ylfeng/ReF-Decompile-dataset huggingface dataset:
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    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x30(%rbp)
  movl   $0x6,-0x2c(%rbp)
  movl   $0x7,-0x28(%rbp)
  movl   $0x8,-0x24(%rbp)
  movl   $0x9,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x2,-0x14(%rbp)
  movl   $0x3,-0x10(%rbp)
  movl   $0x4,-0xc(%rbp)
  lea    -0x30(%rbp),%rax
  mov    $0xa,%edx
  mov    %rax,%rsi
  mov    $0x8,%edi
  call   <searchRotatedArray@plt>
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

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

tool: D9: 
 %d 


assistant: ```c
int main() {
  int ar[10] = {5, 6, 7, 8, 9, 0, 1, 2, 3, 4};
  printf("\n %d \n", searchRotatedArray(8, ar, 10));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x5,(%rsp)
  movl   $0x6,0x4(%rsp)
  movl   $0x7,0x8(%rsp)
  movl   $0x8,0xc(%rsp)
  movl   $0x9,0x10(%rsp)
  movl   $0x0,0x14(%rsp)
  movl   $0x1,0x18(%rsp)
  movl   $0x2,0x1c(%rsp)
  movl   $0x3,0x20(%rsp)
  movl   $0x4,0x24(%rsp)
  mov    %rsp,%rsi
  mov    $0xa,%edx
  mov    $0x8,%edi
  call   <searchRotatedArray@plt>
  mov    %eax,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x38,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type_guessed: string

tool: D2: 
 %d 


assistant: ```c
int main() {
  int ar[10] = {5, 6, 7, 8, 9, 0, 1, 2, 3, 4};
  printf("\n %d \n", searchRotatedArray(8, ar, 10));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    $0x8,%edi
  mov    $0xa,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x600000005,%rax
  mov    %rsp,%rsi
  movq   $0x9,0x10(%rsp)
  mov    %rax,(%rsp)
  movabs $0x800000007,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x200000001,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x400000003,%rax
  mov    %rax,0x20(%rsp)
  call   <searchRotatedArray@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
type_infered: string

tool: D1: 
 %d 


assistant: ```c
int main() {
  int ar[10] = {5, 6, 7, 8, 9, 0, 1, 2, 3, 4};
  printf("\n %d \n", searchRotatedArray(8, ar, 10));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    $0x8,%edi
  mov    $0xa,%edx
  movdqa D10(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D34(%rip),%rax
  mov    %rsp,%rsi
  movaps %xmm0,(%rsp)
  movdqa D39(%rip),%xmm0
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <searchRotatedArray@plt>
  lea    D14(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D10
datatype: byte[16]
name: extract_symbolic_data
arguments: label: D34
datatype: byte[8]
name: extract_symbolic_data
arguments: label: D39
datatype: byte[16]
name: extract_symbolic_data
arguments: label: D14
datatype: string

tool: {"D10": [0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: {"D34": [0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D39": [0x09,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

tool: D14: 
 %d 


assistant: ```c
int main() {
  int ar[10] = {5, 6, 7, 8, 9, 0, 1, 2, 3, 4};
  printf("\n %d \n", searchRotatedArray(8, ar, 10));
}
```
user: What is the c source code of the assembly code below:

```asm
<indexOf_shift>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xc(%rbp)
  jl     L8
  mov    -0x24(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L38
L8:
  movl   $0xffffffff,-0x10(%rbp)
  jmp    L0
L38:
  cmpl   $0x0,-0x24(%rbp)
  jns    L29
  movl   $0x0,-0x24(%rbp)
L29:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strstr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L19
  movl   $0xffffffff,-0x10(%rbp)
  jmp    L0
L19:
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %eax,-0x10(%rbp)
L0:
  mov    -0x10(%rbp),%eax
  leave
  ret
```

assistant: ```c
int indexOf_shift(char *base, char *str, int startIndex) {
  int result;
  int baselen = strlen(base);
  if (strlen(str) > baselen || startIndex > baselen) {
    result = -1;
  } else {
    if (startIndex < 0) {
      startIndex = 0;
    }
    char *pos = strstr(base + startIndex, str);
    if (pos == NULL) {
      result = -1;
    } else {
      result = pos - base;
    }
  }
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<indexOf_shift>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %edx,%ebx
  call   <strlen@plt>
  mov    %eax,%r13d
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  cmovl  %ebx,%eax
  cmp    %eax,%r13d
  jl     L17
  test   %ebx,%ebx
  mov    $0x0,%eax
  cmovns %ebx,%eax
  movslq %eax,%rdi
  add    %rbp,%rdi
  mov    %r12,%rsi
  call   <strstr@plt>
  test   %rax,%rax
  je     L6
  sub    %ebp,%eax
L25:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  mov    $0xffffffff,%eax
  jmp    L25
L6:
  mov    $0xffffffff,%eax
  jmp    L25
```

