ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<__put_unaligned_le32>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  lea    0x8(%rsi),%rsi
  push   %rbp
  mov    %edi,%ebp
  sar    $0x10,%edi
  sub    $0x8,%rsp
  call   <__put_unaligned_le16@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1050 <__put_unaligned_le16@plt>
```

assistant: ```c
void __put_unaligned_le32(u_int32_t val, u_char *p) {
  __put_unaligned_le16(val >> 16, p + 2);
  __put_unaligned_le16(val, p);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<csr_fd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D7(%rip),%rax
  pop    %rbp
  ret
```

assistant: ```c
int *csr_fd(void) {
  static int csrsafe = 0;
  return &csrsafe;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<csr_fd>:
  endbr64
  lea    D0(%rip),%rax
  ret
```

assistant: ```c
int *csr_fd(void) {
  static int csrsafe = 0;
  return &csrsafe;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<csr_fd>:
  endbr64
  lea    D4(%rip),%rax
  ret
```

assistant: ```c
int *csr_fd(void) {
  static int csrsafe = 0;
  return &csrsafe;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<csr_fd>:
  endbr64
  lea    D0(%rip),%rax
  ret
```

assistant: ```c
int *csr_fd(void) {
  static int csrsafe = 0;
  return &csrsafe;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x1,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  je     L161
L15:
  mov    $0x1,%eax
L128:
  add    $0x8,%rsp
  ret
L161:
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  xor    %esi,%esi
  mov    $0x3,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  mov    $0x2,%esi
  mov    $0x4,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  xor    %esi,%esi
  mov    $0x5,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  mov    $0x1,%esi
  mov    $0xa,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  mov    $0x1c,%esi
  mov    $0x10000000,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  mov    $0x1d,%esi
  mov    $0x20000000,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  mov    $0x1d,%esi
  mov    $0x60000000,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  mov    $0x1f,%esi
  mov    $0x80000000,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  xor    %edi,%edi
  mov    $0x28,%esi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  xor    %edi,%edi
  mov    $0x2b,%esi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  xor    %edi,%edi
  mov    $0x39,%esi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  xor    %edi,%edi
  mov    $0x3f,%esi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  xor    %esi,%esi
  mov    $0xa0000000,%edi
  call   <make_ti@plt>
  mov    $0x5d,%esi
  mov    %eax,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  xor    %esi,%esi
  xor    %edi,%edi
  call   <make_ti@plt>
  mov    $0x7c,%esi
  mov    %eax,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L15
  xor    %esi,%esi
  xor    %edi,%edi
  call   <make_ti@plt>
  mov    $0x7f,%esi
  mov    %eax,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L128
```

