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
<readval>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    $0x0,%r13d
  mov    $0x101,%ebx
  mov    $0x100,%r12d
  jmp    L155
L8:
  mov    %esi,%ebx
  jmp    L155
L36:
  lea    D2(%rip),%rdi
  call   <puts@plt>
  mov    $0xfffffffffffffffd,%r13
  jmp    L49
L94:
  lea    D22(%rip),%rdi
  call   <puts@plt>
  mov    $0xfffffffffffffffc,%r13
  jmp    L49
L114:
  shl    $0x8,%r13
  movslq %esi,%rsi
  add    %rsi,%r13
  sub    $0x1,%ebx
L155:
  mov    $0x3,%esi
  mov    %ebp,%edi
  call   <com_getc@plt>
  mov    %eax,%esi
  cmp    $0xffffffff,%eax
  je     L36
  cmp    $0xfffffffe,%eax
  je     L94
  cmp    $0x100,%ebx
  je     L8
  cmp    $0x100,%ebx
  jg     L145
  cmp    $0x1,%ebx
  je     L81
  lea    -0x2(%rbx),%eax
  cmp    $0x2,%eax
  jbe    L114
L92:
  mov    %r13,%rcx
  mov    %ebx,%edx
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xfffffffffffffffe,%r13
L49:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L145:
  cmp    $0x101,%ebx
  jne    L92
  cmp    $0xab,%eax
  je     L0
  cmp    $0xa8,%eax
  cmove  %r12d,%ebx
  jmp    L155
L81:
  cmp    $0xaa,%eax
  jne    L155
  jmp    L49
L0:
  mov    $0xfffffffffffffffe,%r13
  jmp    L49
```

tool_calls: name: unpack_data
arguments: tag: D2
data_type: string
name: unpack_data
arguments: tag: D22
data_type: string
name: unpack_data
arguments: tag: D10
data_type: string

tool: D2: readval: ...Device does not answer!

tool: D22: readval: ...Device disconnected!

tool: D10: 
Error: readval, i = %i, j = %i, val = %li


assistant: ```c
long readval(int fd) {
  int i;
  int j = 257;
  long val = 0;
  while (1) {
    i = com_getc(fd, 3);
    if (i == -1) {
      printf("readval: ...Device does not answer!\n");
      return -3;
    } else if (i == -2) {
      printf("readval: ...Device disconnected!\n");
      return -4;
    }
    switch (j) {
    case 1:
      if (i == 0xAA) {
        return val;
      }
      break;
    case 2:
    case 3:
    case 4:
      val = val * 256 + i;
      j--;
      break;
    case 256:
      j = i;
      break;
    case 257:
      if (i == 0xAB) {
        return -2;
      } else if (i == 0xA8) {
        j = 256;
      }
      break;
    default:
      printf("\nError: readval, i = %i, j = %i, val = %li\n", i, j, val);
      return -2;
    }
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<readval>:
  endbr64
  push   %r13
  mov    $0x101,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbx
  mov    %edi,%ebx
L187:
  mov    $0x3,%esi
  mov    %ebx,%edi
  call   <com_getc@plt>
  cmp    $0xffffffff,%eax
  je     L1
  cmp    $0xfffffffe,%eax
  je     L30
  cmp    $0x4,%r13d
  jg     L57
  cmp    $0x1,%r13d
  jg     L134
  je     L90
L176:
  mov    %r12,%rcx
  mov    %eax,%esi
  mov    %r13d,%edx
  xor    %eax,%eax
  lea    D10(%rip),%rdi
  mov    $0xfffffffffffffffe,%r12
  call   <printf@plt>
L86:
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  cmp    $0xfffffffe,%eax
  je     L30
L90:
  cmp    $0xaa,%eax
  je     L86
  mov    $0x3,%esi
  mov    %ebx,%edi
  call   <com_getc@plt>
  cmp    $0xffffffff,%eax
  jne    L0
  nopl   (%rax)
L1:
  lea    D1(%rip),%rdi
  mov    $0xfffffffffffffffd,%r12
  call   <puts@plt>
  jmp    L86
  nopl   (%rax)
L57:
  cmp    $0x100,%r13d
  je     L16
  cmp    $0x101,%r13d
  jne    L176
L158:
  cmp    $0xab,%eax
  je     L117
  mov    $0x3,%esi
  mov    %ebx,%edi
  cmp    $0xa8,%eax
  je     L40
  call   <com_getc@plt>
  cmp    $0xffffffff,%eax
  je     L1
  cmp    $0xfffffffe,%eax
  jne    L158
  xchg   %ax,%ax
L30:
  lea    D16(%rip),%rdi
  mov    $0xfffffffffffffffc,%r12
  call   <puts@plt>
  jmp    L86
  nopl   0x0(%rax,%rax,1)
L134:
  movslq %eax,%rsi
  shl    $0x8,%r12
  lea    -0x1(%r13),%eax
  add    %rsi,%r12
L16:
  mov    %eax,%r13d
  jmp    L187
  cs nopw 0x0(%rax,%rax,1)
L40:
  call   <com_getc@plt>
  cmp    $0xffffffff,%eax
  je     L1
  cmp    $0xfffffffe,%eax
  je     L30
  mov    %eax,%r13d
  jmp    L187
L117:
  mov    $0xfffffffffffffffe,%r12
  jmp    L86
```

