ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<compute_reaction_forces>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8d,-0x34(%rbp)
  mov    %r9,-0x40(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L10
L15:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L26
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movq   D4(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  movl   $0x1,-0x4(%rbp)
  jmp    L12
L0:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm2,%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jle    L0
L26:
  addl   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jle    L15
  nop
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
datatype: byte[8]

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
void compute_reaction_forces(double *R, double *F, double **K, double *D, int DoF, int *r) {
  int i, j;
  for (i = 1; i <= DoF; i++) {
    R[i] = 0;
    if (r[i]) {
      R[i] = -F[i];
      for (j = 1; j <= DoF; j++)
        R[i] += K[i][j] * D[j];
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<compute_reaction_forces>:
  endbr64
  test   %r8d,%r8d
  jle    L24
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdx,%r10
  mov    %rcx,%r11
  lea    0x8(%rdi),%rsi
  lea    0x1(%r8),%ebp
  lea    0x0(,%rbp,8),%r8
  mov    $0x1,%edx
  movq   D8(%rip),%xmm1
  jmp    L35
L25:
  add    $0x1,%rdx
  add    $0x8,%rsi
  cmp    %rbp,%rdx
  je     L8
L35:
  mov    %rsi,%rcx
  movq   $0x0,(%rsi)
  cmpl   $0x0,(%r9,%rdx,4)
  je     L25
  movsd  (%rbx,%rdx,8),%xmm0
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,(%rsi)
  mov    $0x8,%eax
L57:
  mov    (%r10,%rdx,8),%rdi
  movsd  (%rdi,%rax,1),%xmm0
  mulsd  (%r11,%rax,1),%xmm0
  addsd  (%rcx),%xmm0
  movsd  %xmm0,(%rcx)
  add    $0x8,%rax
  cmp    %r8,%rax
  jne    L57
  jmp    L25
L8:
  pop    %rbx
  pop    %rbp
  ret
L24:
  ret
```

tool_calls: name: parse_data
arguments: tag: D8
data_type: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
void compute_reaction_forces(double *R, double *F, double **K, double *D, int DoF, int *r) {
  int i, j;
  for (i = 1; i <= DoF; i++) {
    R[i] = 0;
    if (r[i]) {
      R[i] = -F[i];
      for (j = 1; j <= DoF; j++)
        R[i] += K[i][j] * D[j];
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<compute_reaction_forces>:
  endbr64
  mov    %rsi,%r11
  mov    %rdx,%r10
  lea    0x1(%r8),%esi
  mov    $0x1,%edx
  test   %r8d,%r8d
  jg     L39
  jmp    L59
  nopw   0x0(%rax,%rax,1)
L5:
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  je     L55
L39:
  mov    (%r9,%rdx,4),%eax
  movq   $0x0,(%rdi,%rdx,8)
  test   %eax,%eax
  je     L5
  movsd  (%r11,%rdx,8),%xmm1
  mov    (%r10,%rdx,8),%r8
  mov    $0x1,%eax
  xorpd  D8(%rip),%xmm1
  movsd  %xmm1,(%rdi,%rdx,8)
  nopl   (%rax)
L20:
  movsd  (%r8,%rax,8),%xmm0
  mulsd  (%rcx,%rax,8),%xmm0
  add    $0x1,%rax
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdi,%rdx,8)
  cmp    %rax,%rsi
  jne    L20
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L39
L55:
  ret
L59:
  ret
```

