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
<dyaddown_2D_keep_even_col>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    %r9d,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L22
L19:
  movl   $0x0,-0x8(%rbp)
  jmp    L9
L6:
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  imul   -0x1c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  imul   -0x1c(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%eax
  add    %ecx,%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x8(%rbp)
L9:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L6
  addl   $0x1,-0x4(%rbp)
L22:
  mov    -0x4(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L19
  nop
  pop    %rbp
  ret
```

assistant: ```c
void dyaddown_2D_keep_even_col(double *matrixIn, int matrixInRow, int matrixInCol, double *matrixOut, int matrixOutRow, int matrixOutCol) {
  int row, col;
  for (col = 0; col < matrixOutCol; col++) {
    for (row = 0; row < matrixInRow; row++)
      matrixOut[row + col * matrixInRow] = matrixIn[row + (2 * col + 1) * matrixInRow];
  }
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dyaddown_2D_keep_even_col>:
  endbr64
  test   %r9d,%r9d
  jle    L44
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r10d
  mov    $0x0,%r8d
  mov    $0x0,%r11d
  mov    %esi,%edi
  jmp    L31
L15:
  lea    (%r10,%r8,2),%eax
  cltq
  lea    (%rbx,%rax,8),%rsi
  movslq %r8d,%rax
  lea    (%rcx,%rax,8),%rdx
  mov    $0x0,%eax
L29:
  movsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L29
L25:
  add    $0x1,%r11d
  add    %r10d,%r8d
  cmp    %r11d,%r9d
  je     L28
L31:
  test   %r10d,%r10d
  jg     L15
  jmp    L25
L28:
  pop    %rbx
  ret
L44:
  ret
```

assistant: ```c
void dyaddown_2D_keep_even_col(double *matrixIn, int matrixInRow, int matrixInCol, double *matrixOut, int matrixOutRow, int matrixOutCol) {
  int row, col;
  for (col = 0; col < matrixOutCol; col++) {
    for (row = 0; row < matrixInRow; row++)
      matrixOut[row + col * matrixInRow] = matrixIn[row + (2 * col + 1) * matrixInRow];
  }
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<dyaddown_2D_keep_even_col>:
  endbr64
  test   %r9d,%r9d
  jle    L34
  push   %rbp
  mov    %esi,%r10d
  mov    %rcx,%rbp
  xor    %r8d,%r8d
  push   %rbx
  xor    %r11d,%r11d
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%edi
  nopl   (%rax)
L25:
  test   %r10d,%r10d
  jle    L20
  lea    (%r10,%r8,2),%eax
  cltq
  lea    (%rbx,%rax,8),%rsi
  movslq %r8d,%rax
  lea    0x0(%rbp,%rax,8),%rcx
  xor    %eax,%eax
  nopl   0x0(%rax)
L13:
  movsd  (%rsi,%rax,8),%xmm0
  mov    %rax,%rdx
  movsd  %xmm0,(%rcx,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rdi
  jne    L13
L20:
  add    $0x1,%r11d
  add    %r10d,%r8d
  cmp    %r11d,%r9d
  jne    L25
  pop    %rbx
  pop    %rbp
  ret
L34:
  ret
```

assistant: ```c
void dyaddown_2D_keep_even_col(double *matrixIn, int matrixInRow, int matrixInCol, double *matrixOut, int matrixOutRow, int matrixOutCol) {
  int row, col;
  for (col = 0; col < matrixOutCol; col++) {
    for (row = 0; row < matrixInRow; row++)
      matrixOut[row + col * matrixInRow] = matrixIn[row + (2 * col + 1) * matrixInRow];
  }
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<dyaddown_2D_keep_even_col>:
  endbr64
  test   %r9d,%r9d
  jle    L59
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  test   %esi,%esi
  jle    L20
  mov    %ebx,%eax
  mov    %r9d,%r13d
  lea    (%rsi,%rsi,1),%r14d
  movslq %esi,%r15
  shr    %eax
  mov    %rcx,%r10
  movslq %r14d,%r14
  shl    $0x3,%r15
  lea    -0x1(%rax),%r9d
  lea    -0x1(%rbx),%eax
  mov    %rcx,%rsi
  mov    %ebx,%ecx
  mov    %eax,-0xc(%rsp)
  add    $0x1,%r9
  mov    %rdi,%r11
  shl    $0x3,%r14
  lea    (%rdi,%r15,1),%rdi
  shl    $0x4,%r9
  and    $0xfffffffe,%ecx
  xor    %r8d,%r8d
  xor    %ebp,%ebp
  mov    %ebx,%r12d
  nopw   0x0(%rax,%rax,1)
L79:
  lea    0x8(%rdi),%rdx
  lea    (%rbx,%r8,2),%eax
  cmp    %rdx,%rsi
  je     L44
  xor    %edx,%edx
  cmpl   $0x2,-0xc(%rsp)
  jbe    L44
  cs nopw 0x0(%rax,%rax,1)
L28:
  movupd (%rdi,%rdx,1),%xmm1
  movups %xmm1,(%rsi,%rdx,1)
  add    $0x10,%rdx
  cmp    %r9,%rdx
  jne    L28
  cmp    %ebx,%ecx
  je     L46
  add    %ecx,%eax
  cltq
  movsd  (%r11,%rax,8),%xmm0
  lea    (%r8,%rcx,1),%eax
  cltq
  movsd  %xmm0,(%r10,%rax,8)
L46:
  add    $0x1,%ebp
  add    %r14,%rdi
  add    %r15,%rsi
  add    %ebx,%r8d
  cmp    %ebp,%r13d
  jne    L79
L20:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L44:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L39:
  movsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%r12
  jne    L39
  jmp    L46
L59:
  ret
```

