ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<omb_utils_dir_exists>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <opendir@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L2
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <closedir@plt>
  mov    $0x1,%eax
  jmp    L13
L2:
  mov    $0x0,%eax
L13:
  leave
  ret
```

assistant: ```c
int omb_utils_dir_exists(const char *folder) {
  DIR *fd = opendir(folder);
  if (fd) {
    closedir(fd);
    return 1;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<omb_utils_dir_exists>:
  endbr64
  sub    $0x8,%rsp
  call   <opendir@plt>
  mov    %rax,%rdi
  mov    $0x0,%eax
  test   %rdi,%rdi
  jne    L3
L10:
  add    $0x8,%rsp
  ret
L3:
  call   <closedir@plt>
  mov    $0x1,%eax
  jmp    L10
```

assistant: ```c
int omb_utils_dir_exists(const char *folder) {
  DIR *fd = opendir(folder);
  if (fd) {
    closedir(fd);
    return 1;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<omb_utils_dir_exists>:
  endbr64
  sub    $0x8,%rsp
  call   <opendir@plt>
  mov    %rax,%rdi
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L6
  call   <closedir@plt>
  mov    $0x1,%eax
L6:
  add    $0x8,%rsp
  ret
```

assistant: ```c
int omb_utils_dir_exists(const char *folder) {
  DIR *fd = opendir(folder);
  if (fd) {
    closedir(fd);
    return 1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<omb_utils_dir_exists>:
  endbr64
  sub    $0x8,%rsp
  call   <opendir@plt>
  mov    %rax,%rdi
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L0
  call   <closedir@plt>
  mov    $0x1,%eax
L0:
  add    $0x8,%rsp
  ret
```

assistant: ```c
int omb_utils_dir_exists(const char *folder) {
  DIR *fd = opendir(folder);
  if (fd) {
    closedir(fd);
    return 1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<readMatrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  jmp    L5
L23:
  movl   $0x0,-0xc(%rbp)
  jmp    L14
L31:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0x11(%rbp),%rax
  mov    %rax,%rdx
  mov    %rcx,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movzbl -0x11(%rbp),%eax
  cmp    $0xa,%al
  je     L17
  addl   $0x1,-0xc(%rbp)
L14:
  cmpl   $0x4,-0xc(%rbp)
  jle    L31
  addl   $0x1,-0x10(%rbp)
L5:
  cmpl   $0x4,-0x10(%rbp)
  jle    L23
  jmp    L3
L17:
  nop
L3:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L69
  call   <__stack_chk_fail@plt>
L69:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
data_type: string

tool: D0: %d%c

assistant: ```c
void readMatrix(int matrix[5][5]) {
  int row, column;
  char space_character;
  for (row = 0; row < 5; row++) {
    for (column = 0; column < 5; column++) {
      scanf("%d%c", &matrix[row][column], &space_character);
      if (space_character == '\n') {
        return;
      }
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<readMatrix>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x14(%rdi),%rbp
  lea    0x78(%rdi),%r13
  lea    D6(%rip),%r12
  jmp    L34
L18:
  add    $0x14,%rbp
  cmp    %r13,%rbp
  je     L57
L34:
  lea    -0x14(%rbp),%rbx
L59:
  lea    0x7(%rsp),%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmpb   $0xa,0x7(%rsp)
  je     L57
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L59
  jmp    L18
L57:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L50:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D6
datatype: string

