ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L61:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x71,%eax
  je     L54
L30:
  add    $0x1,%r12d
  cmp    $0xa,%eax
  je     L8
  cmp    $0x20,%eax
  je     L24
  cmp    $0x9,%eax
  je     L24
  cmp    $0x2,%ebx
  je     L35
  xor    %eax,%eax
  mov    $0x1,%ebx
  call   <getchar@plt>
  cmp    $0x71,%eax
  jne    L30
L54:
  mov    %r12d,%ecx
  mov    %r13d,%edx
  lea    D6(%rip),%rdi
  mov    %ebp,%esi
  call   <printf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L8:
  add    $0x1,%ebp
  mov    $0x2,%ebx
  jmp    L61
  nopw   0x0(%rax,%rax,1)
L24:
  mov    $0x2,%ebx
  jmp    L61
  nopw   0x0(%rax,%rax,1)
L35:
  add    $0x1,%r13d
  mov    $0x1,%ebx
  jmp    L61
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
data_type: string

tool: D6: %d %d %d


assistant: ```c
main() {
  int c, nl, nw, nc, state;
  state = 2;
  nl = nw = nc = 0;
  while ((c = getchar()) != 'q') {
    nc++;
    if (c == '\n')
      nl++;
    if (c == ' ' || c == '\n' || c == '\t')
      state = 2;
    else if (state == 2) {
      state = 1;
      nw++;
    }
  }
  printf("%d %d %d\n", nl, nw, nc);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<uniquePathsWithObstacles1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L15
L26:
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,(%rdx)
  movl   $0x0,-0xc(%rbp)
  jmp    L35
L13:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  movl   $0xffffffff,(%rax)
  addl   $0x1,-0xc(%rbp)
L35:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L13
  addl   $0x1,-0x10(%rbp)
L15:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L26
  mov    -0x20(%rbp),%eax
  lea    -0x1(%rax),%edi
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%esi
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rcx
  mov    %edi,%edx
  mov    %rax,%rdi
  call   <traverse@plt>
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%rax
  sub    $0x4,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  leave
  ret
```

assistant: ```c
int uniquePathsWithObstacles1(int **grid, int rSize, int cSize) {
  int **arrs = (int **)malloc(sizeof(int *) * rSize);
  for (int i = 0; i < rSize; i++) {
    arrs[i] = (int *)malloc(sizeof(int) * cSize);
    for (int j = 0; j < cSize; j++)
      arrs[i][j] = -1;
  }
  traverse(grid, rSize - 1, cSize - 1, arrs);
  return arrs[rSize - 1][cSize - 1];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<uniquePathsWithObstacles1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %esi,%r13d
  mov    %edx,%r12d
  movslq %esi,%rax
  mov    %rax,0x8(%rsp)
  lea    0x0(,%rsi,8),%edi
  call   <malloc@plt>
  mov    %rax,0x10(%rsp)
  test   %r13d,%r13d
  jle    L13
  mov    %rax,%rsi
  lea    0x0(,%r12,4),%r15d
  mov    %rax,%rbp
  mov    %r13d,%eax
  lea    (%rsi,%rax,8),%r14
  lea    -0x1(%r12),%eax
  lea    0x4(,%rax,4),%rbx
L18:
  mov    %r15d,%edi
  call   <malloc@plt>
  mov    %rbp,%rcx
  mov    %rax,0x0(%rbp)
  test   %r12d,%r12d
  jle    L33
  mov    $0x0,%eax
L17:
  mov    (%rcx),%rdx
  movl   $0xffffffff,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rbx,%rax
  jne    L17
L33:
  add    $0x8,%rbp
  cmp    %r14,%rbp
  jne    L18
L13:
  lea    -0x1(%r12),%edx
  lea    -0x1(%r13),%esi
  mov    0x10(%rsp),%rbx
  mov    %rbx,%rcx
  mov    0x18(%rsp),%rdi
  call   <traverse@plt>
  movslq %r12d,%r12
  mov    0x8(%rsp),%rsi
  mov    -0x8(%rbx,%rsi,8),%rax
  mov    -0x4(%rax,%r12,4),%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int uniquePathsWithObstacles1(int **grid, int rSize, int cSize) {
  int **arrs = (int **)malloc(sizeof(int *) * rSize);
