ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D0(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rdx
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    (%rsp),%esi
  cmp    0x4(%rsp),%esi
  jge    L12
  mov    %esi,%eax
  mov    $0x2,%ecx
  cltd
  idiv   %ecx
  mov    %eax,%edi
  cmp    $0x3,%esi
  jle    L69
  test   $0x1,%sil
  jne    L47
  jmp    L12
  cs nopw 0x0(%rax,%rax,1)
L5:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L12
L47:
  add    $0x1,%ecx
  cmp    %edi,%ecx
  jle    L5
L69:
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xchg   %ax,%ax
L12:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L38:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D0
datatype: string
name: unpack_data
arguments: tag: D13
datatype: string
name: unpack_data
arguments: tag: D11
datatype: string
name: unpack_data
arguments: tag: D3
datatype: string

tool: D0: Enter the range

tool: D13: %d%d

tool: D11: Prime numbers=

tool: D3: %d


assistant: ```c
int main() {
  int l, h, i, flag;
  printf("Enter the range\n");
  scanf("%d%d", &l, &h);
  printf("Prime numbers=\n");
  while (l < h) {
    flag = 0;
    for (i = 2; i <= l / 2; ++i) {
      if (l % i == 0) {
        flag = 1;
        break;
      }
    }
    if (flag == 0)
      printf("%d\n", l);
    ++l;
    return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<str_trim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L47
L15:
  addq   $0x1,-0x8(%rbp)
L47:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L15
  jmp    L59
L34:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L28
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
L59:
  subq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jae    L34
  jmp    L9
L28:
  nop
L9:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void str_trim(buf) char *buf;
{
  char *p = buf;
  while (*p)
    p++;
  while (--p >= buf) {
    if (*p == ' ')
      *p = '\0';
    else
      break;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<str_trim>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L25
  mov    %rdi,%rax
L27:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L27
L14:
  sub    $0x1,%rax
  cmp    %rax,%rdi
  ja     L0
  sub    $0x1,%rdi
L12:
  cmpb   $0x20,(%rax)
  jne    L0
  movb   $0x0,(%rax)
  sub    $0x1,%rax
  cmp    %rdi,%rax
  jne    L12
L0:
  ret
L25:
  mov    %rdi,%rax
  jmp    L14
```

assistant: ```c
void str_trim(buf) char *buf;
{
  char *p = buf;
  while (*p)
    p++;
  while (--p >= buf) {
    if (*p == ' ')
      *p = '\0';
    else
      break;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<str_trim>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L43
  mov    %rdi,%rdx
  nopl   0x0(%rax)
L24:
  mov    %rdx,%rax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdx)
  jne    L24
  cmp    %rax,%rdi
  ja     L18
  nopl   0x0(%rax)
L11:
  cmpb   $0x20,(%rax)
  jne    L18
  movb   $0x0,(%rax)
  sub    $0x1,%rax
L47:
  cmp    %rax,%rdi
  jbe    L11
L18:
  ret
  nopw   0x0(%rax,%rax,1)
L43:
  lea    -0x1(%rdi),%rax
  jmp    L47
```

assistant: ```c
void str_trim(buf) char *buf;
{
  char *p = buf;
  while (*p)
    p++;
  while (--p >= buf) {
    if (*p == ' ')
      *p = '\0';
    else
      break;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<str_trim>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L37
  mov    %rdi,%rdx
  nopl   0x0(%rax)
L58:
  mov    %rdx,%rax
  add    $0x1,%rdx
  cmpb   $0x0,(%rdx)
  jne    L58
L14:
  cmp    %rdi,%rax
  jae    L59
  jmp    L34
  nopl   0x0(%rax,%rax,1)
L4:
  movb   $0x0,(%rax)
  sub    $0x1,%rax
  cmp    %rax,%rdi
  ja     L34
L59:
  cmpb   $0x20,(%rax)
  je     L4
L34:
  ret
  nopw   0x0(%rax,%rax,1)
L37:
  lea    -0x1(%rdi),%rax
  jmp    L14
```

assistant: ```c
void str_trim(buf) char *buf;
{
  char *p = buf;
  while (*p)
    p++;