assistant: ```c
void dyaddown_2D_keep_even_col(double *matrixIn, int matrixInRow, int matrixInCol, double *matrixOut, int matrixOutRow, int matrixOutCol) {
  int row, col;
  for (col = 0; col < matrixOutCol; col++) {
    for (row = 0; row < matrixInRow; row++)
      matrixOut[row + col * matrixInRow] = matrixIn[row + (2 * col + 1) * matrixInRow];
  }
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x2a,%edi
  call   <test@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
type_guessed: string

tool: D2: test-

assistant: ```c
int main(int argc, char *argv[]) {
  printf("test-");
  test(42);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x2a,%edi
  call   <test@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D1
data_type: string

tool: D1: test-

assistant: ```c
int main(int argc, char *argv[]) {
  printf("test-");
  test(42);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x2a,%edi
  call   <test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type_guessed: string

tool: D4: test-

assistant: ```c
int main(int argc, char *argv[]) {
  printf("test-");
  test(42);
}
```
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
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x2a,%edi
  call   <test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
type: string

tool: D1: test-

assistant: ```c
int main(int argc, char *argv[]) {
  printf("test-");
  test(42);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sumSelfPowers>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L10
L5:
  mov    -0x14(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <power@plt>
  add    %rax,-0x8(%rbp)
  subl   $0x1,-0x14(%rbp)
L10:
  cmpl   $0x0,-0x14(%rbp)
  jg     L5
  mov    -0x8(%rbp),%rcx
  movabs $0xdbe6fecebdedd5bf,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x21,%rax
  movabs $0x2540be400,%rdx
  imul   %rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  leave
  ret
```

