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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x2,-0x8(%rbp)
  jmp    L12
L35:
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <doprime@plt>
  addl   $0x1,-0x8(%rbp)
L12:
  cmpl   $0x32,-0x8(%rbp)
  jle    L35
  movl   $0x4eefb,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0xf,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L52
L13:
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <doprime@plt>
  addl   $0x1,-0x8(%rbp)
L52:
  mov    -0x4(%rbp),%eax
  add    $0xf,%eax
  cmp    %eax,-0x8(%rbp)
  jle    L13
  movl   $0xbac79,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0xf,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L39
L47:
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <doprime@plt>
  addl   $0x1,-0x8(%rbp)
L39:
  mov    -0x4(%rbp),%eax
  add    $0xf,%eax
  cmp    %eax,-0x8(%rbp)
  jle    L47
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 2; i <= 50; i++) {
    doprime(i);
  }
  int j = 7 * 11 * 13 * 17 * 19;
  for (i = j - 15; i <= j + 15; i++) {
    doprime(i);
  }
  j = 23 * 29 * 31 * 37;
  for (i = j - 15; i <= j + 15; i++) {
    doprime(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   %rbx
  mov    $0x2,%ebx
L23:
  mov    %ebx,%edi
  call   <doprime@plt>
  add    $0x1,%ebx
  cmp    $0x33,%ebx
  jne    L23
  mov    $0x4eeec,%ebx
L14:
  mov    %ebx,%edi
  call   <doprime@plt>
  add    $0x1,%ebx
  cmp    $0x4ef0b,%ebx
  jne    L14
  mov    $0xbac6a,%ebx
L13:
  mov    %ebx,%edi
  call   <doprime@plt>
  add    $0x1,%ebx
  cmp    $0xbac89,%ebx
  jne    L13
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 2; i <= 50; i++) {
    doprime(i);
  }
  int j = 7 * 11 * 13 * 17 * 19;
  for (i = j - 15; i <= j + 15; i++) {
    doprime(i);
  }
  j = 23 * 29 * 31 * 37;
  for (i = j - 15; i <= j + 15; i++) {
    doprime(i);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2,%ebx
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <doprime@plt>
  cmp    $0x33,%ebx
  jne    L11
  mov    $0x4eeec,%ebx
  nopl   0x0(%rax)
L21:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <doprime@plt>
  cmp    $0x4ef0b,%ebx
  jne    L21
  mov    $0xbac6a,%ebx
  nop
L10:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <doprime@plt>
  cmp    $0xbac89,%ebx
  jne    L10
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 2; i <= 50; i++) {
    doprime(i);
  }
  int j = 7 * 11 * 13 * 17 * 19;
  for (i = j - 15; i <= j + 15; i++) {
    doprime(i);
  }
  j = 23 * 29 * 31 * 37;
  for (i = j - 15; i <= j + 15; i++) {
    doprime(i);
  }
  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   %rbx
  mov    $0x2,%ebx
  nopw   0x0(%rax,%rax,1)
L25:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <doprime@plt>
  cmp    $0x33,%ebx
  jne    L25
  mov    $0x4eeec,%ebx
  nopl   0x0(%rax)
L26:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <doprime@plt>
  cmp    $0x4ef0b,%ebx
  jne    L26
  mov    $0xbac6a,%ebx
  nop
L27:
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <doprime@plt>
  cmp    $0xbac89,%ebx
  jne    L27
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 2; i <= 50; i++) {
    doprime(i);
  }
  int j = 7 * 11 * 13 * 17 * 19;
  for (i = j - 15; i <= j + 15; i++) {
    doprime(i);
  }
  j = 23 * 29 * 31 * 37;
  for (i = j - 15; i <= j + 15; i++) {
    doprime(i);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<run_tests>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <test_link_traversal@plt>
  mov    $0x0,%eax
  call   <test_readlink@plt>
  mov    $0x0,%eax
  call   <test_realpath@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <test_statfs@plt>
  nop
  leave
  ret
```

assistant: ```c
void run_tests(bool nosymfollow) {
  test_link_traversal(nosymfollow);
  test_readlink();
  test_realpath();
  test_statfs(nosymfollow);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<run_tests>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <test_link_traversal@plt>
  mov    $0x0,%eax
  call   <test_readlink@plt>
  mov    $0x0,%eax
  call   <test_realpath@plt>
  mov    %ebx,%edi
  call   <test_statfs@plt>
  pop    %rbx
  ret
```