  while (--p >= buf) {
    if (*p == ' ')
      *p = '\0';
    else
      break;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<SHA512_Init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x20(%rax),%rbx
  mov    $0xf3bcc908,%edi
  call   <U64@plt>
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  mov    0x20(%rax),%rax
  lea    0x8(%rax),%rbx
  mov    $0x84caa73b,%edi
  call   <U64@plt>
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  mov    0x20(%rax),%rax
  lea    0x10(%rax),%rbx
  mov    $0xfe94f82b,%edi
  call   <U64@plt>
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  mov    0x20(%rax),%rax
  lea    0x18(%rax),%rbx
  mov    $0x5f1d36f1,%edi
  call   <U64@plt>
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  mov    0x20(%rax),%rax
  lea    0x20(%rax),%rbx
  mov    $0xade682d1,%edi
  call   <U64@plt>
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  mov    0x20(%rax),%rax
  lea    0x28(%rax),%rbx
  mov    $0x2b3e6c1f,%edi
  call   <U64@plt>
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  mov    0x20(%rax),%rax
  lea    0x30(%rax),%rbx
  mov    $0xfb41bd6b,%edi
  call   <U64@plt>
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  mov    0x20(%rax),%rax
  lea    0x38(%rax),%rbx
  mov    $0x137e2179,%edi
  call   <U64@plt>
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  movq   $0x0,0x18(%rax)
  mov    -0x18(%rbp),%rax
  movq   $0x0,0x10(%rax)
  mov    -0x18(%rbp),%rax
  movq   $0x0,0x8(%rax)
  mov    D0(%rip),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int SHA512_Init(SHA512_CTX *c) {
  c->h[0] = U64(0x6a09e667f3bcc908);
  c->h[1] = U64(0xbb67ae8584caa73b);
  c->h[2] = U64(0x3c6ef372fe94f82b);
  c->h[3] = U64(0xa54ff53a5f1d36f1);
  c->h[4] = U64(0x510e527fade682d1);
  c->h[5] = U64(0x9b05688c2b3e6c1f);
  c->h[6] = U64(0x1f83d9abfb41bd6b);
  c->h[7] = U64(0x5be0cd19137e2179);
  c->Nl = 0;
  c->Nh = 0;
  c->num = 0;
  c->md_len = SHA512_DIGEST_LENGTH;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<SHA512_Init>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    0x20(%rdi),%rbp
  mov    $0xf3bcc908,%edi
  call   <U64@plt>
  mov    %rax,0x0(%rbp)
  mov    0x20(%rbx),%rbp
  mov    $0x84caa73b,%edi
  call   <U64@plt>
  mov    %rax,0x8(%rbp)
  mov    0x20(%rbx),%rbp
  mov    $0xfe94f82b,%edi
  call   <U64@plt>
  mov    %rax,0x10(%rbp)
  mov    0x20(%rbx),%rbp
  mov    $0x5f1d36f1,%edi
  call   <U64@plt>
  mov    %rax,0x18(%rbp)
  mov    0x20(%rbx),%rbp
  mov    $0xade682d1,%edi
  call   <U64@plt>
  mov    %rax,0x20(%rbp)
  mov    0x20(%rbx),%rbp
  mov    $0x2b3e6c1f,%edi
  call   <U64@plt>
  mov    %rax,0x28(%rbp)
  mov    0x20(%rbx),%rbp
  mov    $0xfb41bd6b,%edi
  call   <U64@plt>
  mov    %rax,0x30(%rbp)
  mov    0x20(%rbx),%rbp
  mov    $0x137e2179,%edi
  call   <U64@plt>
  mov    %rax,0x38(%rbp)
  movq   $0x0,0x18(%rbx)
  movq   $0x0,0x10(%rbx)
  movq   $0x0,0x8(%rbx)
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,(%rbx)
  mov    $0x1,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int SHA512_Init(SHA512_CTX *c) {
  c->h[0] = U64(0x6a09e667f3bcc908);
  c->h[1] = U64(0xbb67ae8584caa73b);
  c->h[2] = U64(0x3c6ef372fe94f82b);
  c->h[3] = U64(0xa54ff53a5f1d36f1);
  c->h[4] = U64(0x510e527fade682d1);
  c->h[5] = U64(0x9b05688c2b3e6c1f);
  c->h[6] = U64(0x1f83d9abfb41bd6b);
  c->h[7] = U64(0x5be0cd19137e2179);
  c->Nl = 0;
  c->Nh = 0;
  c->num = 0;
  c->md_len = SHA512_DIGEST_LENGTH;
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<SHA512_Init>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    0x20(%rdi),%rbp
  mov    $0xf3bcc908,%edi
  call   <U64@plt>
  mov    $0x84caa73b,%edi
  mov    %rax,0x0(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0xfe94f82b,%edi
  mov    %rax,0x8(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0x5f1d36f1,%edi