tool: D6: %d%c

assistant: ```c
void readMatrix(int matrix[5][5]) {
  int row, column;
  char space_character;
  for (row = 0; row < 5; row++) {
    for (column = 0; column < 5; column++) {
      scanf("%d%c", &matrix[row][column], &space_character);
      if (space_character == '\n') {
        return;
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readMatrix>:
  endbr64
  push   %r14
  lea    0x78(%rdi),%r14
  push   %r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  lea    0x14(%rdi),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%r13
L26:
  lea    -0x14(%rbp),%rbx
L10:
  mov    %r13,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmpb   $0xa,0x7(%rsp)
  je     L32
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L10
  add    $0x14,%rbp
  cmp    %r14,%rbp
  jne    L26
L32:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L19:
  call   <__stack_chk_fail@plt>
```

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

tool: D0: %d%c

assistant: ```c
void readMatrix(int matrix[5][5]) {
  int row, column;
  char space_character;
  for (row = 0; row < 5; row++) {
    for (column = 0; column < 5; column++) {
      scanf("%d%c", &matrix[row][column], &space_character);
      if (space_character == '\n') {
        return;
      }
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<readMatrix>:
  endbr64
  push   %r14
  lea    0x78(%rdi),%r14
  push   %r13
  push   %r12
  lea    D1(%rip),%r12
  push   %rbp
  lea    0x14(%rdi),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%r13
L11:
  lea    -0x14(%rbp),%rbx
L1:
  mov    %r13,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmpb   $0xa,0x7(%rsp)
  je     L21
  add    $0x4,%rbx
  cmp    %rbx,%rbp
  jne    L1
  add    $0x14,%rbp
  cmp    %r14,%rbp
  jne    L11
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D1
type_infered: string

tool: D1: %d%c

assistant: ```c
void readMatrix(int matrix[5][5]) {
  int row, column;
  char space_character;
  for (row = 0; row < 5; row++) {
    for (column = 0; column < 5; column++) {
      scanf("%d%c", &matrix[row][column], &space_character);
      if (space_character == '\n') {
        return;
      }
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xf80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x1f70(%rbp)
  jmp    L63
L31:
  mov    -0x1f70(%rbp),%rax
  movq   $0x0,-0x1f50(%rbp,%rax,8)
  addq   $0x1,-0x1f70(%rbp)
L63:
  cmpq   $0x3e7,-0x1f70(%rbp)
  jle    L31
  movq   $0x1,-0x18(%rbp)
  movq   $0x0,-0x1f60(%rbp)
  movq   $0x0,-0x1f78(%rbp)
  movq   $0x64,-0x1f68(%rbp)
  jmp    L65
L19:
  movq   $0x3e7,-0x1f70(%rbp)
  jmp    L16
L36:
  mov    -0x1f70(%rbp),%rax
  mov    -0x1f50(%rbp,%rax,8),%rdx
  mov    -0x1f68(%rbp),%rax
  imul   %rax,%rdx
  mov    -0x1f70(%rbp),%rax
  mov    %rdx,-0x1f50(%rbp,%rax,8)
  mov    -0x1f70(%rbp),%rax
  mov    -0x1f50(%rbp,%rax,8),%rdx
  mov    -0x1f78(%rbp),%rax
  add    %rax,%rdx
  mov    -0x1f70(%rbp),%rax
  mov    %rdx,-0x1f50(%rbp,%rax,8)
  mov    -0x1f70(%rbp),%rax
  mov    -0x1f50(%rbp,%rax,8),%rax
  mov    %rax,-0x1f58(%rbp)
  mov    -0x1f58(%rbp),%rcx
  movabs $0xcccccccccccccccd,%rdx
  mov    %rcx,%rax
  mul    %rdx
  shr    $0x3,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    -0x1f70(%rbp),%rax
  mov    %rdx,-0x1f50(%rbp,%rax,8)
  mov    -0x1f58(%rbp),%rax
  movabs $0xcccccccccccccccd,%rdx
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x3,%rax
  mov    %rax,-0x1f78(%rbp)
  subq   $0x1,-0x1f70(%rbp)
L16:
  cmpq   $0x0,-0x1f70(%rbp)
  jns    L36
  subq   $0x1,-0x1f68(%rbp)
L65:
  cmpq   $0x0,-0x1f68(%rbp)
  jg     L19
  mov    -0x1f60(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D6
type_infered: string

tool: D6: 
Sum is : %llu

assistant: ```c
int main() {
  unsigned long long int carry, num[1000], sum, temp;
  long long int i, j;
  for (i = 0; i < 1000; i++)
    num[i] = 0;
  num[1000 - 1] = 1;
  sum = 0;
  carry = 0;
  for (j = 100; j > 0; j--) {
    for (i = 1000 - 1; i >= 0; i--) {
      num[i] *= j;
      num[i] += carry;
      temp = num[i];
      num[i] = temp % 10;
      carry = temp / 10;
    }
  }
  printf("\nSum is : %llu", sum);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3e8,%eax
L27:
  sub    $0x1,%rax
  jne    L27
  mov    $0x64,%edx
L4:
  mov    $0x3e8,%eax
L11:
  sub    $0x1,%rax
  jne    L11
  sub    $0x1,%rdx
  jne    L4
  mov    $0x0,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D2
type_infered: string

tool: D2: 
Sum is : %llu

assistant: ```c
int main() {
  unsigned long long int carry, num[1000], sum, temp;
  long long int i, j;
  for (i = 0; i < 1000; i++)
    num[i] = 0;
  num[1000 - 1] = 1;
  sum = 0;
  carry = 0;
  for (j = 100; j > 0; j--) {
    for (i = 1000 - 1; i >= 0; i--) {
      num[i] *= j;
      num[i] += carry;
      temp = num[i];
      num[i] = temp % 10;
      carry = temp / 10;
    }
  }
  printf("\nSum is : %llu", sum);
  return 0;
}
```
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
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_infered: string

