ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ChebyDivRem>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movapd %xmm0,%xmm5
  mov    %rsi,%r8
  mov    %edx,%r9d
  mov    %edx,%r10d
  shr    $0x1f,%r10d
  add    %edx,%r10d
  sar    %r10d
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  lea    -0x1(%r10),%eax
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  add    %eax,%ecx
  sar    %ecx
  lea    0x3(%rdx),%esi
  test   %edx,%edx
  cmovns %edx,%esi
  sar    $0x2,%esi
  cmp    $0x5,%edx
  jle    L10
  movslq %r10d,%r13
  lea    (%rdi,%r13,8),%rbx
  mov    %rbx,%rdx
  movslq %esi,%rsi
  lea    0x0(,%rsi,8),%r11
  lea    (%rdi,%r11,1),%r12
  lea    0x0(%r13,%rsi,1),%rax
  lea    (%rdi,%rax,8),%rbp
  add    %r8,%r11
  mov    $0x0,%eax
  movq   D2(%rip),%xmm3
  movapd %xmm2,%xmm4
  xorpd  %xmm3,%xmm4
  sub    %r13,%rsi
L0:
  movapd %xmm4,%xmm0
  mulsd  (%rdi,%rax,8),%xmm0
  subsd  -0x10(%rdx),%xmm0
  addsd  (%rbx,%rax,8),%xmm0
  movsd  %xmm0,(%r8,%rax,8)
  movsd  -0x10(%rdx,%rsi,8),%xmm0
  xorpd  %xmm3,%xmm0
  movapd %xmm2,%xmm1
  mulsd  (%r12,%rax,8),%xmm1
  subsd  %xmm1,%xmm0
  addsd  0x0(%rbp,%rax,8),%xmm0
  movsd  %xmm0,(%r11,%rax,8)
  add    $0x1,%rax
  sub    $0x8,%rdx
  cmp    %eax,%ecx
  jg     L0
L10:
  movslq %ecx,%rax
  movsd  (%rdi,%rax,8),%xmm0
  add    %r10d,%ecx
  movslq %ecx,%rcx
  movsd  (%rdi,%rcx,8),%xmm1
  subsd  %xmm0,%xmm1
  mulsd  %xmm2,%xmm0
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%r8,%rax,8)
  movslq %r10d,%r10
  movslq %r9d,%r9
  mulsd  -0x8(%rdi,%r10,8),%xmm5
  movsd  -0x8(%rdi,%r9,8),%xmm1
  subsd  %xmm5,%xmm1
  movsd  %xmm1,-0x8(%r8,%r10,8)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type_infered: byte[8]