assistant: ```c
unsigned long long sumSelfPowers(int x) {
  unsigned long long power(int x, int y);
  unsigned long long lastTen = 0;
  while (x > 0) {
    lastTen += power(x, x);
    x--;
  }
  return lastTen % 10000000000;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sumSelfPowers>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edi,%edi
  jle    L22
  mov    %edi,%ebx
  mov    $0x0,%ebp
L5:
  mov    %ebx,%esi
  mov    %ebx,%edi
  call   <power@plt>
  add    %rax,%rbp
  sub    $0x1,%ebx
  jne    L5
L26:
  movabs $0xdbe6fecebdedd5bf,%rdx
  mov    %rbp,%rax
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x21,%rax
  movabs $0x2540be400,%rdx
  imul   %rax,%rdx
  mov    %rbp,%rax
  sub    %rdx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L22:
  mov    $0x0,%ebp
  jmp    L26
```

assistant: ```c
unsigned long long sumSelfPowers(int x) {
  unsigned long long power(int x, int y);
  unsigned long long lastTen = 0;
  while (x > 0) {
    lastTen += power(x, x);
    x--;
  }
  return lastTen % 10000000000;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sumSelfPowers>:
  endbr64
  test   %edi,%edi
  jle    L11
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %ebx,%esi
  mov    %ebx,%edi
  call   <power@plt>
  add    %rax,%rbp
  sub    $0x1,%ebx
  jne    L13
  mov    %rbp,%rax
  add    $0x8,%rsp
  movabs $0xdbe6fecebdedd5bf,%rdx
  mul    %rdx
  pop    %rbx
  mov    %rdx,%rax
  movabs $0x2540be400,%rdx
  shr    $0x21,%rax
  imul   %rax,%rdx
  mov    %rbp,%rax
  pop    %rbp
  sub    %rdx,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned long long sumSelfPowers(int x) {
  unsigned long long power(int x, int y);
  unsigned long long lastTen = 0;
  while (x > 0) {
    lastTen += power(x, x);
    x--;
  }
  return lastTen % 10000000000;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sumSelfPowers>:
  endbr64
  test   %edi,%edi
  jle    L7
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %ebx,%esi
  mov    %ebx,%edi
  call   <power@plt>
  add    %rax,%rbp
  sub    $0x1,%ebx
  jne    L1
  mov    %rbp,%rax
  add    $0x8,%rsp
  movabs $0xdbe6fecebdedd5bf,%rdx
  mul    %rdx
  pop    %rbx
  mov    %rdx,%rax
  movabs $0x2540be400,%rdx
  shr    $0x21,%rax
  imul   %rax,%rdx
  mov    %rbp,%rax
  pop    %rbp
  sub    %rdx,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned long long sumSelfPowers(int x) {
  unsigned long long power(int x, int y);
  unsigned long long lastTen = 0;
  while (x > 0) {
    lastTen += power(x, x);
    x--;
  }
  return lastTen % 10000000000;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xa,-0xc(%rbp)
  movl   $0x5,-0x8(%rbp)
  movb   $0xa,-0xd(%rbp)
  mov    -0xc(%rbp),%eax
  imul   -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movsbl -0xd(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D14
type_infered: string

tool: D1: Value of area : %d

tool: D14: %c

assistant: ```c
int main() {
  const int LENGTH = 10;
  const int WIDTH = 5;
  const char NEWLINE = '\n';
  int area;
  area = LENGTH * WIDTH;
  printf("Value of area : %d", area);
  printf("%c", NEWLINE);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x32,%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D10
type_infered: string
name: extract_symbolic_data
arguments: label: D9
type_infered: string

tool: D10: Value of area : %d

tool: D9: %c

assistant: ```c
int main() {
  const int LENGTH = 10;
  const int WIDTH = 5;
  const char NEWLINE = '\n';
  int area;
  area = LENGTH * WIDTH;
  printf("Value of area : %d", area);
  printf("%c", NEWLINE);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x32,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
data_type: string
name: disassemble_data
arguments: tag: D5
data_type: string

tool: D4: Value of area : %d

tool: D5: %c

assistant: ```c
int main() {
  const int LENGTH = 10;
  const int WIDTH = 5;
  const char NEWLINE = '\n';
  int area;
