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
<binearest_interp>:
  endbr64
  push   %r15
  movd   %xmm2,%r15d
  push   %r14
  movd   %xmm0,%r14d
  movaps %xmm4,%xmm0
  push   %rbp
  movd   %xmm1,%ebp
  push   %rbx
  movd   %xmm3,%ebx
  sub    $0x18,%rsp
  movss  %xmm5,0xc(%rsp)
  call   <roundf@plt>
  movss  0xc(%rsp),%xmm5
  movss  %xmm0,0x8(%rsp)
  movaps %xmm5,%xmm0
  call   <roundf@plt>
  movss  0x8(%rsp),%xmm1
  pxor   %xmm2,%xmm2
  mov    $0x1,%edx
  mov    $0x0,%ecx
  ucomiss %xmm2,%xmm1
  setp   %al
  cmovne %edx,%eax
  ucomiss %xmm2,%xmm0
  setnp  %dl
  cmovne %ecx,%edx
  test   %al,%al
  je     L47
  movd   %ebx,%xmm3
  test   %dl,%dl
  je     L47
L20:
  add    $0x18,%rsp
  movaps %xmm3,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L47:
  ucomiss %xmm2,%xmm1
  mov    $0x0,%esi
  mov    $0x1,%edi
  setnp  %cl
  cmovne %esi,%ecx
  ucomiss %xmm2,%xmm0
  setp   %sil
  cmovne %edi,%esi
  test   %cl,%cl
  je     L61
  movd   %r14d,%xmm3
  test   %sil,%sil
  jne    L20
L61:
  test   %dl,%dl
  je     L42
  movd   %r15d,%xmm3
  test   %cl,%cl
  jne    L20
L42:
  test   %al,%al
  je     L12
  movd   %ebp,%xmm3
  test   %sil,%sil
  jne    L20
L12:
  add    $0x18,%rsp
  movaps %xmm2,%xmm3
  pop    %rbx
  movaps %xmm3,%xmm0
  pop    %rbp
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
float binearest_interp(float tl, float tr, float bl, float br, float x_amount, float y_amount) {
  x_amount = roundf(x_amount);
  y_amount = roundf(y_amount);
  if (x_amount && !y_amount) {
    return br;
  }
  if (!x_amount && y_amount) {
    return tl;
  }
  if (!x_amount && !y_amount) {
    return bl;
  }
  if (x_amount && y_amount) {
    return tr;
  }
  return 0.0f;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    D14(%rip),%r12
  lea    0x4(%rsp),%rbp
  lea    D0(%rip),%rbx
L0:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r9d
  test   %r9d,%r9d
  jle    L0
  movslq %r9d,%rsi
  imul   $0x66666667,%rsi,%rax
  mov    %rax,%rbx
  sar    $0x21,%rbx
  mov    %rbx,%rcx
  mov    %r9d,%edx
  sar    $0x1f,%edx
  sub    %edx,%ecx
  lea    (%rcx,%rcx,4),%edi
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %r9d,%eax
  sub    %ecx,%eax
  movslq %eax,%r8
  imul   $0x66666667,%r8,%r8
  sar    $0x21,%r8
  sar    $0x1f,%eax
  imul   $0x51eb851f,%rsi,%rsi
  mov    %rsi,%rbx
  sar    $0x24,%rbx
  mov    %rbx,%rcx
  sub    %edx,%ecx
  imul   $0x32,%ecx,%ecx
  mov    %r9d,%r10d
  sub    %ecx,%r10d
  movslq %r10d,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  sar    $0x1f,%r10d
  sub    %r10d,%ecx
  sar    $0x25,%rsi
  sub    %edx,%esi
  imul   $0x64,%esi,%edx
  mov    %r9d,%r10d
  sub    %edx,%r10d
  movslq %r10d,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x24,%rdx
  sar    $0x1f,%r10d
  sub    %r10d,%edx
  sub    %edi,%r9d
  sub    %eax,%r8d
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D14
data_type: string
name: retrieve_data
arguments: tag: D0
data_type: string
name: retrieve_data
arguments: tag: D2
data_type: string

tool: D14: Amount (R$): 

