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

```asm
<mrotate>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  lea    0x8(%rsp),%rdi
  mov    %rsp,%rsi
  call   <sincos@plt>
  movsd  (%rbx),%xmm2
  movsd  (%rsp),%xmm3
  movsd  D9(%rip),%xmm0
  movsd  0x8(%rsp),%xmm1
  mulsd  %xmm2,%xmm2
  subsd  %xmm3,%xmm0
  mulsd  %xmm0,%xmm2
  addsd  %xmm3,%xmm2
  movsd  %xmm2,0x0(%rbp)
  movsd  (%rbx),%xmm2
  movsd  0x10(%rbx),%xmm4
  mulsd  %xmm0,%xmm2
  mulsd  0x8(%rbx),%xmm2
  mulsd  %xmm1,%xmm4
  subsd  %xmm4,%xmm2
  movsd  %xmm2,0x8(%rbp)
  movsd  (%rbx),%xmm2
  movsd  0x8(%rbx),%xmm4
  mulsd  %xmm0,%xmm2
  mulsd  0x10(%rbx),%xmm2
  mulsd  %xmm1,%xmm4
  addsd  %xmm4,%xmm2
  movsd  %xmm2,0x10(%rbp)
  movsd  (%rbx),%xmm2
  movsd  0x10(%rbx),%xmm4
  mulsd  %xmm0,%xmm2
  mulsd  0x8(%rbx),%xmm2
  mulsd  %xmm1,%xmm4
  addsd  %xmm4,%xmm2
  movsd  %xmm2,0x18(%rbp)
  movsd  0x8(%rbx),%xmm4
  movapd %xmm0,%xmm2
  mulsd  %xmm4,%xmm2
  mulsd  %xmm4,%xmm2
  addsd  %xmm3,%xmm2
  movsd  %xmm2,0x20(%rbp)
  movsd  0x10(%rbx),%xmm2
  movsd  (%rbx),%xmm4
  mulsd  %xmm0,%xmm2
  mulsd  0x8(%rbx),%xmm2
  mulsd  %xmm1,%xmm4
  subsd  %xmm4,%xmm2
  movsd  %xmm2,0x28(%rbp)
  movsd  (%rbx),%xmm2
  movsd  0x8(%rbx),%xmm4
  mulsd  %xmm0,%xmm2
  mulsd  0x10(%rbx),%xmm2
  mulsd  %xmm1,%xmm4
  subsd  %xmm4,%xmm2
  movsd  %xmm2,0x30(%rbp)
  movsd  0x8(%rbx),%xmm2
  mulsd  (%rbx),%xmm1
  mulsd  %xmm0,%xmm2
  mulsd  0x10(%rbx),%xmm2
  addsd  %xmm1,%xmm2
  movsd  %xmm2,0x38(%rbp)
  movsd  0x10(%rbx),%xmm1
  mulsd  %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  addsd  %xmm3,%xmm0
  movsd  %xmm0,0x40(%rbp)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
type_infered: byte[8]

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
void mrotate(double theta, double axis[3], double m[3][3]) {
  double c = cos(theta);
  double s = sin(theta);
  m[0][0] = c + (1 - c) * ((axis[0]) * (axis[0]));
  m[0][1] = (1 - c) * axis[0] * axis[1] - s * axis[2];
  m[0][2] = (1 - c) * axis[0] * axis[2] + s * axis[1];
  m[1][0] = (1 - c) * axis[0] * axis[1] + s * axis[2];
  m[1][1] = c + (1 - c) * axis[1] * axis[1];
  m[1][2] = (1 - c) * axis[2] * axis[1] - s * axis[0];
  m[2][0] = (1 - c) * axis[0] * axis[2] - s * axis[1];
  m[2][1] = (1 - c) * axis[1] * axis[2] + s * axis[0];
  m[2][2] = c + (1 - c) * axis[2] * axis[2];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mrotate>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  lea    0x8(%rsp),%rdi
  mov    %rsp,%rsi
  call   <sincos@plt>
  movsd  (%rbx),%xmm2
  movsd  (%rsp),%xmm3
  movsd  D5(%rip),%xmm0
  movsd  0x8(%rsp),%xmm1
  mulsd  %xmm2,%xmm2
  subsd  %xmm3,%xmm0
  mulsd  %xmm0,%xmm2
  addsd  %xmm3,%xmm2
  movsd  %xmm2,0x0(%rbp)
  movsd  (%rbx),%xmm2
  movsd  0x10(%rbx),%xmm4
  mulsd  %xmm0,%xmm2