tool_calls: name: read_data
arguments: label: D10
datatype: string
name: read_data
arguments: label: D1
datatype: string
name: read_data
arguments: label: D16
datatype: string

tool: D10: 
Error: readval, i = %i, j = %i, val = %li


tool: D1: readval: ...Device does not answer!

tool: D16: readval: ...Device disconnected!

assistant: ```c
long readval(int fd) {
  int i;
  int j = 257;
  long val = 0;
  while (1) {
    i = com_getc(fd, 3);
    if (i == -1) {
      printf("readval: ...Device does not answer!\n");
      return -3;
    } else if (i == -2) {
      printf("readval: ...Device disconnected!\n");
      return -4;
    }
    switch (j) {
    case 1:
      if (i == 0xAA) {
        return val;
      }
      break;
    case 2:
    case 3:
    case 4:
      val = val * 256 + i;
      j--;
      break;
    case 256:
      j = i;
      break;
    case 257:
      if (i == 0xAB) {
        return -2;
      } else if (i == 0xA8) {
        j = 256;
      }
      break;
    default:
      printf("\nError: readval, i = %i, j = %i, val = %li\n", i, j, val);
      return -2;
    }
  }
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<readval>:
  endbr64
  push   %r13
  mov    $0x101,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbx
  mov    %edi,%ebx
L8:
  mov    $0x3,%esi
  mov    %ebx,%edi
  call   <com_getc@plt>
  cmp    $0xffffffff,%eax
  je     L1
  cmp    $0xfffffffe,%eax
  je     L200
  cmp    $0x4,%r13d
  jg     L142
  cmp    $0x1,%r13d
  jg     L196
  je     L107
L35:
  mov    %r12,%rcx
  mov    %eax,%esi
  mov    %r13d,%edx
  xor    %eax,%eax
  lea    D18(%rip),%rdi
  mov    $0xfffffffffffffffe,%r12
  call   <printf@plt>
L41:
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L158:
  cmp    $0xfffffffe,%eax
  je     L200
L107:
  cmp    $0xaa,%eax
  je     L41
  mov    $0x3,%esi
  mov    %ebx,%edi
  call   <com_getc@plt>
  cmp    $0xffffffff,%eax
  jne    L158
  nopl   (%rax)
L1:
  lea    D1(%rip),%rdi
  mov    $0xfffffffffffffffd,%r12
  call   <puts@plt>
  jmp    L41
  nopl   (%rax)
L142:
  cmp    $0x100,%r13d
  je     L157
  cmp    $0x101,%r13d
  jne    L35
L195:
  cmp    $0xab,%eax
  je     L151
  mov    $0x3,%esi
  mov    %ebx,%edi
  cmp    $0xa8,%eax
  je     L7
  call   <com_getc@plt>
  cmp    $0xffffffff,%eax
  je     L1
  cmp    $0xfffffffe,%eax
  jne    L195
  xchg   %ax,%ax
L200:
  lea    D19(%rip),%rdi
  mov    $0xfffffffffffffffc,%r12
  call   <puts@plt>
  jmp    L41
  nopl   0x0(%rax,%rax,1)
L196:
  movslq %eax,%rsi
  shl    $0x8,%r12
  lea    -0x1(%r13),%eax
  add    %rsi,%r12
L157:
  mov    %eax,%r13d
  jmp    L8
  cs nopw 0x0(%rax,%rax,1)
L7:
  call   <com_getc@plt>
  cmp    $0xffffffff,%eax
  je     L1
  cmp    $0xfffffffe,%eax
  je     L200
  mov    %eax,%r13d
  jmp    L8
L151:
  mov    $0xfffffffffffffffe,%r12
  jmp    L41
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D18
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D1
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D19
type_guessed: string