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

assistant: ```c
void ChebyDivRem(double *dividend, double m0, double *remres, int n) {
  int rlim, dd;
  int tmpfloor;
  double tm0;
  int halfrlim;
  rlim = n / 2;
  tm0 = 2.0 * m0;
  tmpfloor = (rlim - 1) / 2;
  halfrlim = rlim / 2;
  for (dd = 0; dd < tmpfloor; dd++) {
    remres[dd] = -tm0 * dividend[dd] - dividend[rlim - dd - 2] + dividend[rlim + dd];
    remres[dd + halfrlim] = -dividend[halfrlim - 2 - dd] - tm0 * dividend[dd + halfrlim] + dividend[rlim + dd + halfrlim];
  }
  remres[tmpfloor] = dividend[tmpfloor + rlim] - dividend[tmpfloor] - (tm0 * dividend[tmpfloor]);
  remres[rlim - 1] = dividend[n - 1] - (m0 * dividend[rlim - 1]);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ChebyDivRem>:
  endbr64
  movslq %edx,%r9
  mov    %rsi,%r8
  movapd %xmm0,%xmm2
  push   %r13
  mov    %r9d,%r10d
  lea    0x3(%r9),%ecx
  push   %r12
  addsd  %xmm0,%xmm2
  shr    $0x1f,%r10d
  push   %rbp
  movapd %xmm0,%xmm5
  add    %r9d,%r10d
  push   %rbx
  sar    %r10d
  lea    -0x1(%r10),%eax
  mov    %eax,%esi
  shr    $0x1f,%esi
  add    %eax,%esi
  sar    %esi
  test   %r9d,%r9d
  cmovns %r9d,%ecx
  sar    $0x2,%ecx
  cmp    $0x5,%r9d
  jle    L10
  movslq %r10d,%r13
  movslq %ecx,%rcx
  movq   D2(%rip),%xmm3
  movapd %xmm2,%xmm4
  lea    0x0(,%rcx,8),%r11
  lea    (%rcx,%r13,1),%rax
  sub    %r13,%rcx
  lea    (%rdi,%r13,8),%rbx
  lea    (%rdi,%r11,1),%r12
  xorpd  %xmm3,%xmm4
  add    %r8,%r11
  lea    (%rdi,%rax,8),%rbp
  mov    %rbx,%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L1:
  movsd  (%rdi,%rax,8),%xmm0
  mulsd  %xmm4,%xmm0
  subsd  -0x10(%rdx),%xmm0
  addsd  (%rbx,%rax,8),%xmm0
  movsd  %xmm0,(%r8,%rax,8)
  movsd  (%r12,%rax,8),%xmm1
  movsd  -0x10(%rdx,%rcx,8),%xmm0
  sub    $0x8,%rdx
  mulsd  %xmm2,%xmm1
  xorpd  %xmm3,%xmm0
  subsd  %xmm1,%xmm0
  addsd  0x0(%rbp,%rax,8),%xmm0
  movsd  %xmm0,(%r11,%rax,8)
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L1
L23:
  add    %esi,%r10d
  movslq %esi,%rax
  movslq %r10d,%r10
  movsd  (%rdi,%rax,8),%xmm1
  movsd  (%rdi,%r10,8),%xmm0
  subsd  %xmm1,%xmm0
  mulsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%r8,%rax,8)
  mulsd  -0x8(%rdi,%r13,8),%xmm5
  movsd  -0x8(%rdi,%r9,8),%xmm0
  subsd  %xmm5,%xmm0
  movsd  %xmm0,-0x8(%r8,%r13,8)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L10:
  movslq %r10d,%r13
  jmp    L23
```

tool_calls: name: unpack_data
arguments: tag: D2