  mulsd  0x8(%rbx),%xmm2
  mulsd  %xmm1,%xmm4
  subsd  %xmm4,%xmm2
  movsd  %xmm2,0x8(%rbp)
  movsd  (%rbx),%xmm2
  movsd  0x8(%rbx),%xmm4
  mulsd  %xmm0,%xmm2
  mulsd  0x10(%rbx),%xmm2
  mulsd  %xmm1,%xmm4
  addsd  %xmm4,%xmm2
  movsd  %xmm2,0x10(%rbp)
  movsd  (%rbx),%xmm2
  movsd  0x10(%rbx),%xmm4
  mulsd  %xmm0,%xmm2
  mulsd  0x8(%rbx),%xmm2
  mulsd  %xmm1,%xmm4
  addsd  %xmm4,%xmm2
  movsd  %xmm2,0x18(%rbp)
  movsd  0x8(%rbx),%xmm4
  movapd %xmm0,%xmm2
  mulsd  %xmm4,%xmm2
  mulsd  %xmm4,%xmm2
  addsd  %xmm3,%xmm2
  movsd  %xmm2,0x20(%rbp)
  movsd  0x10(%rbx),%xmm2
  movsd  (%rbx),%xmm4
  mulsd  %xmm0,%xmm2
  mulsd  0x8(%rbx),%xmm2
  mulsd  %xmm1,%xmm4
  subsd  %xmm4,%xmm2
  movsd  %xmm2,0x28(%rbp)
  movsd  (%rbx),%xmm2
  movsd  0x8(%rbx),%xmm4
  mulsd  %xmm0,%xmm2
  mulsd  0x10(%rbx),%xmm2
  mulsd  %xmm1,%xmm4
  subsd  %xmm4,%xmm2
  movsd  %xmm2,0x30(%rbp)
  movsd  0x8(%rbx),%xmm2
  mulsd  (%rbx),%xmm1
  mulsd  %xmm0,%xmm2
  mulsd  0x10(%rbx),%xmm2
  addsd  %xmm1,%xmm2
  movsd  %xmm2,0x38(%rbp)
  movsd  0x10(%rbx),%xmm1
  mulsd  %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  addsd  %xmm3,%xmm0
  movsd  %xmm0,0x40(%rbp)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
data_type: byte[8]

tool: {"D5": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
void mrotate(double theta, double axis[3], double m[3][3]) {
  double c = cos(theta);
  double s = sin(theta);
  m[0][0] = c + (1 - c) * ((axis[0]) * (axis[0]));
  m[0][1] = (1 - c) * axis[0] * axis[1] - s * axis[2];
  m[0][2] = (1 - c) * axis[0] * axis[2] + s * axis[1];
  m[1][0] = (1 - c) * axis[0] * axis[1] + s * axis[2];
  m[1][1] = c + (1 - c) * axis[1] * axis[1];
  m[1][2] = (1 - c) * axis[2] * axis[1] - s * axis[0];
  m[2][0] = (1 - c) * axis[0] * axis[2] - s * axis[1];
  m[2][1] = (1 - c) * axis[1] * axis[2] + s * axis[0];
  m[2][2] = c + (1 - c) * axis[2] * axis[2];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<wskip>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L34
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L39
L34:
  mov    -0x8(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <warning@plt>
  mov    -0x8(%rbp),%rax
  jmp    L24
L20:
  addq   $0x1,-0x8(%rbp)
L39:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L23
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L20
L23:
  mov    -0x8(%rbp),%rax
L24:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D5
type: string

tool: D5: expected a space

assistant: ```c
char *wskip(char *cptr) {
  if (!*cptr || !isspace((unsigned char)*cptr)) {
    warning(cptr, "expected a space");
    return (cptr);
  }
  while (*cptr && isspace((unsigned char)*cptr))
    ++cptr;
  return (cptr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<wskip>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L69