assistant: ```c
void run_tests(bool nosymfollow) {
  test_link_traversal(nosymfollow);
  test_readlink();
  test_realpath();
  test_statfs(nosymfollow);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<run_tests>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  call   <test_link_traversal@plt>
  xor    %eax,%eax
  call   <test_readlink@plt>
  xor    %eax,%eax
  call   <test_realpath@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    10b0 <test_statfs@plt>
```

assistant: ```c
void run_tests(bool nosymfollow) {
  test_link_traversal(nosymfollow);
  test_readlink();
  test_realpath();
  test_statfs(nosymfollow);
}
```
user: What is the c source code of the assembly code below:

```asm
<run_tests>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  call   <test_link_traversal@plt>
  xor    %eax,%eax
  call   <test_readlink@plt>
  xor    %eax,%eax
  call   <test_realpath@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    10b0 <test_statfs@plt>
```

assistant: ```c
void run_tests(bool nosymfollow) {
  test_link_traversal(nosymfollow);
  test_readlink();
  test_realpath();
  test_statfs(nosymfollow);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0xb71b00,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  lea    0xb71b00(%rax),%rdx
L69:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L69
  mov    $0x1,%r13d
  jmp    L29
L31:
  sub    0x7a1200(%r12,%rax,1),%ebp
  cmp    %r14d,%ebp
  jle    L97
  mov    %ebx,%r14d
  mov    %ebx,%r15d
L74:
  mov    %r14d,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  cmp    %r14d,%r15d
  cmovg  %r14d,%r15d
  mov    %r14d,%edi
  call   <sum_prop_divisors@plt>
  mov    %eax,%r14d
  cmp    %eax,%ebx
  jne    L74
  mov    %eax,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%r14d
  jmp    L97
L73:
  cmp    %r13d,0x3d0900(%r12,%rax,1)
  je     L31
L97:
  add    $0x1,%r13d
  cmp    $0xf4240,%r13d
  je     L98
L29:
  mov    %r13d,%ebx
  mov    $0x0,%ebp
L47:
  movslq %ebx,%rax
  shl    $0x2,%rax
  lea    (%r12,%rax,1),%rdx
  cmpl   $0x0,(%rdx)
  jne    L73
  movl   $0x1,(%rdx)
  mov    %r13d,0x3d0900(%r12,%rax,1)
  mov    %ebp,0x7a1200(%r12,%rax,1)
  add    $0x1,%ebp
  mov    %ebx,%edi
  call   <sum_prop_divisors@plt>
  mov    %eax,%ebx
  cmp    $0xf423f,%eax
  jle    L47
  jmp    L97
L98:
  mov    %r15d,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: label: D11
data_type: string
name: unpack_data
arguments: label: D0
data_type: string
name: unpack_data
arguments: label: D4
data_type: string