type_guessed: byte[8]

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

assistant: ```c
void ChebyDivRem(double *dividend, double m0, double *remres, int n) {
  int rlim, dd;
  int tmpfloor;
  double tm0;
  int halfrlim;
  rlim = n / 2;
  tm0 = 2.0 * m0;
  tmpfloor = (rlim - 1) / 2;
  halfrlim = rlim / 2;
  for (dd = 0; dd < tmpfloor; dd++) {
    remres[dd] = -tm0 * dividend[dd] - dividend[rlim - dd - 2] + dividend[rlim + dd];
    remres[dd + halfrlim] = -dividend[halfrlim - 2 - dd] - tm0 * dividend[dd + halfrlim] + dividend[rlim + dd + halfrlim];
  }
  remres[tmpfloor] = dividend[tmpfloor + rlim] - dividend[tmpfloor] - (tm0 * dividend[tmpfloor]);
  remres[rlim - 1] = dividend[n - 1] - (m0 * dividend[rlim - 1]);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ChebyDivRem>:
  endbr64
  movslq %edx,%r9
  mov    %rsi,%r8
  movapd %xmm0,%xmm2
  push   %r13
  mov    %r9d,%r10d
  lea    0x3(%r9),%ecx
  push   %r12
  addsd  %xmm0,%xmm2
  shr    $0x1f,%r10d
  push   %rbp
  movapd %xmm0,%xmm5
  add    %r9d,%r10d
  push   %rbx
  sar    %r10d
  lea    -0x1(%r10),%eax
  mov    %eax,%esi
  shr    $0x1f,%esi
  add    %eax,%esi
  sar    %esi
  test   %r9d,%r9d
  cmovns %r9d,%ecx
  sar    $0x2,%ecx
  cmp    $0x5,%r9d
  jle    L27
  movslq %r10d,%r13
  movslq %ecx,%rcx
  movq   D1(%rip),%xmm3
  movapd %xmm2,%xmm4
  lea    0x0(,%rcx,8),%r11
  lea    (%rcx,%r13,1),%rax
  sub    %r13,%rcx
  lea    (%rdi,%r13,8),%rbx
  lea    (%rdi,%r11,1),%r12
  xorpd  %xmm3,%xmm4
  add    %r8,%r11
  lea    (%rdi,%rax,8),%rbp
  mov    %rbx,%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L9:
  movsd  (%rdi,%rax,8),%xmm0
  mulsd  %xmm4,%xmm0
  subsd  -0x10(%rdx),%xmm0
  addsd  (%rbx,%rax,8),%xmm0
  movsd  %xmm0,(%r8,%rax,8)
  movsd  (%r12,%rax,8),%xmm1
  movsd  -0x10(%rdx,%rcx,8),%xmm0
  sub    $0x8,%rdx
  mulsd  %xmm2,%xmm1
  xorpd  %xmm3,%xmm0
  subsd  %xmm1,%xmm0
  addsd  0x0(%rbp,%rax,8),%xmm0
  movsd  %xmm0,(%r11,%rax,8)
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L9
L10:
  add    %esi,%r10d
  movslq %esi,%rax
  movslq %r10d,%r10
  movsd  (%rdi,%rax,8),%xmm1
  movsd  (%rdi,%r10,8),%xmm0
  subsd  %xmm1,%xmm0
  mulsd  %xmm2,%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%r8,%rax,8)
  mulsd  -0x8(%rdi,%r13,8),%xmm5
  movsd  -0x8(%rdi,%r9,8),%xmm0
  subsd  %xmm5,%xmm0
  movsd  %xmm0,-0x8(%r8,%r13,8)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L27:
  movslq %r10d,%r13
  jmp    L10
```