  movzbl %dil,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L69
  movzbl (%rbx),%edi
  mov    %rbx,%rax
  test   %dil,%dil
  je     L35
L52:
  movzbl %dil,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L0
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L52
  mov    %rbx,%rax
  jmp    L35
L69:
  lea    D2(%rip),%rsi
  mov    %rbx,%rdi
  call   <warning@plt>
  mov    %rbx,%rax
L35:
  pop    %rbx
  ret
L0:
  mov    %rbx,%rax
  jmp    L35
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
type: string

tool: D2: expected a space

assistant: ```c
char *wskip(char *cptr) {
  if (!*cptr || !isspace((unsigned char)*cptr)) {
    warning(cptr, "expected a space");
    return (cptr);
  }
  while (*cptr && isspace((unsigned char)*cptr))
    ++cptr;
  return (cptr);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<wskip>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L35
  call   <isspace@plt>
  test   %rax,%rax
  jne    L20
L35:
  lea    D1(%rip),%rsi
  mov    %r12,%rdi
  call   <warning@plt>
L33:
  mov    %r12,%rax
  pop    %r12
  ret
L19:
  call   <isspace@plt>
  test   %rax,%rax
  je     L33
  add    $0x1,%r12
L20:
  movzbl (%r12),%edi
  test   %dil,%dil
  jne    L19
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type: string

tool: D1: expected a space

assistant: ```c
char *wskip(char *cptr) {
  if (!*cptr || !isspace((unsigned char)*cptr)) {
    warning(cptr, "expected a space");
    return (cptr);
  }
  while (*cptr && isspace((unsigned char)*cptr))
    ++cptr;
  return (cptr);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<wskip>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L27
  call   <isspace@plt>
  test   %rax,%rax
  jne    L33
L27:
  lea    D7(%rip),%rsi
  mov    %r12,%rdi
  call   <warning@plt>
L34:
  mov    %r12,%rax
  pop    %r12
  ret
L18:
  call   <isspace@plt>
  test   %rax,%rax
  je     L34
  add    $0x1,%r12
L33:
  movzbl (%r12),%edi
  test   %dil,%dil
  jne    L18
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_label: D7
datatype: string

tool: D7: expected a space

assistant: ```c
char *wskip(char *cptr) {
  if (!*cptr || !isspace((unsigned char)*cptr)) {
    warning(cptr, "expected a space");
    return (cptr);
  }
  while (*cptr && isspace((unsigned char)*cptr))
    ++cptr;
  return (cptr);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x9,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movl   $0x8,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  movl   $0x7,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
int bar(int *a) {
  *(a + 0) = 9;
  *(a + 1) = 8;
  *(a + 2) = 7;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bar>:
  endbr64
  movl   $0x9,(%rdi)
  movl   $0x8,0x4(%rdi)
  movl   $0x7,0x8(%rdi)
  ret
```