  for (int i = 0; i < rSize; i++) {
    arrs[i] = (int *)malloc(sizeof(int) * cSize);
    for (int j = 0; j < cSize; j++)
      arrs[i][j] = -1;
  }
  traverse(grid, rSize - 1, cSize - 1, arrs);
  return arrs[rSize - 1][cSize - 1];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<uniquePathsWithObstacles1>:
  endbr64
  push   %r15
  movslq %esi,%rax
  push   %r14
  push   %r13
  push   %r12
  mov    %rax,%r12
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x38,%rsp
  mov    %rdi,0x20(%rsp)
  lea    0x0(,%rax,8),%edi
  mov    %rax,0x10(%rsp)
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  movslq %ebx,%rax
  mov    %rax,0x18(%rsp)
  test   %r12d,%r12d
  jle    L29
  mov    0x8(%rsp),%rcx
  lea    -0x1(%rbx),%ebp
  mov    %r12d,%eax
  lea    0x0(,%rbx,4),%r14d
  mov    %ebp,0x2c(%rsp)
  add    $0x1,%rbp
  mov    %rcx,%r15
  lea    (%rcx,%rax,8),%r13
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %r14d,%edi
  call   <malloc@plt>
  mov    %rax,(%r15)
  lea    (%rax,%rbp,4),%rdx
  test   %ebx,%ebx
  jle    L4
  nopl   0x0(%rax,%rax,1)
L1:
  movl   $0xffffffff,(%rax)
  add    $0x4,%rax
  cmp    %rax,%rdx
  jne    L1
L4:
  add    $0x8,%r15
  cmp    %r13,%r15
  jne    L10
L33:
  mov    0x8(%rsp),%rbx
  mov    0x2c(%rsp),%edx
  lea    -0x1(%r12),%esi
  mov    0x20(%rsp),%rdi
  mov    %rbx,%rcx
  call   <traverse@plt>
  mov    0x10(%rsp),%rcx
  mov    0x18(%rsp),%rsi
  mov    -0x8(%rbx,%rcx,8),%rax
  mov    -0x4(%rax,%rsi,4),%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L29:
  lea    -0x1(%rbx),%eax
  mov    %eax,0x2c(%rsp)
  jmp    L33
```

assistant: ```c
int uniquePathsWithObstacles1(int **grid, int rSize, int cSize) {
  int **arrs = (int **)malloc(sizeof(int *) * rSize);
  for (int i = 0; i < rSize; i++) {
    arrs[i] = (int *)malloc(sizeof(int) * cSize);
    for (int j = 0; j < cSize; j++)
      arrs[i][j] = -1;
  }
  traverse(grid, rSize - 1, cSize - 1, arrs);
  return arrs[rSize - 1][cSize - 1];
}
```
user: What is the c source code of the assembly code below:

```asm
<uniquePathsWithObstacles1>:
  endbr64
  push   %r15
  movslq %esi,%rax
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  lea    -0x1(%r12),%r13d
  push   %rbx
  mov    %rax,%rbx
  sub    $0x38,%rsp
  mov    %rdi,0x28(%rsp)
  lea    0x0(,%rax,8),%edi
  mov    %eax,0x14(%rsp)
  mov    %rax,0x18(%rsp)
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  movslq %r12d,%rax
  mov    %rax,0x20(%rsp)
  test   %ebx,%ebx
  jle    L47
  mov    0x8(%rsp),%rsi
  mov    0x14(%rsp),%eax
  mov    %r12d,%ebx
  lea    0x0(,%r12,4),%r15d
  lea    -0x1(%r12),%r13d
  and    $0xfffffffc,%ebx
  lea    (%rsi,%rax,8),%rax
  mov    %rsi,%r14
  mov    %rax,(%rsp)
  mov    %r12d,%eax
  shr    $0x2,%eax
  lea    -0x1(%rax),%ebp
  add    $0x1,%rbp
  shl    $0x4,%rbp
  mov    %rbp,%rax
  mov    %r12d,%ebp
  mov    %rax,%r12
L6:
  mov    %r15d,%edi
  call   <malloc@plt>
  test   %ebp,%ebp
  pcmpeqd %xmm0,%xmm0
  mov    %rax,(%r14)
  jle    L84
  cmp    $0x2,%r13d
  jbe    L56
  mov    %rax,%rdx
  lea    (%r12,%rax,1),%rcx
  nopl   0x0(%rax)
L78:
  movups %xmm0,(%rdx)
  add    $0x10,%rdx
  cmp    %rcx,%rdx
  jne    L78
  mov    %ebx,%edx
  cmp    %ebx,%ebp
  je     L84
L30:
  movslq %edx,%rcx
  movl   $0xffffffff,(%rax,%rcx,4)
  lea    0x0(,%rcx,4),%rdi
  lea    0x1(%rdx),%ecx
  cmp    %ecx,%ebp
  jle    L84
  add    $0x2,%edx
  movl   $0xffffffff,0x4(%rax,%rdi,1)
  cmp    %edx,%ebp
  jle    L84
  movl   $0xffffffff,0x8(%rax,%rdi,1)
L84:
  add    $0x8,%r14
  cmp    (%rsp),%r14
  jne    L6
L47:
  mov    0x8(%rsp),%rbx
  mov    0x14(%rsp),%esi
  mov    %r13d,%edx