tool_calls: name: disassemble_data
arguments: label: D1
data_type: byte[8]

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
void ChebyDivRem(double *dividend, double m0, double *remres, int n) {
  int rlim, dd;
  int tmpfloor;
  double tm0;
  int halfrlim;
  rlim = n / 2;
  tm0 = 2.0 * m0;
  tmpfloor = (rlim - 1) / 2;
  halfrlim = rlim / 2;
  for (dd = 0; dd < tmpfloor; dd++) {
    remres[dd] = -tm0 * dividend[dd] - dividend[rlim - dd - 2] + dividend[rlim + dd];
    remres[dd + halfrlim] = -dividend[halfrlim - 2 - dd] - tm0 * dividend[dd + halfrlim] + dividend[rlim + dd + halfrlim];
  }
  remres[tmpfloor] = dividend[tmpfloor + rlim] - dividend[tmpfloor] - (tm0 * dividend[tmpfloor]);
  remres[rlim - 1] = dividend[n - 1] - (m0 * dividend[rlim - 1]);
}
```
user: Translate the below assembly code into C programming language:

```asm
<_cstr_clone>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L25
  movq   $0x0,-0x18(%rbp)
L25:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L13
  mov    $0x0,%eax
  jmp    L9
L13:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
L9:
  leave
  ret
```

assistant: ```c
char *_cstr_clone(char *in_string) {
  if (!in_string)
    in_string = NULL;
  long len = strlen(in_string);
  char *result = malloc(len + 1);
  if (!result)
    return NULL;
  strcpy(result, in_string);
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<_cstr_clone>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L8
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L8:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *_cstr_clone(char *in_string) {
  if (!in_string)
    in_string = NULL;
  long len = strlen(in_string);
  char *result = malloc(len + 1);
  if (!result)
    return NULL;
  strcpy(result, in_string);
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_cstr_clone>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L0
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L0:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *_cstr_clone(char *in_string) {
  if (!in_string)
    in_string = NULL;
  long len = strlen(in_string);
  char *result = malloc(len + 1);
  if (!result)
    return NULL;
  strcpy(result, in_string);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_cstr_clone>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L1
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L1:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *_cstr_clone(char *in_string) {
  if (!in_string)
    in_string = NULL;
  long len = strlen(in_string);
  char *result = malloc(len + 1);
  if (!result)
    return NULL;
  strcpy(result, in_string);
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<File_Exist>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <access@plt>
  test   %rax,%rax
  jne    L1
  mov    $0x1,%eax
  jmp    L18
L1:
  mov    $0x0,%eax
L18:
  leave
  ret
```

assistant: ```c
int File_Exist(const char *FileName) {
  if (access(FileName, F_OK) == 0)
    return 1;
  else
    return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<File_Exist>:
  endbr64
  sub    $0x8,%rsp
  mov    D1(%rip),%rax
  mov    (%rax),%esi
  call   <access@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int File_Exist(const char *FileName) {
  if (access(FileName, F_OK) == 0)
    return 1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<File_Exist>:
  endbr64
  sub    $0x8,%rsp
  mov    D2(%rip),%rax
  mov    (%rax),%esi
  call   <access@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int File_Exist(const char *FileName) {
  if (access(FileName, F_OK) == 0)
    return 1;
  else
    return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<File_Exist>:
  endbr64
  sub    $0x8,%rsp
  mov    D9(%rip),%rax
  mov    (%rax),%esi
  call   <access@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int File_Exist(const char *FileName) {
  if (access(FileName, F_OK) == 0)
    return 1;
  else
    return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x510,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x500(%rbp),%r8
  lea    -0x504(%rbp),%rdi
  lea    -0x4f0(%rbp),%rcx
  lea    -0x330(%rbp),%rdx
  lea    -0x1a0(%rbp),%rsi
  lea    -0x4c0(%rbp),%rax
  mov    %r8,%r9
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <input@plt>
  mov    -0x500(%rbp),%edi
  mov    -0x504(%rbp),%ecx
  lea    -0x4f0(%rbp),%rdx
  lea    -0x1a0(%rbp),%rsi
  lea    -0x4c0(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <banker@plt>
  mov    %eax,-0x4f4(%rbp)
  cmpl   $0x0,-0x4f4(%rbp)
  je     L7
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x4fc(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x4fc(%rbp),%eax
  cmp    $0x1,%eax
  jne    L7
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x4f8(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x500(%rbp),%edi
  mov    -0x4f8(%rbp),%eax
  lea    -0x1(%rax),%ecx
  lea    -0x4f0(%rbp),%rdx
  lea    -0x1a0(%rbp),%rsi
  lea    -0x4c0(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <res_request@plt>
  mov    -0x500(%rbp),%edi
  mov    -0x504(%rbp),%ecx
  lea    -0x4f0(%rbp),%rdx
  lea    -0x1a0(%rbp),%rsi
  lea    -0x4c0(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <banker@plt>
  mov    %eax,-0x4f4(%rbp)
L7:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D37
datatype: string
name: unpack_data
arguments: data_tag: D35
datatype: string
name: unpack_data
arguments: data_tag: D34
datatype: string

tool: D37: 
Make an additional request? (1/0): 

tool: D35: %d

tool: D34: 
Which process: 