tool: D6: 
Sum is : %llu

assistant: ```c
int main() {
  unsigned long long int carry, num[1000], sum, temp;
  long long int i, j;
  for (i = 0; i < 1000; i++)
    num[i] = 0;
  num[1000 - 1] = 1;
  sum = 0;
  carry = 0;
  for (j = 100; j > 0; j--) {
    for (i = 1000 - 1; i >= 0; i--) {
      num[i] *= j;
      num[i] += carry;
      temp = num[i];
      num[i] = temp % 10;
      carry = temp / 10;
    }
  }
  printf("\nSum is : %llu", sum);
  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
  xor    %esi,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D3
type: string

tool: D3: 
Sum is : %llu

assistant: ```c
int main() {
  unsigned long long int carry, num[1000], sum, temp;
  long long int i, j;
  for (i = 0; i < 1000; i++)
    num[i] = 0;
  num[1000 - 1] = 1;
  sum = 0;
  carry = 0;
  for (j = 100; j > 0; j--) {
    for (i = 1000 - 1; i >= 0; i--) {
      num[i] *= j;
      num[i] += carry;
      temp = num[i];
      num[i] = temp % 10;
      carry = temp / 10;
    }
  }
  printf("\nSum is : %llu", sum);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<func_cross>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movss  %xmm1,-0x1c(%rbp)
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x20(%rbp)
  mov    %edx,%eax
  mov    %al,-0x2c(%rbp)
  fldl   -0x18(%rbp)
  fldt   0x10(%rbp)
  faddp  %st,%st(1)
  flds   -0x1c(%rbp)
  faddp  %st,%st(1)
  fstpt  -0x10(%rbp)
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x38(%rbp)
  fildll -0x38(%rbp)
  fldt   -0x10(%rbp)
  fmulp  %st,%st(1)
  fstpt  -0x10(%rbp)
  cmpb   $0x0,-0x2c(%rbp)
  jle    L8
  movzbl -0x2c(%rbp),%eax
  neg    %eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%esi
  mov    -0x28(%rbp),%rcx
  movss  -0x1c(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  push   0x18(%rbp)
  push   0x10(%rbp)
  mov    %rcx,%rdi
  movaps %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <func_cross>
  add    $0x10,%rsp
  fldt   -0x10(%rbp)
  faddp  %st,%st(1)
  fstpt  -0x10(%rbp)
L8:
  fldt   -0x10(%rbp)
  leave
  ret
```

assistant: ```c
long double func_cross(long double a, double b, float c, long d, int e, char f) {
  long double ret;
  ret = a + (long double)b + (long double)c;
  ret *= (long double)(d + (long)e);
  if (f > 0)
    ret += func_cross(a, b, c, d, e, -f);
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<func_cross>:
  endbr64
  sub    $0x28,%rsp
  fldt   0x30(%rsp)
  movsd  %xmm0,0x10(%rsp)
  movss  %xmm1,0x1c(%rsp)
  fld    %st(0)
  faddl  0x10(%rsp)
  fadds  0x1c(%rsp)
  movslq %esi,%rax
  add    %rdi,%rax
  mov    %rax,(%rsp)
  fildll (%rsp)
  fmulp  %st,%st(1)
  fstpt  (%rsp)
  test   %dl,%dl
  jg     L17
  fstp   %st(0)
L3:
  fldt   (%rsp)
  add    $0x28,%rsp
  ret
L17:
  neg    %edx
  movsbl %dl,%edx
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  call   <func_cross>
  add    $0x10,%rsp
  fldt   (%rsp)
  faddp  %st,%st(1)
  fstpt  (%rsp)
  jmp    L3
```