assistant: ```c
int main() {
  if (test__ctzti2(0x00000001, 0))
    return 1;
  if (test__ctzti2(0x00000002, 1))
    return 1;
  if (test__ctzti2(0x00000003, 0))
    return 1;
  if (test__ctzti2(0x00000004, 2))
    return 1;
  if (test__ctzti2(0x00000005, 0))
    return 1;
  if (test__ctzti2(0x0000000A, 1))
    return 1;
  if (test__ctzti2(0x10000000, 28))
    return 1;
  if (test__ctzti2(0x20000000, 29))
    return 1;
  if (test__ctzti2(0x60000000, 29))
    return 1;
  if (test__ctzti2(0x80000000uLL, 31))
    return 1;
  if (test__ctzti2(0x0000050000000000uLL, 40))
    return 1;
  if (test__ctzti2(0x0200080000000000uLL, 43))
    return 1;
  if (test__ctzti2(0x7200000000000000uLL, 57))
    return 1;
  if (test__ctzti2(0x8000000000000000uLL, 63))
    return 1;
  if (test__ctzti2(make_ti(0x00000000A0000000LL, 0x0000000000000000LL), 93))
    return 1;
  if (test__ctzti2(make_ti(0xF000000000000000LL, 0x0000000000000000LL), 124))
    return 1;
  if (test__ctzti2(make_ti(0x8000000000000000LL, 0x0000000000000000LL), 127))
    return 1;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x1,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  je     L127
L5:
  mov    $0x1,%eax
L144:
  add    $0x8,%rsp
  ret
L127:
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  xor    %esi,%esi
  mov    $0x3,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  mov    $0x2,%esi
  mov    $0x4,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  xor    %esi,%esi
  mov    $0x5,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  mov    $0x1,%esi
  mov    $0xa,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  mov    $0x1c,%esi
  mov    $0x10000000,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  mov    $0x1d,%esi
  mov    $0x20000000,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  mov    $0x1d,%esi
  mov    $0x60000000,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  mov    $0x1f,%esi
  mov    $0x80000000,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  xor    %edi,%edi
  mov    $0x28,%esi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  xor    %edi,%edi
  mov    $0x2b,%esi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  xor    %edi,%edi
  mov    $0x39,%esi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  xor    %edi,%edi
  mov    $0x3f,%esi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  xor    %esi,%esi
  mov    $0xa0000000,%edi
  call   <make_ti@plt>
  mov    $0x5d,%esi
  mov    %eax,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  xor    %esi,%esi
  xor    %edi,%edi
  call   <make_ti@plt>
  mov    $0x7c,%esi
  mov    %eax,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  jne    L5
  xor    %esi,%esi
  xor    %edi,%edi
  call   <make_ti@plt>
  mov    $0x7f,%esi
  mov    %eax,%edi
  call   <test__ctzti2@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L144
```

assistant: ```c
int main() {
  if (test__ctzti2(0x00000001, 0))
    return 1;
  if (test__ctzti2(0x00000002, 1))
    return 1;
  if (test__ctzti2(0x00000003, 0))
    return 1;
  if (test__ctzti2(0x00000004, 2))
    return 1;
  if (test__ctzti2(0x00000005, 0))
    return 1;
  if (test__ctzti2(0x0000000A, 1))
    return 1;
  if (test__ctzti2(0x10000000, 28))
    return 1;
  if (test__ctzti2(0x20000000, 29))
    return 1;
  if (test__ctzti2(0x60000000, 29))
    return 1;
  if (test__ctzti2(0x80000000uLL, 31))
    return 1;
  if (test__ctzti2(0x0000050000000000uLL, 40))
    return 1;
  if (test__ctzti2(0x0200080000000000uLL, 43))
    return 1;
  if (test__ctzti2(0x7200000000000000uLL, 57))
    return 1;
  if (test__ctzti2(0x8000000000000000uLL, 63))
    return 1;
  if (test__ctzti2(make_ti(0x00000000A0000000LL, 0x0000000000000000LL), 93))
    return 1;
  if (test__ctzti2(make_ti(0xF000000000000000LL, 0x0000000000000000LL), 124))
    return 1;
  if (test__ctzti2(make_ti(0x8000000000000000LL, 0x0000000000000000LL), 127))
    return 1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

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

tool: D5: cl_dx9_media_sharing_intel.h standalone test PASSED.


assistant: ```c
int main(void) {
  printf("cl_dx9_media_sharing_intel.h standalone test PASSED.\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type: string

tool: D0: cl_dx9_media_sharing_intel.h standalone test PASSED.


assistant: ```c
int main(void) {
  printf("cl_dx9_media_sharing_intel.h standalone test PASSED.\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D0: cl_dx9_media_sharing_intel.h standalone test PASSED.


assistant: ```c
int main(void) {
  printf("cl_dx9_media_sharing_intel.h standalone test PASSED.\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D5: cl_dx9_media_sharing_intel.h standalone test PASSED.


assistant: ```c
int main(void) {
  printf("cl_dx9_media_sharing_intel.h standalone test PASSED.\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<addNumbers>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L2
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <addNumbers>
  mov    -0x4(%rbp),%edx
  add    %edx,%eax
  jmp    L8
L2:
  mov    -0x4(%rbp),%eax
L8:
  leave
  ret
```

assistant: ```c
int addNumbers(int n) {
  if (n != 0)
    return n + addNumbers(n - 1);
  else
    return n;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<addNumbers>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  test   %edi,%edi
  jne    L12
L5:
  mov    %ebx,%eax
  pop    %rbx
  ret
L12:
  lea    -0x1(%rdi),%edi
  call   <addNumbers>
  add    %eax,%ebx
  jmp    L5
```

assistant: ```c
int addNumbers(int n) {
  if (n != 0)
    return n + addNumbers(n - 1);
  else
    return n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<addNumbers>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  je     L9
  nopw   0x0(%rax,%rax,1)
L7:
  add    %edi,%eax
  sub    $0x1,%edi
  jne    L7
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  ret
```

