi
  call   <printf@plt>
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
L37:
  add    $0x1,%ebp
  jmp    L18
  nopl   (%rax)
L14:
  add    $0x1,%r13d
  jmp    L18
```

tool_calls: name: retrieve_data
arguments: label: D10
datatype: string
name: retrieve_data
arguments: label: D0
datatype: string

tool: D10: Please enter something (ends with #)

tool: D0: Space: %d
Line:%d
Other:%d


assistant: ```c
int main(void) {
  char c;
  int spaces, lines, other;
  spaces = lines = other = 0;
  printf("Please enter something (ends with #)\n");
  while ((c = getchar()) != '#') {
    if (' ' == c)
      spaces++;
    else if ('\n' == c)
      lines++;
    else
      other++;
  }
  printf("Space: %d\nLine:%d\nOther:%d\n", spaces, lines, other);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<track_bbox_width>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <blob_in_bounds@plt>
  test   %eax,%eax
  jne    L2
  mov    $0xffffffff,%eax
  jmp    L7
L2:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <track_bbox_right@plt>
  mov    %eax,%ebx
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <track_bbox_left@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  add    $0x1,%eax
L7:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int track_bbox_width(int ch, int i) {
  if (!blob_in_bounds(ch, i))
    return -1;
  return track_bbox_right(ch, i) - track_bbox_left(ch, i) + 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<track_bbox_width>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %esi,%r12d
  call   <blob_in_bounds@plt>
  test   %eax,%eax
  je     L10
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <track_bbox_right@plt>
  mov    %eax,%ebx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <track_bbox_left@plt>
  sub    %eax,%ebx
  lea    0x1(%rbx),%eax
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  mov    $0xffffffff,%eax
  jmp    L13
```

assistant: ```c
int track_bbox_width(int ch, int i) {
  if (!blob_in_bounds(ch, i))
    return -1;
  return track_bbox_right(ch, i) - track_bbox_left(ch, i) + 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<track_bbox_width>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  call   <blob_in_bounds@plt>
  test   %eax,%eax
  je     L5
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <track_bbox_right@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <track_bbox_left@plt>
  sub    %eax,%ebx
  lea    0x1(%rbx),%eax
L0:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L5:
  mov    $0xffffffff,%eax
  jmp    L0
```

assistant: ```c
int track_bbox_width(int ch, int i) {
  if (!blob_in_bounds(ch, i))
    return -1;
  return track_bbox_right(ch, i) - track_bbox_left(ch, i) + 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<track_bbox_width>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  call   <blob_in_bounds@plt>
  test   %eax,%eax
  je     L15
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <track_bbox_right@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <track_bbox_left@plt>
  sub    %eax,%ebx
  lea    0x1(%rbx),%eax
L7:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    $0xffffffff,%eax
  jmp    L7
```

assistant: ```c
int track_bbox_width(int ch, int i) {
  if (!blob_in_bounds(ch, i))
    return -1;
  return track_bbox_right(ch, i) - track_bbox_left(ch, i) + 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<digip>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movl   $0x0,(%rax)
  jmp    L16
L5:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x1,-0x8(%rbp)
L16:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L5
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *digip(char *p, int *n) {
  *n = 0;
  while (isdigit(*p)) {
    *n *= 10;
    *n += (int)(*p) - '0';
    ++p;
  }
  return p;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<digip>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    $0x0,%eax
  jmp    L7
L0:
  mov    0x0(%rbp),%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  mov    %eax,0x0(%rbp)
  movsbl (%rbx),%edx
  lea    -0x30(%rax,%rdx,1),%eax
  add    $0x1,%rbx
L7:
  mov    %eax,0x0(%rbp)
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L0
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *digip(char *p, int *n) {
  *n = 0;
  while (isdigit(*p)) {
    *n *= 10;
    *n += (int)(*p) - '0';
    ++p;
  }
  return p;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<digip>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rdi,%r12
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  jmp    L10
  nopl   (%rax)
L15:
  mov    (%rbx),%eax
  add    $0x1,%r12
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  mov    %eax,(%rbx)
  movsbl -0x1(%r12),%edx
  lea    -0x30(%rax,%rdx,1),%eax
L10:
  mov    %eax,(%rbx)
  movsbl (%r12),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L15
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *digip(char *p, int *n) {
  *n = 0;
  while (isdigit(*p)) {
    *n *= 10;
    *n += (int)(*p) - '0';
    ++p;
  }
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<digip>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rdi,%r12
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  jmp    L11
  nopl   (%rax)
L3:
  mov    (%rbx),%eax
  add    $0x1,%r12
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  mov    %eax,(%rbx)
  movsbl -0x1(%r12),%edx
  lea    -0x30(%rax,%rdx,1),%eax
L11:
  mov    %eax,(%rbx)
  movsbl (%r12),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L3
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *digip(char *p, int *n) {
  *n = 0;
  while (isdigit(*p)) {
    *n *= 10;
    *n += (int)(*p) - '0';
    ++p;
  }
  return p;
}
```
user: What is the c source code of the assembly code below:

```asm
<glmCompMult3fs_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movss  %xmm0,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L8
L1:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mulss  -0x1c(%rbp),%xmm0
  movss  %xmm0,(%rax)
  addq   $0x1,-0x8(%rbp)
L8:
  cmpq   $0x2,-0x8(%rbp)
  jbe    L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void glmCompMult3fs_(float x[3], float y) {
  for (size_t i = 0; i < 3; ++i)
    x[i] *= y;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<glmCompMult3fs_>:
  endbr64
  movaps %xmm0,%xmm1
  mulss  (%rdi),%xmm1
  movss  %xmm1,(%rdi)
  movaps %xmm0,%xmm1
  mulss  0x4(%rdi),%xmm1
  movss  %xmm1,0x4(%rdi)
  mulss  0x8(%rdi),%xmm0
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void glmCompMult3fs_(float x[3], float y) {
  for (size_t i = 0; i < 3; ++i)
    x[i] *= y;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<glmCompMult3fs_>:
  endbr64
  movss  (%rdi),%xmm1
  mulss  %xmm0,%xmm1
  movss  %xmm1,(%rdi)
  movss  0x4(%rdi),%xmm1
  mulss  %xmm0,%xmm1
  mulss  0x8(%rdi),%xmm0
  movss  %xmm1,0x4(%rdi)
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void glmCompMult3fs_(float x[3], float y) {
  for (size_t i = 0; i < 3; ++i)
    x[i] *= y;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<glmCompMult3fs_>:
  endbr64
  movq   (%rdi),%xmm1
  movaps %xmm0,%xmm2
  shufps $0xe0,%xmm2,%xmm2
  mulss  0x8(%rdi),%xmm0
  mulps  %xmm2,%xmm1
  movss  %xmm0,0x8(%rdi)
  movlps %xmm1,(%rdi)
  ret
```

assistant: ```c
void glmCompMult3fs_(float x[3], float y) {
  for (size_t i = 0; i < 3; ++i)
    x[i] *= y;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0xdeadbeef,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <test_api@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L15:
  mov    $0x0,%eax
  call   <test_shuffle@plt>
  mov    $0x0,%eax
  call   <test_map@plt>
  addl   $0x1,-0x4(%rbp)
L8:
  cmpl   $0x3,-0x4(%rbp)
  jbe    L15
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  unsigned int i;
  srand(0xdeadbeef);
  test_api();
  for (i = 0; i < 4; ++i) {
    test_shuffle();
    test_map();
  }
  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    $0xdeadbeef,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <test_api@plt>
  mov    $0x0,%eax
  call   <test_shuffle@plt>
  mov    $0x0,%eax
  call   <test_map@plt>
  mov    $0x0,%eax
  call   <test_shuffle@plt>
  mov    $0x0,%eax
  call   <test_map@plt>
  mov    $0x0,%eax
  call   <test_shuffle@plt>
  mov    $0x0,%eax
  call   <test_map@plt>
  mov    $0x0,%eax
  call   <test_shuffle@plt>
  mov    $0x0,%eax
  call   <test_map@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  unsigned int i;
  srand(0xdeadbeef);
  test_api();
  for (i = 0; i < 4; ++i) {
    test_shuffle();
    test_map();
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xdeadbeef,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <test_api@plt>
