ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<ft_strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L2
  mov    $0x1,%eax
  xchg   %ax,%ax
L18:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L18
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int ft_strlen(char *str) {
  int n;
  n = 0;
  while (str[n]) {
    n++;
  }
  return (n);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x5,-0x4(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int alloc(int *p) {
  int tmp = 5;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<alloc>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int alloc(int *p) {
  int tmp = 5;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<alloc>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int alloc(int *p) {
  int tmp = 5;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<alloc>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int alloc(int *p) {
  int tmp = 5;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ichol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <icholu@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int ichol(double *A, int N) {
  int sc;
  sc = icholu(A, N);
  return sc;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ichol>:
  endbr64
  sub    $0x8,%rsp
  call   <icholu@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int ichol(double *A, int N) {
  int sc;
  sc = icholu(A, N);
  return sc;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ichol>:
  endbr64
  jmp    1050 <icholu@plt>
```

assistant: ```c
int ichol(double *A, int N) {
  int sc;
  sc = icholu(A, N);
  return sc;
}
```
user: What is the c source code of the assembly code below:

```asm
<ichol>:
  endbr64
  jmp    1050 <icholu@plt>
```

assistant: ```c
int ichol(double *A, int N) {
  int sc;
  sc = icholu(A, N);
  return sc;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<test_bit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x4(%rbp)
  mov    %ax,-0x8(%rbp)
  movswl -0x8(%rbp),%eax
  cltq
  lea    D4(%rip),%rdx
  movzbl (%rax,%rdx,1),%eax
  and    -0x4(%rbp),%al
  pop    %rbp
  ret
```

assistant: ```c
unsigned char test_bit(byte, bitpos)
unsigned char byte;
short bitpos;
{
  static unsigned char mask[9] = {0x1, 0x2, 0x4, 0x8, 0x10, 0x20, 0x40, 0x80};
  return (byte & mask[bitpos]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<classify_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L10
  mov    $0x1,%eax
  jmp    L13
L10:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <isalnum@plt>
  test   %rax,%rax
  jne    L4
  cmpl   $0x5f,-0x4(%rbp)
  jne    L38
L4:
  mov    $0x2,%eax
  jmp    L13
L38:
  mov    $0x3,%eax
L13:
  leave
  ret
```

assistant: ```c
int classify_char(int c) {
  if (isspace(c))
    return 1;
  if (isalnum(c) || (c == '_'))
    return 2;
  return 3;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<classify_char>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <isspace@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  je     L5
L10:
  pop    %rbx
  ret
L5:
  mov    %ebx,%edi
  call   <isalnum@plt>
  cmp    $0x5f,%ebx
  sete   %dl
  test   %rax,%rax
  setne  %al
  or     %eax,%edx
  movzbl %dl,%edx
  mov    $0x3,%eax
  sub    %edx,%eax
  jmp    L10
```

assistant: ```c
int classify_char(int c) {
  if (isspace(c))
    return 1;
  if (isalnum(c) || (c == '_'))
    return 2;
  return 3;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<classify_char>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <isspace@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L5
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %ebx,%edi
  call   <isalnum@plt>
  cmp    $0x5f,%ebx
  pop    %rbx
  sete   %dl
  test   %rax,%rax
  setne  %al
  or     %eax,%edx
  mov    $0x3,%eax
  movzbl %dl,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int classify_char(int c) {
  if (isspace(c))
    return 1;
  if (isalnum(c) || (c == '_'))
    return 2;
  return 3;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<classify_char>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <isspace@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L1
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %ebx,%edi
  call   <isalnum@plt>
  cmp    $0x5f,%ebx
  pop    %rbx
  sete   %dl
  test   %rax,%rax
  setne  %al