assistant: ```c
int bar(int *a) {
  *(a + 0) = 9;
  *(a + 1) = 8;
  *(a + 2) = 7;
}
```
user: What is the c source code of the assembly code below:

```asm
<bar>:
  endbr64
  movabs $0x800000009,%rax
  movl   $0x7,0x8(%rdi)
  mov    %rax,(%rdi)
  ret
```

assistant: ```c
int bar(int *a) {
  *(a + 0) = 9;
  *(a + 1) = 8;
  *(a + 2) = 7;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bar>:
  endbr64
  mov    D6(%rip),%rax
  movl   $0x7,0x8(%rdi)
  mov    %rax,(%rdi)
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type_guessed: byte[8]

tool: {"D6": [0x09,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

assistant: ```c
int bar(int *a) {
  *(a + 0) = 9;
  *(a + 1) = 8;
  *(a + 2) = 7;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<compile_stdin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x8,%rsp
  jmp    L14
L17:
  mov    %rbx,%rdi
  call   <eval@plt>
L14:
  mov    $0x0,%eax
  call   <lread@plt>
  mov    %rax,%rbx
  test   %rbx,%rbx
  jne    L17
  mov    $0x1,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
compile_stdin() {
  register struct lnode *lp;
  while ((lp = lread()) != ((struct lnode *)0))
    eval(lp);
  return (1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<compile_stdin>:
  endbr64
  sub    $0x8,%rsp
  jmp    L19
L9:
  call   <eval@plt>
L19:
  mov    $0x0,%eax
  call   <lread@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L9
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
compile_stdin() {
  register struct lnode *lp;
  while ((lp = lread()) != ((struct lnode *)0))
    eval(lp);
  return (1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<compile_stdin>:
  endbr64
  sub    $0x8,%rsp
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L16:
  call   <eval@plt>
L0:
  xor    %eax,%eax
  call   <lread@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L16
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
compile_stdin() {
  register struct lnode *lp;
  while ((lp = lread()) != ((struct lnode *)0))
    eval(lp);
  return (1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<compile_stdin>:
  endbr64
  sub    $0x8,%rsp
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L6:
  call   <eval@plt>
L1:
  xor    %eax,%eax
  call   <lread@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L6
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
compile_stdin() {
  register struct lnode *lp;
  while ((lp = lread()) != ((struct lnode *)0))
    eval(lp);
  return (1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<force>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  movl   $0x1,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
L22:
  mov    -0x8(%rbp),%eax
  sub    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffffb21642c9,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x4,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x17,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  test   %edx,%edx
  jne    L13
  mov    -0x8(%rbp),%eax
  sub    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x4,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1c,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  test   %edx,%edx
  jne    L13
  mov    -0x8(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x3e0f83e1,%rax,%rax
  shr    $0x20,%rax
  sar    $0x3,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x5,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L12
L13:
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  jmp    L22
L12:
  nop
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int force(int p, int e, int i, int d) {
  int next;
  int step = 1;
  next = d + 1;
  while (1) {
    if ((next - p) % 23 == 0 && (next - e) % 28 == 0 && (next - i) % 33 == 0)
      break;
    next += step;
  }
  return next;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<force>:
  endbr64
  mov    %edx,%eax
  lea    0x1(%rcx),%edx
  sub    %edi,%edx
  mov    %edi,%r8d
  sub    %esi,%r8d
  mov    %edi,%esi
  sub    %eax,%esi
L6:
  lea    (%rdx,%rdi,1),%ecx
  lea    (%r8,%rdx,1),%r9d
  movslq %r9d,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %r9d,%eax
  sar    $0x4,%eax
  mov    %r9d,%r10d
  sar    $0x1f,%r10d
  sub    %r10d,%eax
  imul   $0x1c,%eax,%eax
  mov    %r9d,%r10d
  sub    %eax,%r10d
  lea    (%rsi,%rdx,1),%r9d
  movslq %r9d,%rax
  imul   $0x3e0f83e1,%rax,%rax
  sar    $0x23,%rax
  mov    %r9d,%r11d
  sar    $0x1f,%r11d
  sub    %r11d,%eax
  mov    %eax,%r11d
  shl    $0x5,%r11d
  add    %r11d,%eax
  sub    %eax,%r9d
  or     %r10d,%r9d
  movslq %edx,%rax
  imul   $0xffffffffb21642c9,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x4,%eax
  mov    %edx,%r10d
  sar    $0x1f,%r10d
  sub    %r10d,%eax
  imul   $0x17,%eax,%eax
  mov    %edx,%r10d
  sub    %eax,%r10d
  add    $0x1,%edx
  or     %r10d,%r9d
  jne    L6
  mov    %ecx,%eax
  ret
```