assistant: ```c
int main(void) {
  int n;
  int m;
  int allocation[10][10];
  int maximum[10][10];
  int need[10][10];
  int available[10];
  input(allocation, need, maximum, available, &n, &m);
  int res = banker(allocation, need, available, n, m);
  if (res != 0) {
    int choice;
    printf("\nMake an additional request? (1/0): ");
    scanf("%d", &choice);
    if (choice == 1) {
      int pid;
      printf("\nWhich process: ");
      scanf("%d", &pid);
      res_request(allocation, need, available, pid - 1, m);
      res = banker(allocation, need, available, 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   %r12
  push   %rbp
  push   %rbx
  sub    $0x500,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4f8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  lea    0x1d0(%rsp),%rdx
  lea    0x360(%rsp),%rbp
  lea    0x40(%rsp),%rbx
  lea    0x4(%rsp),%r9
  mov    %rsp,%r8
  mov    %r12,%rcx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <input@plt>
  mov    0x4(%rsp),%r8d
  mov    (%rsp),%ecx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <banker@plt>
  test   %eax,%eax
  jne    L18
L28:
  mov    0x4f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  mov    $0x0,%eax
  add    $0x500,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D38(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x1,0x8(%rsp)
  jne    L28
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D38(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%ecx
  mov    0x4(%rsp),%r8d
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <res_request@plt>
  mov    0x4(%rsp),%r8d
  mov    (%rsp),%ecx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <banker@plt>
  jmp    L28
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_guessed: string
name: disassemble_data
arguments: tag: D38
type_guessed: string
name: disassemble_data
arguments: tag: D9
type_guessed: string

tool: D6: 
Make an additional request? (1/0): 

tool: D38: %d

tool: D9: 
Which process: 

assistant: ```c
int main(void) {
  int n;
  int m;
  int allocation[10][10];
  int maximum[10][10];
  int need[10][10];
  int available[10];
  input(allocation, need, maximum, available, &n, &m);
  int res = banker(allocation, need, available, n, m);
  if (res != 0) {
    int choice;
    printf("\nMake an additional request? (1/0): ");
    scanf("%d", &choice);
    if (choice == 1) {
      int pid;
      printf("\nWhich process: ");
      scanf("%d", &pid);
      res_request(allocation, need, available, pid - 1, m);
      res = banker(allocation, need, available, 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   %r14
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x508,%rsp
  lea    0x10(%rsp),%r13
  lea    0x40(%rsp),%rbp
  mov    %rsp,%r8
  lea    0x360(%rsp),%r12
  mov    %r13,%rcx
  lea    0x4(%rsp),%r9
  mov    %rbp,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x4f8(%rsp)
  xor    %eax,%eax
  lea    0x1d0(%rsp),%rdx
  mov    %r12,%rsi
  call   <input@plt>
  mov    0x4(%rsp),%r8d
  mov    (%rsp),%ecx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <banker@plt>
  test   %eax,%eax
  jne    L37
L11:
  mov    0x4f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x508,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L37:
  lea    D3(%rip),%rdi
  lea    D5(%rip),%r14
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  cmpl   $0x1,0x8(%rsp)
  jne    L11
  lea    D6(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    0x4(%rsp),%r8d
  mov    %rbp,%rdi
  lea    -0x1(%rax),%ecx
  call   <res_request@plt>
  mov    0x4(%rsp),%r8d
  mov    (%rsp),%ecx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <banker@plt>
  jmp    L11
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D3
type_infered: string
name: disassemble_data
arguments: label: D5
type_infered: string
name: disassemble_data
arguments: label: D6
type_infered: string

tool: D3: 
Make an additional request? (1/0): 

tool: D5: %d

tool: D6: 
Which process: 