assistant: ```c
long double func_cross(long double a, double b, float c, long d, int e, char f) {
  long double ret;
  ret = a + (long double)b + (long double)c;
  ret *= (long double)(d + (long)e);
  if (f > 0)
    ret += func_cross(a, b, c, d, e, -f);
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<func_cross>:
  endbr64
  movsd  %xmm0,-0x18(%rsp)
  fldt   0x8(%rsp)
  faddl  -0x18(%rsp)
  movslq %esi,%rsi
  movss  %xmm1,-0xc(%rsp)
  add    %rdi,%rsi
  mov    %rsi,-0x18(%rsp)
  fadds  -0xc(%rsp)
  fildll -0x18(%rsp)
  fmulp  %st,%st(1)
  test   %dl,%dl
  jg     L7
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  fadd   %st(0),%st
  ret
```

assistant: ```c
long double func_cross(long double a, double b, float c, long d, int e, char f) {
  long double ret;
  ret = a + (long double)b + (long double)c;
  ret *= (long double)(d + (long)e);
  if (f > 0)
    ret += func_cross(a, b, c, d, e, -f);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<func_cross>:
  endbr64
  movsd  %xmm0,-0x18(%rsp)
  fldt   0x8(%rsp)
  faddl  -0x18(%rsp)
  movslq %esi,%rsi
  movss  %xmm1,-0xc(%rsp)
  add    %rdi,%rsi
  mov    %rsi,-0x18(%rsp)
  fadds  -0xc(%rsp)
  fildll -0x18(%rsp)
  fmulp  %st,%st(1)
  test   %dl,%dl
  jg     L6
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  fadd   %st(0),%st
  ret
```

assistant: ```c
long double func_cross(long double a, double b, float c, long d, int e, char f) {
  long double ret;
  ret = a + (long double)b + (long double)c;
  ret *= (long double)(d + (long)e);
  if (f > 0)
    ret += func_cross(a, b, c, d, e, -f);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<r>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D119(%rip),%eax
  shl    $0xb,%eax
  mov    %eax,%edx
  mov    D119(%rip),%eax
  xor    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    D70(%rip),%eax
  mov    %eax,D58(%rip)
  mov    D58(%rip),%eax
  mov    %eax,D54(%rip)
  mov    D54(%rip),%eax
  mov    %eax,D64(%rip)
  mov    D54(%rip),%eax
  shr    $0x13,%eax
  mov    %eax,%edx
  mov    D54(%rip),%eax
  xor    %edx,%eax
  xor    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  xor    %edx,%eax
  mov    %eax,D64(%rip)
  mov    D64(%rip),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned r(void) {
  static unsigned x = 0xdeadbeefU;
  static unsigned y = 0x12347654U;
  static unsigned z = 0x1a2b3c4dU;
  static unsigned w = 0x87654321U;
  unsigned t = x ^ (x << 11);
  x = y;
  y = z;
  z = w;
  w = w ^ (w >> 19) ^ t ^ (t >> 8);
  return w;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<r>:
  endbr64
  mov    D34(%rip),%eax
  mov    %eax,%edx
  shl    $0xb,%edx
  xor    %eax,%edx
  mov    D1(%rip),%eax
  mov    %eax,D50(%rip)
  mov    D50(%rip),%eax
  mov    %eax,D5(%rip)
  mov    D5(%rip),%ecx
  mov    %ecx,D53(%rip)
  mov    %ecx,%eax
  shr    $0x13,%eax
  xor    %ecx,%eax
  xor    %edx,%eax
  shr    $0x8,%edx
  xor    %edx,%eax
  mov    %eax,D53(%rip)
  ret
```

assistant: ```c
unsigned r(void) {
  static unsigned x = 0xdeadbeefU;
  static unsigned y = 0x12347654U;
  static unsigned z = 0x1a2b3c4dU;
  static unsigned w = 0x87654321U;
  unsigned t = x ^ (x << 11);
  x = y;
  y = z;
  z = w;
  w = w ^ (w >> 19) ^ t ^ (t >> 8);
  return w;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<r>:
  endbr64
  mov    D8(%rip),%eax
  mov    D19(%rip),%ecx
  mov    %eax,%edx
  shl    $0xb,%edx
  xor    %eax,%edx
  mov    D73(%rip),%eax
  mov    %eax,D46(%rip)
  mov    D46(%rip),%eax
  mov    %ecx,D1(%rip)
  mov    %eax,D1(%rip)
  mov    %ecx,%eax
  shr    $0x13,%eax
  xor    %ecx,%eax
  xor    %edx,%eax
  shr    $0x8,%edx
  xor    %edx,%eax
  mov    %eax,D1(%rip)
  ret
```