  or     %eax,%edx
  mov    $0x3,%eax
  movzbl %dl,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int classify_char(int c) {
  if (isspace(c))
    return 1;
  if (isalnum(c) || (c == '_'))
    return 2;
  return 3;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sig_restore>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  lea    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    %rax,%rdi
  call   <xsigemptyset@plt>
  movq   $0x0,-0x20(%rbp)
  mov    D22(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rcx
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <xsigaction@plt>
  mov    D29(%rip),%rax
  mov    (%rax),%eax
  mov    -0x28(%rbp),%rcx
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <xsigaction@plt>
  mov    D10(%rip),%rax
  mov    (%rax),%eax
  mov    -0x30(%rbp),%rcx
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <xsigaction@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
void sig_restore(const struct sigaction *old_int, const struct sigaction *old_term) {
  {
    struct sigaction sa;
    sa.sa_handler = SIG_DFL;
    xsigemptyset(&sa.sa_mask);
    sa.sa_flags = 0;
    xsigaction(SIGCONT, &sa, 0);
  }
  xsigaction(SIGINT, old_int, 0);
  xsigaction(SIGTERM, old_term, 0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sig_restore>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D17(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0xc(%rsp)
  mov    %rsp,%r12
  lea    0x8(%rsp),%rdi
  call   <xsigemptyset@plt>
  movq   $0x0,(%rsp)
  mov    $0x0,%edx
  mov    %r12,%rsi
  mov    D3(%rip),%rax
  mov    (%rax),%edi
  call   <xsigaction@plt>
  mov    $0x0,%edx
  mov    %rbp,%rsi
  mov    D15(%rip),%rax
  mov    (%rax),%edi
  call   <xsigaction@plt>
  mov    $0x0,%edx
  mov    %rbx,%rsi
  mov    D13(%rip),%rax
  mov    (%rax),%edi
  call   <xsigaction@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sig_restore(const struct sigaction *old_int, const struct sigaction *old_term) {
  {
    struct sigaction sa;
    sa.sa_handler = SIG_DFL;
    xsigemptyset(&sa.sa_mask);
    sa.sa_flags = 0;
    xsigaction(SIGCONT, &sa, 0);
  }
  xsigaction(SIGINT, old_int, 0);
  xsigaction(SIGTERM, old_term, 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sig_restore>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    D32(%rip),%rax
  lea    0x8(%rsp),%rdi
  mov    %rsp,%r13
  mov    (%rax),%eax
  mov    %eax,0xc(%rsp)
  call   <xsigemptyset@plt>
  mov    D28(%rip),%rax
  xor    %edx,%edx
  mov    %r13,%rsi
  movq   $0x0,(%rsp)
  mov    (%rax),%edi
  call   <xsigaction@plt>
  mov    D3(%rip),%rax
  xor    %edx,%edx
  mov    %rbp,%rsi
  mov    (%rax),%edi
  call   <xsigaction@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    D17(%rip),%rax
  mov    %r12,%rsi
  xor    %edx,%edx
  mov    (%rax),%edi
  add    $0x20,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1090 <xsigaction@plt>
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sig_restore(const struct sigaction *old_int, const struct sigaction *old_term) {
  {
    struct sigaction sa;
    sa.sa_handler = SIG_DFL;
    xsigemptyset(&sa.sa_mask);
    sa.sa_flags = 0;
    xsigaction(SIGCONT, &sa, 0);
  }
  xsigaction(SIGINT, old_int, 0);
  xsigaction(SIGTERM, old_term, 0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<sig_restore>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    D16(%rip),%rax
  lea    0x8(%rsp),%rdi
  mov    %rsp,%r13
  mov    (%rax),%eax
  mov    %eax,0xc(%rsp)
  call   <xsigemptyset@plt>
  mov    D27(%rip),%rax
  xor    %edx,%edx
  mov    %r13,%rsi
  movq   $0x0,(%rsp)
  mov    (%rax),%edi
  call   <xsigaction@plt>
  mov    D9(%rip),%rax
  xor    %edx,%edx
  mov    %rbp,%rsi
  mov    (%rax),%edi
  call   <xsigaction@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    D0(%rip),%rax
  mov    %r12,%rsi
  xor    %edx,%edx
  mov    (%rax),%edi
  add    $0x20,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1090 <xsigaction@plt>
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sig_restore(const struct sigaction *old_int, const struct sigaction *old_term) {
  {
    struct sigaction sa;
    sa.sa_handler = SIG_DFL;
    xsigemptyset(&sa.sa_mask);
    sa.sa_flags = 0;
    xsigaction(SIGCONT, &sa, 0);
  }
  xsigaction(SIGINT, old_int, 0);
