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
<eval_blow_effect>:
  endbr64
  mov    %esi,%r8d
  cmp    $0x97,%edi
  ja     L3
  lea    D8(%rip),%rsi
  mov    %edi,%edi
  movslq (%rsi,%rdi,4),%rcx
  add    %rsi,%rcx
  notrack jmp *%rcx
  nopw   0x0(%rax,%rax,1)
L3:
  movslq %r8d,%rax
  ret
  nopl   0x0(%rax)
  lea    0x14(%r8),%eax
  cltq
  ret
  nopw   0x0(%rax,%rax,1)
  lea    0xa(%r8),%eax
  cltq
  ret
  nopw   0x0(%rax,%rax,1)
  lea    0x1e(%r8),%eax
  cltq
  ret
  nopw   0x0(%rax,%rax,1)
  lea    0x5(%r8),%eax
  cltq
  ret
  nopw   0x0(%rax,%rax,1)
  mov    $0x7d0,%eax
  lea    0x1(%rdx),%ecx
  cltd
  idiv   %ecx
  add    %r8d,%eax
  cltq
  ret
  nopl   0x0(%rax)
  lea    (%r8,%r8,4),%edx
  test   %edx,%edx
  lea    0x3(%rdx),%eax
  cmovns %edx,%eax
  sar    $0x2,%eax
  add    $0x5,%eax
  cltq
  ret
  nopl   (%rax)
  mov    $0x3e8,%eax
  lea    0x1(%rdx),%ecx
  cltd
  idiv   %ecx
  add    %r8d,%eax
  cltq
  ret
  nopl   0x0(%rax)
  lea    0x28(%r8),%eax
  cltq
  ret
  lea    0x12c(%r8),%eax
  cltq
  ret
  lea    0x32(%r8),%eax
  cltq
  ret
  mov    $0x5,%eax
  ret
```

assistant: ```c
long eval_blow_effect(int effect, int atk_dam, int rlev) {
  switch (effect) {
  case 65:
  case 66:
  case 67:
  case 68:
  case 82:
  case 2: {
    atk_dam += 5;
    break;
  }
  case 101:
  case 20:
  case 11:
  case 27:
  case 151: {
    atk_dam *= 5;
    atk_dam /= 4;
    atk_dam += 5;
    break;
  }
  case 75:
  case 17:
  case 19:
  case 18:
  case 100:
  case 145:
  case 22:
  case 23:
  case 74:
  case 110: {
    atk_dam += 10;
    break;
  }
  case 9:
  case 16:
  case 24:
  case 26:
  case 28:
  case 73:
  case 62:
  case 25:
  case 77:
  case 78:
  case 79:
  case 80:
  case 94:
  case 112: {
    atk_dam += 20;
    break;
  }
  case 12:
  case 13:
  case 14:
  case 63:
  case 64:
  case 81:
  case 21:
  case 40:
  case 50: {
    atk_dam += 30;
    break;
  }
  case 76:
  case 83: {
    atk_dam += 40;
    break;
  }
  case 31: {
    atk_dam += 50;
    break;
  }
  case 85:
  case 86: {
    atk_dam += 1000 / (rlev + 1);
    break;
  }
  case 29:
  case 30:
  case 15:
  case 87:
  case 88: {
    atk_dam += 2000 / (rlev + 1);
    break;
  }
  case 84: {
    atk_dam += 300;
    break;
  }
  case 6:
  case 7:
  case 8:
  case 93:
  case 105: {
    atk_dam = 5;
  }
  default:
    break;
  }
  return (atk_dam);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<eval_blow_effect>:
  endbr64
  mov    %esi,%r8d
  cmp    $0x97,%edi
  ja     L8
  lea    D7(%rip),%rsi
  mov    %edi,%edi
  movslq (%rsi,%rdi,4),%rcx
  add    %rsi,%rcx
  notrack jmp *%rcx
  nopw   0x0(%rax,%rax,1)
L8:
  movslq %r8d,%rax
  ret
  nopl   0x0(%rax)
  lea    0x14(%r8),%eax
  cltq
  ret
  nopw   0x0(%rax,%rax,1)
  lea    0xa(%r8),%eax
  cltq
  ret
  nopw   0x0(%rax,%rax,1)
  lea    0x1e(%r8),%eax
  cltq
  ret
  nopw   0x0(%rax,%rax,1)
  lea    0x5(%r8),%eax
  cltq
  ret
  nopw   0x0(%rax,%rax,1)
  mov    $0x7d0,%eax
  lea    0x1(%rdx),%ecx
  cltd
  idiv   %ecx
  add    %r8d,%eax
  cltq
  ret
  nopl   0x0(%rax)
  lea    (%r8,%r8,4),%edx
  test   %edx,%edx
  lea    0x3(%rdx),%eax
  cmovns %edx,%eax
  sar    $0x2,%eax
  add    $0x5,%eax
  cltq
  ret
  nopl   (%rax)
  mov    $0x3e8,%eax
  lea    0x1(%rdx),%ecx
  cltd
  idiv   %ecx
  add    %r8d,%eax
  cltq
  ret
  nopl   0x0(%rax)
  lea    0x28(%r8),%eax
  cltq
  ret
  lea    0x12c(%r8),%eax
  cltq
  ret
  lea    0x32(%r8),%eax
  cltq
  ret
  mov    $0x5,%eax
  ret
```

