ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的：

```asm
<string_printable>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    $0x400,%edx
  mov    $0x0,%esi
  lea    D15(%rip),%rdi
  call   <memset@plt>
  movzbl (%rbx),%eax
  test   %al,%al
  je     L82
  lea    D15(%rip),%rbp
  lea    D5(%rip),%r12
  jmp    L45
L26:
  cmp    $0x7,%al
  je     L6
  cmp    $0x8,%al
  je     L39
  cmp    $0xa,%al
  je     L84
  cmp    $0xd,%al
  jne    L43
  movb   $0x5c,0x0(%rbp)
  movb   $0x72,0x1(%rbp)
  lea    0x2(%rbp),%rbp
  jmp    L125
L6:
  movb   $0x5c,0x0(%rbp)
  movb   $0x61,0x1(%rbp)
  lea    0x2(%rbp),%rbp
L125:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L82
L45:
  lea    -0x20(%rax),%edx
  cmp    $0x5e,%dl
  ja     L26
  mov    %al,0x0(%rbp)
  lea    0x1(%rbp),%rbp
  jmp    L125
L39:
  movb   $0x5c,0x0(%rbp)
  movb   $0x62,0x1(%rbp)
  lea    0x2(%rbp),%rbp
  jmp    L125
L84:
  movb   $0x5c,0x0(%rbp)
  movb   $0x6e,0x1(%rbp)
  lea    0x2(%rbp),%rbp
  jmp    L125
L43:
  movzbl %al,%ecx
  mov    %rsp,%r13
  mov    %r12,%rdx
  mov    $0x400,%esi
  mov    %r13,%rdi
  call   <snprintf@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <strncat@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  cltq
  add    %rax,%rbp
  jmp    L125
L82:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L86
  lea    D15(%rip),%rax
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L86:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D5
data_type: string

tool: D5: \%03o

assistant: ```c
char *string_printable(char *in) {
  static char out[1024];
  char octal[1024];
  char *ret;
  memset(out, '\0', 1024);
  ret = out;
  while (*in) {
    if (*in >= 32 && *in != 127) {
      *ret++ = *in;
    } else if (*in == '\\') {
      *ret++ = '\\';
      *ret++ = '\\';
    } else if (*in == '\a') {
      *ret++ = '\\';
      *ret++ = 'a';
    } else if (*in == '\b') {
      *ret++ = '\\';
      *ret++ = 'b';
    } else if (*in == '\n') {
      *ret++ = '\\';
      *ret++ = 'n';
    } else if (*in == '\r') {
      *ret++ = '\\';
      *ret++ = 'r';
    } else {
      snprintf(octal, 1024, "\\%03o", (unsigned char)*in);
      strncat(ret, octal, strlen(octal));
      ret += strlen(octal);
    }
    in++;
  }
  return out;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<string_printable>:
  endbr64
  push   %r14
  mov    $0x400,%edx
  xor    %esi,%esi
  push   %r13
  push   %r12
  lea    D2(%rip),%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %r12,%rdi
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <memset@plt>
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L83
  mov    %r12,%rbp
  lea    D15(%rip),%r14
  jmp    L10
  nopw   0x0(%rax,%rax,1)
L86:
  mov    %cl,0x0(%rbp)
  add    $0x1,%rbp
L4:
  movzbl 0x1(%rbx),%ecx
  add    $0x1,%rbx
  test   %cl,%cl
  je     L83
L10:
  lea    -0x20(%rcx),%eax
  cmp    $0x5e,%al
  jbe    L86
  cmp    $0x7,%cl
  je     L12
  cmp    $0x8,%cl
  je     L77
  cmp    $0xa,%cl
  je     L48
  cmp    $0xd,%cl
  jne    L2
  mov    $0x725c,%eax
  add    $0x1,%rbx
  add    $0x2,%rbp
  mov    %ax,-0x2(%rbp)
  movzbl (%rbx),%ecx
  test   %cl,%cl
  jne    L10
  cs nopw 0x0(%rax,%rax,1)
L83:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0x410,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  mov    $0x615c,%esi
  add    $0x2,%rbp
  mov    %si,-0x2(%rbp)
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L77:
  mov    $0x625c,%ecx
  add    $0x2,%rbp
  mov    %cx,-0x2(%rbp)
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L48:
  mov    $0x6e5c,%edx
  add    $0x2,%rbp
  mov    %dx,-0x2(%rbp)
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L2:
  mov    %rsp,%r13
  mov    %r14,%rdx
  mov    $0x400,%esi
  mov    %r13,%rdi
  call   <snprintf@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %r13,%rsi
  mov    %eax,%edx
  call   <strncat@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  cltq
  add    %rax,%rbp
  jmp    L4
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
type_infered: string