assistant: ```c
int force(int p, int e, int i, int d) {
  int next;
  int step = 1;
  next = d + 1;
  while (1) {
    if ((next - p) % 23 == 0 && (next - e) % 28 == 0 && (next - i) % 33 == 0)
      break;
    next += step;
  }
  return next;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<force>:
  endbr64
  mov    %esi,%r9d
  mov    %edx,%r8d
  lea    0x1(%rcx),%edx
  mov    %edi,%esi
  mov    %edi,%ecx
  sub    %edi,%edx
  sub    %r9d,%esi
  sub    %r8d,%ecx
  nopl   0x0(%rax)
L1:
  lea    (%rsi,%rdx,1),%r9d
  lea    (%rdi,%rdx,1),%r8d
  movslq %r9d,%rax
  mov    %r9d,%r10d
  imul   $0xffffffff92492493,%rax,%rax
  sar    $0x1f,%r10d
  shr    $0x20,%rax
  add    %r9d,%eax
  sar    $0x4,%eax
  sub    %r10d,%eax
  mov    %r9d,%r10d
  lea    (%rcx,%rdx,1),%r9d
  imul   $0x1c,%eax,%eax
  mov    %r9d,%r11d
  sar    $0x1f,%r11d
  sub    %eax,%r10d
  movslq %r9d,%rax
  imul   $0x3e0f83e1,%rax,%rax
  sar    $0x23,%rax
  sub    %r11d,%eax
  mov    %eax,%r11d
  shl    $0x5,%r11d
  add    %r11d,%eax
  sub    %eax,%r9d
  movslq %edx,%rax
  imul   $0xffffffffb21642c9,%rax,%rax
  or     %r10d,%r9d
  mov    %edx,%r10d
  sar    $0x1f,%r10d
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x4,%eax
  sub    %r10d,%eax
  mov    %edx,%r10d
  add    $0x1,%edx
  imul   $0x17,%eax,%eax
  sub    %eax,%r10d
  or     %r10d,%r9d
  jne    L1
  mov    %r8d,%eax
  ret
```

assistant: ```c
int force(int p, int e, int i, int d) {
  int next;
  int step = 1;
  next = d + 1;
  while (1) {
    if ((next - p) % 23 == 0 && (next - e) % 28 == 0 && (next - i) % 33 == 0)
      break;
    next += step;
  }
  return next;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<force>:
  endbr64
  lea    0x1(%rcx),%r8d
  mov    %edx,%r9d
  mov    %ecx,%edx
  mov    %esi,%r10d
  mov    %r8d,%ecx
  sub    %esi,%ecx
  movslq %ecx,%rax
  mov    %ecx,%esi
  imul   $0xffffffff92492493,%rax,%rax
  sar    $0x1f,%esi
  shr    $0x20,%rax
  add    %ecx,%eax
  sar    $0x4,%eax
  sub    %esi,%eax
  mov    %r8d,%esi
  imul   $0x1c,%eax,%eax
  sub    %edi,%esi
  mov    %esi,%r11d
  sar    $0x1f,%r11d
  sub    %eax,%ecx
  movslq %esi,%rax
  imul   $0xffffffffb21642c9,%rax,%rax
  shr    $0x20,%rax
  add    %esi,%eax
  sar    $0x4,%eax
  sub    %r11d,%eax
  imul   $0x17,%eax,%eax
  sub    %eax,%esi
  or     %esi,%ecx
  mov    %r8d,%esi
  sub    %r9d,%esi
  movslq %esi,%rax
  mov    %esi,%r11d
  imul   $0x3e0f83e1,%rax,%rax
  sar    $0x1f,%r11d
  sar    $0x23,%rax
  sub    %r11d,%eax
  mov    %eax,%r11d
  shl    $0x5,%r11d
  add    %r11d,%eax
  sub    %eax,%esi
  or     %esi,%ecx
  je     L16
  add    $0x2,%edx
  mov    %edi,%esi
  mov    %edi,%ecx
  sub    %edi,%edx
  sub    %r10d,%esi
  sub    %r9d,%ecx
  xchg   %ax,%ax
L18:
  lea    (%rsi,%rdx,1),%r9d
  lea    (%rdx,%rdi,1),%r8d
  movslq %r9d,%rax
  mov    %r9d,%r10d
  imul   $0xffffffff92492493,%rax,%rax
  sar    $0x1f,%r10d
  shr    $0x20,%rax
  add    %r9d,%eax
  sar    $0x4,%eax
  sub    %r10d,%eax
  mov    %r9d,%r10d
  lea    (%rcx,%rdx,1),%r9d
  imul   $0x1c,%eax,%eax
  mov    %r9d,%r11d
  sar    $0x1f,%r11d
  sub    %eax,%r10d
  movslq %r9d,%rax
  imul   $0x3e0f83e1,%rax,%rax
  sar    $0x23,%rax
  sub    %r11d,%eax
  mov    %eax,%r11d
  shl    $0x5,%r11d
  add    %r11d,%eax
  sub    %eax,%r9d
  movslq %edx,%rax
  imul   $0xffffffffb21642c9,%rax,%rax
  or     %r10d,%r9d
  mov    %edx,%r10d
  sar    $0x1f,%r10d
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x4,%eax
  sub    %r10d,%eax
  mov    %edx,%r10d
  add    $0x1,%edx
  imul   $0x17,%eax,%eax
  sub    %eax,%r10d
  or     %r10d,%r9d
  jne    L18
L16:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int force(int p, int e, int i, int d) {
  int next;
  int step = 1;
  next = d + 1;
  while (1) {
    if ((next - p) % 23 == 0 && (next - e) % 28 == 0 && (next - i) % 33 == 0)
      break;
    next += step;
  }
  return next;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jg     L16
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L37
L16:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jg     L20
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L37
L20:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <factorize@plt>
  mov    $0x0,%eax
L37:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D6
type_infered: string
name: read_data
arguments: data_label: D14
type_infered: string

tool: D6: Usage: ./factorize N
Please input a value to factorize!


tool: D14: You must input a value higher than 2


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("Usage: ./factorize N\nPlease input a value to factorize!\n");
    return -1;
  }
  int n = atoi(argv[1]);
  if (n < 2) {
    printf("You must input a value higher than 2\n");
    return -1;
  }
  factorize(n);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L38
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  cmp    $0x1,%eax
  jle    L5
  call   <factorize@plt>
  mov    $0x0,%eax
L1:
  add    $0x8,%rsp
  ret
L38:
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L1
L5:
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L1
```