assistant: ```c
unsigned r(void) {
  static unsigned x = 0xdeadbeefU;
  static unsigned y = 0x12347654U;
  static unsigned z = 0x1a2b3c4dU;
  static unsigned w = 0x87654321U;
  unsigned t = x ^ (x << 11);
  x = y;
  y = z;
  z = w;
  w = w ^ (w >> 19) ^ t ^ (t >> 8);
  return w;
}
```
user: What is the c source code of the assembly code below:

```asm
<r>:
  endbr64
  mov    D20(%rip),%eax
  mov    D51(%rip),%ecx
  mov    %eax,%edx
  shl    $0xb,%edx
  xor    %eax,%edx
  mov    D52(%rip),%eax
  mov    %eax,D29(%rip)
  mov    D29(%rip),%eax
  mov    %ecx,D66(%rip)
  mov    %eax,D66(%rip)
  mov    %ecx,%eax
  shr    $0x13,%eax
  xor    %ecx,%eax
  xor    %edx,%eax
  shr    $0x8,%edx
  xor    %edx,%eax
  mov    %eax,D66(%rip)
  ret
```

assistant: ```c
unsigned r(void) {
  static unsigned x = 0xdeadbeefU;
  static unsigned y = 0x12347654U;
  static unsigned z = 0x1a2b3c4dU;
  static unsigned w = 0x87654321U;
  unsigned t = x ^ (x << 11);
  x = y;
  y = z;
  z = w;
  w = w ^ (w >> 19) ^ t ^ (t >> 8);
  return w;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<copyBytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L7:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L7
  nop
  pop    %rbp
  ret
```

assistant: ```c
unsigned char *copyBytes(unsigned char *sourceStr, unsigned char *destinationStr, int len) {
  int i;
  for (i = 0; i < len; i++) {
    destinationStr[i] = sourceStr[i];
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<copyBytes>:
  endbr64
  test   %edx,%edx
  jle    L18
  mov    %edx,%edx
  mov    $0x0,%eax
L8:
  movzbl (%rdi,%rax,1),%ecx
  mov    %cl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L8
L18:
  ret
```

assistant: ```c
unsigned char *copyBytes(unsigned char *sourceStr, unsigned char *destinationStr, int len) {
  int i;
  for (i = 0; i < len; i++) {
    destinationStr[i] = sourceStr[i];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<copyBytes>:
  endbr64
  test   %edx,%edx
  jle    L7
  movslq %edx,%rdx
  xor    %eax,%eax
  nopl   (%rax)
L8:
  movzbl (%rdi,%rax,1),%ecx
  mov    %cl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L8
L7:
  ret
```

