ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x3,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int foo(int xxx) {
  int yyy = xxx + 3;
  int z = yyy + xxx;
  return z + xxx;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<foo>:
  endbr64
  lea    0x3(%rdi,%rdi,2),%eax
  ret
```

assistant: ```c
int foo(int xxx) {
  int yyy = xxx + 3;
  int z = yyy + xxx;
  return z + xxx;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<foo>:
  endbr64
  lea    0x3(%rdi,%rdi,2),%eax
  ret
```

assistant: ```c
int foo(int xxx) {
  int yyy = xxx + 3;
  int z = yyy + xxx;
  return z + xxx;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<foo>:
  endbr64
  lea    0x3(%rdi,%rdi,2),%eax
  ret
```

assistant: ```c
int foo(int xxx) {
  int yyy = xxx + 3;
  int z = yyy + xxx;
  return z + xxx;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <test_local_continuation@plt>
  mov    $0x0,%eax
  call   <test_empty@plt>
  mov    $0x0,%eax
  call   <test_wait@plt>
  mov    $0x0,%eax
  call   <test_yield@plt>
  mov    $0x0,%eax
  call   <test_exit@plt>
  mov    $0x0,%eax
  call   <test_loop@plt>
  mov    $0x0,%eax
  call   <test_queues@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  test_local_continuation();
  test_empty();
  test_wait();
  test_yield();
  test_exit();
  test_loop();
  test_queues();
  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
  mov    $0x0,%eax
  call   <test_local_continuation@plt>
  mov    $0x0,%eax
  call   <test_empty@plt>
  mov    $0x0,%eax
  call   <test_wait@plt>
  mov    $0x0,%eax
  call   <test_yield@plt>
  mov    $0x0,%eax
  call   <test_exit@plt>
  mov    $0x0,%eax
  call   <test_loop@plt>
  mov    $0x0,%eax
  call   <test_queues@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  test_local_continuation();
  test_empty();
  test_wait();
  test_yield();
  test_exit();
  test_loop();
  test_queues();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_local_continuation@plt>
  xor    %eax,%eax
  call   <test_empty@plt>
  xor    %eax,%eax
  call   <test_wait@plt>
  xor    %eax,%eax
  call   <test_yield@plt>
  xor    %eax,%eax
  call   <test_exit@plt>
  xor    %eax,%eax
  call   <test_loop@plt>
  xor    %eax,%eax
  call   <test_queues@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  test_local_continuation();
  test_empty();
  test_wait();
  test_yield();
  test_exit();
  test_loop();
  test_queues();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_local_continuation@plt>
  xor    %eax,%eax
  call   <test_empty@plt>
  xor    %eax,%eax
  call   <test_wait@plt>
  xor    %eax,%eax
  call   <test_yield@plt>
  xor    %eax,%eax
  call   <test_exit@plt>
  xor    %eax,%eax
  call   <test_loop@plt>
  xor    %eax,%eax
  call   <test_queues@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  test_local_continuation();
  test_empty();
  test_wait();
  test_yield();
  test_exit();
  test_loop();
  test_queues();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<multiply372>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L26
L24:
  movl   $0x0,-0x3c(%rbp)
  jmp    L19
L11:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L19:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L11
  addl   $0x1,-0x40(%rbp)
L26:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L24
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply372(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<multiply372>:
  endbr64
  test   %edi,%edi
  jle    L7
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L20:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L3:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L3
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L20
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  ret
```

