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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  xor    %eax,%eax
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = 54;
  void *p = &x;
  int *p2 = p;
  assert(*p2 == 54);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  xor    %eax,%eax
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x = 54;
  void *p = &x;
  int *p2 = p;
  assert(*p2 == 54);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x3c(%rbp)
  jmp    L67
L121:
  movl   $0x0,-0x38(%rbp)
  jmp    L149
L119:
  movl   $0x0,-0x34(%rbp)
  jmp    L7
L10:
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x34(%rbp),%eax
  cltq
  mov    %rdx,-0x30(%rbp,%rax,8)
  addl   $0x1,-0x34(%rbp)
L7:
  cmpl   $0x4,-0x34(%rbp)
  jle    L10
  mov    -0x38(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  lea    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x3c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  lea    -0x30(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <foo@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L6
L47:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,8),%rdx
  mov    -0x34(%rbp),%eax
  cltq
  cmp    %rax,%rdx
  je     L97
  mov    $0x0,%eax
  call   <abort@plt>
L97:
  addl   $0x1,-0x34(%rbp)
L6:
  mov    -0x34(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jle    L47
  mov    -0x3c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x34(%rbp)
  jmp    L111
L2:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,8),%rdx
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  cltq
  cmp    %rax,%rdx
  je     L33
  mov    $0x0,%eax
  call   <abort@plt>
L33:
  addl   $0x1,-0x34(%rbp)
L111:
  mov    -0x34(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jle    L2
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x34(%rbp)
  jmp    L131
L72:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,8),%rdx
  mov    -0x34(%rbp),%eax
  cltq
  cmp    %rax,%rdx
  je     L125
  mov    $0x0,%eax
  call   <abort@plt>
L125:
  addl   $0x1,-0x34(%rbp)
L131:
  cmpl   $0x4,-0x34(%rbp)
  jle    L72
  addl   $0x1,-0x38(%rbp)
L149:
  cmpl   $0x4,-0x38(%rbp)
  jle    L119
  addl   $0x1,-0x3c(%rbp)
L67:
  cmpl   $0x4,-0x3c(%rbp)
  jle    L121
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L88
  call   <__stack_chk_fail@plt>
L88:
  leave
  ret
```

assistant: ```c
int main(void) {
  unsigned long a[5];
  int start, end, k;
  for (start = 0; start < 5; start++)
    for (end = 0; end < 5; end++) {
      for (k = 0; k < 5; k++)
        a[k] = k;
      foo(a + start, a + end);
      for (k = 0; k <= start; k++)
        if (a[k] != k)
          abort();
      for (k = start + 1; k <= end; k++)
        if (a[k] != k - 1)
          abort();
      for (k = end + 1; k < 5; k++)
        if (a[k] != k)
          abort();
    }
  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
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    $0x1,%r15d
  movq   $0x0,(%rsp)
  lea    0x10(%rsp),%rbp
  jmp    L110
L160:
  add    $0x1,%rbx
  cmp    %ebx,%r14d
  jl     L74
L91:
  cmp    %rbx,0x0(%rbp,%rbx,8)
  je     L160
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L160
L120:
  add    $0x1,%rbx
  cmp    %ebx,%r13d
  jl     L88
L24:
  lea    -0x1(%rbx),%rax
  cmp    %rax,0x0(%rbp,%rbx,8)
  je     L120
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L120
L88:
  cmp    $0x4,%r12
  je     L72
  add    $0x1,%r12
  mov    %r12,%rbx
  jmp    L51
L8:
  add    $0x1,%rbx
  cmp    $0x4,%ebx
  jg     L12
L51:
  cmp    %rbx,0x0(%rbp,%rbx,8)
  je     L8
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L8
L148:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L82