assistant: ```c
unsigned char *copyBytes(unsigned char *sourceStr, unsigned char *destinationStr, int len) {
  int i;
  for (i = 0; i < len; i++) {
    destinationStr[i] = sourceStr[i];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copyBytes>:
  endbr64
  mov    %rdi,%rcx
  test   %edx,%edx
  jle    L29
  lea    0x1(%rdi),%r9
  mov    %rsi,%rdi
  lea    -0x1(%rdx),%eax
  mov    %edx,%r8d
  sub    %r9,%rdi
  cmp    $0xe,%rdi
  jbe    L54
  cmp    $0x6,%eax
  jbe    L54
  cmp    $0xe,%eax
  jbe    L1
  mov    %edx,%edi
  xor    %eax,%eax
  shr    $0x4,%edi
  shl    $0x4,%rdi
  cs nopw 0x0(%rax,%rax,1)
L146:
  movdqu (%rcx,%rax,1),%xmm0
  movups %xmm0,(%rsi,%rax,1)
  add    $0x10,%rax
  cmp    %rdi,%rax
  jne    L146
  mov    %edx,%edi
  and    $0xfffffff0,%edi
  mov    %edi,%eax
  cmp    %edi,%edx
  je     L29
  mov    %edx,%r8d
  sub    %edi,%r8d
  lea    -0x1(%r8),%r9d
  cmp    $0x6,%r9d
  jbe    L77
L51:
  mov    (%rcx,%rdi,1),%r9
  mov    %r9,(%rsi,%rdi,1)
  mov    %r8d,%edi
  and    $0xfffffff8,%edi
  add    %edi,%eax
  cmp    %edi,%r8d
  je     L29
L77:
  movslq %eax,%rdi
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x1(%rax),%edi
  cmp    %edi,%edx
  jle    L29
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x2(%rax),%edi
  cmp    %edi,%edx
  jle    L29
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x3(%rax),%edi
  cmp    %edi,%edx
  jle    L29
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x4(%rax),%edi
  cmp    %edi,%edx
  jle    L29
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x5(%rax),%edi
  cmp    %edi,%edx
  jle    L29
  movslq %edi,%rdi
  add    $0x6,%eax
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  cmp    %eax,%edx
  jle    L29
  cltq
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,(%rsi,%rax,1)
  ret
  nopw   0x0(%rax,%rax,1)
L54:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L98:
  movzbl (%rcx,%rax,1),%edi
  mov    %dil,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L98
L29:
  ret
L1:
  xor    %edi,%edi
  xor    %eax,%eax
  jmp    L51
```

assistant: ```c
unsigned char *copyBytes(unsigned char *sourceStr, unsigned char *destinationStr, int len) {
  int i;
  for (i = 0; i < len; i++) {
    destinationStr[i] = sourceStr[i];
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<x23_teardown>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x110(%rbp),%rax
  mov    $0x1,%edx
  lea    D8(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x110(%rbp),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type_guessed: string

tool: D8: iptables -D INPUT -p tcp -m multiport --sport 53,80 --tcp-flags SYN,ACK SYN,ACK -j NFQUEUE --queue-num %d

assistant: ```c
int x23_teardown() {
  char cmd[256];
  sprintf(cmd, "iptables -D INPUT -p tcp -m multiport --sport 53,80 --tcp-flags SYN,ACK SYN,ACK -j NFQUEUE --queue-num %d", 1);
  system(cmd);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<x23_teardown>:
  endbr64
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    $0x1,%edx
  lea    D7(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rdi
  call   <system@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x110,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D7
datatype: string

tool: D7: iptables -D INPUT -p tcp -m multiport --sport 53,80 --tcp-flags SYN,ACK SYN,ACK -j NFQUEUE --queue-num %d

assistant: ```c
int x23_teardown() {
  char cmd[256];
  sprintf(cmd, "iptables -D INPUT -p tcp -m multiport --sport 53,80 --tcp-flags SYN,ACK SYN,ACK -j NFQUEUE --queue-num %d", 1);
  system(cmd);
  return 0;
}
```