tool: D11: %d --> 

tool: D0: %d



tool: D4: answer: %d


assistant: ```c
int main() {
  int curr_num, i, j, branch, curr_chain_len, big_chain_len, temp, STOP;
  int *arr = (int *)malloc(sizeof(int) * 3 * 1000000);
  for (i = 0; i < 3 * 1000000; i++) {
    arr[i] = 0;
  }
  branch = 1;
  curr_chain_len = 0;
  for (i = 1; i < 1000000; i++) {
    temp = i;
    STOP = 0;
    while (!arr[i]) {
      arr[i] = 1;
      arr[1000000 + i] = branch;
      arr[(2 * 1000000) + i] = curr_chain_len;
      curr_chain_len++;
      i = sum_prop_divisors(i);
      if (i >= 1000000) {
        STOP = 1;
        break;
      }
    }
    if (STOP) {
      i = temp;
      curr_chain_len = 0;
      branch++;
      continue;
    }
    if (branch == arr[1000000 + i]) {
      if ((curr_chain_len - arr[(2 * 1000000) + i]) > big_chain_len) {
        big_chain_len = curr_chain_len - arr[(2 * 1000000) + i];
        curr_num = i;
        j = i;
        do {
          printf("%d --> ", j);
          if (j < curr_num) {
            curr_num = j;
          }
          j = sum_prop_divisors(j);
        } while (j != i);
        printf("%d\n\n", j);
      }
    }
    curr_chain_len = 0;
    branch++;
    i = temp;
  }
  printf("answer: %d\n", curr_num);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  mov    $0xb71b00,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x2,%ebp
  push   %rbx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    $0xb71b00,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r15
  call   <memset@plt>
  nopl   0x0(%rax,%rax,1)
L21:
  lea    -0x1(%rbp),%ebx
  xor    %r14d,%r14d
  mov    %ebx,%r13d
  jmp    L46
  nopl   0x0(%rax,%rax,1)
L65:
  movl   $0x1,(%rsi)
  mov    %ebx,(%rdi)
  mov    %r13d,%edi
  mov    %r14d,0x7a1200(%r15,%rax,1)
  add    $0x1,%r14d
  call   <sum_prop_divisors@plt>
  mov    %eax,%r13d
  cmp    $0xf423f,%eax
  jg     L53
L46:
  movslq %r13d,%rax
  shl    $0x2,%rax
  lea    (%r15,%rax,1),%rsi
  lea    0x3d0900(%r15,%rax,1),%rdi
  mov    (%rsi),%edx
  test   %edx,%edx
  je     L65
  cmp    (%rdi),%ebx
  je     L63
L53:
  add    $0x1,%ebp
  cmp    $0xf4241,%ebp
  jne    L21
  mov    %r12d,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L63:
  mov    %r14d,%ebx
  sub    0x7a1200(%r15,%rax,1),%ebx
  cmp    0xc(%rsp),%ebx
  jle    L53
  mov    %r13d,%r14d
  mov    %r13d,%r12d
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %r14d,%esi
  lea    D24(%rip),%rdi
  call   <printf@plt>
  cmp    %r14d,%r12d
  mov    %r14d,%edi
  cmovg  %r14d,%r12d
  call   <sum_prop_divisors@plt>
  mov    %eax,%r14d
  cmp    %eax,%r13d
  jne    L13
  mov    %r13d,%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,0xc(%rsp)
  jmp    L53
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
data_type: string
name: extract_symbolic_data
arguments: data_tag: D24
data_type: string
name: extract_symbolic_data
arguments: data_tag: D10
data_type: string