tool: D18: 
Error: readval, i = %i, j = %i, val = %li


tool: D1: readval: ...Device does not answer!

tool: D19: readval: ...Device disconnected!

assistant: ```c
long readval(int fd) {
  int i;
  int j = 257;
  long val = 0;
  while (1) {
    i = com_getc(fd, 3);
    if (i == -1) {
      printf("readval: ...Device does not answer!\n");
      return -3;
    } else if (i == -2) {
      printf("readval: ...Device disconnected!\n");
      return -4;
    }
    switch (j) {
    case 1:
      if (i == 0xAA) {
        return val;
      }
      break;
    case 2:
    case 3:
    case 4:
      val = val * 256 + i;
      j--;
      break;
    case 256:
      j = i;
      break;
    case 257:
      if (i == 0xAB) {
        return -2;
      } else if (i == 0xA8) {
        j = 256;
      }
      break;
    default:
      printf("\nError: readval, i = %i, j = %i, val = %li\n", i, j, val);
      return -2;
    }
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<do_cmd_lua_script>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x666f645f656d6f74,%rax
  movabs $0x77796e615f656c69,%rdx
  mov    %rax,-0x60(%rbp)
  mov    %rdx,-0x58(%rbp)
  movabs $0x474e412865726568,%rax
  movabs $0x5249445f444e4142,%rdx
  mov    %rax,-0x50(%rbp)
  mov    %rdx,-0x48(%rbp)
  movabs $0x27202c45524f435f,%rax
  movabs $0x2e7864695f6e6567,%rdx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  movabs $0x292761756c,%rax
  mov    $0x0,%edx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  lea    -0x60(%rbp),%rax
  mov    $0x50,%edx
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <get_string@plt>
  test   %eax,%eax
  je     L15
  lea    -0x60(%rbp),%rax
  mov    %rax,%rdi
  call   <exec_lua@plt>
  jmp    L25
L15:
  nop
L25:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

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

tool: D6: Script:

assistant: ```c
void do_cmd_lua_script() {
  char script[80] = "tome_dofile_anywhere(ANGBAND_DIR_CORE, 'gen_idx.lua')";
  if (!get_string("Script:", script, 80))
    return;
  exec_lua(script);
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<do_cmd_lua_script>:
  endbr64
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movabs $0x666f645f656d6f74,%rax
  movabs $0x77796e615f656c69,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movabs $0x474e412865726568,%rax
  movabs $0x5249445f444e4142,%rdx
  mov    %rax,0x10(%rsp)
  mov    %rdx,0x18(%rsp)
  movabs $0x27202c45524f435f,%rax
  movabs $0x2e7864695f6e6567,%rdx
  mov    %rax,0x20(%rsp)
  mov    %rdx,0x28(%rsp)
  movabs $0x292761756c,%rax
  mov    $0x0,%edx
  mov    %rax,0x30(%rsp)
  mov    %rdx,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  mov    %rsp,%rsi
  mov    $0x50,%edx
  lea    D1(%rip),%rdi
  call   <get_string@plt>
  test   %eax,%eax
  jne    L4
L24:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x68,%rsp
  ret
L4:
  mov    %rsp,%rdi
  call   <exec_lua@plt>
  jmp    L24
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D1
datatype: string