assistant: ```c
int addNumbers(int n) {
  if (n != 0)
    return n + addNumbers(n - 1);
  else
    return n;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<addNumbers>:
  endbr64
  test   %edi,%edi
  je     L121
  lea    -0x1(%rdi),%edx
  mov    %edi,%ecx
  cmp    $0x9,%edx
  jbe    L94
  movd   %edi,%xmm4
  mov    %edi,%edx
  movdqa D18(%rip),%xmm3
  xor    %eax,%eax
  pshufd $0x0,%xmm4,%xmm1
  paddd  D17(%rip),%xmm1
  shr    $0x2,%edx
  pxor   %xmm0,%xmm0
  xchg   %ax,%ax
L115:
  movdqa %xmm1,%xmm2
  add    $0x1,%eax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %eax,%edx
  jne    L115
  movdqa %xmm0,%xmm1
  mov    %edi,%edx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%edx
  paddd  %xmm1,%xmm0
  sub    %edx,%edi
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %edx,%ecx
  je     L114
  lea    -0x1(%rdi),%edx
L9:
  add    %edi,%eax
  test   %edx,%edx
  je     L114
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x2,%edx
  je     L114
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x3,%edx
  je     L114
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x4,%edx
  je     L114
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x5,%edx
  je     L114
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x6,%edx
  je     L114
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x7,%edx
  je     L114
  add    %edx,%eax
  mov    %edi,%edx
  sub    $0x8,%edx
  je     L114
  add    %eax,%edx
  lea    -0x9(%rdi,%rdx,1),%eax
  ret
  nopl   0x0(%rax,%rax,1)
L121:
  xor    %eax,%eax
L114:
  ret
L94:
  xor    %eax,%eax
  jmp    L9
```

tool_calls: name: read_data
arguments: label: D18
datatype: byte[16]
name: read_data
arguments: label: D17
datatype: byte[16]

tool: {"D18": [0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff]}

tool: {"D17": [0x00,0x00,0x00,0x00,0xff,0xff,0xff,0xff,0xfe,0xff,0xff,0xff,0xfd,0xff,0xff,0xff]}

assistant: ```c
int addNumbers(int n) {
  if (n != 0)
    return n + addNumbers(n - 1);
  else
    return n;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<tail>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L23
L20:
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x30(%rbp)
  mov    %rdx,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L14
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x30(%rbp)
  mov    %rdx,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
L14:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    $0x0,%edx
  divq   -0x30(%rbp)
  mov    %rdx,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,(%rdx)
L23:
  mov    -0x38(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <readline@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L20
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  divq   -0x30(%rbp)
  mov    %rdx,%rax
  mov    %eax,%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
char **tail(FILE *fp, size_t n, size_t buffsize, int *pos) {
  char **lines;
  size_t current = 0;
  lines = (char **)calloc(n, sizeof(char *));
  char *newline;
  while ((newline = readline(fp, buffsize)) != NULL) {
    if (lines[current % n] != NULL) {
      free(lines[current % n]);
    }
    lines[current++ % n] = newline;
  }
  *pos = current % n;
  return lines;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tail>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  mov    %rsi,%r13
  mov    %rdx,(%rsp)
  mov    %rcx,0x8(%rsp)
  mov    $0x8,%esi
  mov    %r13,%rdi
  call   <calloc@plt>
  mov    %rax,%r14
  mov    $0x0,%ebp
  jmp    L24
L12:
  add    $0x1,%rbp
  mov    %rbx,(%r12)
L24:
  mov    (%rsp),%rsi
  mov    %r15,%rdi
  call   <readline@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L4
  mov    %rbp,%rax
  mov    $0x0,%edx
  div    %r13
  lea    (%r14,%rdx,8),%r12
  mov    (%r12),%rdi
  test   %rdi,%rdi
  je     L12
  call   <free@plt>
  jmp    L12
L4:
  mov    %rbp,%rax
  mov    $0x0,%edx
  div    %r13
  mov    0x8(%rsp),%rax
  mov    %edx,(%rax)
  mov    %r14,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char **tail(FILE *fp, size_t n, size_t buffsize, int *pos) {
  char **lines;
  size_t current = 0;
  lines = (char **)calloc(n, sizeof(char *));
  char *newline;
  while ((newline = readline(fp, buffsize)) != NULL) {
    if (lines[current % n] != NULL) {
      free(lines[current % n]);
    }
    lines[current++ % n] = newline;
  }
  *pos = current % n;
  return lines;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tail>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x8,%esi
  push   %rbx
  mov    %rbp,%rdi
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %rcx,0x8(%rsp)
  call   <calloc@plt>
  mov    %rax,%r12
  jmp    L5
  nopl   0x0(%rax)
L23:
  lea    (%r12,%rdx,8),%rdx
  mov    (%rdx),%rdi
  test   %rdi,%rdi
  je     L13
  mov    %rdx,(%rsp)
  call   <free@plt>
  mov    (%rsp),%rdx
L13:
  mov    %r15,(%rdx)
  add    $0x1,%rbx
L5:
  mov    %r13,%rsi
  mov    %r14,%rdi
  call   <readline@plt>
  xor    %edx,%edx
  mov    %rax,%r15
  mov    %rbx,%rax
  div    %rbp
  test   %r15,%r15
  jne    L23
  mov    0x8(%rsp),%rax
  mov    %edx,(%rax)
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char **tail(FILE *fp, size_t n, size_t buffsize, int *pos) {
  char **lines;
  size_t current = 0;
  lines = (char **)calloc(n, sizeof(char *));
  char *newline;
  while ((newline = readline(fp, buffsize)) != NULL) {
    if (lines[current % n] != NULL) {
      free(lines[current % n]);
    }
    lines[current++ % n] = newline;
  }
  *pos = current % n;
  return lines;
}
```
user: What is the c source code of the assembly code below?

```asm
<tail>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x8,%esi
  push   %rbx
  mov    %rbp,%rdi
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %rcx,0x8(%rsp)
  call   <calloc@plt>
  mov    %rax,%r12
  jmp    L17
  nopl   0x0(%rax)
L26:
  lea    (%r12,%rdx,8),%rdx
  mov    (%rdx),%rdi
  test   %rdi,%rdi
  je     L23
  mov    %rdx,(%rsp)
  call   <free@plt>
  mov    (%rsp),%rdx
L23:
  mov    %r15,(%rdx)
  add    $0x1,%rbx
L17:
  mov    %r13,%rsi
  mov    %r14,%rdi
  call   <readline@plt>
  xor    %edx,%edx
  mov    %rax,%r15
  mov    %rbx,%rax
  div    %rbp
  test   %r15,%r15
  jne    L26
  mov    0x8(%rsp),%rax
  mov    %edx,(%rax)
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char **tail(FILE *fp, size_t n, size_t buffsize, int *pos) {
  char **lines;
  size_t current = 0;
  lines = (char **)calloc(n, sizeof(char *));
  char *newline;
  while ((newline = readline(fp, buffsize)) != NULL) {
    if (lines[current % n] != NULL) {
      free(lines[current % n]);
    }
    lines[current++ % n] = newline;
  }
  *pos = current % n;
  return lines;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<_tc_vsnprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    %r9d,-0x20(%rbp)
  mov    -0x20(%rbp),%ecx
  mov    -0x38(%rbp),%rdx
  mov    -0x30(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  mov    %eax,-0x4(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%edi
  mov    -0x4(%rbp),%ecx
  mov    -0x30(%rbp),%rdx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <tc_test_string@plt>
  test   %rax,%rax
  jne    L19
  mov    -0x4(%rbp),%eax
  jmp    L5
L19:
  mov    $0xffffffff,%eax
L5:
  leave
  ret
```