assistant: ```c
int indexOf_shift(char *base, char *str, int startIndex) {
  int result;
  int baselen = strlen(base);
  if (strlen(str) > baselen || startIndex > baselen) {
    result = -1;
  } else {
    if (startIndex < 0) {
      startIndex = 0;
    }
    char *pos = strstr(base + startIndex, str);
    if (pos == NULL) {
      result = -1;
    } else {
      result = pos - base;
    }
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<indexOf_shift>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  cmp    %ebx,%eax
  cmovl  %ebx,%eax
  cmp    %eax,%r13d
  jl     L26
  xor    %eax,%eax
  test   %ebx,%ebx
  mov    %r12,%rsi
  cmovns %ebx,%eax
  movslq %eax,%rdi
  add    %rbp,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L26
  sub    %ebp,%eax
L25:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  mov    $0xffffffff,%eax
  jmp    L25
```

assistant: ```c
int indexOf_shift(char *base, char *str, int startIndex) {
  int result;
  int baselen = strlen(base);
  if (strlen(str) > baselen || startIndex > baselen) {
    result = -1;
  } else {
    if (startIndex < 0) {
      startIndex = 0;
    }
    char *pos = strstr(base + startIndex, str);
    if (pos == NULL) {
      result = -1;
    } else {
      result = pos - base;
    }
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<indexOf_shift>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  cmp    %ebx,%eax
  cmovl  %ebx,%eax
  cmp    %eax,%r13d
  jl     L14
  xor    %eax,%eax
  test   %ebx,%ebx
  mov    %r12,%rsi
  cmovns %ebx,%eax
  movslq %eax,%rdi
  add    %rbp,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L14
  sub    %ebp,%eax
L28:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  mov    $0xffffffff,%eax
  jmp    L28
```