tool: D0: %d

tool: D2: R$ 100: %d
 R$ 50: %d
 R$ 10: %d
  R$ 5: %d
  R$ 1: %d


assistant: ```c
int main() {
  int amount = 0;
  while (amount <= 0) {
    printf("Amount (R$): ");
    scanf("%d", &amount);
  }
  int n100, n50, n10, n5, n1;
  n100 = amount / 100;
  n50 = amount % 100 / 50;
  n10 = amount % 50 / 10;
  n5 = amount % 10 / 5;
  n1 = amount % 5;
  printf("R$ 100: %d\n R$ 50: %d\n R$ 10: %d\n  R$ 5: %d\n  R$ 1: %d\n", n100, n50, n10, n5, n1);
  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   %r12
  lea    D9(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D14(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rbp
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r9d
  test   %r9d,%r9d
  jle    L7
  movslq %r9d,%rsi
  mov    %r9d,%edx
  mov    %r9d,%r8d
  mov    $0xcccccccd,%r10d
  imul   $0x66666667,%rsi,%rax
  sar    $0x1f,%edx
  imul   $0x51eb851f,%rsi,%rsi
  mov    %rax,%rbx
  sar    $0x22,%rax
  sub    %edx,%eax
  sar    $0x21,%rbx
  lea    (%rax,%rax,4),%eax
  mov    %rbx,%rcx
  add    %eax,%eax
  sub    %edx,%ecx
  sub    %eax,%r8d
  mov    %rsi,%rax
  sar    $0x25,%rsi
  lea    (%rcx,%rcx,4),%edi
  sar    $0x24,%rax
  imul   %r10,%r8
  sub    %edx,%esi
  mov    %r9d,%ecx
  sub    %edx,%eax
  mov    %r9d,%edx
  sub    %edi,%r9d
  imul   $0x32,%eax,%eax
  lea    D24(%rip),%rdi
  shr    $0x22,%r8
  sub    %eax,%ecx
  imul   $0x64,%esi,%eax
  imul   %r10,%rcx
  sub    %eax,%edx
  xor    %eax,%eax
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x23,%rcx
  shr    $0x24,%rdx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
type_guessed: string
name: disassemble_data
arguments: label: D14
type_guessed: string
name: disassemble_data
arguments: label: D24
type_guessed: string

tool: D9: Amount (R$): 

tool: D14: %d

tool: D24: R$ 100: %d
 R$ 50: %d
 R$ 10: %d
  R$ 5: %d
  R$ 1: %d


assistant: ```c
int main() {
  int amount = 0;
  while (amount <= 0) {
    printf("Amount (R$): ");
    scanf("%d", &amount);
  }
  int n100, n50, n10, n5, n1;
  n100 = amount / 100;
  n50 = amount % 100 / 50;
  n10 = amount % 50 / 10;
  n5 = amount % 10 / 5;
  n1 = amount % 5;
  printf("R$ 100: %d\n R$ 50: %d\n R$ 10: %d\n  R$ 5: %d\n  R$ 1: %d\n", n100, n50, n10, n5, n1);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D21(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rbp
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r9d
  test   %r9d,%r9d
  jle    L18
  movslq %r9d,%rsi
  mov    %r9d,%edx
  mov    %r9d,%r8d
  mov    $0xcccccccd,%r10d
  imul   $0x66666667,%rsi,%rax
  sar    $0x1f,%edx
  imul   $0x51eb851f,%rsi,%rsi
  mov    %rax,%rbx
  sar    $0x22,%rax
  sub    %edx,%eax
  sar    $0x21,%rbx
  lea    (%rax,%rax,4),%eax
  mov    %rbx,%rcx
  add    %eax,%eax
  sub    %edx,%ecx
  sub    %eax,%r8d
  mov    %rsi,%rax
  sar    $0x25,%rsi
  lea    (%rcx,%rcx,4),%edi
  sar    $0x24,%rax
  imul   %r10,%r8
  sub    %edx,%esi
  mov    %r9d,%ecx
  sub    %edx,%eax
  mov    %r9d,%edx
  sub    %edi,%r9d
  imul   $0x32,%eax,%eax
  lea    D11(%rip),%rdi
  shr    $0x22,%r8
  sub    %eax,%ecx
  imul   $0x64,%esi,%eax
  imul   %r10,%rcx
  sub    %eax,%edx
  xor    %eax,%eax
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x23,%rcx
  shr    $0x24,%rdx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D4
type_infered: string
name: unpack_data
arguments: data_label: D21
type_infered: string
name: unpack_data
arguments: data_label: D11
type_infered: string

tool: D4: Amount (R$): 