assistant: ```c
void multiply372(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<multiply372>:
  endbr64
  test   %edi,%edi
  jle    L13
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L17:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L17
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L8
  pop    %rbx
  ret
L13:
  ret
```

assistant: ```c
void multiply372(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<multiply372>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L23
  mov    %rbp,%r11
  mov    %esi,%ebx
  lea    0x4(%rdx),%r10
  mov    %r8,%rax
  mov    %r11d,%edx
  mov    %r11d,%esi
  lea    0x4(%rcx),%r9
  shl    $0x2,%rbp
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  movd   %ebx,%xmm4
  xor    %r12d,%r12d
  lea    0x1(%rsi),%ecx
  lea    0x2(%rsi),%edi
  mov    %esi,-0x24(%rsp)
  movslq %esi,%r13
  lea    -0x1(%rdx),%r8d
  mov    %r11d,%esi
  lea    -0x1(%r11),%edx
  mov    %ecx,-0x20(%rsp)
  pshufd $0x0,%xmm4,%xmm2
  mov    %edi,-0x1c(%rsp)
  add    $0x1,%r8
  movslq %ecx,%r14
  mov    %edx,-0x28(%rsp)
  movdqa %xmm2,%xmm3
  shl    $0x4,%r8
  movslq %edi,%r15
  mov    %rsi,-0x10(%rsp)
  psrlq  $0x20,%xmm3
  nopl   (%rax)
L6:
  mov    %rax,%rdx
  lea    -0x4(%r10),%rcx
  lea    -0x4(%r9),%rsi
  sub    %r10,%rdx
  cmp    $0x8,%rdx
  mov    %rax,%rdx
  seta   %dil
  sub    %r9,%rdx
  cmp    $0x8,%rdx
  seta   %dl
  test   %dl,%dil
  je     L42
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L42
  nopl   0x0(%rax)
L27:
  movdqu (%rcx,%rdx,1),%xmm0
  movdqu (%rcx,%rdx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqu (%rsi,%rdx,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %r8,%rdx
  jne    L27
  cmp    -0x24(%rsp),%r11d
  je     L7
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L7
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L7
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L7:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L6
L23:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L42:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L78:
  mov    (%rcx,%rdx,4),%edi
  imul   %ebx,%edi
  add    (%rsi,%rdx,4),%edi
  mov    %edi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r8
  jne    L78
  mov    -0x18(%rsp),%r8
  jmp    L7
```

assistant: ```c
void multiply372(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
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
  finit
  mov    $0x0,%eax
  call   <do_fprem@plt>
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <do_fprem1@plt>
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <do_fsincos@plt>
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
datatype: string
name: retrieve_data
arguments: data_tag: D22
datatype: string
name: retrieve_data
arguments: data_tag: D5
datatype: string

tool: D6: fprem   %f


tool: D22: fprem1  %f


tool: D5: fsincos %f


assistant: ```c
int main(void) {
  __asm__ __volatile__("finit");
  printf("fprem   %f\n", do_fprem());
  printf("fprem1  %f\n", do_fprem1());
  printf("fsincos %f\n", do_fsincos());
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  finit
  mov    $0x0,%eax
  call   <do_fprem@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <do_fprem1@plt>
  mov    %eax,%esi
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <do_fsincos@plt>
  mov    %eax,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D3
datatype: string
name: read_data
arguments: data_tag: D12
datatype: string
name: read_data
arguments: data_tag: D6
datatype: string

tool: D3: fprem   %f


tool: D12: fprem1  %f


tool: D6: fsincos %f


assistant: ```c
int main(void) {
  __asm__ __volatile__("finit");
  printf("fprem   %f\n", do_fprem());
  printf("fprem1  %f\n", do_fprem1());
  printf("fsincos %f\n", do_fsincos());
  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
  finit
  xor    %eax,%eax
  call   <do_fprem@plt>
  lea    D8(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <do_fprem1@plt>
  lea    D24(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <do_fsincos@plt>
  lea    D19(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
type_guessed: string
name: read_data
arguments: data_label: D24
type_guessed: string
name: read_data
arguments: data_label: D19
type_guessed: string

tool: D8: fprem   %f


tool: D24: fprem1  %f


tool: D19: fsincos %f


assistant: ```c
int main(void) {
  __asm__ __volatile__("finit");
  printf("fprem   %f\n", do_fprem());
  printf("fprem1  %f\n", do_fprem1());
  printf("fsincos %f\n", do_fsincos());
  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
  sub    $0x8,%rsp
  finit
  xor    %eax,%eax
  call   <do_fprem@plt>
  lea    D12(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <do_fprem1@plt>
  lea    D20(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <do_fsincos@plt>
  lea    D13(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D12
data_type: string
name: retrieve_data
arguments: tag: D20
data_type: string
name: retrieve_data
arguments: tag: D13
data_type: string