  mov    %rax,0x10(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0xade682d1,%edi
  mov    %rax,0x18(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0x2b3e6c1f,%edi
  mov    %rax,0x20(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0xfb41bd6b,%edi
  mov    %rax,0x28(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0x137e2179,%edi
  mov    %rax,0x30(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    %rax,0x38(%rbp)
  mov    D1(%rip),%rax
  movq   $0x0,0x18(%rbx)
  mov    (%rax),%eax
  movq   $0x0,0x10(%rbx)
  movq   $0x0,0x8(%rbx)
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int SHA512_Init(SHA512_CTX *c) {
  c->h[0] = U64(0x6a09e667f3bcc908);
  c->h[1] = U64(0xbb67ae8584caa73b);
  c->h[2] = U64(0x3c6ef372fe94f82b);
  c->h[3] = U64(0xa54ff53a5f1d36f1);
  c->h[4] = U64(0x510e527fade682d1);
  c->h[5] = U64(0x9b05688c2b3e6c1f);
  c->h[6] = U64(0x1f83d9abfb41bd6b);
  c->h[7] = U64(0x5be0cd19137e2179);
  c->Nl = 0;
  c->Nh = 0;
  c->num = 0;
  c->md_len = SHA512_DIGEST_LENGTH;
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<SHA512_Init>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    0x20(%rdi),%rbp
  mov    $0xf3bcc908,%edi
  call   <U64@plt>
  mov    $0x84caa73b,%edi
  mov    %rax,0x0(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0xfe94f82b,%edi
  mov    %rax,0x8(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0x5f1d36f1,%edi
  mov    %rax,0x10(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0xade682d1,%edi
  mov    %rax,0x18(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0x2b3e6c1f,%edi
  mov    %rax,0x20(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0xfb41bd6b,%edi
  mov    %rax,0x28(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  mov    $0x137e2179,%edi
  mov    %rax,0x30(%rbp)
  mov    0x20(%rbx),%rbp
  call   <U64@plt>
  pxor   %xmm0,%xmm0
  mov    %rax,0x38(%rbp)
  mov    D9(%rip),%rax
  movq   $0x0,0x18(%rbx)
  mov    (%rax),%eax
  movups %xmm0,0x8(%rbx)
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int SHA512_Init(SHA512_CTX *c) {
  c->h[0] = U64(0x6a09e667f3bcc908);
  c->h[1] = U64(0xbb67ae8584caa73b);
  c->h[2] = U64(0x3c6ef372fe94f82b);
  c->h[3] = U64(0xa54ff53a5f1d36f1);
  c->h[4] = U64(0x510e527fade682d1);
  c->h[5] = U64(0x9b05688c2b3e6c1f);
  c->h[6] = U64(0x1f83d9abfb41bd6b);
  c->h[7] = U64(0x5be0cd19137e2179);
  c->Nl = 0;
  c->Nh = 0;
  c->num = 0;
  c->md_len = SHA512_DIGEST_LENGTH;
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L13:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <factorial@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L2:
  cmpl   $0xc,-0x4(%rbp)
  jle    L13
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D6: %d! = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 0; i < 13; i++) {
    printf("%d! = %d\n", i, factorial(i));
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebx
  lea    D3(%rip),%rbp
L8:
  mov    %ebx,%edi
  call   <factorial@plt>
  mov    %eax,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0xd,%ebx
  jne    L8
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
type: string

tool: D3: %d! = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 0; i < 13; i++) {
    printf("%d! = %d\n", i, factorial(i));
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %ebx,%edi
  call   <factorial@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  mov    %eax,%edx
  call   <printf@plt>
  cmp    $0xd,%ebx
  jne    L8
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
type: string

tool: D5: %d! = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 0; i < 13; i++) {
    printf("%d! = %d\n", i, factorial(i));
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %ebx,%edi
  call   <factorial@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  mov    %eax,%edx
  call   <printf@plt>
  cmp    $0xd,%ebx