  xor    %eax,%eax
  call   <test_shuffle@plt>
  xor    %eax,%eax
  call   <test_map@plt>
  xor    %eax,%eax
  call   <test_shuffle@plt>
  xor    %eax,%eax
  call   <test_map@plt>
  xor    %eax,%eax
  call   <test_shuffle@plt>
  xor    %eax,%eax
  call   <test_map@plt>
  xor    %eax,%eax
  call   <test_shuffle@plt>
  xor    %eax,%eax
  call   <test_map@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  unsigned int i;
  srand(0xdeadbeef);
  test_api();
  for (i = 0; i < 4; ++i) {
    test_shuffle();
    test_map();
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xdeadbeef,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <test_api@plt>
  xor    %eax,%eax
  call   <test_shuffle@plt>
  xor    %eax,%eax
  call   <test_map@plt>
  xor    %eax,%eax
  call   <test_shuffle@plt>
  xor    %eax,%eax
  call   <test_map@plt>
  xor    %eax,%eax
  call   <test_shuffle@plt>
  xor    %eax,%eax
  call   <test_map@plt>
  xor    %eax,%eax
  call   <test_shuffle@plt>
  xor    %eax,%eax
  call   <test_map@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  unsigned int i;
  srand(0xdeadbeef);
  test_api();
  for (i = 0; i < 4; ++i) {
    test_shuffle();
    test_map();
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<parse_human_number>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <atol@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  nop
L8:
  addq   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L8
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L17
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L17
  mov    -0x8(%rbp),%rax
  sub    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  lea    D13(%rip),%rax
  mov    (%rdx,%rax,1),%rax
  mov    -0x20(%rbp),%rdx
  imul   %rdx,%rax
  mov    %rax,-0x20(%rbp)
L17:
  mov    -0x20(%rbp),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D18
type_infered: string
name: read_data
arguments: data_tag: D13
type_infered: string

tool: D18: KMG

tool: 

assistant: ```c
unsigned long parse_human_number(const char *num) {
  unsigned long ret = 0;
  static const unsigned long mul[] = {1024, 1024 * 1024, 1024 * 1024 * 1024};
  const char *kmg = "KMG";
  const char *kmgind, *p;
  ret = atol(num);
  p = num;
  while (isdigit(*(++p)))
    ;
  if (*p && (kmgind = strchr(kmg, *p)))
    ret *= mul[kmgind - kmg];
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<parse_human_number>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <atol@plt>
  mov    %rax,%rbp
L6:
  add    $0x1,%rbx
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L6
  movzbl (%rbx),%esi
  test   %sil,%sil
  jne    L2
L9:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  movsbl %sil,%esi
  lea    D17(%rip),%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L9
  lea    D17(%rip),%rdx
  sub    %rdx,%rax
  lea    D26(%rip),%rdx
  imul   (%rdx,%rax,8),%rbp
  jmp    L9
```

