ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<mp_squh>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  mov    %rcx,-0x18(%rbp)
  mov    %r8d,-0x1c(%rbp)
  mov    %r9,-0x28(%rbp)
  mov    -0x28(%rbp),%rdi
  mov    -0x10(%rbp),%rcx
  mov    -0x1c(%rbp),%edx
  mov    -0x8(%rbp),%esi
  mov    -0x4(%rbp),%eax
  mov    %rdi,%r9
  mov    %rcx,%r8
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <mp_mul_i2d@plt>
  mov    0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rsi
  mov    -0x1c(%rbp),%eax
  mov    %rdx,%rcx
  mov    0x10(%rbp),%rdx
  mov    %eax,%edi
  call   <mp_mul_csqu@plt>
  mov    -0x18(%rbp),%rdi
  mov    -0x28(%rbp),%rcx
  mov    -0x1c(%rbp),%edx
  mov    -0x8(%rbp),%esi
  mov    -0x4(%rbp),%eax
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <mp_mul_d2i@plt>
  nop
  leave
  ret
```

assistant: ```c
void mp_squh(int n, int radix, int in[], int out[], int nfft, double outfft[], int ip[], double w[]) {
  void mp_mul_i2d(int n, int radix, int nfft, int shift, int in[], double dout[]);
  void mp_mul_csqu(int nfft, double d1[], int ip[], double w[]);
  void mp_mul_d2i(int n, int radix, int nfft, double din[], int out[]);
  mp_mul_i2d(n, radix, nfft, 0, in, outfft);
  mp_mul_csqu(nfft, outfft, ip, w);
  mp_mul_d2i(n, radix, nfft, outfft, out);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mp_squh>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %esi,%r13d
  mov    %rcx,%r14
  mov    %r8d,%ebx
  mov    %r9,%rbp
  mov    %rdx,%r8
  mov    $0x0,%ecx
  mov    %ebx,%edx
  call   <mp_mul_i2d@plt>
  mov    0x38(%rsp),%rcx
  mov    0x30(%rsp),%rdx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <mp_mul_csqu@plt>
  mov    %r14,%r8
  mov    %rbp,%rcx
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %r12d,%edi
  call   <mp_mul_d2i@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void mp_squh(int n, int radix, int in[], int out[], int nfft, double outfft[], int ip[], double w[]) {
  void mp_mul_i2d(int n, int radix, int nfft, int shift, int in[], double dout[]);
  void mp_mul_csqu(int nfft, double d1[], int ip[], double w[]);
  void mp_mul_d2i(int n, int radix, int nfft, double din[], int out[]);
  mp_mul_i2d(n, radix, nfft, 0, in, outfft);
  mp_mul_csqu(nfft, outfft, ip, w);
  mp_mul_d2i(n, radix, nfft, outfft, out);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<mp_squh>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %r9,%r13
  push   %r12
  mov    %r8d,%r12d
  mov    %rdx,%r8
  push   %rbp
  mov    %r12d,%edx
  mov    %edi,%ebp
  push   %rbx
  mov    %rcx,%rbx
  xor    %ecx,%ecx
  sub    $0x18,%rsp
  mov    0x58(%rsp),%r10
  mov    0x50(%rsp),%r15
  mov    %r10,0x8(%rsp)
  call   <mp_mul_i2d@plt>
  mov    0x8(%rsp),%rcx
  mov    %r15,%rdx
  mov    %r13,%rsi
  mov    %r12d,%edi
  call   <mp_mul_csqu@plt>
  add    $0x18,%rsp
  mov    %rbx,%r8
  mov    %r13,%rcx
  pop    %rbx
  mov    %r12d,%edx
  mov    %r14d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <mp_mul_d2i@plt>
```

assistant: ```c
void mp_squh(int n, int radix, int in[], int out[], int nfft, double outfft[], int ip[], double w[]) {
  void mp_mul_i2d(int n, int radix, int nfft, int shift, int in[], double dout[]);
  void mp_mul_csqu(int nfft, double d1[], int ip[], double w[]);
  void mp_mul_d2i(int n, int radix, int nfft, double din[], int out[]);
  mp_mul_i2d(n, radix, nfft, 0, in, outfft);
  mp_mul_csqu(nfft, outfft, ip, w);
  mp_mul_d2i(n, radix, nfft, outfft, out);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<mp_squh>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %r9,%r13
  push   %r12
  mov    %r8d,%r12d
  mov    %rdx,%r8
  push   %rbp
  mov    %r12d,%edx
  mov    %edi,%ebp
  push   %rbx
  mov    %rcx,%rbx
  xor    %ecx,%ecx
  sub    $0x18,%rsp
  mov    0x58(%rsp),%r10
  mov    0x50(%rsp),%r15
  mov    %r10,0x8(%rsp)
  call   <mp_mul_i2d@plt>
  mov    0x8(%rsp),%rcx
  mov    %r15,%rdx
  mov    %r13,%rsi
  mov    %r12d,%edi
  call   <mp_mul_csqu@plt>
  add    $0x18,%rsp
  mov    %rbx,%r8
  mov    %r13,%rcx
  pop    %rbx
  mov    %r12d,%edx
  mov    %r14d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <mp_mul_d2i@plt>
```

assistant: ```c
void mp_squh(int n, int radix, int in[], int out[], int nfft, double outfft[], int ip[], double w[]) {
  void mp_mul_i2d(int n, int radix, int nfft, int shift, int in[], double dout[]);
  void mp_mul_csqu(int nfft, double d1[], int ip[], double w[]);
  void mp_mul_d2i(int n, int radix, int nfft, double din[], int out[]);