  jne    L8
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
type_infered: string

tool: D3: %d! = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 0; i < 13; i++) {
    printf("%d! = %d\n", i, factorial(i));
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<transpose>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L30
L24:
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L36:
  mov    -0x4(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  add    %ecx,%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L36
  addl   $0x1,-0x8(%rbp)
L30:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L24
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void transpose(int i_matdim, const double *pd_B, double *B) {
  int i, j;
  for (i = 0; i < i_matdim; i++)
    for (j = 0; j < i_matdim; j++) {
      B[i * i_matdim + j] = pd_B[j * i_matdim + i];
    }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<transpose>:
  endbr64
  mov    %edi,%r9d
  test   %edi,%edi
  jle    L4
  movslq %edi,%rdi
  shl    $0x3,%rdi
  lea    -0x1(%r9),%eax
  lea    0x8(%rdx,%rax,8),%rcx
  mov    $0x0,%r8d
  mov    %r9d,%r10d
  neg    %r10
  shl    $0x3,%r10
L19:
  lea    (%r10,%rcx,1),%rax
  mov    %rsi,%rdx
L23:
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  add    %rdi,%rdx
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L23
  add    $0x1,%r8d
  add    $0x8,%rsi
  add    %rdi,%rcx
  cmp    %r8d,%r9d
  jne    L19
L4:
  ret
```

assistant: ```c
void transpose(int i_matdim, const double *pd_B, double *B) {
  int i, j;
  for (i = 0; i < i_matdim; i++)
    for (j = 0; j < i_matdim; j++) {
      B[i * i_matdim + j] = pd_B[j * i_matdim + i];
    }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<transpose>:
  endbr64
  mov    %edi,%r9d
  test   %edi,%edi
  jle    L0
  lea    -0x1(%r9),%eax
  movslq %edi,%rdi
  mov    $0xfffffffffffffff8,%r10
  xor    %r8d,%r8d
  shl    $0x3,%rax
  shl    $0x3,%rdi
  lea    0x8(%rdx,%rax,1),%rcx
  sub    %rax,%r10
  nopl   0x0(%rax)
L28:
  lea    (%r10,%rcx,1),%rax
  mov    %rsi,%rdx
  nopw   0x0(%rax,%rax,1)
L22:
  movsd  (%rdx),%xmm0
  add    $0x8,%rax
  add    %rdi,%rdx
  movsd  %xmm0,-0x8(%rax)
  cmp    %rcx,%rax
  jne    L22
  add    $0x1,%r8d
  add    $0x8,%rsi
  add    %rdi,%rcx
  cmp    %r8d,%r9d
  jne    L28
L0:
  ret
```

assistant: ```c
void transpose(int i_matdim, const double *pd_B, double *B) {
  int i, j;
  for (i = 0; i < i_matdim; i++)
    for (j = 0; j < i_matdim; j++) {
      B[i * i_matdim + j] = pd_B[j * i_matdim + i];
    }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<transpose>:
  endbr64
  mov    %edi,%r9d
  test   %edi,%edi
  jle    L23
  lea    -0x1(%r9),%eax
  movslq %edi,%rdi
  mov    $0xfffffffffffffff8,%r10
  xor    %r8d,%r8d
  shl    $0x3,%rax
  shl    $0x3,%rdi
  lea    0x8(%rdx,%rax,1),%rcx
  sub    %rax,%r10
  nopl   0x0(%rax)
L9:
  lea    (%r10,%rcx,1),%rax
  mov    %rsi,%rdx
  nopw   0x0(%rax,%rax,1)
L28:
  movsd  (%rdx),%xmm0
  add    $0x8,%rax
  add    %rdi,%rdx
  movsd  %xmm0,-0x8(%rax)
  cmp    %rcx,%rax
  jne    L28
  add    $0x1,%r8d
  add    $0x8,%rsi
  add    %rdi,%rcx
  cmp    %r8d,%r9d
  jne    L9
L23:
  ret
```

assistant: ```c
void transpose(int i_matdim, const double *pd_B, double *B) {
  int i, j;
  for (i = 0; i < i_matdim; i++)
    for (j = 0; j < i_matdim; j++) {
      B[i * i_matdim + j] = pd_B[j * i_matdim + i];
    }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D27(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    D17(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    D21(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x2,-0x24(%rbp)
  jle    L15
  mov    -0x30(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x1c(%rbp)
L49:
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <get_process_id@plt>