tool: D15: \%03o

assistant: ```c
char *string_printable(char *in) {
  static char out[1024];
  char octal[1024];
  char *ret;
  memset(out, '\0', 1024);
  ret = out;
  while (*in) {
    if (*in >= 32 && *in != 127) {
      *ret++ = *in;
    } else if (*in == '\\') {
      *ret++ = '\\';
      *ret++ = '\\';
    } else if (*in == '\a') {
      *ret++ = '\\';
      *ret++ = 'a';
    } else if (*in == '\b') {
      *ret++ = '\\';
      *ret++ = 'b';
    } else if (*in == '\n') {
      *ret++ = '\\';
      *ret++ = 'n';
    } else if (*in == '\r') {
      *ret++ = '\\';
      *ret++ = 'r';
    } else {
      snprintf(octal, 1024, "\\%03o", (unsigned char)*in);
      strncat(ret, octal, strlen(octal));
      ret += strlen(octal);
    }
    in++;
  }
  return out;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<string_printable>:
  endbr64
  push   %r14
  mov    $0x400,%edx
  xor    %esi,%esi
  push   %r13
  push   %r12
  lea    D8(%rip),%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %r12,%rdi
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <memset@plt>
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L115
  mov    %r12,%rbp
  lea    D12(%rip),%r14
  jmp    L90
  nopw   0x0(%rax,%rax,1)
L137:
  mov    %cl,0x0(%rbp)
  add    $0x1,%rbp
L112:
  movzbl 0x1(%rbx),%ecx
  add    $0x1,%rbx
  test   %cl,%cl
  je     L115
L90:
  lea    -0x20(%rcx),%eax
  cmp    $0x5e,%al
  jbe    L137
  cmp    $0x7,%cl
  je     L76
  cmp    $0x8,%cl
  je     L39
  cmp    $0xa,%cl
  je     L94
  cmp    $0xd,%cl
  jne    L109
  mov    $0x725c,%eax
  add    $0x1,%rbx
  add    $0x2,%rbp
  mov    %ax,-0x2(%rbp)
  movzbl (%rbx),%ecx
  test   %cl,%cl
  jne    L90
  cs nopw 0x0(%rax,%rax,1)
L115:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L107
  add    $0x410,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L76:
  mov    $0x615c,%esi
  add    $0x2,%rbp
  mov    %si,-0x2(%rbp)
  jmp    L112
  nopw   0x0(%rax,%rax,1)
L39:
  mov    $0x625c,%ecx
  add    $0x2,%rbp
  mov    %cx,-0x2(%rbp)
  jmp    L112
  nopw   0x0(%rax,%rax,1)
L94:
  mov    $0x6e5c,%edx
  add    $0x2,%rbp
  mov    %dx,-0x2(%rbp)
  jmp    L112
  nopw   0x0(%rax,%rax,1)
L109:
  mov    %rsp,%r13
  mov    %r14,%rdx
  mov    $0x400,%esi
  mov    %r13,%rdi
  call   <snprintf@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %r13,%rsi
  mov    %eax,%edx
  call   <strncat@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  cltq
  add    %rax,%rbp
  jmp    L112
L107:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D12
type: string

tool: D12: \%03o

assistant: ```c
char *string_printable(char *in) {
  static char out[1024];
  char octal[1024];
  char *ret;
  memset(out, '\0', 1024);
  ret = out;
  while (*in) {
    if (*in >= 32 && *in != 127) {
      *ret++ = *in;
    } else if (*in == '\\') {
      *ret++ = '\\';
      *ret++ = '\\';
    } else if (*in == '\a') {
      *ret++ = '\\';
      *ret++ = 'a';
    } else if (*in == '\b') {
      *ret++ = '\\';
      *ret++ = 'b';
    } else if (*in == '\n') {
      *ret++ = '\\';
      *ret++ = 'n';
    } else if (*in == '\r') {
      *ret++ = '\\';
      *ret++ = 'r';
    } else {
      snprintf(octal, 1024, "\\%03o", (unsigned char)*in);
      strncat(ret, octal, strlen(octal));
      ret += strlen(octal);
    }
    in++;
  }
  return out;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<file_removepackage>:
  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),%rdx
  lea    -0x90(%rbp),%rax
  mov    $0x80,%esi
  mov    %rax,%rdi
  call   <file_getlocalpackagepath@plt>
  lea    -0x90(%rbp),%rax
  mov    %rax,%rdi
  call   <file_remove@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
unsigned int file_removepackage(const char *name) {
  char remotedatapath[128];
  file_getlocalpackagepath(remotedatapath, 128, name);
  return file_remove(remotedatapath);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<file_removepackage>:
  endbr64
  push   %rbx
  sub    $0x90,%rsp
  mov    %rdi,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    $0x80,%esi