tool: D1: Script:

assistant: ```c
void do_cmd_lua_script() {
  char script[80] = "tome_dofile_anywhere(ANGBAND_DIR_CORE, 'gen_idx.lua')";
  if (!get_string("Script:", script, 80))
    return;
  exec_lua(script);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<do_cmd_lua_script>:
  endbr64
  push   %rbp
  xor    %edx,%edx
  lea    D5(%rip),%rdi
  sub    $0x60,%rsp
  movdqa D18(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rdx,0x38(%rsp)
  mov    $0x50,%edx
  movaps %xmm0,(%rsp)
  movdqa D26(%rip),%xmm0
  mov    %rbp,%rsi
  movabs $0x292761756c,%rax
  mov    %rax,0x30(%rsp)
  movaps %xmm0,0x10(%rsp)
  movdqa D24(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  pxor   %xmm0,%xmm0
  movaps %xmm0,0x40(%rsp)
  call   <get_string@plt>
  test   %eax,%eax
  je     L6
  mov    %rbp,%rdi
  call   <exec_lua@plt>
L6:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x60,%rsp
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D5
data_type: string
name: parse_data
arguments: label: D18
data_type: byte[16]
name: parse_data
arguments: label: D26
data_type: byte[16]
name: parse_data
arguments: label: D24
data_type: byte[16]

tool: D5: Script:

tool: {"D18": [0x74,0x6f,0x6d,0x65,0x5f,0x64,0x6f,0x66,0x69,0x6c,0x65,0x5f,0x61,0x6e,0x79,0x77]}

tool: {"D26": [0x68,0x65,0x72,0x65,0x28,0x41,0x4e,0x47,0x42,0x41,0x4e,0x44,0x5f,0x44,0x49,0x52]}

tool: {"D24": [0x5f,0x43,0x4f,0x52,0x45,0x2c,0x20,0x27,0x67,0x65,0x6e,0x5f,0x69,0x64,0x78,0x2e]}

assistant: ```c
void do_cmd_lua_script() {
  char script[80] = "tome_dofile_anywhere(ANGBAND_DIR_CORE, 'gen_idx.lua')";
  if (!get_string("Script:", script, 80))
    return;
  exec_lua(script);
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<do_cmd_lua_script>:
  endbr64
  push   %rbp
  xor    %edx,%edx
  lea    D37(%rip),%rdi
  sub    $0x60,%rsp
  movdqa D1(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rdx,0x38(%rsp)
  mov    $0x50,%edx
  movaps %xmm0,(%rsp)
  movdqa D12(%rip),%xmm0
  mov    %rbp,%rsi
  movabs $0x292761756c,%rax
  mov    %rax,0x30(%rsp)
  movaps %xmm0,0x10(%rsp)
  movdqa D9(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  pxor   %xmm0,%xmm0
  movaps %xmm0,0x40(%rsp)
  call   <get_string@plt>
  test   %eax,%eax
  je     L16
  mov    %rbp,%rdi
  call   <exec_lua@plt>
L16:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x60,%rsp
  pop    %rbp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D37
type_guessed: string
name: unpack_data
arguments: data_label: D1
type_guessed: byte[16]
name: unpack_data
arguments: data_label: D12
type_guessed: byte[16]
name: unpack_data
arguments: data_label: D9
type_guessed: byte[16]

tool: D37: Script:

tool: {"D1": [0x74,0x6f,0x6d,0x65,0x5f,0x64,0x6f,0x66,0x69,0x6c,0x65,0x5f,0x61,0x6e,0x79,0x77]}

tool: {"D12": [0x68,0x65,0x72,0x65,0x28,0x41,0x4e,0x47,0x42,0x41,0x4e,0x44,0x5f,0x44,0x49,0x52]}

tool: {"D9": [0x5f,0x43,0x4f,0x52,0x45,0x2c,0x20,0x27,0x67,0x65,0x6e,0x5f,0x69,0x64,0x78,0x2e]}