assistant: ```c
int indexOf_shift(char *base, char *str, int startIndex) {
  int result;
  int baselen = strlen(base);
  if (strlen(str) > baselen || startIndex > baselen) {
    result = -1;
  } else {
    if (startIndex < 0) {
      startIndex = 0;
    }
    char *pos = strstr(base + startIndex, str);
    if (pos == NULL) {
      result = -1;
    } else {
      result = pos - base;
    }
  }
  return result;
}
```
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    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <getGcm@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D10
data_type: string
name: parse_data
arguments: label: D18
data_type: string

tool: D10: %lld %lld

tool: D18: %lld


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long long num1, num2;
  unsigned long long gcm;
  scanf("%lld %lld", &num1, &num2);
  gcm = getGcm(num1, num2);
  printf("%lld\n", gcm);
  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    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%rsi
  mov    0x8(%rsp),%rdi
  call   <getGcm@plt>
  mov    %rax,%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
type: string
name: disassemble_data
arguments: label: D3
type: string

tool: D9: %lld %lld

tool: D3: %lld


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long long num1, num2;
  unsigned long long gcm;
  scanf("%lld %lld", &num1, &num2);
  gcm = getGcm(num1, num2);
  printf("%lld\n", gcm);
  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    $0x28,%rsp
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x10(%rsp),%rsi
  mov    0x8(%rsp),%rdi
  call   <getGcm@plt>
  lea    D10(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D10
type_infered: string

tool: D5: %lld %lld

tool: D10: %lld


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long long num1, num2;
  unsigned long long gcm;
  scanf("%lld %lld", &num1, &num2);
  gcm = getGcm(num1, num2);
  printf("%lld\n", gcm);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D9(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x10(%rsp),%rsi
  mov    0x8(%rsp),%rdi
  call   <getGcm@plt>
  lea    D6(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D9
type_guessed: string
name: unpack_data
arguments: data_label: D6
type_guessed: string

tool: D9: %lld %lld

tool: D6: %lld


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long long num1, num2;
  unsigned long long gcm;
  scanf("%lld %lld", &num1, &num2);
  gcm = getGcm(num1, num2);
  printf("%lld\n", gcm);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%r8
  lea    -0x14(%rbp),%rdi
  lea    -0x18(%rbp),%rcx
  lea    -0x1c(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  sub    $0x8,%rsp
  lea    -0xc(%rbp),%rsi
  push   %rsi
  mov    %r8,%r9
  mov    %rdi,%r8
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  add    $0x10,%rsp
  movss  -0x14(%rbp),%xmm0
  movss  -0x1c(%rbp),%xmm1
  subss  %xmm1,%xmm0
  movss  -0x18(%rbp),%xmm1
  movss  -0x20(%rbp),%xmm2
  subss  %xmm2,%xmm1
  divss  %xmm1,%xmm0
  movss  -0xc(%rbp),%xmm1
  movss  -0x14(%rbp),%xmm2
  subss  %xmm2,%xmm1
  movss  -0x10(%rbp),%xmm2
  movss  -0x18(%rbp),%xmm3
  subss  %xmm3,%xmm2
  divss  %xmm2,%xmm1
  ucomiss %xmm1,%xmm0
  jp     L32
  ucomiss %xmm1,%xmm0
  jne    L32
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L28
L32:
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L28:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D14
type: string
name: parse_data
arguments: tag: D28
type: string
name: parse_data
arguments: tag: D1
type: string

tool: D14: %d %d %d %d %d %d

tool: D28: All the points are on same line.

tool: D1: All the points are not on same line.

assistant: ```c
int main() {
  float x1, y1, x2, y2, x3, y3;
  scanf("%d %d %d %d %d %d", &x1, &y1, &x2, &y2, &x3, &y3);
  if (((y2 - y1) / (x2 - x1)) == ((y3 - y2) / (x3 - x2))) {
    printf("All the points are on same line.");
  } else {
    printf("All the points are not on same line.");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rcx
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  sub    $0x8,%rsp
  lea    0x1c(%rsp),%rax
  push   %rax
  lea    0x20(%rsp),%r9
  lea    0x1c(%rsp),%r8
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  movss  0x1c(%rsp),%xmm2
  movss  0x18(%rsp),%xmm3
  movaps %xmm2,%xmm1
  subss  0x14(%rsp),%xmm1
  movaps %xmm3,%xmm0
  subss  0x10(%rsp),%xmm0
  divss  %xmm0,%xmm1
  movss  0x24(%rsp),%xmm0
  subss  %xmm2,%xmm0
  movss  0x20(%rsp),%xmm2
  subss  %xmm3,%xmm2
  divss  %xmm2,%xmm0
  add    $0x10,%rsp
  ucomiss %xmm0,%xmm1
  jp     L12
  jne    L12
  lea    D8(%rip),%rdi
  call   <printf@plt>
L30:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L12:
  lea    D13(%rip),%rdi
  call   <printf@plt>
  jmp    L30
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D25
data_type: string
name: extract_symbolic_data
arguments: label: D8
data_type: string
name: extract_symbolic_data
arguments: label: D13
data_type: string

tool: D25: %d %d %d %d %d %d

tool: D8: All the points are on same line.

tool: D13: All the points are not on same line.

assistant: ```c
int main() {
  float x1, y1, x2, y2, x3, y3;
  scanf("%d %d %d %d %d %d", &x1, &y1, &x2, &y2, &x3, &y3);
  if (((y2 - y1) / (x2 - x1)) == ((y3 - y2) / (x3 - x2))) {
    printf("All the points are on same line.");
  } else {
    printf("All the points are not on same line.");
  }
  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    $0x28,%rsp
  lea    D15(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  lea    0x8(%rsp),%rcx
  mov    %rsp,%rsi
  sub    $0x8,%rsp
  lea    0x1c(%rsp),%rax
  push   %rax
  lea    0x20(%rsp),%r9
  lea    0x1c(%rsp),%r8
  call   <scanf@plt>
  movss  0x1c(%rsp),%xmm2
  movss  0x18(%rsp),%xmm3
  movaps %xmm2,%xmm1
  movaps %xmm3,%xmm0
  subss  0x14(%rsp),%xmm1
  subss  0x10(%rsp),%xmm0
  divss  %xmm0,%xmm1
  movss  0x24(%rsp),%xmm0
  subss  %xmm2,%xmm0
  movss  0x20(%rsp),%xmm2
  pop    %rax
  pop    %rdx
  subss  %xmm3,%xmm2
  divss  %xmm2,%xmm0
  ucomiss %xmm0,%xmm1
  jp     L15
  jne    L15
  lea    D12(%rip),%rdi
  call   <printf@plt>
L26:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L15:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L26
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D15
type_infered: string
name: read_data
arguments: data_tag: D12
type_infered: string
name: read_data
arguments: data_tag: D10
type_infered: string

tool: D15: %d %d %d %d %d %d

tool: D12: All the points are on same line.

tool: D10: All the points are not on same line.

assistant: ```c
int main() {
  float x1, y1, x2, y2, x3, y3;
  scanf("%d %d %d %d %d %d", &x1, &y1, &x2, &y2, &x3, &y3);
  if (((y2 - y1) / (x2 - x1)) == ((y3 - y2) / (x3 - x2))) {
    printf("All the points are on same line.");
  } else {
    printf("All the points are not on same line.");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  lea    0x8(%rsp),%rcx
  mov    %rsp,%rsi
  sub    $0x8,%rsp
  lea    0x1c(%rsp),%rax
  push   %rax
  lea    0x20(%rsp),%r9
  lea    0x1c(%rsp),%r8
  call   <scanf@plt>
  movss  0x1c(%rsp),%xmm2
  movss  0x18(%rsp),%xmm3
  movaps %xmm2,%xmm1
  movaps %xmm3,%xmm0
  subss  0x14(%rsp),%xmm1
  subss  0x10(%rsp),%xmm0
  divss  %xmm0,%xmm1
  movss  0x24(%rsp),%xmm0
  subss  %xmm2,%xmm0
  movss  0x20(%rsp),%xmm2
  pop    %rax
  pop    %rdx
  subss  %xmm3,%xmm2
  divss  %xmm2,%xmm0
  ucomiss %xmm0,%xmm1
  jp     L37
  jne    L37
  lea    D11(%rip),%rdi
  call   <printf@plt>
L6:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L37:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  jmp    L6
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D14
datatype: string
name: parse_data
arguments: data_tag: D11
datatype: string
name: parse_data
arguments: data_tag: D9
datatype: string

tool: D14: %d %d %d %d %d %d

tool: D11: All the points are on same line.

tool: D9: All the points are not on same line.

assistant: ```c
int main() {
  float x1, y1, x2, y2, x3, y3;
  scanf("%d %d %d %d %d %d", &x1, &y1, &x2, &y2, &x3, &y3);
  if (((y2 - y1) / (x2 - x1)) == ((y3 - y2) / (x3 - x2))) {
    printf("All the points are on same line.");
  } else {
    printf("All the points are not on same line.");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<multi_instance_dump>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x3,%al
  je     L5
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L12
L5:
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L12:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
type_infered: string
name: retrieve_data
arguments: tag: D15
type_infered: string

tool: D7: Impossible command %x


tool: D15: class %x has %d instance(s)


assistant: ```c
void multi_instance_dump(void *data, void *payload, int len) {
  unsigned char *pp = (unsigned char *)payload;
  if (pp[0] != 0x03) {
    printf("Impossible command %x\n", pp[0]);
    return;
  }
  printf("class %x has %d instance(s)\n", pp[1], pp[2]);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<multi_instance_dump>:
  endbr64
  sub    $0x8,%rsp
  movzbl (%rsi),%eax
  cmp    $0x3,%al
  je     L8
  movzbl %al,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L12:
  add    $0x8,%rsp
  ret
L8:
  movzbl 0x2(%rsi),%edx
  movzbl 0x1(%rsi),%esi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L12
```