assistant: ```c
long eval_blow_effect(int effect, int atk_dam, int rlev) {
  switch (effect) {
  case 65:
  case 66:
  case 67:
  case 68:
  case 82:
  case 2: {
    atk_dam += 5;
    break;
  }
  case 101:
  case 20:
  case 11:
  case 27:
  case 151: {
    atk_dam *= 5;
    atk_dam /= 4;
    atk_dam += 5;
    break;
  }
  case 75:
  case 17:
  case 19:
  case 18:
  case 100:
  case 145:
  case 22:
  case 23:
  case 74:
  case 110: {
    atk_dam += 10;
    break;
  }
  case 9:
  case 16:
  case 24:
  case 26:
  case 28:
  case 73:
  case 62:
  case 25:
  case 77:
  case 78:
  case 79:
  case 80:
  case 94:
  case 112: {
    atk_dam += 20;
    break;
  }
  case 12:
  case 13:
  case 14:
  case 63:
  case 64:
  case 81:
  case 21:
  case 40:
  case 50: {
    atk_dam += 30;
    break;
  }
  case 76:
  case 83: {
    atk_dam += 40;
    break;
  }
  case 31: {
    atk_dam += 50;
    break;
  }
  case 85:
  case 86: {
    atk_dam += 1000 / (rlev + 1);
    break;
  }
  case 29:
  case 30:
  case 15:
  case 87:
  case 88: {
    atk_dam += 2000 / (rlev + 1);
    break;
  }
  case 84: {
    atk_dam += 300;
    break;
  }
  case 6:
  case 7:
  case 8:
  case 93:
  case 105: {
    atk_dam = 5;
  }
  default:
    break;
  }
  return (atk_dam);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<losequote>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %rdi,-0x98(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x98(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x22,%al
  jne    L5
  mov    -0x98(%rbp),%rax
  lea    0x1(%rax),%rdx
  lea    -0x90(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x90(%rbp),%rdx
  mov    -0x98(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L5:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
void losequote(char *str) {
  char tmp[128];
  if (*str == '"') {
    strcpy(tmp, str + 1);
    strcpy(str, tmp);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<losequote>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  cmpb   $0x22,(%rdi)
  je     L29
L16:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  ret
L29:
  mov    %rdi,%rbx
  lea    0x1(%rdi),%rsi
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  jmp    L16
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void losequote(char *str) {
  char tmp[128];
  if (*str == '"') {
    strcpy(tmp, str + 1);
    strcpy(str, tmp);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<losequote>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  cmpb   $0x22,(%rdi)
  je     L23
L12:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x98,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L23:
  mov    %rsp,%r12
  mov    %rdi,%rbp
  lea    0x1(%rdi),%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  jmp    L12
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void losequote(char *str) {
  char tmp[128];
  if (*str == '"') {
    strcpy(tmp, str + 1);
    strcpy(str, tmp);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<losequote>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  cmpb   $0x22,(%rdi)
  je     L25
L12:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x98,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L25:
  mov    %rsp,%r12
  mov    %rdi,%rbp
  lea    0x1(%rdi),%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  jmp    L12
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void losequote(char *str) {
  char tmp[128];
  if (*str == '"') {
    strcpy(tmp, str + 1);
    strcpy(str, tmp);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <error@plt>
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <error@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type_infered: string