assistant: ```c
void do_cmd_lua_script() {
  char script[80] = "tome_dofile_anywhere(ANGBAND_DIR_CORE, 'gen_idx.lua')";
  if (!get_string("Script:", script, 80))
    return;
  exec_lua(script);
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rotateMatrix>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%edx
  shr    $0x1f,%edx
  add    %edi,%edx
  sar    %edx
  mov    %edx,-0x2c(%rsp)
  cmp    $0x1,%edi
  jle    L4
  mov    %edi,%r14d
  mov    %rsi,%r12
  movslq %edi,%rax
  lea    -0x1(%rdi),%ebx
  lea    0x1(%rax),%rdx
  lea    0x0(,%rdx,4),%rdi
  mov    %rdi,-0x28(%rsp)
  mov    %rsi,%r15
  mov    $0x1,%ecx
  sub    %rax,%rcx
  lea    0x0(,%rcx,4),%rsi
  mov    %rsi,-0x20(%rsp)
  lea    -0x1(%rax),%rcx
  lea    -0x4(%rdi),%r13
  imul   %rcx,%r13
  add    %r12,%r13
  mov    %rdx,%rsi
  neg    %rsi
  lea    0x0(,%rsi,4),%rdi
  mov    %rdi,-0x10(%rsp)
  imul   %rcx,%rdx
  lea    (%r12,%rdx,4),%rbp
  shl    $0x2,%rcx
  mov    %rcx,-0x18(%rsp)
  add    %rcx,%r12
  lea    0x4(%rdi),%r10
  lea    0x0(,%rax,4),%r9
  imul   %rax,%rax
  sub    %rdx,%rax
  mov    $0x0,%r11d
  lea    -0x4(,%rax,4),%rax
  mov    %rax,-0x8(%rsp)
L17:
  cmp    %r11d,%ebx
  jle    L11
  mov    -0x8(%rsp),%rax
  lea    -0x4(%rbp,%rax,1),%r8
  lea    (%rbx,%rbx,1),%eax
  sub    %r14d,%eax
  shl    $0x2,%rax
  sub    %rax,%r8
  mov    %r12,%rsi
  mov    %rbp,%rax
  mov    %r13,%rcx
  mov    %r15,%rdx
  mov    %r11d,-0x30(%rsp)
L27:
  mov    (%rdx),%edi
  mov    (%rcx),%r11d
  mov    %r11d,(%rdx)
  mov    (%rax),%r11d
  mov    %r11d,(%rcx)
  mov    (%rsi),%r11d
  mov    %r11d,(%rax)
  mov    %edi,(%rsi)
  add    $0x4,%rdx
  add    %r10,%rcx
  sub    $0x4,%rax
  add    %r9,%rsi
  cmp    %r8,%rax
  jne    L27
  mov    -0x30(%rsp),%r11d
L11:
  add    $0x1,%r11d
  sub    $0x1,%ebx
  add    -0x28(%rsp),%r15
  add    -0x20(%rsp),%r13
  add    -0x10(%rsp),%rbp
  add    -0x18(%rsp),%r12
  cmp    %r11d,-0x2c(%rsp)
  jg     L17
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void rotateMatrix(int n, int matrix[n][n]) {
  int layers = n / 2;
  for (int layer = 0; layer < layers; ++layer) {
    int first = layer;
    int last = n - 1 - layer;
    for (int i = first; i < last; ++i) {
      int offset = last - i + first;
      int top = matrix[first][i];
      matrix[first][i] = matrix[offset][first];
      matrix[offset][first] = matrix[last][offset];
      matrix[last][offset] = matrix[i][last];
      matrix[i][last] = top;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rotateMatrix>:
  endbr64
  movslq %edi,%rax
  push   %r15
  mov    %eax,%edx
  push   %r14
  shr    $0x1f,%edx
  push   %r13
  push   %r12
  add    %eax,%edx
  push   %rbp
  sar    %edx
  push   %rbx
  mov    %edx,-0x1c(%rsp)
  cmp    $0x1,%eax
  jle    L18
  mov    $0x1,%ecx
  lea    -0x1(%rax),%r14d
  mov    %rsi,%r12
  mov    %rax,%rbx
  sub    %rax,%rcx
  lea    0x1(%rax),%rdx
  mov    $0x4,%r10d
  sub    $0x2,%ebx
  imul   %rax,%rax
  lea    0x0(,%rcx,4),%rsi
  movslq %r14d,%rcx
  xor    %r11d,%r11d
  lea    0x0(,%rdx,4),%rdi
  mov    %rsi,-0x10(%rsp)
  imul   %rcx,%rdx
  lea    -0x4(%rdi),%r13
  mov    %rdi,-0x18(%rsp)