assistant: ```c
int _tc_vsnprintf(const char *file, int line, char *buf, size_t limit, const char *format, va_list args) {
  int res = vsnprintf(buf, limit, format, args);
  return tc_test_string(file, line, limit, res, errno) ? -1 : res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<_tc_vsnprintf>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %rdx,%rdi
  mov    %rcx,%rbp
  mov    %r8,%rdx
  mov    %r9d,%ecx
  mov    %rbp,%rsi
  call   <vsnprintf@plt>
  mov    %eax,%ebx
  mov    D3(%rip),%rax
  mov    (%rax),%r8d
  mov    %ebx,%ecx
  mov    %rbp,%rdx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <tc_test_string@plt>
  test   %rax,%rax
  mov    $0xffffffff,%eax
  cmovne %eax,%ebx
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int _tc_vsnprintf(const char *file, int line, char *buf, size_t limit, const char *format, va_list args) {
  int res = vsnprintf(buf, limit, format, args);
  return tc_test_string(file, line, limit, res, errno) ? -1 : res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<_tc_vsnprintf>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rcx,%r13
  mov    %r9d,%ecx
  push   %r12
  mov    %r13,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %rdx,%rdi
  mov    %r8,%rdx
  sub    $0x8,%rsp
  call   <vsnprintf@plt>
  mov    %r13,%rdx
  mov    %r14d,%esi
  mov    %rbp,%rdi
  mov    %eax,%r12d
  mov    D2(%rip),%rax
  mov    %r12d,%ecx
  mov    (%rax),%r8d
  call   <tc_test_string@plt>
  test   %rax,%rax
  mov    $0xffffffff,%eax
  cmovne %eax,%r12d
  add    $0x8,%rsp
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int _tc_vsnprintf(const char *file, int line, char *buf, size_t limit, const char *format, va_list args) {
  int res = vsnprintf(buf, limit, format, args);
  return tc_test_string(file, line, limit, res, errno) ? -1 : res;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<_tc_vsnprintf>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rcx,%r13
  mov    %r9d,%ecx
  push   %r12
  mov    %r13,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %rdx,%rdi
  mov    %r8,%rdx
  sub    $0x8,%rsp
  call   <vsnprintf@plt>
  mov    %r13,%rdx
  mov    %r14d,%esi
  mov    %rbp,%rdi
  mov    %eax,%r12d
  mov    D7(%rip),%rax
  mov    %r12d,%ecx
  mov    (%rax),%r8d
  call   <tc_test_string@plt>
  test   %rax,%rax
  mov    $0xffffffff,%eax
  cmovne %eax,%r12d
  add    $0x8,%rsp
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int _tc_vsnprintf(const char *file, int line, char *buf, size_t limit, const char *format, va_list args) {
  int res = vsnprintf(buf, limit, format, args);
  return tc_test_string(file, line, limit, res, errno) ? -1 : res;
}
```
user: What is the c source code of the assembly code below?