  mp_mul_i2d(n, radix, nfft, 0, in, outfft);
  mp_mul_csqu(nfft, outfft, ip, w);
  mp_mul_d2i(n, radix, nfft, outfft, out);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ts_eit_test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_test1@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_dump@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_free@plt>
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_test2@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_dump@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_free@plt>
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_test3@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_dump@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_free@plt>
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_test4@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_dump@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ts_eit_free@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void ts_eit_test(void) {
  struct ts_eit *eit;
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test1(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test2(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test3(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test4(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ts_eit_test>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test1@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <ts_eit_free@plt>
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test2@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbx,%rdi
  call   <ts_eit_free@plt>
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test3@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbx,%rdi
  call   <ts_eit_free@plt>
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test4@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbx,%rdi
  call   <ts_eit_free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  pop    %rbx
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ts_eit_test(void) {
  struct ts_eit *eit;
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test1(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test2(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test3(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test4(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ts_eit_test>:
  endbr64
  push   %rbp
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test1@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbp,%rdi
  call   <ts_eit_free@plt>
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test2@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbp,%rdi
  call   <ts_eit_free@plt>
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test3@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbp,%rdi
  call   <ts_eit_free@plt>
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test4@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbp,%rdi
  call   <ts_eit_free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ts_eit_test(void) {
  struct ts_eit *eit;
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test1(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test2(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test3(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test4(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ts_eit_test>:
  endbr64
  push   %rbp
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test1@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbp,%rdi
  call   <ts_eit_free@plt>
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test2@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbp,%rdi
  call   <ts_eit_free@plt>
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test3@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbp,%rdi
  call   <ts_eit_free@plt>
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <ts_eit_alloc_init@plt>
  mov    %rax,%rdi
  mov    %rax,(%rsp)
  call   <ts_eit_test4@plt>
  mov    (%rsp),%rdi
  call   <ts_eit_dump@plt>
  mov    %rbp,%rdi
  call   <ts_eit_free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void ts_eit_test(void) {
  struct ts_eit *eit;
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test1(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test2(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test3(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
  eit = ts_eit_alloc_init(1, 2, 3, 1, 0, 0);