  mov    %rbx,%rdi
  call   <file_getlocalpackagepath@plt>
  mov    %rbx,%rdi
  call   <file_remove@plt>
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x90,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int file_removepackage(const char *name) {
  char remotedatapath[128];
  file_getlocalpackagepath(remotedatapath, 128, name);
  return file_remove(remotedatapath);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<file_removepackage>:
  endbr64
  push   %rbp
  mov    %rdi,%rdx
  mov    $0x80,%esi
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <file_getlocalpackagepath@plt>
  mov    %rbp,%rdi
  call   <file_remove@plt>
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x90,%rsp
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int file_removepackage(const char *name) {
  char remotedatapath[128];
  file_getlocalpackagepath(remotedatapath, 128, name);
  return file_remove(remotedatapath);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<file_removepackage>:
  endbr64
  push   %rbp
  mov    %rdi,%rdx
  mov    $0x80,%esi
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <file_getlocalpackagepath@plt>
  mov    %rbp,%rdi
  call   <file_remove@plt>
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x90,%rsp
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int file_removepackage(const char *name) {
  char remotedatapath[128];
  file_getlocalpackagepath(remotedatapath, 128, name);
  return file_remove(remotedatapath);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<computeArea>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %edx,-0x3c(%rbp)
  mov    %ecx,-0x40(%rbp)
  mov    %r8d,-0x44(%rbp)
  mov    %r9d,-0x48(%rbp)
  mov    -0x3c(%rbp),%eax
  sub    -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    -0x40(%rbp),%eax
  sub    -0x38(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x24(%rbp)
  mov    0x10(%rbp),%eax
  sub    -0x44(%rbp),%eax
  mov    %eax,%edx
  mov    0x18(%rbp),%eax
  sub    -0x48(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x44(%rbp),%edx
  mov    -0x34(%rbp),%eax
  cmp    %eax,%edx
  cmovge %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    0x10(%rbp),%edx
  mov    -0x3c(%rbp),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x48(%rbp),%edx
  mov    -0x38(%rbp),%eax
  cmp    %eax,%edx
  cmovge %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    0x18(%rbp),%edx
  mov    -0x40(%rbp),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L22
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  jmp    L21
L22:
  mov    $0x0,%eax
L21:
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L15
  mov    -0x10(%rbp),%eax
  sub    -0x14(%rbp),%eax
  jmp    L35
L15:
  mov    $0x0,%eax
L35:
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  imul   -0x8(%rbp),%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    -0x20(%rbp),%eax
  add    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
  int area1 = (C - A) * (D - B);
  int area2 = (G - E) * (H - F);
  int inL = A > E ? A : E;
  int inR = C < G ? C : G;
  int inB = B > F ? B : F;
  int inT = D < H ? D : H;
  int inW = inR > inL ? inR - inL : 0;
  int inH = inT > inB ? inT - inB : 0;
  int sum = area1 - inW * inH + area2;
  return sum;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<computeArea>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edi,%r10d
  mov    %esi,%edi
  mov    %edx,%esi
  mov    %ecx,%edx
  mov    0x18(%rsp),%ebx
  mov    0x20(%rsp),%ecx
  mov    %esi,%eax
  sub    %r10d,%eax
  mov    %edx,%r11d
  sub    %edi,%r11d
  imul   %r11d,%eax
  mov    %ebx,%r11d
  sub    %r8d,%r11d
  mov    %ecx,%ebp
  sub    %r9d,%ebp
  imul   %ebp,%r11d
  cmp    %r8d,%r10d
  cmovl  %r8d,%r10d
  cmp    %ebx,%esi
  cmovg  %ebx,%esi
  cmp    %r9d,%edi
  cmovl  %r9d,%edi
  cmp    %ecx,%edx
  cmovg  %ecx,%edx
  mov    %esi,%ecx
  sub    %r10d,%ecx
  cmp    %esi,%r10d
  mov    $0x0,%esi
  cmovge %esi,%ecx
  mov    %edx,%esi
  sub    %edi,%esi
  cmp    %edx,%edi
  mov    $0x0,%edx