tool: D21: %d

tool: D11: R$ 100: %d
 R$ 50: %d
 R$ 10: %d
  R$ 5: %d
  R$ 1: %d


assistant: ```c
int main() {
  int amount = 0;
  while (amount <= 0) {
    printf("Amount (R$): ");
    scanf("%d", &amount);
  }
  int n100, n50, n10, n5, n1;
  n100 = amount / 100;
  n50 = amount % 100 / 50;
  n10 = amount % 50 / 10;
  n5 = amount % 10 / 5;
  n1 = amount % 5;
  printf("R$ 100: %d\n R$ 50: %d\n R$ 10: %d\n  R$ 5: %d\n  R$ 1: %d\n", n100, n50, n10, n5, n1);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<REVERSE>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  lea    D7(%rip),%rax
  movzbl (%rax),%eax
  cmp    $0x1,%al
  je     L31
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L24
L30:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x1c(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x8(%rbp)
L24:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L30
  jmp    L29
L31:
  nop
L29:
  pop    %rbp
  ret
```

assistant: ```c
void REVERSE(char *intp, int l) {
  int i, j;
  char save;
  static int one = 1;
  if (((unsigned char *)&one)[sizeof(one) - 1] == 1)
    return;
  j = l / 2;
  for (i = 0; i < j; i++) {
    save = intp[i];
    intp[i] = intp[l - i - 1];
    intp[l - i - 1] = save;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<REVERSE>:
  endbr64
  mov    %esi,%r8d
  shr    $0x1f,%r8d
  add    %esi,%r8d
  sar    %r8d
  cmp    $0x1,%esi
  jle    L7
  movslq %esi,%rsi
  lea    -0x1(%rdi,%rsi,1),%rdx
  mov    $0x0,%eax
L14:
  movzbl (%rdi,%rax,1),%ecx
  movzbl (%rdx),%esi
  mov    %sil,(%rdi,%rax,1)
  mov    %cl,(%rdx)
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%r8d
  jg     L14
L7:
  ret
```

assistant: ```c
void REVERSE(char *intp, int l) {
  int i, j;
  char save;
  static int one = 1;
  if (((unsigned char *)&one)[sizeof(one) - 1] == 1)
    return;
  j = l / 2;
  for (i = 0; i < j; i++) {
    save = intp[i];
    intp[i] = intp[l - i - 1];
    intp[l - i - 1] = save;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<REVERSE>:
  endbr64
  mov    %esi,%r8d
  shr    $0x1f,%r8d
  add    %esi,%r8d
  sar    %r8d
  cmp    $0x1,%esi
  jle    L8
  movslq %esi,%rsi
  xor    %eax,%eax
  lea    -0x1(%rdi,%rsi,1),%rdx
L5:
  movzbl (%rdx),%esi
  movzbl (%rdi,%rax,1),%ecx
  sub    $0x1,%rdx
  mov    %sil,(%rdi,%rax,1)
  add    $0x1,%rax
  mov    %cl,0x1(%rdx)
  cmp    %eax,%r8d
  jg     L5
L8:
  ret
```

assistant: ```c
void REVERSE(char *intp, int l) {
  int i, j;
  char save;
  static int one = 1;
  if (((unsigned char *)&one)[sizeof(one) - 1] == 1)
    return;
  j = l / 2;
  for (i = 0; i < j; i++) {
    save = intp[i];
    intp[i] = intp[l - i - 1];
    intp[l - i - 1] = save;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<REVERSE>:
  endbr64
  mov    %esi,%r8d
  shr    $0x1f,%r8d
  add    %esi,%r8d
  sar    %r8d
  cmp    $0x1,%esi
  jle    L15
  movslq %esi,%rsi
  xor    %eax,%eax
  lea    -0x1(%rdi,%rsi,1),%rdx
L10:
  movzbl (%rdx),%esi
  movzbl (%rdi,%rax,1),%ecx
  sub    $0x1,%rdx
  mov    %sil,(%rdi,%rax,1)
  add    $0x1,%rax
  mov    %cl,0x1(%rdx)
  cmp    %eax,%r8d
  jg     L10
L15:
  ret
```