assistant: ```c
int main(void) {
  int n;
  int m;
  int allocation[10][10];
  int maximum[10][10];
  int need[10][10];
  int available[10];
  input(allocation, need, maximum, available, &n, &m);
  int res = banker(allocation, need, available, n, m);
  if (res != 0) {
    int choice;
    printf("\nMake an additional request? (1/0): ");
    scanf("%d", &choice);
    if (choice == 1) {
      int pid;
      printf("\nWhich process: ");
      scanf("%d", &pid);
      res_request(allocation, need, available, pid - 1, m);
      res = banker(allocation, need, available, n, m);
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x508,%rsp
  lea    0x10(%rsp),%r13
  lea    0x40(%rsp),%rbp
  mov    %rsp,%r8
  lea    0x360(%rsp),%r12
  mov    %r13,%rcx
  lea    0x4(%rsp),%r9
  mov    %rbp,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x4f8(%rsp)
  xor    %eax,%eax
  lea    0x1d0(%rsp),%rdx
  mov    %r12,%rsi
  call   <input@plt>
  mov    0x4(%rsp),%r8d
  mov    (%rsp),%ecx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <banker@plt>
  test   %eax,%eax
  jne    L33
L15:
  mov    0x4f8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x508,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L33:
  lea    D8(%rip),%rdi
  lea    D14(%rip),%r14
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  cmpl   $0x1,0x8(%rsp)
  jne    L15
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    0x4(%rsp),%r8d
  mov    %rbp,%rdi
  lea    -0x1(%rax),%ecx
  call   <res_request@plt>
  mov    0x4(%rsp),%r8d
  mov    (%rsp),%ecx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <banker@plt>
  jmp    L15
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D8
type_infered: string
name: parse_data
arguments: tag: D14
type_infered: string
name: parse_data
arguments: tag: D11
type_infered: string

tool: D8: 
Make an additional request? (1/0): 

tool: D14: %d

tool: D11: 
Which process: 

assistant: ```c
int main(void) {
  int n;
  int m;
  int allocation[10][10];
  int maximum[10][10];
  int need[10][10];
  int available[10];
  input(allocation, need, maximum, available, &n, &m);
  int res = banker(allocation, need, available, n, m);
  if (res != 0) {
    int choice;
    printf("\nMake an additional request? (1/0): ");
    scanf("%d", &choice);
    if (choice == 1) {
      int pid;
      printf("\nWhich process: ");
      scanf("%d", &pid);
      res_request(allocation, need, available, pid - 1, m);
      res = banker(allocation, need, available, n, m);
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  jmp    L9
L13:
  movl   $0x0,-0x4(%rbp)
  jmp    L29
L17:
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <doit@plt>
  addl   $0x1,-0x4(%rbp)
L29:
  cmpl   $0x1b57,-0x4(%rbp)
  jle    L17
  addl   $0x1,-0x8(%rbp)
L9:
  cmpl   $0x1b57,-0x8(%rbp)
  jle    L13
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int i, j;
  for (i = 0; i < 7000; i++)
    for (j = 0; j < 7000; j++)
      doit(i, j);
  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
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebp
L11:
  mov    $0x0,%ebx
L10:
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <doit@plt>
  add    $0x1,%ebx
  cmp    $0x1b58,%ebx
  jne    L10
  add    $0x1,%ebp
  cmp    $0x1b58,%ebp
  jne    L11
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int i, j;
  for (i = 0; i < 7000; i++)
    for (j = 0; j < 7000; j++)
      doit(i, j);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x8,%rsp
L2:
  xor    %ebx,%ebx
  xchg   %ax,%ax
L17:
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%ebx
  call   <doit@plt>
  cmp    $0x1b58,%ebx
  jne    L17
  add    $0x1,%ebp
  cmp    $0x1b58,%ebp
  jne    L2
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int i, j;
  for (i = 0; i < 7000; i++)
    for (j = 0; j < 7000; j++)
      doit(i, j);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x8,%rsp
L0:
  xor    %ebx,%ebx
  xchg   %ax,%ax
L9:
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%ebx
  call   <doit@plt>
  cmp    $0x1b58,%ebx
  jne    L9
  add    $0x1,%ebp
  cmp    $0x1b58,%ebp
  jne    L0
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int i, j;
  for (i = 0; i < 7000; i++)
    for (j = 0; j < 7000; j++)
      doit(i, j);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<power>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L13
L10:
  mov    -0x8(%rbp),%rax
  imul   -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L13:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L10
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long power(long number, int power) {
  long tmp = number;
  for (int i = 1; i < power; i++) {
    tmp *= number;
  }
  return tmp;
}
```