tool_calls: name: parse_data
arguments: tag: D8
type: byte[16]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void compute_reaction_forces(double *R, double *F, double **K, double *D, int DoF, int *r) {
  int i, j;
  for (i = 1; i <= DoF; i++) {
    R[i] = 0;
    if (r[i]) {
      R[i] = -F[i];
      for (j = 1; j <= DoF; j++)
        R[i] += K[i][j] * D[j];
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<compute_reaction_forces>:
  endbr64
  mov    %rsi,%r11
  mov    %rdx,%r10
  lea    0x1(%r8),%esi
  mov    $0x1,%edx
  test   %r8d,%r8d
  jg     L53
  jmp    L10
  nopw   0x0(%rax,%rax,1)
L27:
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  je     L24
L53:
  mov    (%r9,%rdx,4),%eax
  movq   $0x0,(%rdi,%rdx,8)
  test   %eax,%eax
  je     L27
  movsd  (%r11,%rdx,8),%xmm1
  mov    (%r10,%rdx,8),%r8
  mov    $0x1,%eax
  xorpd  D2(%rip),%xmm1
  movsd  %xmm1,(%rdi,%rdx,8)
  nopl   (%rax)
L28:
  movsd  (%r8,%rax,8),%xmm0
  mulsd  (%rcx,%rax,8),%xmm0
  add    $0x1,%rax
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdi,%rdx,8)
  cmp    %rax,%rsi
  jne    L28
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L53
L24:
  ret
L10:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type_infered: byte[16]

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void compute_reaction_forces(double *R, double *F, double **K, double *D, int DoF, int *r) {
  int i, j;
  for (i = 1; i <= DoF; i++) {
    R[i] = 0;
    if (r[i]) {
      R[i] = -F[i];
      for (j = 1; j <= DoF; j++)
        R[i] += K[i][j] * D[j];
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type_guessed: string

tool: D3: sleep 30

assistant: ```c
int main() {
  int return_value;
  return_value = system("sleep 30");
  return (return_value);
}
```
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
  lea    D8(%rip),%rdi
  call   <system@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
datatype: string

tool: D8: sleep 30

assistant: ```c
int main() {
  int return_value;
  return_value = system("sleep 30");
  return (return_value);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  lea    D9(%rip),%rdi
  jmp    1050 <system@plt>
```

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

tool: D9: sleep 30

assistant: ```c
int main() {
  int return_value;
  return_value = system("sleep 30");
  return (return_value);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  lea    D1(%rip),%rdi
  jmp    1050 <system@plt>
```

tool_calls: name: parse_data
arguments: label: D1
data_type: string

tool: D1: sleep 30

assistant: ```c
int main() {
  int return_value;
  return_value = system("sleep 30");
  return (return_value);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<y_memdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L7
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *y_memdup(const void *addr, int n) {
  void *new_chunk = malloc(n);
  if (new_chunk)
    memcpy(new_chunk, addr, n);
  return new_chunk;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<y_memdup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %esi,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L2
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L2:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void *y_memdup(const void *addr, int n) {
  void *new_chunk = malloc(n);
  if (new_chunk)
    memcpy(new_chunk, addr, n);
  return new_chunk;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<y_memdup>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %esi,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L0
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L0:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void *y_memdup(const void *addr, int n) {
  void *new_chunk = malloc(n);
  if (new_chunk)
    memcpy(new_chunk, addr, n);
  return new_chunk;
}
```
user: What is the c source code of the assembly code below:

```asm
<y_memdup>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %esi,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L8
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L8:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void *y_memdup(const void *addr, int n) {
  void *new_chunk = malloc(n);
  if (new_chunk)
    memcpy(new_chunk, addr, n);
  return new_chunk;
}
```
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    $0x420,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D8(%rip),%rax
  mov    %rax,-0x418(%rbp)
  lea    -0x410(%rbp),%rdx
  mov    -0x418(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <expand@plt>
  lea    -0x410(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@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: read_data
arguments: data_tag: D8
data_type: string
name: read_data
arguments: data_tag: D19
data_type: string

tool: D8: a-z0-9A-Z---

tool: D19: %s


assistant: ```c
int main() {
  char result[1024];
  char *s1 = "a-z0-9A-Z---";
  expand(s1, result);
  printf("%s\n", result);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D11(%rip),%rdi
  call   <expand@plt>
  mov    %rbx,%rsi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x410,%rsp
  pop    %rbx
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D11
type: string
name: retrieve_data
arguments: data_label: D17
type: string