  ts_eit_test4(eit);
  ts_eit_dump(eit);
  ts_eit_free(&eit);
}
```
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    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x94(%rbp)
  movl   $0x0,-0x90(%rbp)
  movl   $0x65,-0x8c(%rbp)
  jmp    L50
L3:
  movl   $0x3,-0x88(%rbp)
  jmp    L78
L7:
  mov    -0x8c(%rbp),%eax
  cltd
  idivl  -0x88(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L62
  addl   $0x1,-0x94(%rbp)
L62:
  addl   $0x1,-0x88(%rbp)
L78:
  mov    -0x8c(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  add    $0x1,%eax
  cmp    %eax,-0x88(%rbp)
  jle    L7
  cmpl   $0x0,-0x94(%rbp)
  jne    L70
  mov    -0x90(%rbp),%eax
  cltq
  mov    -0x8c(%rbp),%edx
  mov    %edx,-0x80(%rbp,%rax,4)
  addl   $0x1,-0x90(%rbp)
L70:
  movl   $0x0,-0x94(%rbp)
  addl   $0x2,-0x8c(%rbp)
L50:
  cmpl   $0x1e,-0x90(%rbp)
  jne    L3
  movl   $0x0,-0x84(%rbp)
  jmp    L57
L38:
  mov    -0x84(%rbp),%eax
  cltq
  mov    -0x80(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x84(%rbp)
L57:
  cmpl   $0x1d,-0x84(%rbp)
  jle    L38
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L89
  call   <__stack_chk_fail@plt>
L89:
  leave
  ret
```

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

tool: D6: %d


assistant: ```c
void main() {
  int primo = 0, vzs = 0, X[30];
  for (int i = 101; vzs != 30; i += 2) {
    for (int j = 3; j <= (int)sqrt(i) + 1; j++) {
      if (i % j == 0) {
        primo++;
      }
    }
    if (primo == 0) {
      X[vzs] = i;
      vzs++;
    }
    primo = 0;
  }
  for (int i = 0; i < 30; i++) {
    printf("%d\n", X[i]);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    $0x65,%ebp
  mov    $0x0,%r13d
  jmp    L68
L16:
  mov    %ebp,%eax
  cltd
  idiv   %ebx
  cmp    $0x1,%edx
  adc    $0x0,%r12d
  add    $0x1,%ebx
L36:
  mov    %ebp,%edi
  call   <sqrt@plt>
  add    $0x1,%eax
  cmp    %ebx,%eax
  jge    L16
  test   %r12d,%r12d
  jne    L54
  movslq %r13d,%rax
  mov    %ebp,(%rsp,%rax,4)
  add    $0x1,%r13d
L54:
  add    $0x2,%ebp
  cmp    $0x1e,%r13d
  je     L8
L68:
  mov    $0x3,%ebx
  mov    $0x0,%r12d
  jmp    L36
L8:
  mov    %rsp,%rbx
  lea    0x78(%rsp),%r12
  lea    D6(%rip),%rbp
L12:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L12
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
type_guessed: string

tool: D6: %d


assistant: ```c
void main() {
  int primo = 0, vzs = 0, X[30];
  for (int i = 101; vzs != 30; i += 2) {
    for (int j = 3; j <= (int)sqrt(i) + 1; j++) {
      if (i % j == 0) {
        primo++;
      }
    }
    if (primo == 0) {
      X[vzs] = i;
      vzs++;
    }
    primo = 0;
  }
  for (int i = 0; i < 30; i++) {
    printf("%d\n", X[i]);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0x65,%r12d
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L18:
  mov    $0x3,%ebx
  xor    %ebp,%ebp
  jmp    L46
  nopl   0x0(%rax)
L54:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  cmp    $0x1,%edx
  adc    $0x0,%ebp
  add    $0x1,%ebx
L46:
  mov    %r12d,%edi
  call   <sqrt@plt>
  add    $0x1,%eax
  cmp    %ebx,%eax
  jge    L54
  test   %ebp,%ebp
  jne    L27
  movslq %r13d,%rax
  add    $0x1,%r13d
  mov    %r12d,(%rsp,%rax,4)
L27:
  add    $0x2,%r12d
  cmp    $0x1e,%r13d
  jne    L18
  mov    %rsp,%rbx
  lea    0x78(%rsp),%r12
  lea    D2(%rip),%rbp
  cs nopw 0x0(%rax,%rax,1)
L52:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L52
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D2
type: string

tool: D2: %d


assistant: ```c
void main() {
  int primo = 0, vzs = 0, X[30];
  for (int i = 101; vzs != 30; i += 2) {
    for (int j = 3; j <= (int)sqrt(i) + 1; j++) {
      if (i % j == 0) {
        primo++;
      }
    }
    if (primo == 0) {
      X[vzs] = i;
      vzs++;
    }
    primo = 0;
  }
  for (int i = 0; i < 30; i++) {
    printf("%d\n", X[i]);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0x65,%r12d
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L13:
  mov    $0x3,%ebx
  xor    %ebp,%ebp
  jmp    L15
  nopl   0x0(%rax)
L16:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  cmp    $0x1,%edx
  adc    $0x0,%ebp
  add    $0x1,%ebx
L15:
  mov    %r12d,%edi
  call   <sqrt@plt>
  add    $0x1,%eax
  cmp    %ebx,%eax
  jge    L16
  test   %ebp,%ebp
  jne    L74
  movslq %r13d,%rax
  add    $0x1,%r13d
  mov    %r12d,(%rsp,%rax,4)
  add    $0x2,%r12d
  cmp    $0x1e,%r13d
  jne    L13
L56:
  mov    %rsp,%rbx
  lea    0x78(%rsp),%r12
  lea    D9(%rip),%rbp