```asm
<read_bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <check_bytes@plt>
  movq   $0x0,-0x8(%rbp)
  jmp    L39
L19:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    -0x8(%rbp),%rdx
  cmp    %dl,%al
  je     L34
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L16
L34:
  addq   $0x1,-0x8(%rbp)
L39:
  cmpq   $0xfffffff,-0x8(%rbp)
  jbe    L19
  nop
L16:
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D7
type: string

tool: D7: Mismatch at %lu


assistant: ```c
void read_bytes(char *addr) {
  unsigned long i;
  check_bytes(addr);
  for (i = 0; i < (256UL * 1024 * 1024); i++)
    if (*(addr + i) != (char)i) {
      printf("Mismatch at %lu\n", i);
      break;
    }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<read_bytes>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <check_bytes@plt>
  mov    $0x0,%esi
L23:
  cmp    %sil,(%rbx,%rsi,1)
  jne    L26
  add    $0x1,%rsi
  cmp    $0x10000000,%rsi
  jne    L23
  jmp    L3
L26:
  lea    D2(%rip),%rdi
  call   <printf@plt>
L3:
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: label: D2
type_guessed: string

tool: D2: Mismatch at %lu


assistant: ```c
void read_bytes(char *addr) {
  unsigned long i;
  check_bytes(addr);
  for (i = 0; i < (256UL * 1024 * 1024); i++)
    if (*(addr + i) != (char)i) {
      printf("Mismatch at %lu\n", i);
      break;
    }
}
```