  sub    %rdi,%r10
  imul   %rcx,%r13
  shl    $0x2,%rcx
  lea    (%r12,%rcx,1),%r15
  lea    0x4(%rcx),%r9
  sub    %rdx,%rax
  lea    (%r12,%rdx,4),%rbp
  lea    -0x4(,%rax,4),%rax
  add    %r12,%r13
  mov    %rax,-0x8(%rsp)
  nopl   0x0(%rax)
L28:
  cmp    %r11d,%r14d
  jle    L11
  mov    -0x8(%rsp),%rax
  mov    %r11d,-0x20(%rsp)
  mov    %r15,%rsi
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    -0x4(%rbp,%rax,1),%r8
  mov    %ebx,%eax
  shl    $0x2,%rax
  sub    %rax,%r8
  mov    %rbp,%rax
  nopl   0x0(%rax)
L13:
  mov    (%rcx),%r11d
  mov    (%rdx),%edi
  sub    $0x4,%rax
  add    $0x4,%rdx
  mov    %r11d,-0x4(%rdx)
  mov    0x4(%rax),%r11d
  mov    %r11d,(%rcx)
  mov    (%rsi),%r11d
  add    %r10,%rcx
  mov    %r11d,0x4(%rax)
  mov    %edi,(%rsi)
  add    %r9,%rsi
  cmp    %r8,%rax
  jne    L13
  mov    -0x20(%rsp),%r11d
L11:
  mov    -0x18(%rsp),%rdi
  add    $0x1,%r11d
  sub    $0x1,%r14d
  sub    $0x2,%ebx
  mov    -0x10(%rsp),%rax
  add    %rdi,%r12
  sub    %rdi,%rbp
  add    %rax,%r13
  sub    %rax,%r15
  cmp    %r11d,-0x1c(%rsp)
  jg     L28
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void rotateMatrix(int n, int matrix[n][n]) {
  int layers = n / 2;
  for (int layer = 0; layer < layers; ++layer) {
    int first = layer;
    int last = n - 1 - layer;
    for (int i = first; i < last; ++i) {
      int offset = last - i + first;
      int top = matrix[first][i];
      matrix[first][i] = matrix[offset][first];
      matrix[offset][first] = matrix[last][offset];
      matrix[last][offset] = matrix[i][last];
      matrix[i][last] = top;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rotateMatrix>:
  endbr64
  movslq %edi,%rax
  push   %r15
  mov    %eax,%edx
  push   %r14
  shr    $0x1f,%edx
  push   %r13
  push   %r12
  add    %eax,%edx
  push   %rbp
  sar    %edx
  push   %rbx
  mov    %eax,-0xc(%rsp)
  mov    %edx,-0x10(%rsp)
  cmp    $0x1,%eax
  jle    L6
  mov    %rax,%rbx
  mov    $0x1,%ecx
  lea    0x1(%rax),%rdx
  mov    %rsi,%r12
  sub    %rax,%rcx
  sub    $0x1,%ebx
  mov    $0x4,%r10d
  xor    %r11d,%r11d
  lea    0x0(,%rcx,4),%rdi
  movslq %ebx,%rcx
  lea    0x0(,%rdx,4),%r15
  imul   %rax,%rax
  lea    -0x4(%r15),%r13
  mov    %rdi,-0x18(%rsp)
  sub    %r15,%r10
  imul   %rcx,%rdx
  imul   %rcx,%r13
  shl    $0x2,%rcx
  lea    (%rsi,%rcx,1),%r14
  lea    0x4(%rcx),%r9
  sub    %rdx,%rax
  lea    (%rsi,%rdx,4),%rbp
  lea    -0x4(,%rax,4),%rax
  add    %rsi,%r13
  mov    %rax,-0x8(%rsp)
  nopw   0x0(%rax,%rax,1)
L7:
  cmp    %r11d,%ebx
  jle    L4
  mov    -0x8(%rsp),%rax
  mov    %r11d,-0x1c(%rsp)
  mov    %r14,%rsi
  mov    %r13,%rcx
  mov    %r12,%rdx
  lea    -0x4(%rbp,%rax,1),%r8
  lea    (%rbx,%rbx,1),%eax
  sub    -0xc(%rsp),%eax
  shl    $0x2,%rax
  sub    %rax,%r8
  mov    %rbp,%rax
  xchg   %ax,%ax
L5:
  mov    (%rcx),%r11d
  mov    (%rdx),%edi
  sub    $0x4,%rax
  add    $0x4,%rdx
  mov    %r11d,-0x4(%rdx)
  mov    0x4(%rax),%r11d
  mov    %r11d,(%rcx)
  mov    (%rsi),%r11d
  add    %r10,%rcx
  mov    %r11d,0x4(%rax)
  mov    %edi,(%rsi)
  add    %r9,%rsi
  cmp    %r8,%rax
  jne    L5
  mov    -0x1c(%rsp),%r11d
L4:
  mov    -0x18(%rsp),%rax
  add    $0x1,%r11d
  sub    $0x1,%ebx
  add    %r15,%r12
  sub    %r15,%rbp
  add    %rax,%r13
  sub    %rax,%r14
  cmp    %r11d,-0x10(%rsp)
  jg     L7
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void rotateMatrix(int n, int matrix[n][n]) {
  int layers = n / 2;
  for (int layer = 0; layer < layers; ++layer) {
    int first = layer;
    int last = n - 1 - layer;
    for (int i = first; i < last; ++i) {
      int offset = last - i + first;
      int top = matrix[first][i];
      matrix[first][i] = matrix[offset][first];
      matrix[offset][first] = matrix[last][offset];
      matrix[last][offset] = matrix[i][last];
      matrix[i][last] = top;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<example_sub_var>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    $0xa,%edi
  call   <compute@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int example_sub_var(int *array, int array_size) {
  compute(10);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<example_sub_var>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <compute@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int example_sub_var(int *array, int array_size) {
  compute(10);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<example_sub_var>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <compute@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int example_sub_var(int *array, int array_size) {
  compute(10);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<example_sub_var>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%edi
  call   <compute@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int example_sub_var(int *array, int array_size) {
  compute(10);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<absf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  comisd -0x18(%rbp),%xmm0
  jbe    L8
  movsd  -0x18(%rbp),%xmm0
  movq   D4(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  jmp    L10
L8:
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
L10:
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D4
type: f32[2]

tool: {"D4": [0.0,-0.0]}

assistant: ```c
double absf(double a) {
  double ret = 0.0;
  if (a < 0) {
    ret = -1 * a;
  } else {
    ret = a;
  }
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<absf>:
  endbr64
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  ja     L6
L10:
  ret
L6:
  xorpd  D5(%rip),%xmm0
  jmp    L10
```

tool_calls: name: read_data
arguments: data_tag: D5
type_guessed: f32[4]

tool: {"D5": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double absf(double a) {
  double ret = 0.0;
  if (a < 0) {
    ret = -1 * a;
  } else {
    ret = a;
  }
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<absf>:
  endbr64
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  ja     L0
  ret
  nop
L0:
  xorpd  D2(%rip),%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type_infered: f32[4]

tool: {"D2": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double absf(double a) {
  double ret = 0.0;
  if (a < 0) {
    ret = -1 * a;
  } else {
    ret = a;
  }
  return ret;
}
```