tool_calls: name: unpack_data
arguments: data_tag: D16
type_infered: string
name: unpack_data
arguments: data_tag: D17
type_infered: string

tool: D16: Usage: ./factorize N
Please input a value to factorize!


tool: D17: You must input a value higher than 2


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("Usage: ./factorize N\nPlease input a value to factorize!\n");
    return -1;
  }
  int n = atoi(argv[1]);
  if (n < 2) {
    printf("You must input a value higher than 2\n");
    return -1;
  }
  factorize(n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L7
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  cmp    $0x1,%eax
  jle    L4
  call   <factorize@plt>
  xor    %eax,%eax
L9:
  add    $0x8,%rsp
  ret
L7:
  lea    D15(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L9
L4:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L9
```

tool_calls: name: read_data
arguments: data_label: D15
type: string
name: read_data
arguments: data_label: D6
type: string

tool: D15: Usage: ./factorize N
Please input a value to factorize!


tool: D6: You must input a value higher than 2


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("Usage: ./factorize N\nPlease input a value to factorize!\n");
    return -1;
  }
  int n = atoi(argv[1]);
  if (n < 2) {
    printf("You must input a value higher than 2\n");
    return -1;
  }
  factorize(n);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L8
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  cmp    $0x1,%eax
  jle    L24
  call   <factorize@plt>
  xor    %eax,%eax
L3:
  add    $0x8,%rsp
  ret
L8:
  lea    D19(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L3
L24:
  lea    D18(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L3
```

tool_calls: name: disassemble_data
arguments: data_tag: D19
data_type: string
name: disassemble_data
arguments: data_tag: D18
data_type: string

tool: D19: Usage: ./factorize N
Please input a value to factorize!


tool: D18: You must input a value higher than 2


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("Usage: ./factorize N\nPlease input a value to factorize!\n");
    return -1;
  }
  int n = atoi(argv[1]);
  if (n < 2) {
    printf("You must input a value higher than 2\n");
    return -1;
  }
  factorize(n);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<sysfs_write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xffffffff,-0xc(%rbp)
  mov    -0x20(%rbp),%rax
  lea    D7(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L16
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x1,%edx
  mov    %rax,%rdi
  call   <fwrite@plt>
  cmp    $0xffffffff,%eax
  je     L37
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L37:
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  jmp    L3
L16:
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
L3:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0xc(%rbp),%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D7
datatype: string
name: read_data
arguments: tag: D25
datatype: string
name: read_data
arguments: tag: D23
datatype: string

tool: D7: w

tool: D25: fwrite

tool: D23: fopen

assistant: ```c
int sysfs_write(char *str, char *file) {
  FILE *f;
  int ret = -1;
  f = fopen(file, "w");
  if (f != NULL) {
    if (-1 != fwrite(str, strlen(str), 1, f)) {
      ret = 0;
    } else {
      perror("fwrite");
    }
  } else {
    perror("fopen");
  }
  fclose(f);
  return ret;
}
```