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L72:
  addq   $0x1,(%rsp)
  mov    (%rsp),%rax
  add    $0x1,%r15
  cmp    $0x5,%rax
  je     L148
L110:
  mov    (%rsp),%rax
  lea    0x0(%rbp,%rax,8),%rdx
  mov    %rdx,0x8(%rsp)
  mov    $0x0,%r12d
  mov    %eax,%r14d
L12:
  movq   $0x0,0x10(%rsp)
  movq   $0x1,0x18(%rsp)
  movq   $0x2,0x20(%rsp)
  movq   $0x3,0x28(%rsp)
  movq   $0x4,0x30(%rsp)
  lea    0x0(%rbp,%r12,8),%rsi
  mov    0x8(%rsp),%rdi
  call   <foo@plt>
  mov    $0x0,%ebx
  cmpl   $0x0,(%rsp)
  jns    L91
L74:
  mov    %r12d,%r13d
  cmp    %r15d,%r12d
  jl     L88
  mov    %r15,%rbx
  jmp    L24
L82:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  unsigned long a[5];
  int start, end, k;
  for (start = 0; start < 5; start++)
    for (end = 0; end < 5; end++) {
      for (k = 0; k < 5; k++)
        a[k] = k;
      foo(a + start, a + end);
      for (k = 0; k <= start; k++)
        if (a[k] != k)
          abort();
      for (k = start + 1; k <= end; k++)
        if (a[k] != k - 1)
          abort();
      for (k = end + 1; k < 5; k++)
        if (a[k] != k)
          abort();
    }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  mov    $0x1,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
L83:
  lea    0x0(%rbp,%r13,8),%rax
  xor    %r12d,%r12d
  mov    %rax,0x8(%rsp)
L39:
  mov    0x8(%rsp),%rdi
  lea    0x0(%rbp,%r12,8),%rsi
  xor    %ebx,%ebx
  movq   $0x0,0x10(%rsp)
  movq   $0x1,0x18(%rsp)
  movq   $0x2,0x20(%rsp)
  movq   $0x3,0x28(%rsp)
  movq   $0x4,0x30(%rsp)
  call   <foo@plt>
L13:
  cmp    0x0(%rbp,%rbx,8),%rbx
  je     L71
  xor    %eax,%eax
  call   <abort@plt>
L71:
  add    $0x1,%rbx
  cmp    %r13d,%ebx
  jle    L13
  mov    %r12d,%r15d
  mov    %r14,%rbx
  cmp    %r14d,%r12d
  jl     L12
L68:
  lea    -0x1(%rbx),%rax
  cmp    %rax,0x0(%rbp,%rbx,8)
  je     L4
  xor    %eax,%eax
  call   <abort@plt>
L4:
  add    $0x1,%rbx
  cmp    %ebx,%r15d
  jge    L68
L12:
  cmp    $0x4,%r12
  je     L15
  add    $0x1,%r12
  mov    %r12,%rbx
L92:
  cmp    %rbx,0x0(%rbp,%rbx,8)
  je     L108
  xor    %eax,%eax
  call   <abort@plt>
L108:
  add    $0x1,%rbx
  cmp    $0x5,%ebx
  jne    L92
  jmp    L39
  nopl   0x0(%rax)
L15:
  add    $0x1,%r14
  add    $0x1,%r13
  cmp    $0x6,%r14
  jne    L83
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  unsigned long a[5];
  int start, end, k;
  for (start = 0; start < 5; start++)
    for (end = 0; end < 5; end++) {
      for (k = 0; k < 5; k++)
        a[k] = k;
      foo(a + start, a + end);
      for (k = 0; k <= start; k++)
        if (a[k] != k)
          abort();
      for (k = start + 1; k <= end; k++)
        if (a[k] != k - 1)
          abort();
      for (k = end + 1; k < 5; k++)
        if (a[k] != k)
          abort();
    }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<CountDivisibility>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %ax,-0x14(%rbp)
  movw   $0x0,-0x2(%rbp)
  movzwl -0x14(%rbp),%ecx
  mov    $0x3e7,%eax
  cltd
  idiv   %ecx
  mov    %ax,-0x2(%rbp)
  movzwl -0x2(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned short CountDivisibility(unsigned short divisor) {
  unsigned short cnt = 0;
  cnt = (1000 - 1) / divisor;
  return cnt;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<CountDivisibility>:
  endbr64
  movzwl %di,%edi
  mov    $0x3e7,%eax
  cltd
  idiv   %edi
  ret
```

assistant: ```c
unsigned short CountDivisibility(unsigned short divisor) {
  unsigned short cnt = 0;
  cnt = (1000 - 1) / divisor;
  return cnt;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<CountDivisibility>:
  endbr64
  mov    $0x3e7,%eax
  movzwl %di,%edi
  cltd
  idiv   %edi
  ret
```

assistant: ```c