tool_calls: name: retrieve_data
arguments: label: D6
datatype: string
name: retrieve_data
arguments: label: D5
datatype: string

tool: D6: Impossible command %x


tool: D5: class %x has %d instance(s)


assistant: ```c
void multi_instance_dump(void *data, void *payload, int len) {
  unsigned char *pp = (unsigned char *)payload;
  if (pp[0] != 0x03) {
    printf("Impossible command %x\n", pp[0]);
    return;
  }
  printf("class %x has %d instance(s)\n", pp[1], pp[2]);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<multi_instance_dump>:
  endbr64
  movzbl (%rsi),%eax
  cmp    $0x3,%al
  je     L9
  movzbl %al,%esi
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L9:
  movzbl 0x2(%rsi),%edx
  movzbl 0x1(%rsi),%esi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: data_label: D5
type: string
name: parse_data
arguments: data_label: D7
type: string

tool: D5: Impossible command %x


tool: D7: class %x has %d instance(s)


assistant: ```c
void multi_instance_dump(void *data, void *payload, int len) {
  unsigned char *pp = (unsigned char *)payload;
  if (pp[0] != 0x03) {
    printf("Impossible command %x\n", pp[0]);
    return;
  }
  printf("class %x has %d instance(s)\n", pp[1], pp[2]);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<multi_instance_dump>:
  endbr64
  movzbl (%rsi),%eax
  cmp    $0x3,%al
  je     L2
  movzbl %al,%esi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L2:
  movzbl 0x2(%rsi),%edx
  movzbl 0x1(%rsi),%esi
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D13
datatype: string
name: disassemble_data
arguments: tag: D10
datatype: string

tool: D13: Impossible command %x


tool: D10: class %x has %d instance(s)


assistant: ```c
void multi_instance_dump(void *data, void *payload, int len) {
  unsigned char *pp = (unsigned char *)payload;
  if (pp[0] != 0x03) {
    printf("Impossible command %x\n", pp[0]);
    return;
  }
  printf("class %x has %d instance(s)\n", pp[1], pp[2]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mkstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *mkstr(const char *s, int len) {
  char *result = (char *)malloc(len + 1);
  memcpy(result, s, len);
  result[len] = '\0';
  return result;
}
```