  cmovl  %esi,%edx
  imul   %ecx,%edx
  sub    %edx,%eax
  add    %r11d,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
  int area1 = (C - A) * (D - B);
  int area2 = (G - E) * (H - F);
  int inL = A > E ? A : E;
  int inR = C < G ? C : G;
  int inB = B > F ? B : F;
  int inT = D < H ? D : H;
  int inW = inR > inL ? inR - inL : 0;
  int inH = inT > inB ? inT - inB : 0;
  int sum = area1 - inW * inH + area2;
  return sum;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<computeArea>:
  endbr64
  mov    %esi,%r10d
  mov    %edx,%esi
  mov    %ecx,%edx
  push   %rbp
  mov    %esi,%ecx
  push   %rbx
  mov    %edx,%eax
  mov    0x18(%rsp),%ebx
  mov    0x20(%rsp),%r11d
  sub    %r10d,%eax
  sub    %edi,%ecx
  imul   %eax,%ecx
  mov    %ebx,%eax
  mov    %r11d,%ebp
  sub    %r8d,%eax
  sub    %r9d,%ebp
  imul   %ebp,%eax
  cmp    %r8d,%edi
  cmovl  %r8d,%edi
  cmp    %ebx,%esi
  cmovg  %ebx,%esi
  cmp    %r9d,%r10d
  cmovl  %r9d,%r10d
  cmp    %r11d,%edx
  cmovg  %r11d,%edx
  mov    %esi,%r8d
  sub    %edi,%r8d
  cmp    %esi,%edi
  mov    $0x0,%esi
  cmovge %esi,%r8d
  cmp    %edx,%r10d
  jge    L2
  sub    %r10d,%edx
  imul   %r8d,%edx
  sub    %edx,%ecx
L2:
  add    %ecx,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
  int area1 = (C - A) * (D - B);
  int area2 = (G - E) * (H - F);
  int inL = A > E ? A : E;
  int inR = C < G ? C : G;
  int inB = B > F ? B : F;
  int inT = D < H ? D : H;
  int inW = inR > inL ? inR - inL : 0;
  int inH = inT > inB ? inT - inB : 0;
  int sum = area1 - inW * inH + area2;
  return sum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<computeArea>:
  endbr64
  mov    %esi,%r10d
  mov    %edx,%esi
  mov    %ecx,%edx
  push   %rbp
  mov    %esi,%ecx
  push   %rbx
  mov    %edx,%eax
  mov    0x18(%rsp),%ebx
  mov    0x20(%rsp),%r11d
  sub    %r10d,%eax
  sub    %edi,%ecx
  imul   %eax,%ecx
  mov    %ebx,%eax
  mov    %r11d,%ebp
  sub    %r8d,%eax
  sub    %r9d,%ebp
  imul   %ebp,%eax
  cmp    %r8d,%edi
  cmovl  %r8d,%edi
  cmp    %ebx,%esi
  cmovg  %ebx,%esi
  cmp    %r9d,%r10d
  cmovl  %r9d,%r10d
  cmp    %r11d,%edx
  cmovg  %r11d,%edx
  mov    %esi,%r8d
  sub    %edi,%r8d
  cmp    %esi,%edi
  mov    $0x0,%esi
  cmovge %esi,%r8d
  cmp    %edx,%r10d
  jge    L3
  sub    %r10d,%edx
  imul   %r8d,%edx
  sub    %edx,%ecx
L3:
  add    %ecx,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
  int area1 = (C - A) * (D - B);
  int area2 = (G - E) * (H - F);
  int inL = A > E ? A : E;
  int inR = C < G ? C : G;
  int inB = B > F ? B : F;
  int inT = D < H ? D : H;
  int inW = inR > inL ? inR - inL : 0;
  int inH = inT > inB ? inT - inB : 0;
  int sum = area1 - inW * inH + area2;
  return sum;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <Joseph@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D26
datatype: string
name: disassemble_data
arguments: data_tag: D40
datatype: string
name: disassemble_data
arguments: data_tag: D34
datatype: string
name: disassemble_data
arguments: data_tag: D35
datatype: string

tool: D26: Input amount: 

tool: D40: %d

tool: D34: 
Input upper limit: 

tool: D35: 


assistant: ```c
int main(int argc, const char *argv[]) {
  int n, m;
  printf("Input amount: ");
  scanf("%d", &n);
  printf("\nInput upper limit: ");
  scanf("%d", &m);
  printf("\n");
  Joseph(n, m);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D7(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D27(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D32(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edi
  call   <Joseph@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type_infered: string
name: retrieve_data
arguments: data_tag: D7
type_infered: string
name: retrieve_data
arguments: data_tag: D27
type_infered: string
name: retrieve_data
arguments: data_tag: D32
type_infered: string

tool: D9: Input amount: 

tool: D7: %d

tool: D27: 
Input upper limit: 

tool: D32: 


assistant: ```c
int main(int argc, const char *argv[]) {
  int n, m;
  printf("Input amount: ");
  scanf("%d", &n);