unsigned short CountDivisibility(unsigned short divisor) {
  unsigned short cnt = 0;
  cnt = (1000 - 1) / divisor;
  return cnt;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<CountDivisibility>:
  endbr64
  mov    $0x3e7,%eax
  movzwl %di,%edi
  cltd
  idiv   %edi
  ret
```

assistant: ```c
unsigned short CountDivisibility(unsigned short divisor) {
  unsigned short cnt = 0;
  cnt = (1000 - 1) / divisor;
  return cnt;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<setp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  jmp    L14
L31:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <findprop@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L10
  mov    -0x1c(%rbp),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <putofield@plt>
  jmp    L4
L10:
  mov    -0x14(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <getofield@plt>
  mov    %eax,-0x14(%rbp)
L14:
  cmpl   $0x0,-0x14(%rbp)
  jne    L31
  mov    $0x0,%eax
L4:
  leave
  ret
```

assistant: ```c
int setp(int obj, int prop, int val) {
  int p;
  for (; obj; obj = getofield(obj, ((int)0)))
    if (p = findprop(obj, prop))
      return (putofield(obj, p, val));
  return (0L);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<setp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %edx,%r12d
  test   %edi,%edi
  je     L29
L40:
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <findprop@plt>
  test   %eax,%eax
  jne    L28
  mov    $0x0,%esi
  mov    %ebx,%edi
  call   <getofield@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L40
  jmp    L48
L28:
  mov    %r12d,%edx
  mov    %eax,%esi
  mov    %ebx,%edi
  call   <putofield@plt>
L48:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  mov    %edi,%eax
  jmp    L48
```

assistant: ```c
int setp(int obj, int prop, int val) {
  int p;
  for (; obj; obj = getofield(obj, ((int)0)))
    if (p = findprop(obj, prop))
      return (putofield(obj, p, val));
  return (0L);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<setp>:
  endbr64
  test   %edi,%edi
  je     L6
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  jmp    L9
  nopl   (%rax)
L20:
  mov    %ebp,%edi
  xor    %esi,%esi
  call   <getofield@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  je     L24
L9:
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <findprop@plt>
  test   %eax,%eax
  je     L20
  pop    %rbx
  mov    %r12d,%edx
  mov    %ebp,%edi
  mov    %eax,%esi
  pop    %rbp
  pop    %r12
  jmp    1070 <putofield@plt>
  nopl   0x0(%rax)
L24:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L6:
  xor    %eax,%eax
  ret
```

assistant: ```c
int setp(int obj, int prop, int val) {
  int p;
  for (; obj; obj = getofield(obj, ((int)0)))
    if (p = findprop(obj, prop))
      return (putofield(obj, p, val));
  return (0L);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<setp>:
  endbr64
  test   %edi,%edi
  je     L14
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  jmp    L19
  nopl   (%rax)
L33:
  mov    %ebp,%edi
  xor    %esi,%esi
  call   <getofield@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  je     L23
L19:
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <findprop@plt>
  test   %eax,%eax
  je     L33
  pop    %rbx
  mov    %r12d,%edx
  mov    %ebp,%edi
  mov    %eax,%esi
  pop    %rbp
  pop    %r12
  jmp    1070 <putofield@plt>
  nopl   0x0(%rax)
L23:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L14:
  xor    %eax,%eax
  ret
```

assistant: ```c
int setp(int obj, int prop, int val) {
  int p;
  for (; obj; obj = getofield(obj, ((int)0)))
    if (p = findprop(obj, prop))
      return (putofield(obj, p, val));
  return (0L);
}
```
user: What is the c source code of the assembly code below?