tool: D11: a-z0-9A-Z---

tool: D17: %s


assistant: ```c
int main() {
  char result[1024];
  char *s1 = "a-z0-9A-Z---";
  expand(s1, result);
  printf("%s\n", result);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D15(%rip),%rdi
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <expand@plt>
  mov    %rbp,%rsi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x410,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D15
type_guessed: string
name: parse_data
arguments: data_label: D17
type_guessed: string

tool: D15: a-z0-9A-Z---

tool: D17: %s


assistant: ```c
int main() {
  char result[1024];
  char *s1 = "a-z0-9A-Z---";
  expand(s1, result);
  printf("%s\n", result);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D19(%rip),%rdi
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <expand@plt>
  mov    %rbp,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x410,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D19
data_type: string
name: read_data
arguments: data_label: D8
data_type: string

tool: D19: a-z0-9A-Z---

tool: D8: %s


assistant: ```c
int main() {
  char result[1024];
  char *s1 = "a-z0-9A-Z---";
  expand(s1, result);
  printf("%s\n", result);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <factorial@plt>
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D16
type: string
name: unpack_data
arguments: tag: D19
type: string
name: unpack_data
arguments: tag: D28
type: string

tool: D16: Enter number: 

tool: D19: %d

tool: D28: Result is %d! = %1d


assistant: ```c
int main(void) {
  int n;
  printf("Enter number: \n");
  scanf("%d", &n);
  printf("Result is %d! = %1d\n", n, factorial(n));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <factorial@plt>
  mov    %eax,%edx
  mov    0x4(%rsp),%esi
  lea    D13(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D14
datatype: string
name: parse_data
arguments: data_label: D22
datatype: string
name: parse_data
arguments: data_label: D13
datatype: string

tool: D14: Enter number: 

tool: D22: %d

tool: D13: Result is %d! = %1d


assistant: ```c
int main(void) {
  int n;
  printf("Enter number: \n");
  scanf("%d", &n);
  printf("Result is %d! = %1d\n", n, factorial(n));
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D20(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <factorial@plt>
  mov    0x4(%rsp),%esi
  lea    D15(%rip),%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D14
type_infered: string
name: parse_data
arguments: data_tag: D20
type_infered: string
name: parse_data
arguments: data_tag: D15
type_infered: string

tool: D14: Enter number: 

tool: D20: %d

tool: D15: Result is %d! = %1d


assistant: ```c
int main(void) {
  int n;
  printf("Enter number: \n");
  scanf("%d", &n);
  printf("Result is %d! = %1d\n", n, factorial(n));
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D28(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <factorial@plt>
  mov    0x4(%rsp),%esi
  lea    D19(%rip),%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D28
type_infered: string
name: disassemble_data
arguments: data_label: D29
type_infered: string
name: disassemble_data
arguments: data_label: D19
type_infered: string

tool: D28: Enter number: 