tool: D2: answer: %d


tool: D24: %d --> 

tool: D10: %d



assistant: ```c
int main() {
  int curr_num, i, j, branch, curr_chain_len, big_chain_len, temp, STOP;
  int *arr = (int *)malloc(sizeof(int) * 3 * 1000000);
  for (i = 0; i < 3 * 1000000; i++) {
    arr[i] = 0;
  }
  branch = 1;
  curr_chain_len = 0;
  for (i = 1; i < 1000000; i++) {
    temp = i;
    STOP = 0;
    while (!arr[i]) {
      arr[i] = 1;
      arr[1000000 + i] = branch;
      arr[(2 * 1000000) + i] = curr_chain_len;
      curr_chain_len++;
      i = sum_prop_divisors(i);
      if (i >= 1000000) {
        STOP = 1;
        break;
      }
    }
    if (STOP) {
      i = temp;
      curr_chain_len = 0;
      branch++;
      continue;
    }
    if (branch == arr[1000000 + i]) {
      if ((curr_chain_len - arr[(2 * 1000000) + i]) > big_chain_len) {
        big_chain_len = curr_chain_len - arr[(2 * 1000000) + i];
        curr_num = i;
        j = i;
        do {
          printf("%d --> ", j);
          if (j < curr_num) {
            curr_num = j;
          }
          j = sum_prop_divisors(j);
        } while (j != i);
        printf("%d\n\n", j);
      }
    }
    curr_chain_len = 0;
    branch++;
    i = temp;
  }
  printf("answer: %d\n", curr_num);
  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   %r15
  mov    $0xb71b00,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    $0x2,%ebp
  push   %rbx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    $0xb71b00,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r15
  call   <memset@plt>
  nopl   0x0(%rax,%rax,1)
L51:
  lea    -0x1(%rbp),%ebx
  xor    %r14d,%r14d
  mov    %ebx,%r13d
  jmp    L76
  nopl   0x0(%rax,%rax,1)
L35:
  movl   $0x1,(%rsi)
  mov    %ebx,(%rdi)
  mov    %r13d,%edi
  mov    %r14d,0x7a1200(%r15,%rax,1)
  add    $0x1,%r14d
  call   <sum_prop_divisors@plt>
  mov    %eax,%r13d
  cmp    $0xf423f,%eax
  jg     L59
L76:
  movslq %r13d,%rax
  shl    $0x2,%rax
  lea    (%r15,%rax,1),%rsi
  lea    0x3d0900(%r15,%rax,1),%rdi
  mov    (%rsi),%edx
  test   %edx,%edx
  je     L35
  cmp    (%rdi),%ebx
  je     L67
L59:
  add    $0x1,%ebp
  cmp    $0xf4241,%ebp
  jne    L51
  mov    %r12d,%esi
  lea    D26(%rip),%rdi
  call   <printf@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L67:
  mov    %r14d,%ebx
  sub    0x7a1200(%r15,%rax,1),%ebx
  cmp    0xc(%rsp),%ebx
  jle    L59
  mov    %r13d,%r14d
  mov    %r13d,%r12d
  nopw   0x0(%rax,%rax,1)
L15:
  mov    %r14d,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  cmp    %r14d,%r12d
  mov    %r14d,%edi
  cmovg  %r14d,%r12d
  call   <sum_prop_divisors@plt>
  mov    %eax,%r14d
  cmp    %eax,%r13d
  jne    L15
  mov    %r13d,%esi
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,0xc(%rsp)
  jmp    L59
```

tool_calls: name: retrieve_data
arguments: data_tag: D26
data_type: string
name: retrieve_data
arguments: data_tag: D2
data_type: string
name: retrieve_data
arguments: data_tag: D21
data_type: string

tool: D26: answer: %d


tool: D2: %d --> 

tool: D21: %d



assistant: ```c
int main() {
  int curr_num, i, j, branch, curr_chain_len, big_chain_len, temp, STOP;
  int *arr = (int *)malloc(sizeof(int) * 3 * 1000000);
  for (i = 0; i < 3 * 1000000; i++) {
    arr[i] = 0;
  }
  branch = 1;
  curr_chain_len = 0;
  for (i = 1; i < 1000000; i++) {
    temp = i;
    STOP = 0;
    while (!arr[i]) {
      arr[i] = 1;
      arr[1000000 + i] = branch;
      arr[(2 * 1000000) + i] = curr_chain_len;
      curr_chain_len++;
      i = sum_prop_divisors(i);
      if (i >= 1000000) {
        STOP = 1;
        break;
      }
    }
    if (STOP) {
      i = temp;
      curr_chain_len = 0;
      branch++;
      continue;
    }
    if (branch == arr[1000000 + i]) {
      if ((curr_chain_len - arr[(2 * 1000000) + i]) > big_chain_len) {
        big_chain_len = curr_chain_len - arr[(2 * 1000000) + i];
        curr_num = i;
        j = i;
        do {
          printf("%d --> ", j);
          if (j < curr_num) {
            curr_num = j;
          }
          j = sum_prop_divisors(j);
        } while (j != i);
        printf("%d\n\n", j);
      }
    }
    curr_chain_len = 0;
    branch++;
    i = temp;
  }
  printf("answer: %d\n", curr_num);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<selinux_relabel_path>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  nop
  leave
  ret
```

assistant: ```c
void selinux_relabel_path(const char *path, const char *inside_path) {
  assert(path);
  assert(inside_path);
  (void)path;
  (void)inside_path;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<selinux_relabel_path>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <assert@plt>
  pop    %rbx
  ret
```

assistant: ```c
void selinux_relabel_path(const char *path, const char *inside_path) {
  assert(path);
  assert(inside_path);
  (void)path;
  (void)inside_path;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<selinux_relabel_path>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  call   <assert@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1050 <assert@plt>
```