```asm
<trio_equal_case_max>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  cmpq   $0x0,-0x8(%rbp)
  je     L18
  cmpq   $0x0,-0x18(%rbp)
  je     L18
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L24
L18:
  mov    $0x0,%eax
L24:
  leave
  ret
```

assistant: ```c
__attribute__((visibility("default"))) int trio_equal_case_max(const char *first, size_t max, const char *second) {
  assert(first);
  assert(second);
  if ((first != 0) && (second != 0)) {
    return (0 == strncmp(first, second, max));
  }
  return (1 == 0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<trio_equal_case_max>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <assert@plt>
  test   %rbx,%rbx
  je     L7
  test   %rbp,%rbp
  jne    L6
L7:
  mov    $0x0,%eax
L28:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L28
```

assistant: ```c
__attribute__((visibility("default"))) int trio_equal_case_max(const char *first, size_t max, const char *second) {
  assert(first);
  assert(second);
  if ((first != 0) && (second != 0)) {
    return (0 == strncmp(first, second, max));
  }
  return (1 == 0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<trio_equal_case_max>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <assert@plt>
  mov    %r12,%rdi
  call   <assert@plt>
  test   %rbp,%rbp
  je     L15
  test   %r12,%r12
  jne    L7
L15:
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L7:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  pop    %rbp
  pop    %r12
  test   %rax,%rax
  pop    %r13
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
__attribute__((visibility("default"))) int trio_equal_case_max(const char *first, size_t max, const char *second) {
  assert(first);
  assert(second);
  if ((first != 0) && (second != 0)) {
    return (0 == strncmp(first, second, max));
  }
  return (1 == 0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<trio_equal_case_max>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <assert@plt>
  mov    %r12,%rdi
  call   <assert@plt>
  test   %rbp,%rbp
  je     L7
  test   %r12,%r12
  jne    L8
L7:
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L8:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  pop    %rbp
  pop    %r12
  test   %rax,%rax
  pop    %r13
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
__attribute__((visibility("default"))) int trio_equal_case_max(const char *first, size_t max, const char *second) {
  assert(first);
  assert(second);
  if ((first != 0) && (second != 0)) {
    return (0 == strncmp(first, second, max));
  }
  return (1 == 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<LTC2309_read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    $0x0,%eax
  call   <gpio_iic_start@plt>
  mov    $0x11,%edi
  call   <gpio_iic_write@plt>
  mov    $0x0,%eax
  call   <gpio_iic_Gack@plt>
  test   %rax,%rax
  jne    L23
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <gpio_iic_read@plt>
  mov    $0x0,%eax
  call   <gpio_iic_Pack@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <gpio_iic_read@plt>
  mov    $0x0,%eax
  call   <gpio_iic_Pnack@plt>
  mov    $0x0,%eax
  call   <gpio_iic_stop@plt>
  shll   $0x8,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  add    %eax,-0xc(%rbp)
  sarl   $0x4,-0xc(%rbp)
  jmp    L3
L23:
  movl   $0xf0000,-0xc(%rbp)
L3:
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
uint32_t LTC2309_read() {
  uint32_t rdata;
  uint8_t temp;
  rdata = 0;
  temp = 0;
  gpio_iic_start();
  gpio_iic_write(((0x10) + 1));
  if (gpio_iic_Gack() == 0) {
    gpio_iic_read(&temp);
    gpio_iic_Pack();
    rdata = temp;
    gpio_iic_read(&temp);
    gpio_iic_Pnack();
    gpio_iic_stop();
    rdata <<= 8;
    rdata += temp;
    rdata >>= 4;
  } else {
    rdata = 0xf0000;
  }
  return rdata;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<LTC2309_read>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <gpio_iic_start@plt>
  mov    $0x11,%edi
  call   <gpio_iic_write@plt>
  mov    $0x0,%eax
  call   <gpio_iic_Gack@plt>
  mov    %rax,%rdx
  mov    $0xf0000,%eax
  test   %rdx,%rdx
  je     L28
L19:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L28:
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rdi
  call   <gpio_iic_read@plt>
  mov    $0x0,%eax
  call   <gpio_iic_Pack@plt>
  mov    0x4(%rsp),%ebx
  mov    %rbp,%rdi
  call   <gpio_iic_read@plt>
  mov    $0x0,%eax
  call   <gpio_iic_Pnack@plt>
  mov    $0x0,%eax
  call   <gpio_iic_stop@plt>
  mov    %ebx,%eax
  shl    $0x8,%eax
  add    0x4(%rsp),%eax
  sar    $0x4,%eax
  jmp    L19
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t LTC2309_read() {
  uint32_t rdata;