tool: D29: %d

tool: D19: Result is %d! = %1d


assistant: ```c
int main(void) {
  int n;
  printf("Enter number: \n");
  scanf("%d", &n);
  printf("Result is %d! = %1d\n", n, factorial(n));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<distanciaRuta>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r14d
  mov    %rsi,%r15
  mov    %rdx,%r12
  mov    %rcx,%rbp
  cmp    $0x1,%edi
  jle    L4
  mov    %rsi,%rbx
  lea    -0x2(%rdi),%eax
  lea    0x4(%rsi,%rax,4),%r13
  pxor   %xmm3,%xmm3
  movss  %xmm3,0x4(%rsp)
L28:
  movslq 0x4(%rbx),%rdx
  movslq (%rbx),%rax
  movss  (%r12,%rdx,4),%xmm0
  subss  (%r12,%rax,4),%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %rax,0x8(%rsp)
  movslq 0x4(%rbx),%rdx
  movslq (%rbx),%rax
  movss  0x0(%rbp,%rdx,4),%xmm0
  subss  0x0(%rbp,%rax,4),%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  mov    0x8(%rsp),%rdi
  add    %rax,%rdi
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  addss  0x4(%rsp),%xmm0
  movss  %xmm0,0x4(%rsp)
  add    $0x4,%rbx
  cmp    %r13,%rbx
  jne    L28
L19:
  movslq %r14d,%r14
  lea    -0x4(%r15,%r14,4),%r13
  movslq 0x0(%r13),%rdx
  movslq (%r15),%rax
  movss  (%r12,%rdx,4),%xmm0
  subss  (%r12,%rax,4),%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %rax,%rbx
  movslq 0x0(%r13),%rdx
  movslq (%r15),%rax
  movss  0x0(%rbp,%rdx,4),%xmm0
  subss  0x0(%rbp,%rax,4),%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  lea    (%rbx,%rax,1),%rdi
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  addss  0x4(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  D8(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  pxor   %xmm4,%xmm4
  movss  %xmm4,0x4(%rsp)
  jmp    L19
```

tool_calls: name: unpack_data
arguments: tag: D8
datatype: f64

tool: D8: 111.32

assistant: ```c
float distanciaRuta(int ciudades, int *ruta, float *latitudes, float *longitudes) {
  float distancia = 0.0f;
  for (int i = 1; i < ciudades; i++)
    distancia += sqrt(pow(latitudes[ruta[i]] - latitudes[ruta[i - 1]], 2) + pow(longitudes[ruta[i]] - longitudes[ruta[i - 1]], 2));
  distancia += sqrt(pow(latitudes[ruta[ciudades - 1]] - latitudes[ruta[0]], 2) + pow(longitudes[ruta[ciudades - 1]] - longitudes[ruta[0]], 2));
  return distancia * 111.32;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<distanciaRuta>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  movslq %edi,%r14
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  cmp    $0x1,%r14d
  jle    L13
  lea    -0x2(%r14),%eax
  pxor   %xmm3,%xmm3
  mov    %rsi,%rbx
  lea    0x4(%rsi,%rax,4),%r13
  movss  %xmm3,0x4(%rsp)
L12:
  movslq 0x4(%rbx),%rcx
  movslq (%rbx),%rax
  mov    $0x2,%edi
  add    $0x4,%rbx
  movss  (%r15,%rcx,4),%xmm0
  subss  (%r15,%rax,4),%xmm0
  call   <pow@plt>
  movslq (%rbx),%rcx
  mov    $0x2,%edi
  mov    %rax,%rbp
  movslq -0x4(%rbx),%rax
  movss  (%r12,%rcx,4),%xmm0
  subss  (%r12,%rax,4),%xmm0
  call   <pow@plt>
  lea    0x0(%rbp,%rax,1),%rdi
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  addss  0x4(%rsp),%xmm0
  movss  %xmm0,0x4(%rsp)
  cmp    %rbx,%r13
  jne    L12
L23:
  mov    0x8(%rsp),%rax
  mov    $0x2,%edi
  lea    -0x4(%rax,%r14,4),%rbp
  mov    %rax,%r14
  movslq (%rax),%rax
  movslq 0x0(%rbp),%rcx