  area = LENGTH * WIDTH;
  printf("Value of area : %d", area);
  printf("%c", NEWLINE);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x32,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D0
type_infered: string

tool: D9: Value of area : %d

tool: D0: %c

assistant: ```c
int main() {
  const int LENGTH = 10;
  const int WIDTH = 5;
  const char NEWLINE = '\n';
  int area;
  area = LENGTH * WIDTH;
  printf("Value of area : %d", area);
  printf("%c", NEWLINE);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<artificial1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x47464544434241,%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,%ecx
  shl    $0x3,%ecx
  sub    %edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  movslq %edx,%rax
  movzbl -0x10(%rbp,%rax,1),%eax
  mov    %al,-0x11(%rbp)
  movsbl -0x11(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movsbl -0x11(%rbp),%eax
  mov    %eax,%esi
  mov    $0x5e,%edi
  call   <update@plt>
  mov    $0x0,%eax
  call   <print_board@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D6
type: string

tool: D6: IA chose %c


assistant: ```c
void artificial1(void) {
  char colors[] = "ABCDEFG";
  char color;
  color = colors[rand() % 7];
  printf("IA chose %c\n", color);
  update('^', color);
  print_board();
}
```
user: Translate the below assembly code into C programming language:

```asm
<artificial1>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movabs $0x47464544434241,%rax
  mov    %rax,(%rsp)
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    0x0(,%rdx,8),%ecx
  sub    %edx,%ecx
  sub    %ecx,%eax
  cltq
  movsbl (%rsp,%rax,1),%ebx
  mov    %ebx,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%esi
  mov    $0x5e,%edi
  call   <update@plt>
  mov    $0x0,%eax
  call   <print_board@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D1
datatype: string

tool: D1: IA chose %c


assistant: ```c
void artificial1(void) {
  char colors[] = "ABCDEFG";
  char color;
  color = colors[rand() % 7];
  printf("IA chose %c\n", color);
  update('^', color);
  print_board();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<artificial1>:
  endbr64
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x47464544434241,%rax
  mov    %rax,(%rsp)
  xor    %eax,%eax
  call   <rand@plt>
  lea    D4(%rip),%rdi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0xffffffff92492493,%rdx,%rdx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  sub    %ecx,%edx
  lea    0x0(,%rdx,8),%ecx
  sub    %edx,%ecx
  sub    %ecx,%eax
  cltq
  movsbl (%rsp,%rax,1),%ebp
  mov    %ebp,%esi
  call   <printf@plt>
  mov    %ebp,%esi
  mov    $0x5e,%edi
  call   <update@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  jmp    10c0 <print_board@plt>
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D4
data_type: string

tool: D4: IA chose %c


assistant: ```c
void artificial1(void) {
  char colors[] = "ABCDEFG";
  char color;
  color = colors[rand() % 7];
  printf("IA chose %c\n", color);
  update('^', color);
  print_board();
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<artificial1>:
  endbr64
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x47464544434241,%rax
  mov    %rax,(%rsp)
  xor    %eax,%eax
  call   <rand@plt>
  lea    D6(%rip),%rdi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0xffffffff92492493,%rdx,%rdx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  sub    %ecx,%edx
  lea    0x0(,%rdx,8),%ecx
  sub    %edx,%ecx
  sub    %ecx,%eax
  cltq
  movsbl (%rsp,%rax,1),%ebp
  mov    %ebp,%esi
  call   <printf@plt>
  mov    %ebp,%esi
  mov    $0x5e,%edi
  call   <update@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  jmp    10c0 <print_board@plt>
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
type_guessed: string

tool: D6: IA chose %c


assistant: ```c
void artificial1(void) {
  char colors[] = "ABCDEFG";
  char color;