tool: D3: This isver 0

assistant: ```c
int main() {
  int err = 0, err2 = 0;
  error(err);
  printf("This is"
         "ver 0");
  error(err2);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <error@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <error@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D9
type: string

tool: D9: This isver 0

assistant: ```c
int main() {
  int err = 0, err2 = 0;
  error(err);
  printf("This is"
         "ver 0");
  error(err2);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <error@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <error@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D1
data_type: string

tool: D1: This isver 0

assistant: ```c
int main() {
  int err = 0, err2 = 0;
  error(err);
  printf("This is"
         "ver 0");
  error(err2);
  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
  xor    %edi,%edi
  call   <error@plt>
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <error@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D3
datatype: string

tool: D3: This isver 0

assistant: ```c
int main() {
  int err = 0, err2 = 0;
  error(err);
  printf("This is"
         "ver 0");
  error(err2);
  return 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
  movl   $0x0,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L26
L15:
  cmpl   $0x1,-0x18(%rbp)
  jg     L47
  mov    -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L25
L47:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L25:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x18(%rbp)
L26:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x18(%rbp)
  jl     L15
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D24
type_infered: string
name: unpack_data
arguments: data_label: D10
type_infered: string
name: unpack_data
arguments: data_label: D27
type_infered: string

tool: D24: %d

tool: D10: %d 

tool: D27: %d


assistant: ```c
int main() {
  int n, i, f = 0, s = 1, sum;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    if (i <= 1) {
      sum = i;
    } else {
      sum = f + s;
      f = s;
      s = sum;
    }
    printf("%d ", sum);
  }
  printf("%d\n", f + s);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    $0x1,%ebp
  mov    $0x0,%r12d
  cmpl   $0x1,0x4(%rsp)
  jle    L2
  mov    $0x0,%ebx
  mov    $0x0,%esi
  mov    $0x1,%ebp
  mov    $0x0,%r12d
  lea    D20(%rip),%r13
  jmp    L54
L18:
  lea    (%r12,%rbp,1),%esi
  mov    %ebp,%r12d
  mov    %esi,%ebp
L54:
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jle    L2
  mov    %ebx,%esi
  cmp    $0x1,%ebx
  jg     L18
  jmp    L54
L2:
  lea    (%r12,%rbp,1),%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D16
type_infered: string
name: parse_data
arguments: data_tag: D20
type_infered: string
name: parse_data
arguments: data_tag: D13
type_infered: string

tool: D16: %d

tool: D20: %d 

tool: D13: %d


assistant: ```c
int main() {
  int n, i, f = 0, s = 1, sum;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    if (i <= 1) {
      sum = i;
    } else {
      sum = f + s;
      f = s;
      s = sum;
    }
    printf("%d ", sum);
  }
  printf("%d\n", f + s);
  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    D11(%rip),%rdi
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L10
  xor    %ebx,%ebx
  xor    %esi,%esi
  mov    $0x1,%ebp
  xor    %r12d,%r12d
  lea    D6(%rip),%r13
  nopl   0x0(%rax)
L15:
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jle    L10
  mov    $0x1,%esi
  cmp    $0x1,%ebx
  je     L15
  lea    (%r12,%rbp,1),%esi
  mov    %ebp,%r12d
  mov    %esi,%ebp
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L10:
  lea    (%r12,%rbp,1),%esi
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D11
type: string
name: read_data
arguments: data_tag: D6
type: string
name: read_data
arguments: data_tag: D26
type: string