  cs nopw 0x0(%rax,%rax,1)
L41:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L41
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L74:
  add    $0x2,%r12d
  cmp    $0x1e,%r13d
  jne    L13
  jmp    L56
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
data_type: string

tool: D9: %d


assistant: ```c
void main() {
  int primo = 0, vzs = 0, X[30];
  for (int i = 101; vzs != 30; i += 2) {
    for (int j = 3; j <= (int)sqrt(i) + 1; j++) {
      if (i % j == 0) {
        primo++;
      }
    }
    if (primo == 0) {
      X[vzs] = i;
      vzs++;
    }
    primo = 0;
  }
  for (int i = 0; i < 30; i++) {
    printf("%d\n", X[i]);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<swapfunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %edx,-0x54(%rbp)
  mov    %ecx,-0x58(%rbp)
  cmpl   $0x1,-0x58(%rbp)
  jg     L34
  mov    -0x54(%rbp),%eax
  cltq
  shr    $0x3,%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,-0x28(%rbp)
L4:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x30(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x30(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    (%rdx),%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  subq   $0x1,-0x38(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  jg     L4
  jmp    L16
L34:
  mov    -0x54(%rbp),%eax
  cltq
  mov    %rax,-0x20(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x50(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L38:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x39(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl -0x39(%rbp),%edx
  mov    %dl,(%rax)
  subq   $0x1,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jg     L38
L16:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swapfunc(char *a, char *b, int n, int swaptype) {
  if (swaptype <= 1) {
    long i = (n) / sizeof(long);
    long *pi = (long *)(a);
    long *pj = (long *)(b);
    do {
      long t = *pi;
      *pi++ = *pj;
      *pj++ = t;
    } while (--i > 0);
  } else {
    long i = (n) / sizeof(char);
    char *pi = (char *)(a);
    char *pj = (char *)(b);
    do {
      char t = *pi;
      *pi++ = *pj;
      *pj++ = t;
    } while (--i > 0);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<swapfunc>:
  endbr64
  mov    %rsi,%rax
  mov    %ecx,%esi
  mov    %rdi,%rcx
  movslq %edx,%r8
  add    %rdi,%r8
  cmp    $0x1,%esi
  jle    L18
L22:
  movzbl (%rcx),%edx
  add    $0x1,%rcx
  movzbl (%rax),%esi
  mov    %sil,-0x1(%rcx)
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  mov    %r8,%rdx
  sub    %rcx,%rdx
  test   %rdx,%rdx
  jg     L22
  ret
L18:
  movslq %edx,%rdx
  shr    $0x3,%rdx
L19:
  mov    (%rdi),%rcx
  add    $0x8,%rdi
  mov    (%rax),%rsi
  mov    %rsi,-0x8(%rdi)
  add    $0x8,%rax
  mov    %rcx,-0x8(%rax)
  sub    $0x1,%rdx
  test   %rdx,%rdx
  jg     L19
  ret
```

assistant: ```c
void swapfunc(char *a, char *b, int n, int swaptype) {
  if (swaptype <= 1) {
    long i = (n) / sizeof(long);
    long *pi = (long *)(a);
    long *pj = (long *)(b);
    do {
      long t = *pi;
      *pi++ = *pj;
      *pj++ = t;
    } while (--i > 0);
  } else {
    long i = (n) / sizeof(char);
    char *pi = (char *)(a);
    char *pj = (char *)(b);
    do {
      char t = *pi;
      *pi++ = *pj;
      *pj++ = t;
    } while (--i > 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
<swapfunc>:
  endbr64
  movslq %edx,%rdx
  mov    %rdi,%rax
  lea    (%rdi,%rdx,1),%r8
  cmp    $0x1,%ecx
  jle    L17
  nopl   0x0(%rax,%rax,1)
L15:
  movzbl (%rax),%edx
  movzbl (%rsi),%ecx
  add    $0x1,%rsi
  add    $0x1,%rax
  mov    %cl,-0x1(%rax)
  mov    %dl,-0x1(%rsi)
  mov    %r8,%rdx
  sub    %rax,%rdx
  test   %rdx,%rdx
  jg     L15
  ret
  nopl   0x0(%rax,%rax,1)
L17:
  shr    $0x3,%rdx
  nopl   0x0(%rax)
L14:
  mov    (%rdi),%rax
  mov    (%rsi),%rcx
  sub    $0x1,%rdx
  add    $0x8,%rsi
  add    $0x8,%rdi
  mov    %rcx,-0x8(%rdi)
  mov    %rax,-0x8(%rsi)
  test   %rdx,%rdx
  jg     L14
  ret
```

assistant: ```c
void swapfunc(char *a, char *b, int n, int swaptype) {
  if (swaptype <= 1) {
    long i = (n) / sizeof(long);
    long *pi = (long *)(a);
    long *pj = (long *)(b);
    do {
      long t = *pi;
      *pi++ = *pj;
      *pj++ = t;
    } while (--i > 0);
  } else {
    long i = (n) / sizeof(char);
    char *pi = (char *)(a);
    char *pj = (char *)(b);
    do {
      char t = *pi;
      *pi++ = *pj;
      *pj++ = t;
    } while (--i > 0);
  }
}
```