  uint8_t temp;
  rdata = 0;
  temp = 0;
  gpio_iic_start();
  gpio_iic_write(((0x10) + 1));
  if (gpio_iic_Gack() == 0) {
    gpio_iic_read(&temp);
    gpio_iic_Pack();
    rdata = temp;
    gpio_iic_read(&temp);
    gpio_iic_Pnack();
    gpio_iic_stop();
    rdata <<= 8;
    rdata += temp;
    rdata >>= 4;
  } else {
    rdata = 0xf0000;
  }
  return rdata;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<LTC2309_read>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <gpio_iic_start@plt>
  mov    $0x11,%edi
  call   <gpio_iic_write@plt>
  xor    %eax,%eax
  call   <gpio_iic_Gack@plt>
  mov    %rax,%r8
  mov    $0xf0000,%eax
  test   %r8,%r8
  jne    L5
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rdi
  call   <gpio_iic_read@plt>
  xor    %eax,%eax
  call   <gpio_iic_Pack@plt>
  mov    %rbp,%rdi
  mov    0x4(%rsp),%ebx
  call   <gpio_iic_read@plt>
  xor    %eax,%eax
  call   <gpio_iic_Pnack@plt>
  xor    %eax,%eax
  call   <gpio_iic_stop@plt>
  mov    %ebx,%eax
  shl    $0x8,%eax
  add    0x4(%rsp),%eax
  sar    $0x4,%eax
L5:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t LTC2309_read() {
  uint32_t rdata;
  uint8_t temp;
  rdata = 0;
  temp = 0;
  gpio_iic_start();
  gpio_iic_write(((0x10) + 1));
  if (gpio_iic_Gack() == 0) {
    gpio_iic_read(&temp);
    gpio_iic_Pack();
    rdata = temp;
    gpio_iic_read(&temp);
    gpio_iic_Pnack();
    gpio_iic_stop();
    rdata <<= 8;
    rdata += temp;
    rdata >>= 4;
  } else {
    rdata = 0xf0000;
  }
  return rdata;
}
```
user: Translate the below assembly code into C programming language:

```asm
<LTC2309_read>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  call   <gpio_iic_start@plt>
  mov    $0x11,%edi
  call   <gpio_iic_write@plt>
  xor    %eax,%eax
  call   <gpio_iic_Gack@plt>
  mov    %rax,%r8
  mov    $0xf0000,%eax
  test   %r8,%r8
  jne    L8
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rdi
  call   <gpio_iic_read@plt>
  xor    %eax,%eax
  call   <gpio_iic_Pack@plt>
  mov    %rbp,%rdi
  mov    0x4(%rsp),%ebx
  call   <gpio_iic_read@plt>
  xor    %eax,%eax
  call   <gpio_iic_Pnack@plt>
  xor    %eax,%eax
  call   <gpio_iic_stop@plt>
  mov    %ebx,%eax
  shl    $0x8,%eax
  add    0x4(%rsp),%eax
  sar    $0x4,%eax
L8:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t LTC2309_read() {
  uint32_t rdata;
  uint8_t temp;
  rdata = 0;
  temp = 0;
  gpio_iic_start();
  gpio_iic_write(((0x10) + 1));
  if (gpio_iic_Gack() == 0) {
    gpio_iic_read(&temp);
    gpio_iic_Pack();
    rdata = temp;
    gpio_iic_read(&temp);
    gpio_iic_Pnack();
    gpio_iic_stop();
    rdata <<= 8;
    rdata += temp;
    rdata >>= 4;
  } else {
    rdata = 0xf0000;
  }
  return rdata;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<pthread_mutex_lock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <lksmith_prelock@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L15
  mov    -0x4(%rbp),%eax
  jmp    L18
L15:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <r_pthread_mutex_lock@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <lksmith_postlock@plt>
  mov    -0x4(%rbp),%eax
L18:
  leave
  ret
```

assistant: ```c
int pthread_mutex_lock(pthread_mutex_t *mutex) {
  int ret = lksmith_prelock(mutex, 1);
  if (ret)
    return ret;
  ret = r_pthread_mutex_lock(mutex);
  lksmith_postlock(mutex, ret);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<pthread_mutex_lock>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x1,%esi
  call   <lksmith_prelock@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  je     L5
L3:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  mov    %rbp,%rdi
  call   <r_pthread_mutex_lock@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <lksmith_postlock@plt>
  jmp    L3
```

assistant: ```c
int pthread_mutex_lock(pthread_mutex_t *mutex) {
  int ret = lksmith_prelock(mutex, 1);
  if (ret)
    return ret;
  ret = r_pthread_mutex_lock(mutex);
  lksmith_postlock(mutex, ret);
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<pthread_mutex_lock>:
  endbr64
  push   %r12
  mov    $0x1,%esi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <lksmith_prelock@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L1
  mov    %rbp,%rdi
  call   <r_pthread_mutex_lock@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  mov    %eax,%r12d
  call   <lksmith_postlock@plt>
L1:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int pthread_mutex_lock(pthread_mutex_t *mutex) {
  int ret = lksmith_prelock(mutex, 1);
  if (ret)
    return ret;
  ret = r_pthread_mutex_lock(mutex);
  lksmith_postlock(mutex, ret);
  return ret;
}
```