tool: D11: %d

tool: D6: %d 

tool: D26: %d


assistant: ```c
int main() {
  int n, i, f = 0, s = 1, sum;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    if (i <= 1) {
      sum = i;
    } else {
      sum = f + s;
      f = s;
      s = sum;
    }
    printf("%d ", sum);
  }
  printf("%d\n", f + s);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D15(%rip),%rdi
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L6
  xor    %ebx,%ebx
  xor    %esi,%esi
  mov    $0x1,%ebp
  xor    %r12d,%r12d
  lea    D20(%rip),%r13
  nopl   0x0(%rax)
L49:
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  cmp    %ebx,%eax
  jle    L6
  mov    $0x1,%esi
  cmp    $0x1,%ebx
  je     L49
  lea    (%r12,%rbp,1),%esi
  mov    %ebp,%r12d
  mov    %esi,%ebp
  jmp    L49
  nopl   0x0(%rax,%rax,1)
L6:
  lea    (%r12,%rbp,1),%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D15
type: string
name: retrieve_data
arguments: label: D20
type: string
name: retrieve_data
arguments: label: D10
type: string

tool: D15: %d

tool: D20: %d 

tool: D10: %d


assistant: ```c
int main() {
  int n, i, f = 0, s = 1, sum;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    if (i <= 1) {
      sum = i;
    } else {
      sum = f + s;
      f = s;
      s = sum;
    }
    printf("%d ", sum);
  }
  printf("%d\n", f + s);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<substract>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
type_guessed: string

tool: D0: %d


assistant: ```c
void substract(char *arg1, char *arg2) {
  int a, b;
  a = atoi(arg1);
  b = atoi(arg2);
  printf("%d\n", a - b);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<substract>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  call   <atoi@plt>
  mov    %eax,%ebx
  mov    %rbp,%rdi
  call   <atoi@plt>
  sub    %eax,%ebx
  mov    %ebx,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

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

tool: D5: %d


assistant: ```c
void substract(char *arg1, char *arg2) {
  int a, b;
  a = atoi(arg1);
  b = atoi(arg2);
  printf("%d\n", a - b);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<substract>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <atoi@plt>
  add    $0x8,%rsp
  lea    D6(%rip),%rdi
  sub    %eax,%ebx
  mov    %ebx,%esi
  pop    %rbx
  pop    %rbp
  jmp    1060 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type: string