assistant: ```c
void selinux_relabel_path(const char *path, const char *inside_path) {
  assert(path);
  assert(inside_path);
  (void)path;
  (void)inside_path;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<selinux_relabel_path>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  call   <assert@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1050 <assert@plt>
```

assistant: ```c
void selinux_relabel_path(const char *path, const char *inside_path) {
  assert(path);
  assert(inside_path);
  (void)path;
  (void)inside_path;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  jmp    L48
L37:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0xc(%rbp)
  jne    L28
  mov    -0x8(%rbp),%rax
  jmp    L29
L28:
  addq   $0x1,-0x8(%rbp)
L48:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L37
  cmpl   $0x0,-0xc(%rbp)
  jne    L13
  mov    -0x8(%rbp),%rax
  jmp    L29
L13:
  mov    $0x0,%eax
L29:
  pop    %rbp
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  for (; *s != '\0'; s++)
    if (*s == c)
      return (char *)s;
  if (c == '\0')
    return (char *)s;
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L5
L18:
  movsbl %dl,%edx
  cmp    %esi,%edx
  je     L24
  add    $0x1,%rdi
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L18
L5:
  test   %esi,%esi
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
L24:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  for (; *s != '\0'; s++)
    if (*s == c)
      return (char *)s;
  if (c == '\0')
    return (char *)s;
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strchr>:
  endbr64
  jmp    L3
  cs nopw 0x0(%rax,%rax,1)
L23:
  cmp    %esi,%edx
  je     L29
  add    $0x1,%rdi
L3:
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L23
  xor    %eax,%eax
  test   %esi,%esi
  cmove  %rdi,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L29:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  for (; *s != '\0'; s++)
    if (*s == c)
      return (char *)s;
  if (c == '\0')
    return (char *)s;
  return NULL;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strchr>:
  endbr64
  jmp    L4
  cs nopw 0x0(%rax,%rax,1)
L21:
  cmp    %esi,%edx
  je     L15
  add    $0x1,%rdi
L4:
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L21
  xor    %eax,%eax
  test   %esi,%esi
  cmove  %rdi,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  for (; *s != '\0'; s++)
    if (*s == c)
      return (char *)s;
  if (c == '\0')
    return (char *)s;
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<tl_unserialize>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  mov    %r8d,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <tl_compiler_clear_errors@plt>
  lea    -0x10(%rbp),%rdi
  mov    -0x40(%rbp),%ecx
  mov    -0x3c(%rbp),%edx
  mov    -0x38(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rdi,%r9
  mov    $0x0,%r8d
  mov    %rax,%rdi
  call   <tl_expression_unserialize@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  js     L26
  mov    -0x10(%rbp),%rax
  test   %rax,%rax
  je     L26
  mov    -0x10(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <tl_scheme_object_sbdump@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <tl_scheme_object_free@plt>
L26:
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
int tl_unserialize(struct tl_compiler *C, struct tl_buffer *b, int *input, int ilen, int section_mask) {
  struct tl_scheme_object *O = NULL;
  tl_compiler_clear_errors(C);
  int r = tl_expression_unserialize(C, input, ilen, section_mask, NULL, &O);
  if (r >= 0 && O != NULL) {
    tl_scheme_object_sbdump(b, O);
    tl_scheme_object_free(O);
  }
  return r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tl_unserialize>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r13
  mov    %rdx,%rbp
  mov    %ecx,%r12d
  mov    %r8d,%r14d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  call   <tl_compiler_clear_errors@plt>
  mov    %rsp,%r9
  mov    $0x0,%r8d
  mov    %r14d,%ecx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <tl_expression_unserialize@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  js     L7
  mov    (%rsp),%rsi
  test   %rsi,%rsi
  je     L7
  mov    %r13,%rdi
  call   <tl_scheme_object_sbdump@plt>
  mov    (%rsp),%rdi
  call   <tl_scheme_object_free@plt>
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    %ebx,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int tl_unserialize(struct tl_compiler *C, struct tl_buffer *b, int *input, int ilen, int section_mask) {
  struct tl_scheme_object *O = NULL;
  tl_compiler_clear_errors(C);
  int r = tl_expression_unserialize(C, input, ilen, section_mask, NULL, &O);
  if (r >= 0 && O != NULL) {
    tl_scheme_object_sbdump(b, O);
    tl_scheme_object_free(O);
  }
  return r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tl_unserialize>:
  endbr64
  push   %r15
  mov    %r8d,%r15d
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  call   <tl_compiler_clear_errors@plt>
  mov    %r12,%rsi
  mov    %rsp,%r9
  xor    %r8d,%r8d
  mov    %r15d,%ecx
  mov    %r14d,%edx
  mov    %rbp,%rdi
  call   <tl_expression_unserialize@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  js     L19
  mov    (%rsp),%rsi
  test   %rsi,%rsi
  je     L19
  mov    %r13,%rdi
  call   <tl_scheme_object_sbdump@plt>
  mov    (%rsp),%rdi
  call   <tl_scheme_object_free@plt>
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int tl_unserialize(struct tl_compiler *C, struct tl_buffer *b, int *input, int ilen, int section_mask) {
  struct tl_scheme_object *O = NULL;
  tl_compiler_clear_errors(C);
  int r = tl_expression_unserialize(C, input, ilen, section_mask, NULL, &O);
  if (r >= 0 && O != NULL) {
    tl_scheme_object_sbdump(b, O);
    tl_scheme_object_free(O);
  }
  return r;
}
```