  mov    0x28(%rsp),%rdi
  sub    $0x1,%esi
  mov    %rbx,%rcx
  call   <traverse@plt>
  mov    0x18(%rsp),%rsi
  mov    -0x8(%rbx,%rsi,8),%rax
  mov    0x20(%rsp),%rsi
  mov    -0x4(%rax,%rsi,4),%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L56:
  xor    %edx,%edx
  jmp    L30
```

assistant: ```c
int uniquePathsWithObstacles1(int **grid, int rSize, int cSize) {
  int **arrs = (int **)malloc(sizeof(int *) * rSize);
  for (int i = 0; i < rSize; i++) {
    arrs[i] = (int *)malloc(sizeof(int) * cSize);
    for (int j = 0; j < cSize; j++)
      arrs[i][j] = -1;
  }
  traverse(grid, rSize - 1, cSize - 1, arrs);
  return arrs[rSize - 1][cSize - 1];
}
```
user: What is the c source code of the assembly code below:

```asm
<logger_mkdir>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x440,%rsp
  mov    %rdi,-0x438(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0x420(%rbp)
  movl   $0xffffffff,-0x41c(%rbp)
  movq   $0x0,-0x418(%rbp)
  cmpq   $0x0,-0x438(%rbp)
  je     L22
  mov    -0x438(%rbp),%rdx
  lea    -0x410(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x410(%rbp),%rax
  mov    %rax,-0x418(%rbp)
  jmp    L139
L37:
  mov    -0x418(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L108
  jmp    L153
L58:
  addq   $0x1,-0x418(%rbp)
L153:
  mov    -0x418(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L162
  mov    -0x418(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L162
  mov    -0x418(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L58
L162:
  cmpl   $0x0,-0x420(%rbp)
  jle    L33
  mov    -0x418(%rbp),%rax
  movb   $0x0,(%rax)
  lea    -0x424(%rbp),%rax
  mov    $0x4,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0x424(%rbp),%rdx
  lea    -0x410(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  mov    %eax,-0x41c(%rbp)
  cmpl   $0x0,-0x41c(%rbp)
  jne    L68
  mov    -0x424(%rbp),%eax
  mov    %eax,%edi
  call   <S_ISDIR@plt>
  test   %eax,%eax
  jne    L68
  mov    $0xffffffff,%eax
  jmp    L2
L68:
  cmpl   $0x0,-0x41c(%rbp)
  je     L10
  lea    -0x410(%rbp),%rax
  mov    $0x1ed,%esi
  mov    %rax,%rdi
  call   <mkdir@plt>
  test   %rax,%rax
  je     L10
  mov    $0xffffffff,%eax
  jmp    L2
L10:
  mov    -0x418(%rbp),%rax
  movb   $0x2f,(%rax)
L33:
  addl   $0x1,-0x420(%rbp)
L108:
  addq   $0x1,-0x418(%rbp)
L139:
  mov    -0x418(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L37
  mov    $0x0,%eax
  jmp    L2
L22:
  mov    $0xffffffff,%eax
L2:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

assistant: ```c
int logger_mkdir(char *path) {
  char fullpath[1024];
  int level = -1, ret = -1;
  char *p = NULL;
  struct stat st;
  if (path) {
    strcpy(fullpath, path);
    p = fullpath;
    while (*p != '\0') {
      if (*p == '/') {
        while (*p != '\0' && *p == '/' && *(p + 1) == '/')
          ++p;
        if (level > 0) {
          *p = '\0';
          memset(&st, 0, sizeof(struct stat));
          ret = stat(fullpath, &st);
          if (ret == 0 && !S_ISDIR(st.st_mode))
            return -1;
          if (ret != 0 && mkdir(fullpath, 0755) != 0)
            return -1;
          *p = '/';
        }
        level++;
      }
      ++p;
    }
    return 0;
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<logger_mkdir>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x420,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L5
  mov    %rdi,%rsi
  lea    0x10(%rsp),%rdi
  call   <strcpy@plt>
  movzbl 0x10(%rsp),%edx
  test   %dl,%dl
  je     L60
  lea    0x10(%rsp),%rax
  mov    $0xffffffff,%ebp
  lea    0xc(%rsp),%r12
  jmp    L85
L176:
  mov    %rax,%rbx
L184:
  test   %ebp,%ebp
  jg     L9
L141:
  add    $0x1,%ebp
L64:
  lea    0x1(%rbx),%rax
  movzbl 0x1(%rbx),%edx
  test   %dl,%dl
  je     L109
L85:
  mov    %rax,%rbx
  cmp    $0x2f,%dl
  jne    L64
  cmpb   $0x2f,(%rax)
  jne    L184
L44:
  cmpb   $0x2f,0x1(%rax)
  jne    L176
  add    $0x1,%rax
  cmpb   $0x2f,(%rax)
  je     L44
  mov    %rax,%rbx
  jmp    L184
L9:
  movb   $0x0,(%rbx)
  mov    $0x4,%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <memset@plt>
  lea    0x10(%rsp),%rdi
  mov    %r12,%rsi
  call   <stat@plt>
  test   %eax,%eax
  je     L142
  lea    0x10(%rsp),%rdi
  mov    $0x1ed,%esi
  call   <mkdir@plt>
  test   %rax,%rax
  je     L13
  mov    $0xffffffff,%eax
  jmp    L23
L142:
  mov    0xc(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %eax,%eax
  je     L187
L13:
  movb   $0x2f,(%rbx)
  jmp    L141
L109:
  mov    $0x0,%eax
L23:
  mov    0x418(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L32
  add    $0x420,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  mov    $0xffffffff,%eax
  jmp    L23
L60:
  mov    $0x0,%eax
  jmp    L23
L187:
  mov    $0xffffffff,%eax
  jmp    L23
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int logger_mkdir(char *path) {
  char fullpath[1024];
  int level = -1, ret = -1;
  char *p = NULL;
  struct stat st;
  if (path) {
    strcpy(fullpath, path);
    p = fullpath;
    while (*p != '\0') {
      if (*p == '/') {
        while (*p != '\0' && *p == '/' && *(p + 1) == '/')
          ++p;
        if (level > 0) {
          *p = '\0';
          memset(&st, 0, sizeof(struct stat));
          ret = stat(fullpath, &st);
          if (ret == 0 && !S_ISDIR(st.st_mode))
            return -1;
          if (ret != 0 && mkdir(fullpath, 0755) != 0)
            return -1;
          *p = '/';
        }
        level++;
      }
      ++p;
    }
    return 0;
  }
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<logger_mkdir>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x420,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L62
  lea    0x10(%rsp),%r13
  mov    %rdi,%rsi
  mov    $0xffffffff,%r12d
  mov    %r13,%rdi
  mov    %r13,%rbx
  lea    0xc(%rsp),%r14
  call   <strcpy@plt>
  movzbl 0x10(%rsp),%eax
  test   %al,%al
  jne    L61
  jmp    L129
  nopl   0x0(%rax,%rax,1)
L70:
  movzbl 0x1(%rbp),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L129
L61:
  mov    %rbx,%rbp
  cmp    $0x2f,%al
  jne    L70
  cmpb   $0x2f,(%rbx)
  je     L56
  jmp    L136
  nopw   0x0(%rax,%rax,1)
L49:
  add    $0x1,%rbx
L56:
  cmpb   $0x2f,0x1(%rbx)
  je     L49
  mov    %rbx,%rbp
L136:
  test   %r12d,%r12d
  jg     L150
L139:
  movzbl 0x1(%rbp),%eax
  add    $0x1,%r12d
  add    $0x1,%rbx
  test   %al,%al
  jne    L61
  nopw   0x0(%rax,%rax,1)
L129:
  xor    %eax,%eax
L22:
  mov    0x418(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L75
  add    $0x420,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L150:
  movb   $0x0,(%rbx)
  xor    %esi,%esi
  mov    %r14,%rdi
  mov    $0x4,%edx
  call   <memset@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <stat@plt>
  test   %eax,%eax
  je     L36
  mov    $0x1ed,%esi
  mov    %r13,%rdi
  call   <mkdir@plt>
  test   %rax,%rax
  jne    L62
L128:
  movb   $0x2f,(%rbx)
  jmp    L139
  nopl   0x0(%rax,%rax,1)
L36:
  mov    0xc(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %eax,%eax
  jne    L128
L62:
  mov    $0xffffffff,%eax
  jmp    L22
L75:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int logger_mkdir(char *path) {
  char fullpath[1024];
  int level = -1, ret = -1;
  char *p = NULL;
  struct stat st;
  if (path) {