tool: D6: %d


assistant: ```c
void substract(char *arg1, char *arg2) {
  int a, b;
  a = atoi(arg1);
  b = atoi(arg2);
  printf("%d\n", a - b);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<substract>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <atoi@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <atoi@plt>
  add    $0x8,%rsp
  lea    D3(%rip),%rdi
  sub    %eax,%ebx
  mov    %ebx,%esi
  pop    %rbx
  pop    %rbp
  jmp    1060 <printf@plt>
```

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

tool: D3: %d


assistant: ```c
void substract(char *arg1, char *arg2) {
  int a, b;
  a = atoi(arg1);
  b = atoi(arg2);
  printf("%d\n", a - b);
}
```
user: Translate the below assembly code into C programming language:

```asm
<simple_trsml>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    %r9,-0x30(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L45
L14:
  movl   $0x0,-0x8(%rbp)
  jmp    L59
L1:
  movl   $0x0,-0x4(%rbp)
  jmp    L41
L24:
  mov    -0xc(%rbp),%eax
  imul   0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  imul   0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  imul   -0x24(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  imul   -0x28(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0x4(%rbp),%eax
  imul   0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  imul   0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mulsd  %xmm2,%xmm1
  mov    -0xc(%rbp),%eax
  imul   0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  imul   0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L41:
  mov    -0x4(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L24
  mov    -0xc(%rbp),%eax
  imul   0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  imul   0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  imul   -0x24(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  imul   -0x28(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0xc(%rbp),%eax
  imul   0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  imul   0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x8(%rbp)
L59:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L1
  addl   $0x1,-0xc(%rbp)
L45:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L14
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void simple_trsml(int m, int n, double *a, int a_row_stride, int a_col_stride, double *b, int b_row_stride, int b_col_stride) {
  int i, j, p;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      for (p = 0; p < i; p++) {
        b[(i)*b_row_stride + (j)*b_col_stride] = b[(i)*b_row_stride + (j)*b_col_stride] - a[(i)*a_row_stride + (p)*a_col_stride] * b[(p)*b_row_stride + (j)*b_col_stride];
      }
      b[(i)*b_row_stride + (j)*b_col_stride] = b[(i)*b_row_stride + (j)*b_col_stride] / a[(i)*a_row_stride + (i)*a_col_stride];
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<simple_trsml>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,-0x10(%rsp)
  mov    %rdx,-0x8(%rsp)
  mov    %ecx,-0xc(%rsp)
  test   %edi,%edi
  jle    L70
  mov    %esi,%ebp
  mov    %r8d,%eax
  mov    %r9,%r12
  lea    (%r8,%rcx,1),%ebx
  mov    %ebx,-0x14(%rsp)
  movslq 0x40(%rsp),%rdx
  lea    0x0(,%rdx,8),%rbx
  mov    %rbx,-0x28(%rsp)
  cltq
  lea    0x0(,%rax,8),%r8
  movslq 0x38(%rsp),%r9
  shl    $0x3,%r9
  movl   $0x0,-0x18(%rsp)
  movl   $0x0,-0x1c(%rsp)
  mov    $0x0,%r13d
  mov    $0x0,%edi
  jmp    L38
L88:
  lea    (%r11,%r13,1),%eax
  cltq
  lea    (%r12,%rax,8),%rsi
  movslq %r11d,%rax
  lea    (%r12,%rax,8),%rcx
  mov    %r15,%rdx
  mov    $0x0,%eax
L17:
  movsd  (%rdx),%xmm1
  mulsd  (%rcx),%xmm1
  movsd  (%rsi),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rsi)
  add    $0x1,%eax
  add    %r8,%rdx
  add    %r9,%rcx
  cmp    %edi,%eax
  jne    L17
L79:
  movsd  (%r10),%xmm0
  divsd  (%r14),%xmm0
  movsd  %xmm0,(%r10)
  add    $0x1,%ebx
  add    -0x28(%rsp),%r10
  add    0x40(%rsp),%r11d
  cmp    %ebx,%ebp
  je     L55
L83:
  test   %edi,%edi
  jg     L88
  jmp    L79
L55:
  add    $0x1,%edi
  add    0x38(%rsp),%r13d
  mov    -0x14(%rsp),%ebx
  add    %ebx,-0x1c(%rsp)
  mov    -0xc(%rsp),%ebx
  add    %ebx,-0x18(%rsp)
  cmp    %edi,-0x10(%rsp)
  je     L70
L38:
  test   %ebp,%ebp
  jle    L55
  movslq -0x1c(%rsp),%rax
  mov    -0x8(%rsp),%rbx
  lea    (%rbx,%rax,8),%r14
  movslq %r13d,%rax
  lea    (%r12,%rax,8),%r10
  movslq -0x18(%rsp),%rax
  lea    (%rbx,%rax,8),%r15
  mov    $0x0,%r11d
  mov    $0x0,%ebx
  jmp    L83
L70:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void simple_trsml(int m, int n, double *a, int a_row_stride, int a_col_stride, double *b, int b_row_stride, int b_col_stride) {
  int i, j, p;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      for (p = 0; p < i; p++) {
        b[(i)*b_row_stride + (j)*b_col_stride] = b[(i)*b_row_stride + (j)*b_col_stride] - a[(i)*a_row_stride + (p)*a_col_stride] * b[(p)*b_row_stride + (j)*b_col_stride];
      }
      b[(i)*b_row_stride + (j)*b_col_stride] = b[(i)*b_row_stride + (j)*b_col_stride] / a[(i)*a_row_stride + (i)*a_col_stride];
    }
  }
}
```