assistant: ```c
void REVERSE(char *intp, int l) {
  int i, j;
  char save;
  static int one = 1;
  if (((unsigned char *)&one)[sizeof(one) - 1] == 1)
    return;
  j = l / 2;
  for (i = 0; i < j; i++) {
    save = intp[i];
    intp[i] = intp[l - i - 1];
    intp[l - i - 1] = save;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L16
L19:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x31,%al
  jne    L23
  mov    -0x8(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  add    %eax,-0xc(%rbp)
L23:
  addl   $0x1,-0x8(%rbp)
L16:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L19
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getint(char str[]) {
  int len = strlen(str);
  int res = 0;
  for (int i = 0; i < len; i++) {
    if (str[i] == '1') {
      res = res + (1 << i);
    }
  }
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getint>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L16
  mov    %eax,%eax
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x1,%edi
  jmp    L46
L49:
  add    $0x1,%rcx
  cmp    %rax,%rcx
  je     L28
L46:
  cmpb   $0x31,(%rbx,%rcx,1)
  jne    L49
  mov    %edi,%esi
  shl    %cl,%esi
  add    %esi,%edx
  jmp    L49
L16:
  mov    $0x0,%edx
L28:
  mov    %edx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int getint(char str[]) {
  int len = strlen(str);
  int res = 0;
  for (int i = 0; i < len; i++) {
    if (str[i] == '1') {
      res = res + (1 << i);
    }
  }
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getint>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L26
  cltq
  xor    %ecx,%ecx
  xor    %r8d,%r8d
  mov    $0x1,%esi
  nopl   (%rax)
L27:
  cmpb   $0x31,(%rbx,%rcx,1)
  jne    L6
  mov    %esi,%edx
  shl    %cl,%edx
  add    %edx,%r8d
L6:
  add    $0x1,%rcx
  cmp    %rax,%rcx
  jne    L27
  mov    %r8d,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
```

assistant: ```c
int getint(char str[]) {
  int len = strlen(str);
  int res = 0;
  for (int i = 0; i < len; i++) {
    if (str[i] == '1') {
      res = res + (1 << i);
    }
  }
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<getint>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L10
  cltq
  xor    %ecx,%ecx
  xor    %r8d,%r8d
  mov    $0x1,%esi
  nopl   (%rax)
L18:
  cmpb   $0x31,(%rbx,%rcx,1)
  jne    L0
  mov    %esi,%edx
  shl    %cl,%edx
  add    %edx,%r8d
L0:
  add    $0x1,%rcx
  cmp    %rcx,%rax
  jne    L18
  mov    %r8d,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  ret
```

assistant: ```c
int getint(char str[]) {
  int len = strlen(str);
  int res = 0;
  for (int i = 0; i < len; i++) {
    if (str[i] == '1') {
      res = res + (1 << i);
    }
  }
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<save_file>:
  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
  lea    D1(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L0
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L0:
  mov    -0x8(%rbp),%rcx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fwrite@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D1
data_type: string

tool: D1: wb

assistant: ```c
void save_file(const char *filename, const uint8_t *buf, int buf_len) {
  FILE *f;
  f = fopen(filename, "wb");
  if (!f) {
    perror(filename);
    exit(1);
  }
  fwrite(buf, 1, buf_len, f);
  fclose(f);
}
```
user: What is the c source code of the assembly code below:

```asm
<save_file>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %edx,%r13d
  lea    D0(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L2
L15:
  mov    %rbx,%rcx
  mov    %r13d,%edx
  mov    $0x1,%esi
  mov    %r12,%rdi
  call   <fwrite@plt>
  mov    %rbx,%rdi
  call   <fclose@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  mov    %rbp,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L15
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
type_guessed: string

tool: D0: wb

assistant: ```c
void save_file(const char *filename, const uint8_t *buf, int buf_len) {
  FILE *f;
  f = fopen(filename, "wb");
  if (!f) {
    perror(filename);
    exit(1);
  }
  fwrite(buf, 1, buf_len, f);
  fclose(f);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<save_file>:
  endbr64
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rsi,%r13
  lea    D1(%rip),%rsi
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  sub    $0x8,%rsp
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L1
L9:
  mov    %r13,%rdi
  mov    %rbp,%rcx
  mov    %r14d,%edx
  mov    $0x1,%esi
  call   <fwrite@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <fclose@plt>
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %r12,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L9
```

tool_calls: name: parse_data
arguments: data_label: D1
type_infered: string

tool: D1: wb

assistant: ```c
void save_file(const char *filename, const uint8_t *buf, int buf_len) {
  FILE *f;
  f = fopen(filename, "wb");
  if (!f) {
    perror(filename);
    exit(1);
  }
  fwrite(buf, 1, buf_len, f);
  fclose(f);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<save_file>:
  endbr64
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rsi,%r13
  lea    D9(%rip),%rsi
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  sub    $0x8,%rsp
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L8
L10:
  mov    %r13,%rdi
  mov    %rbp,%rcx
  mov    %r14d,%edx
  mov    $0x1,%esi
  call   <fwrite@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <fclose@plt>
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %r12,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L10
```