tool: D12: fprem   %f


tool: D20: fprem1  %f


tool: D13: fsincos %f


assistant: ```c
int main(void) {
  __asm__ __volatile__("finit");
  printf("fprem   %f\n", do_fprem());
  printf("fprem1  %f\n", do_fprem1());
  printf("fsincos %f\n", do_fsincos());
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<xrop3_218_1_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x18(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  and    -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  notl   -0xc(%rbp)
  mov    -0x10(%rbp),%eax
  and    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  xor    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  not    %eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  not    %eax
  and    %eax,%edx
  mov    -0x18(%rbp),%eax
  and    -0x14(%rbp),%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%eax
  and    -0x10(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_218_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = S & D;
  stk3 = ~stk3;
  stk2 = T & stk3;
  stk1 = D ^ stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<xrop3_218_1_1>:
  endbr64
  movzbl %dil,%edi
  movzbl %sil,%esi
  mov    %esi,%eax
  shl    $0x8,%eax
  or     %esi,%eax
  not    %edi
  mov    %edi,%edx
  and    %eax,%edx
  mov    %edx,%eax
  xor    $0xff,%dl
  xor    %edx,%eax
  ret
```

assistant: ```c
unsigned xrop3_218_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = S & D;
  stk3 = ~stk3;
  stk2 = T & stk3;
  stk1 = D ^ stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xrop3_218_1_1>:
  endbr64
  movzbl %sil,%esi
  movzbl %dil,%edi
  mov    %esi,%eax
  not    %edi
  shl    $0x8,%eax
  mov    %edi,%edx
  or     %esi,%eax
  and    %eax,%edx
  mov    %edx,%eax
  xor    $0xff,%dl
  xor    %edx,%eax
  ret
```

assistant: ```c
unsigned xrop3_218_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = S & D;
  stk3 = ~stk3;
  stk2 = T & stk3;
  stk1 = D ^ stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<xrop3_218_1_1>:
  endbr64
  movzbl %sil,%esi
  movzbl %dil,%edi
  mov    %esi,%eax
  not    %edi
  shl    $0x8,%eax
  mov    %edi,%edx
  or     %esi,%eax
  and    %eax,%edx
  mov    %edx,%eax
  xor    $0xff,%dl
  xor    %edx,%eax
  ret
```

assistant: ```c
unsigned xrop3_218_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = S & D;
  stk3 = ~stk3;
  stk2 = T & stk3;
  stk1 = D ^ stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x18(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L23
L13:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x14(%rbp)
L23:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L13
  mov    -0x1c(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_input@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x1c(%rbp),%edx
  lea    -0x18(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <print_permutations@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  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: read_data
arguments: data_label: D31
data_type: string
name: read_data
arguments: data_label: D39
data_type: string
name: read_data
arguments: data_label: D33
data_type: string
name: read_data
arguments: data_label: D19
data_type: string
name: read_data
arguments: data_label: D34
data_type: string

tool: D31: %d


tool: D39: Given string length: %d


tool: D33: %c

tool: D19: All possible permutations of %s are: 


tool: D34: 
Total of %d permutations printed


assistant: ```c
int main() {
  int i, length, counter = 0;
  char *str;
  scanf("%d\n", &length);
  printf("Given string length: %d\n", length);
  str = (char *)malloc(length * sizeof(char));
  for (i = 0; i < length; i++) {
    scanf("%c", (str + i));
  }
  print_input(str, length);
  printf("All possible permutations of %s are: \n", str);
  print_permutations(str, 0, length, &counter);
  printf("\nTotal of %d permutations printed\n", counter);
  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
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  mov    %rsp,%rsi
  lea    D26(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%esi
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    (%rsp),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    (%rsp),%esi
  test   %esi,%esi
  jle    L5
  mov    $0x0,%ebx
  lea    D15(%rip),%r12
L12:
  lea    0x0(%rbp,%rbx,1),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    (%rsp),%esi
  add    $0x1,%rbx
  cmp    %ebx,%esi
  jg     L12
L5:
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %rbp,%rsi
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rcx
  mov    (%rsp),%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <print_permutations@plt>
  mov    0x4(%rsp),%esi
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D26
type: string
name: extract_symbolic_data
arguments: label: D17
type: string
name: extract_symbolic_data
arguments: label: D15
type: string
name: extract_symbolic_data
arguments: label: D33
type: string
name: extract_symbolic_data
arguments: label: D8
type: string

tool: D26: %d


tool: D17: Given string length: %d


tool: D15: %c

tool: D33: All possible permutations of %s are: 


tool: D8: 
Total of %d permutations printed


assistant: ```c
int main() {
  int i, length, counter = 0;
  char *str;
  scanf("%d\n", &length);
  printf("Given string length: %d\n", length);
  str = (char *)malloc(length * sizeof(char));
  for (i = 0; i < length; i++) {
    scanf("%c", (str + i));
  }
  print_input(str, length);
  printf("All possible permutations of %s are: \n", str);
  print_permutations(str, 0, length, &counter);
  printf("\nTotal of %d permutations printed\n", counter);
  return 0;
}
```