  xsigaction(SIGTERM, old_term, 0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<serial_flush>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    $0x2710,%edi
  call   <usleep@plt>
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <tcflush@plt>
  mov    %eax,-0x4(%rbp)
  mov    $0xf4240,%edi
  call   <usleep@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int serial_flush(int fd) {
  int ret;
  usleep(10000);
  ret = tcflush(fd, TCIOFLUSH);
  usleep(1000000);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<serial_flush>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x2710,%edi
  call   <usleep@plt>
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  mov    %ebx,%edi
  call   <tcflush@plt>
  mov    %eax,%ebx
  mov    $0xf4240,%edi
  call   <usleep@plt>
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int serial_flush(int fd) {
  int ret;
  usleep(10000);
  ret = tcflush(fd, TCIOFLUSH);
  usleep(1000000);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<serial_flush>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  mov    $0x2710,%edi
  call   <usleep@plt>
  mov    D7(%rip),%rax
  mov    %r12d,%edi
  mov    (%rax),%esi
  call   <tcflush@plt>
  mov    $0xf4240,%edi
  mov    %eax,%r12d
  call   <usleep@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int serial_flush(int fd) {
  int ret;
  usleep(10000);
  ret = tcflush(fd, TCIOFLUSH);
  usleep(1000000);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<serial_flush>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  mov    $0x2710,%edi
  call   <usleep@plt>
  mov    D5(%rip),%rax
  mov    %r12d,%edi
  mov    (%rax),%esi
  call   <tcflush@plt>
  mov    $0xf4240,%edi
  mov    %eax,%r12d
  call   <usleep@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int serial_flush(int fd) {
  int ret;
  usleep(10000);
  ret = tcflush(fd, TCIOFLUSH);
  usleep(1000000);
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L9
L8:
  mov    -0x18(%rbp),%edx
  mov    -0xc(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x14(%rbp),%edi
  mov    -0x10(%rbp),%eax
  cltd
  idiv   %edi
  mov    %eax,%esi
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x4,%edx
  sar    $0x1f,%eax
  mov    %eax,%edi
  mov    %edx,%eax
  sub    %edi,%eax
  imul   %eax,%esi
  mov    %esi,%edx
  mov    %ecx,%eax
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L9:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L8
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
datatype: string
name: retrieve_data
arguments: data_tag: D22
datatype: string

tool: D13: %d

tool: D22: 
%d

assistant: ```c
int main() {
  int n, m, k, i, sum;
  sum = 0;
  i = 0;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  for (; i < n; i++) {
    sum = sum + m - (i / k * (2 * m / 100));
  }
  printf("\n%d", sum);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D8(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%r10d
  test   %r10d,%r10d
  jle    L18
  mov    0x10(%rsp),%r9d
  mov    0x14(%rsp),%edi
  movslq %r9d,%r8
  imul   $0x51eb851f,%r8,%r8
  sar    $0x24,%r8
  mov    %r9d,%eax
  sar    $0x1f,%eax
  sub    %eax,%r8d
  mov    $0x0,%esi
  mov    $0x0,%ecx
L15:
  add    %r9d,%esi
  mov    %ecx,%eax
  cltd
  idiv   %edi
  imul   %r8d,%eax
  sub    %eax,%esi
  add    $0x1,%ecx
  cmp    %r10d,%ecx
  jne    L15
L8:
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L18:
  mov    $0x0,%esi
  jmp    L8
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D8
data_type: string
name: retrieve_data
arguments: label: D19
data_type: string

tool: D8: %d

tool: D19: 
%d

assistant: ```c
int main() {
  int n, m, k, i, sum;
  sum = 0;
  i = 0;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  for (; i < n; i++) {
    sum = sum + m - (i / k * (2 * m / 100));
  }
  printf("\n%d", sum);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D1(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%r10d
  test   %r10d,%r10d
  jle    L21
  movslq 0x10(%rsp),%r8
  mov    0x14(%rsp),%edi
  xor    %esi,%esi
  xor    %ecx,%ecx
  mov    %r8,%r9
  imul   $0x51eb851f,%r8,%r8
  mov    %r9d,%eax
  sar    $0x1f,%eax
  sar    $0x24,%r8
  sub    %eax,%r8d
  nopl   (%rax)
L13:
  mov    %ecx,%eax
  add    %r9d,%esi
  add    $0x1,%ecx
  cltd
  idiv   %edi
  imul   %r8d,%eax
  sub    %eax,%esi