  movss  (%r15,%rcx,4),%xmm0
  subss  (%r15,%rax,4),%xmm0
  call   <pow@plt>
  movslq 0x0(%rbp),%rdx
  mov    $0x2,%edi
  mov    %rax,%rbx
  movslq (%r14),%rax
  movss  (%r12,%rdx,4),%xmm0
  subss  (%r12,%rax,4),%xmm0
  call   <pow@plt>
  lea    (%rbx,%rax,1),%rdi
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  addss  0x4(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  D5(%rip),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  cvtsd2ss %xmm0,%xmm0
  ret
  cs nopw 0x0(%rax,%rax,1)
L13:
  pxor   %xmm4,%xmm4
  movss  %xmm4,0x4(%rsp)
  jmp    L23
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_infered: f64

tool: D5: 111.32

assistant: ```c
float distanciaRuta(int ciudades, int *ruta, float *latitudes, float *longitudes) {
  float distancia = 0.0f;
  for (int i = 1; i < ciudades; i++)
    distancia += sqrt(pow(latitudes[ruta[i]] - latitudes[ruta[i - 1]], 2) + pow(longitudes[ruta[i]] - longitudes[ruta[i - 1]], 2));
  distancia += sqrt(pow(latitudes[ruta[ciudades - 1]] - latitudes[ruta[0]], 2) + pow(longitudes[ruta[ciudades - 1]] - longitudes[ruta[0]], 2));
  return distancia * 111.32;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<distanciaRuta>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  movslq %edi,%r14
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  cmp    $0x1,%r14d
  jle    L27
  lea    -0x2(%r14),%eax
  pxor   %xmm3,%xmm3
  mov    %rsi,%rbx
  lea    0x4(%rsi,%rax,4),%r13
  movss  %xmm3,0x4(%rsp)
L1:
  movslq 0x4(%rbx),%rcx
  movslq (%rbx),%rax
  mov    $0x2,%edi
  add    $0x4,%rbx
  movss  (%r15,%rcx,4),%xmm0
  subss  (%r15,%rax,4),%xmm0
  call   <pow@plt>
  movslq (%rbx),%rcx
  mov    $0x2,%edi
  mov    %rax,%rbp
  movslq -0x4(%rbx),%rax
  movss  (%r12,%rcx,4),%xmm0
  subss  (%r12,%rax,4),%xmm0
  call   <pow@plt>
  lea    0x0(%rbp,%rax,1),%rdi
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  addss  0x4(%rsp),%xmm0
  movss  %xmm0,0x4(%rsp)
  cmp    %rbx,%r13
  jne    L1
L2:
  mov    0x8(%rsp),%rax
  mov    $0x2,%edi
  lea    -0x4(%rax,%r14,4),%rbp
  mov    %rax,%r14
  movslq (%rax),%rax
  movslq 0x0(%rbp),%rcx
  movss  (%r15,%rcx,4),%xmm0
  subss  (%r15,%rax,4),%xmm0
  call   <pow@plt>
  movslq 0x0(%rbp),%rdx
  mov    $0x2,%edi
  mov    %rax,%rbx
  movslq (%r14),%rax
  movss  (%r12,%rdx,4),%xmm0
  subss  (%r12,%rax,4),%xmm0
  call   <pow@plt>
  lea    (%rbx,%rax,1),%rdi
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  addss  0x4(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  D5(%rip),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  cvtsd2ss %xmm0,%xmm0
  ret
  cs nopw 0x0(%rax,%rax,1)
L27:
  pxor   %xmm4,%xmm4
  movss  %xmm4,0x4(%rsp)
  jmp    L2
```

tool_calls: name: parse_data
arguments: data_tag: D5
type: f64

tool: D5: 111.32

assistant: ```c
float distanciaRuta(int ciudades, int *ruta, float *latitudes, float *longitudes) {
  float distancia = 0.0f;
  for (int i = 1; i < ciudades; i++)
    distancia += sqrt(pow(latitudes[ruta[i]] - latitudes[ruta[i - 1]], 2) + pow(longitudes[ruta[i]] - longitudes[ruta[i - 1]], 2));
  distancia += sqrt(pow(latitudes[ruta[ciudades - 1]] - latitudes[ruta[0]], 2) + pow(longitudes[ruta[ciudades - 1]] - longitudes[ruta[0]], 2));
  return distancia * 111.32;
}
```