tool_calls: name: disassemble_data
arguments: label: D9
data_type: string

tool: D9: wb

assistant: ```c
void save_file(const char *filename, const uint8_t *buf, int buf_len) {
  FILE *f;
  f = fopen(filename, "wb");
  if (!f) {
    perror(filename);
    exit(1);
  }
  fwrite(buf, 1, buf_len, f);
  fclose(f);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<check_cpuset>:
  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
  mov    $0x0,%eax
  call   <hwloc_bitmap_alloc@plt>
  mov    %eax,-0x14(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    -0x24(%rbp),%edx
  lea    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <hwloc_bitmap_asprintf@plt>
  cmpq   $0x0,-0x30(%rbp)
  je     L11
  mov    -0x10(%rbp),%rax
  mov    -0x30(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L11
  mov    $0x0,%edi
  call   <assert@plt>
L11:
  mov    -0x10(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <hwloc_bitmap_sscanf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <hwloc_bitmap_isequal@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x24(%rbp),%edx
  lea    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <hwloc_bitmap_list_asprintf@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <hwloc_bitmap_list_sscanf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <hwloc_bitmap_isequal@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x24(%rbp),%edx
  lea    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <hwloc_bitmap_taskset_asprintf@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <hwloc_bitmap_taskset_sscanf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <hwloc_bitmap_isequal@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <hwloc_bitmap_free@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void check_cpuset(hwloc_bitmap_t set, const char *expected) {
  hwloc_bitmap_t set2 = hwloc_bitmap_alloc();
  char *string = NULL;
  hwloc_bitmap_asprintf(&string, set);
  if (expected && strcmp(string, expected))
    assert(0);
  hwloc_bitmap_sscanf(set2, string);
  free(string);
  assert(hwloc_bitmap_isequal(set, set2));
  hwloc_bitmap_list_asprintf(&string, set);
  hwloc_bitmap_list_sscanf(set2, string);
  free(string);
  assert(hwloc_bitmap_isequal(set, set2));
  hwloc_bitmap_taskset_asprintf(&string, set);
  hwloc_bitmap_taskset_sscanf(set2, string);
  free(string);
  assert(hwloc_bitmap_isequal(set, set2));
  hwloc_bitmap_free(set2);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<check_cpuset>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <hwloc_bitmap_alloc@plt>
  mov    %eax,%ebx
  movq   $0x0,(%rsp)
  mov    %rsp,%rdi
  mov    %ebp,%esi
  call   <hwloc_bitmap_asprintf@plt>
  test   %r12,%r12
  je     L5
  mov    %r12,%rsi
  mov    (%rsp),%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L6
L5:
  mov    (%rsp),%rsi
  mov    %ebx,%edi
  call   <hwloc_bitmap_sscanf@plt>
  mov    (%rsp),%rdi
  call   <free@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <hwloc_bitmap_isequal@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    %rsp,%r12
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <hwloc_bitmap_list_asprintf@plt>
  mov    (%rsp),%rsi
  mov    %ebx,%edi
  call   <hwloc_bitmap_list_sscanf@plt>
  mov    (%rsp),%rdi
  call   <free@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <hwloc_bitmap_isequal@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <hwloc_bitmap_taskset_asprintf@plt>
  mov    (%rsp),%rsi
  mov    %ebx,%edi
  call   <hwloc_bitmap_taskset_sscanf@plt>
  mov    (%rsp),%rdi
  call   <free@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <hwloc_bitmap_isequal@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    %ebx,%edi
  call   <hwloc_bitmap_free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  mov    $0x0,%edi
  call   <assert@plt>
  jmp    L5
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void check_cpuset(hwloc_bitmap_t set, const char *expected) {
  hwloc_bitmap_t set2 = hwloc_bitmap_alloc();
  char *string = NULL;
  hwloc_bitmap_asprintf(&string, set);
  if (expected && strcmp(string, expected))
    assert(0);
  hwloc_bitmap_sscanf(set2, string);
  free(string);
  assert(hwloc_bitmap_isequal(set, set2));
  hwloc_bitmap_list_asprintf(&string, set);
  hwloc_bitmap_list_sscanf(set2, string);
  free(string);
  assert(hwloc_bitmap_isequal(set, set2));
  hwloc_bitmap_taskset_asprintf(&string, set);
  hwloc_bitmap_taskset_sscanf(set2, string);
  free(string);
  assert(hwloc_bitmap_isequal(set, set2));
  hwloc_bitmap_free(set2);
}
```