    strcpy(fullpath, path);
    p = fullpath;
    while (*p != '\0') {
      if (*p == '/') {
        while (*p != '\0' && *p == '/' && *(p + 1) == '/')
          ++p;
        if (level > 0) {
          *p = '\0';
          memset(&st, 0, sizeof(struct stat));
          ret = stat(fullpath, &st);
          if (ret == 0 && !S_ISDIR(st.st_mode))
            return -1;
          if (ret != 0 && mkdir(fullpath, 0755) != 0)
            return -1;
          *p = '/';
        }
        level++;
      }
      ++p;
    }
    return 0;
  }
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<logger_mkdir>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x420,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L127
  lea    0x10(%rsp),%r13
  mov    %rdi,%rsi
  mov    $0xffffffff,%r12d
  mov    %r13,%rdi
  mov    %r13,%rbx
  lea    0xc(%rsp),%r14
  call   <strcpy@plt>
  movzbl 0x10(%rsp),%eax
  test   %al,%al
  jne    L99
  jmp    L134
  nopl   0x0(%rax,%rax,1)
L97:
  movzbl 0x1(%rbp),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L134
L99:
  mov    %rbx,%rbp
  cmp    $0x2f,%al
  jne    L97
  cmpb   $0x2f,0x1(%rbx)
  jne    L143
  nopl   0x0(%rax)
L44:
  add    $0x1,%rbx
  cmpb   $0x2f,0x1(%rbx)
  je     L44
  mov    %rbx,%rbp
L143:
  test   %r12d,%r12d
  jg     L120
L33:
  movzbl 0x1(%rbp),%eax
  add    $0x1,%r12d
  add    $0x1,%rbx
  test   %al,%al
  jne    L99
  nopw   0x0(%rax,%rax,1)
L134:
  xor    %eax,%eax
L100:
  mov    0x418(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x420,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L120:
  movb   $0x0,(%rbx)
  xor    %esi,%esi
  mov    %r14,%rdi
  mov    $0x4,%edx
  call   <memset@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <stat@plt>
  test   %eax,%eax
  je     L36
  mov    $0x1ed,%esi
  mov    %r13,%rdi
  call   <mkdir@plt>
  test   %rax,%rax
  jne    L127
L133:
  movb   $0x2f,(%rbx)
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L36:
  mov    0xc(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %eax,%eax
  jne    L133
L127:
  mov    $0xffffffff,%eax
  jmp    L100
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int logger_mkdir(char *path) {
  char fullpath[1024];
  int level = -1, ret = -1;
  char *p = NULL;
  struct stat st;
  if (path) {
    strcpy(fullpath, path);
    p = fullpath;
    while (*p != '\0') {
      if (*p == '/') {
        while (*p != '\0' && *p == '/' && *(p + 1) == '/')
          ++p;
        if (level > 0) {
          *p = '\0';
          memset(&st, 0, sizeof(struct stat));
          ret = stat(fullpath, &st);
          if (ret == 0 && !S_ISDIR(st.st_mode))
            return -1;
          if (ret != 0 && mkdir(fullpath, 0755) != 0)
            return -1;
          *p = '/';
        }
        level++;
      }
      ++p;
    }
    return 0;
  }
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<add_f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  %xmm0,-0x4(%rbp)
  movss  %xmm1,-0x8(%rbp)
  movss  -0x8(%rbp),%xmm1
  movss  -0x4(%rbp),%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  pop    %rbp
  ret
```

assistant: ```c
float add_f(float x, float y) {
  __asm__ __volatile__("addss %1, %0" : "+x"(x) : "x"(y));
  return x;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<add_f>:
  endbr64
  addss  %xmm1,%xmm0
  ret
```

assistant: ```c
float add_f(float x, float y) {
  __asm__ __volatile__("addss %1, %0" : "+x"(x) : "x"(y));
  return x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<add_f>:
  endbr64
  addss  %xmm1,%xmm0
  ret
```

assistant: ```c
float add_f(float x, float y) {