  printf("\nInput upper limit: ");
  scanf("%d", &m);
  printf("\n");
  Joseph(n, m);
  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    D5(%rip),%rdi
  lea    D30(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edi
  call   <Joseph@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
type_infered: string
name: disassemble_data
arguments: tag: D30
type_infered: string
name: disassemble_data
arguments: tag: D9
type_infered: string
name: disassemble_data
arguments: tag: D8
type_infered: string

tool: D5: Input amount: 

tool: D30: %d

tool: D9: 
Input upper limit: 

tool: D8: 


assistant: ```c
int main(int argc, const char *argv[]) {
  int n, m;
  printf("Input amount: ");
  scanf("%d", &n);
  printf("\nInput upper limit: ");
  scanf("%d", &m);
  printf("\n");
  Joseph(n, m);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D27(%rip),%rdi
  lea    D1(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edi
  call   <Joseph@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D27
datatype: string
name: unpack_data
arguments: data_label: D1
datatype: string
name: unpack_data
arguments: data_label: D0
datatype: string
name: unpack_data
arguments: data_label: D21
datatype: string

tool: D27: Input amount: 

tool: D1: %d

tool: D0: 
Input upper limit: 

tool: D21: 


assistant: ```c
int main(int argc, const char *argv[]) {
  int n, m;
  printf("Input amount: ");
  scanf("%d", &n);
  printf("\nInput upper limit: ");
  scanf("%d", &m);
  printf("\n");
  Joseph(n, m);
  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
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L21
  lea    0x8(%rsi),%rbx
  lea    -0x2(%rdi),%eax
  lea    0x10(%rsi,%rax,8),%r12
  lea    D32(%rip),%r14
  lea    D18(%rip),%r13
  lea    D7(%rip),%rbp
  jmp    L63
L99:
  mov    %r14,%rdi
  call   <panic@plt>
L20:
  add    $0x8,%rbx
  cmp    %r12,%rbx
  je     L21
L63:
  mov    (%rbx),%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  je     L99
  lea    -0x2a(%rax),%edx
  cmp    $0xf,%dl
  ja     L101
  ja     L101
  movzbl %dl,%eax
  movslq 0x0(%rbp,%rax,4),%rax
  add    %rbp,%rax
  notrack jmp *%rax
  call   <atof@plt>
  call   <push@plt>
  jmp    L20
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %xmm0,%r15
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %r15,%xmm1
  addsd  %xmm1,%xmm0
  call   <push@plt>
  jmp    L20
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %xmm0,%r15
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %r15,%xmm2
  subsd  %xmm2,%xmm0
  call   <push@plt>
  jmp    L20
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %xmm0,%r15
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %r15,%xmm3
  mulsd  %xmm3,%xmm0
  call   <push@plt>
  jmp    L20
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %xmm0,%r15
  mov    $0x0,%eax
  call   <pop@plt>
  movq   %r15,%xmm4
  divsd  %xmm4,%xmm0
  call   <push@plt>
  jmp    L20
L101:
  mov    %r13,%rdi
  call   <panic@plt>
  jmp    L20
L21:
  mov    $0x0,%eax
  call   <pop@plt>
  lea    D33(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: tag: D32
type: string
name: parse_data
arguments: tag: D18
type: string
name: parse_data
arguments: tag: D33
type: string

tool: D32: empty command line argument

tool: D18: unknown operator

tool: D33: %g


assistant: ```c
int main(int argc, char **argv) {
  int i;
  double value;
  for (i = 1; i < argc; ++i) {
    switch (argv[i][0]) {
    case '\0':
      panic("empty command line argument");
      break;
    case '0':
    case '1':
    case '2':
    case '3':
    case '4':
    case '5':
    case '6':
    case '7':
    case '8':
    case '9':
      push(atof(argv[i]));
      break;
    case '+':
      push(pop() + pop());
      break;
    case '-':
      value = pop();
      push(pop() - value);
      break;
    case '*':
      push(pop() * pop());
      break;
    case '/':
      value = pop();
      push(pop() / value);
      break;
    default:
      panic("unknown operator");
      break;
    }
  }
  printf("%g\n", pop());
  return 0;
}
```