  mov    %eax,-0x18(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jle    L15
  cmpl   $0x0,-0x1c(%rbp)
  js     L15
  cmpl   $0x2,-0x1c(%rbp)
  jg     L15
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x14(%rbp,%rax,4),%edx
  mov    -0x18(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <kill@plt>
  jmp    L49
L15:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int pid, idx;
  int sig[] = {SIGUSR1, SIGUSR2, SIGCONT};
  if (argc > 2) {
    idx = atoi(argv[2]);
    while (1) {
      pid = get_process_id(argv[1]);
      if (pid > 0 && idx >= 0 && idx < 3) {
        kill(pid, sig[idx]);
      } else
        break;
    }
  }
  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
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D29(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0xc(%rsp)
  mov    D23(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x10(%rsp)
  mov    D19(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x14(%rsp)
  cmp    $0x2,%edi
  jg     L6
L19:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  mov    %rsi,%rbx
  mov    0x10(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%ebp
  movslq %eax,%r12
L39:
  mov    0x8(%rbx),%rdi
  call   <get_process_id@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jle    L19
  cmp    $0x2,%ebp
  ja     L19
  mov    0xc(%rsp,%r12,4),%esi
  call   <kill@plt>
  jmp    L39
L45:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int pid, idx;
  int sig[] = {SIGUSR1, SIGUSR2, SIGCONT};
  if (argc > 2) {
    idx = atoi(argv[2]);
    while (1) {
      pid = get_process_id(argv[1]);
      if (pid > 0 && idx >= 0 && idx < 3) {
        kill(pid, sig[idx]);
      } else
        break;
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    D27(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0xc(%rsp)
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x10(%rsp)
  mov    D19(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x14(%rsp)
  cmp    $0x2,%edi
  jg     L12
L51:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    0x10(%rsi),%rdi
  mov    %rsi,%rbx
  call   <atoi@plt>
  cmp    $0x2,%eax
  movslq %eax,%rbp
  setbe  %r12b
  jmp    L39
  nopl   0x0(%rax)
L37:
  test   %r12b,%r12b
  je     L51
  mov    0xc(%rsp,%rbp,4),%esi
  call   <kill@plt>
L39:
  mov    0x8(%rbx),%rdi
  call   <get_process_id@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jg     L37
  jmp    L51
L34:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int pid, idx;
  int sig[] = {SIGUSR1, SIGUSR2, SIGCONT};
  if (argc > 2) {
    idx = atoi(argv[2]);
    while (1) {
      pid = get_process_id(argv[1]);
      if (pid > 0 && idx >= 0 && idx < 3) {
        kill(pid, sig[idx]);
      } else
        break;
    }
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    D1(%rip),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    D26(%rip),%rax
  movd   (%rdx),%xmm0
  mov    (%rax),%eax
  movd   %eax,%xmm1
  mov    D3(%rip),%rax
  punpckldq %xmm1,%xmm0
  mov    (%rax),%eax
  movq   %xmm0,0x8(%rsp)
  mov    %eax,0x10(%rsp)
  cmp    $0x2,%edi
  jg     L17
L32:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  mov    0x10(%rsi),%rdi
  mov    %rsi,%rbx
  call   <atoi@plt>
  cmp    $0x2,%eax
  movslq %eax,%rbp
  setbe  %r12b
  jmp    L26
  nopl   0x0(%rax)
L41:
  test   %r12b,%r12b
  je     L32
  mov    0x8(%rsp,%rbp,4),%esi
  call   <kill@plt>
L26:
  mov    0x8(%rbx),%rdi
  call   <get_process_id@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jg     L41
  jmp    L32
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int pid, idx;
  int sig[] = {SIGUSR1, SIGUSR2, SIGCONT};
  if (argc > 2) {
    idx = atoi(argv[2]);
    while (1) {
      pid = get_process_id(argv[1]);
      if (pid > 0 && idx >= 0 && idx < 3) {
        kill(pid, sig[idx]);
      } else
        break;
    }
  }
  return 0;
}
```