  cmp    %r10d,%ecx
  jne    L13
L22:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L21:
  xor    %esi,%esi
  jmp    L22
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D1
datatype: string
name: retrieve_data
arguments: tag: D11
datatype: string

tool: D1: %d

tool: D11: 
%d

assistant: ```c
int main() {
  int n, m, k, i, sum;
  sum = 0;
  i = 0;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  for (; i < n; i++) {
    sum = sum + m - (i / k * (2 * m / 100));
  }
  printf("\n%d", sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D14(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%r10d
  test   %r10d,%r10d
  jle    L6
  movslq 0x10(%rsp),%r8
  mov    0x14(%rsp),%edi
  xor    %esi,%esi
  xor    %ecx,%ecx
  mov    %r8,%r9
  imul   $0x51eb851f,%r8,%r8
  mov    %r9d,%eax
  sar    $0x1f,%eax
  sar    $0x24,%r8
  sub    %eax,%r8d
  nopl   (%rax)
L29:
  mov    %ecx,%eax
  add    %r9d,%esi
  add    $0x1,%ecx
  cltd
  idiv   %edi
  imul   %r8d,%eax
  sub    %eax,%esi
  cmp    %r10d,%ecx
  jne    L29
L20:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  xor    %esi,%esi
  jmp    L20
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D14
type_infered: string
name: disassemble_data
arguments: tag: D4
type_infered: string

tool: D14: %d

tool: D4: 
%d

assistant: ```c
int main() {
  int n, m, k, i, sum;
  sum = 0;
  i = 0;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  for (; i < n; i++) {
    sum = sum + m - (i / k * (2 * m / 100));
  }
  printf("\n%d", sum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  not    %eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  neg    %eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  subl   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  mov    %eax,-0x14(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int main() {
  int x = 1;
  int y;
  y = ~x;
  y = -x;
  y = +x;
  y = x++;
  y = ++x;
  y = --x;
  y = x--;
  int a;
  int *b;
  b = &a;
  a = *(b + 1);
  *b = *b + 1;
  a = *(&a);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int x = 1;
  int y;
  y = ~x;
  y = -x;
  y = +x;
  y = x++;
  y = ++x;
  y = --x;
  y = x--;
  int a;
  int *b;
  b = &a;
  a = *(b + 1);
  *b = *b + 1;
  a = *(&a);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int x = 1;
  int y;
  y = ~x;
  y = -x;
  y = +x;
  y = x++;
  y = ++x;
  y = --x;
  y = x--;
  int a;
  int *b;
  b = &a;
  a = *(b + 1);
  *b = *b + 1;
  a = *(&a);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int x = 1;
  int y;
  y = ~x;
  y = -x;
  y = +x;
  y = x++;
  y = ++x;
  y = --x;
  y = x--;
  int a;
  int *b;
  b = &a;
  a = *(b + 1);
  *b = *b + 1;
  a = *(&a);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <Palindrom@plt>
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D9
data_type: string
name: disassemble_data
arguments: tag: D0
data_type: string

tool: D9: %s

tool: D0: %d 


assistant: ```c
int main() {
  int i, a;
  char Pali[100];
  scanf("%s", Pali);
  printf("%d \n", Palindrom(Pali));
  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   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <Palindrom@plt>
  mov    %eax,%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D6
data_type: string
name: unpack_data
arguments: tag: D15
data_type: string

tool: D6: %s

tool: D15: %d 


assistant: ```c
int main() {
  int i, a;
  char Pali[100];
  scanf("%s", Pali);
  printf("%d \n", Palindrom(Pali));
  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   %rbp
  lea    D5(%rip),%rdi
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <Palindrom@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D5
type_infered: string
name: read_data
arguments: data_label: D6
type_infered: string

tool: D5: %s

tool: D6: %d 


assistant: ```c
int main() {
  int i, a;
  char Pali[100];
  scanf("%s", Pali);
  printf("%d \n", Palindrom(Pali));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rdi
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <Palindrom@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type_infered: string
name: disassemble_data
arguments: data_label: D7
type_infered: string

tool: D3: %s

tool: D7: %d 


assistant: ```c
int main() {
  int i, a;
  char Pali[100];
  scanf("%s", Pali);
  printf("%d \n", Palindrom(Pali));
  return 0;
}
```