  color = colors[rand() % 7];
  printf("IA chose %c\n", color);
  update('^', color);
  print_board();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<hex_to_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L24:
  shll   $0x4,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L10
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L10
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,-0x8(%rbp)
  jmp    L96
L10:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L3
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x46,%al
  jg     L3
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x37,%eax
  add    %eax,-0x8(%rbp)
  jmp    L96
L3:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L96
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x66,%al
  jg     L96
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x57,%eax
  add    %eax,-0x8(%rbp)
L96:
  addl   $0x1,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L83
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L24
L83:
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int hex_to_int(char *hex, int l) {
  int n = 0;
  for (int i = 0; hex[i] && i < l; i++) {
    n <<= 4;
    if (hex[i] >= '0' && hex[i] <= '9') {
      n += hex[i] - '0';
    } else if (hex[i] >= 'A' && hex[i] <= 'F') {
      n += hex[i] - 'A' + 0x0A;
    } else if (hex[i] >= 'a' && hex[i] <= 'f') {
      n += hex[i] - 'a' + 0x0A;
    }
  }
  return n;
}
```
user: What is the c source code of the assembly code below:

```asm
<hex_to_int>:
  endbr64
  mov    %rdi,%r9
  movzbl (%rdi),%eax
  test   %esi,%esi
  jle    L64
  test   %al,%al
  je     L64
  mov    $0x1,%edx
  mov    $0x0,%ecx
  jmp    L44
L55:
  lea    -0x41(%rax),%edi
  cmp    $0x5,%dil
  ja     L74
  movsbl %al,%eax
  lea    -0x37(%rcx,%rax,1),%ecx
L89:
  movzbl (%r9,%rdx,1),%eax
  cmp    %edx,%esi
  setg   %r8b
  test   %al,%al
  setne  %dil
  add    $0x1,%rdx
  test   %dil,%r8b
  je     L24
L44:
  shl    $0x4,%ecx
  mov    %eax,%r8d
  lea    -0x30(%rax),%edi
  cmp    $0x9,%dil
  ja     L55
  movsbl %al,%eax
  lea    -0x30(%rcx,%rax,1),%ecx
  jmp    L89
L74:
  sub    $0x61,%r8d
  cmp    $0x5,%r8b
  ja     L89
  movsbl %al,%eax
  lea    -0x57(%rcx,%rax,1),%ecx
  jmp    L89
L64:
  mov    $0x0,%ecx
L24:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int hex_to_int(char *hex, int l) {
  int n = 0;
  for (int i = 0; hex[i] && i < l; i++) {
    n <<= 4;
    if (hex[i] >= '0' && hex[i] <= '9') {
      n += hex[i] - '0';
    } else if (hex[i] >= 'A' && hex[i] <= 'F') {
      n += hex[i] - 'A' + 0x0A;
    } else if (hex[i] >= 'a' && hex[i] <= 'f') {
      n += hex[i] - 'a' + 0x0A;
    }
  }
  return n;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hex_to_int>:
  endbr64
  movzbl (%rdi),%eax
  mov    %rdi,%r9
  test   %al,%al
  je     L11
  test   %esi,%esi
  jle    L11
  mov    $0x1,%edx
  xor    %r8d,%r8d
  jmp    L76
  nopl   0x0(%rax)
L53:
  movsbl %cl,%ecx
  add    %ecx,%r8d
L19:
  movzbl (%r9,%rdx,1),%eax
  cmp    %edx,%esi
  setg   %dil
  test   %al,%al
  setne  %cl
  add    $0x1,%rdx
  test   %cl,%dil
  je     L62
L76:
  lea    -0x30(%rax),%ecx
  shl    $0x4,%r8d
  cmp    $0x9,%cl
  jbe    L53
  lea    -0x41(%rax),%ecx
  cmp    $0x5,%cl
  ja     L65
  sub    $0x37,%eax
  movsbl %al,%eax
  add    %eax,%r8d
  jmp    L19
  xchg   %ax,%ax
L65:
  lea    -0x61(%rax),%ecx
  cmp    $0x5,%cl
  ja     L19
  sub    $0x57,%eax
  movsbl %al,%eax
  add    %eax,%r8d
  jmp    L19
  nopl   0x0(%rax,%rax,1)
L11:
  xor    %r8d,%r8d
L62:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int hex_to_int(char *hex, int l) {
  int n = 0;
  for (int i = 0; hex[i] && i < l; i++) {
    n <<= 4;
    if (hex[i] >= '0' && hex[i] <= '9') {
      n += hex[i] - '0';
    } else if (hex[i] >= 'A' && hex[i] <= 'F') {
      n += hex[i] - 'A' + 0x0A;
    } else if (hex[i] >= 'a' && hex[i] <= 'f') {
      n += hex[i] - 'a' + 0x0A;
    }
  }
  return n;
}
```