tool_calls: name: unpack_data
arguments: tag: D17
type: string
name: unpack_data
arguments: tag: D26
type: string

tool: D17: KMG

tool: 

assistant: ```c
unsigned long parse_human_number(const char *num) {
  unsigned long ret = 0;
  static const unsigned long mul[] = {1024, 1024 * 1024, 1024 * 1024 * 1024};
  const char *kmg = "KMG";
  const char *kmgind, *p;
  ret = atol(num);
  p = num;
  while (isdigit(*(++p)))
    ;
  if (*p && (kmgind = strchr(kmg, *p)))
    ret *= mul[kmgind - kmg];
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<parse_human_number>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <atol@plt>
  mov    %rax,%r12
  cs nopw 0x0(%rax,%rax,1)
L4:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L4
  movsbl (%rbx),%esi
  test   %sil,%sil
  jne    L29
L8:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   (%rax)
L29:
  lea    D4(%rip),%rbx
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L8
  sub    %rbx,%rax
  lea    D5(%rip),%rdx
  imul   (%rdx,%rax,8),%r12
  add    $0x8,%rsp
  pop    %rbx
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D4
datatype: string
name: parse_data
arguments: tag: D5
datatype: string

tool: D4: KMG

tool: 

assistant: ```c
unsigned long parse_human_number(const char *num) {
  unsigned long ret = 0;
  static const unsigned long mul[] = {1024, 1024 * 1024, 1024 * 1024 * 1024};
  const char *kmg = "KMG";
  const char *kmgind, *p;
  ret = atol(num);
  p = num;
  while (isdigit(*(++p)))
    ;
  if (*p && (kmgind = strchr(kmg, *p)))
    ret *= mul[kmgind - kmg];
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_human_number>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <atol@plt>
  mov    %rax,%r12
  cs nopw 0x0(%rax,%rax,1)
L25:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L25
  movsbl (%rbx),%esi
  test   %sil,%sil
  jne    L27
L0:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   (%rax)
L27:
  lea    D2(%rip),%rbx
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L0
  sub    %rbx,%rax
  lea    D16(%rip),%rdx
  imul   (%rdx,%rax,8),%r12
  add    $0x8,%rsp
  pop    %rbx
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D2
data_type: string
name: disassemble_data
arguments: data_label: D16
data_type: string

tool: D2: KMG

tool: 

assistant: ```c
unsigned long parse_human_number(const char *num) {
  unsigned long ret = 0;
  static const unsigned long mul[] = {1024, 1024 * 1024, 1024 * 1024 * 1024};
  const char *kmg = "KMG";
  const char *kmgind, *p;
  ret = atol(num);
  p = num;
  while (isdigit(*(++p)))
    ;
  if (*p && (kmgind = strchr(kmg, *p)))
    ret *= mul[kmgind - kmg];
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xc,-0x4(%rbp)
  cmpl   $0xc,-0x4(%rbp)
  jne    L3
  mov    -0x4(%rbp),%eax
  cltq
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L3:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
main() {
  int x = 12;
  if (x == 12)
    printf(x);
}
```
user: 请从给定的汇编指令推导出C源代码:

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

assistant: ```c
main() {
  int x = 12;
  if (x == 12)
    printf(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
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xc,%edi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  int x = 12;
  if (x == 12)
    printf(x);
}
```
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    $0xc,%edi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
main() {
  int x = 12;
  if (x == 12)
    printf(x);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<testRotation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D63(%rip),%rax
  movq   %rax,%xmm0
  call   <rotateRobot@plt>
  mov    $0x0,%eax
  call   <stopRobot@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  mov    D63(%rip),%rax
  movq   %rax,%xmm0
  call   <rotateRobot@plt>
  mov    $0x0,%eax
  call   <stopRobot@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  mov    D63(%rip),%rax
  movq   %rax,%xmm0
  call   <rotateRobot@plt>
  mov    $0x0,%eax
  call   <stopRobot@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  mov    D63(%rip),%rax
  movq   %rax,%xmm0
  call   <rotateRobot@plt>
  mov    $0x0,%eax
  call   <stopRobot@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  mov    D72(%rip),%rax
  movq   %rax,%xmm0
  call   <rotateRobot@plt>
  mov    $0x0,%eax
  call   <stopRobot@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  mov    D72(%rip),%rax
  movq   %rax,%xmm0
  call   <rotateRobot@plt>
  mov    $0x0,%eax
  call   <stopRobot@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  mov    D72(%rip),%rax
  movq   %rax,%xmm0
  call   <rotateRobot@plt>
  mov    $0x0,%eax
  call   <stopRobot@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  mov    D72(%rip),%rax
  movq   %rax,%xmm0
  call   <rotateRobot@plt>
  mov    $0x0,%eax
  call   <stopRobot@plt>
  mov    $0x64,%edi
  call   <wait10Msec@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D63
type_infered: byte[8]
name: read_data
arguments: data_tag: D72
type_infered: byte[8]

tool: {"D63": [0x18,0x2d,0x44,0x54,0xfb,0x21,0xf9,0x3f]}

tool: {"D72": [0x18,0x2d,0x44,0x54,0xfb,0x21,0xf9,0xbf]}

assistant: ```c
void testRotation() {
  rotateRobot(3.14159265358979323846 / 2);
  stopRobot();
  wait10Msec(100);
  rotateRobot(3.14159265358979323846 / 2);
  stopRobot();
  wait10Msec(100);
  rotateRobot(3.14159265358979323846 / 2);
  stopRobot();
  wait10Msec(100);
  rotateRobot(3.14159265358979323846 / 2);
  stopRobot();
  wait10Msec(100);
  rotateRobot(-3.14159265358979323846 / 2);
  stopRobot();
  wait10Msec(100);
  rotateRobot(-3.14159265358979323846 / 2);
  stopRobot();
  wait10Msec(100);
  rotateRobot(-3.14159265358979323846 / 2);
  stopRobot();
  wait10Msec(100);
  rotateRobot(-3.14159265358979323846 / 2);
  stopRobot();
  wait10Msec(100);
}
```