  __asm__ __volatile__("addss %1, %0" : "+x"(x) : "x"(y));
  return x;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<add_f>:
  endbr64
  addss  %xmm1,%xmm0
  ret
```

assistant: ```c
float add_f(float x, float y) {
  __asm__ __volatile__("addss %1, %0" : "+x"(x) : "x"(y));
  return x;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<write_all>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
L44:
  mov    -0x18(%rbp),%eax
  sub    -0x8(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rax,%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L7
  mov    D12(%rip),%rax
  mov    (%rax),%rdx
  mov    D1(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L44
L7:
  cmpl   $0x0,-0x4(%rbp)
  jle    L28
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
L28:
  cmpl   $0x0,-0x4(%rbp)
  jle    L8
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L44
L8:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int write_all(int fd, void *buf, int len) {
  int n;
  int bc = 0;
  do {
    do {
      n = write(fd, (char *)buf + bc, len - bc);
    } while (n < 0 && errno == EINTR);
    if (n > 0)
      bc += n;
  } while (n > 0 && bc < len);
  return bc;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<write_all>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  mov    %rsi,%r12
  mov    %edx,%ebp
  mov    $0x0,%ebx
  jmp    L40
L25:
  mov    D0(%rip),%rdx
  mov    D6(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L40
L0:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  add    %eax,%ebx
  cmp    %ebx,%ebp
  jle    L0
L40:
  mov    %ebp,%edx
  sub    %ebx,%edx
  movslq %ebx,%rsi
  add    %r12,%rsi
  mov    %r13d,%edi
  call   <write@plt>
  test   %eax,%eax
  js     L25
  jg     L10
  jmp    L0
```

assistant: ```c
int write_all(int fd, void *buf, int len) {
  int n;
  int bc = 0;
  do {
    do {
      n = write(fd, (char *)buf + bc, len - bc);
    } while (n < 0 && errno == EINTR);
    if (n > 0)
      bc += n;
  } while (n > 0 && bc < len);
  return bc;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<write_all>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %edx,%ebx
  nopl   0x0(%rax)
L25:
  mov    %ebx,%edx
  lea    (%r12,%rbp,1),%rsi
  mov    %r13d,%edi
  sub    %r14d,%edx
  call   <write@plt>
  test   %eax,%eax
  js     L11
  je     L42
  add    %eax,%r14d
  cmp    %r14d,%ebx
  jle    L42
  movslq %r14d,%rbp
  jmp    L25
  nopl   0x0(%rax)
L11:
  mov    D12(%rip),%rax
  mov    D2(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L25
L42:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int write_all(int fd, void *buf, int len) {
  int n;
  int bc = 0;
  do {
    do {
      n = write(fd, (char *)buf + bc, len - bc);
    } while (n < 0 && errno == EINTR);
    if (n > 0)
      bc += n;
  } while (n > 0 && bc < len);
  return bc;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<write_all>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %edx,%ebx
  nopl   0x0(%rax)
L35:
  mov    %ebx,%edx
  lea    (%r12,%rbp,1),%rsi
  mov    %r13d,%edi
  sub    %r14d,%edx
  call   <write@plt>
  test   %eax,%eax
  js     L47
  je     L29
  add    %eax,%r14d
  cmp    %r14d,%ebx
  jle    L29
  movslq %r14d,%rbp
  jmp    L35
  nopl   0x0(%rax)
L47:
  mov    D4(%rip),%rax
  mov    D16(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L35
L29:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int write_all(int fd, void *buf, int len) {
  int n;
  int bc = 0;
  do {
    do {
      n = write(fd, (char *)buf + bc, len - bc);
    } while (n < 0 && errno == EINTR);
    if (n > 0)
      bc += n;
  } while (n > 0 && bc < len);
  return bc;
}
```