ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<bptCountLevels>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L4
L8:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %eax,-0x18(%rbp)
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x18(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L8
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int bptCountLevels(int maxBlockSize, int itemCount) {
  int levels = 1;
  while (itemCount > maxBlockSize) {
    itemCount = (itemCount + maxBlockSize - 1) / maxBlockSize;
    levels += 1;
  }
  return levels;
}
```
user: What is the c source code of the assembly code below?

```asm
<bptCountLevels>:
  endbr64
  cmp    %edi,%esi
  jle    L27
  mov    $0x1,%ecx
L10:
  lea    -0x1(%rdi,%rsi,1),%eax
  cltd
  idiv   %edi
  mov    %eax,%esi
  add    $0x1,%ecx
  cmp    %eax,%edi
  jl     L10
L0:
  mov    %ecx,%eax
  ret
L27:
  mov    $0x1,%ecx
  jmp    L0
```

assistant: ```c
int bptCountLevels(int maxBlockSize, int itemCount) {
  int levels = 1;
  while (itemCount > maxBlockSize) {
    itemCount = (itemCount + maxBlockSize - 1) / maxBlockSize;
    levels += 1;
  }
  return levels;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bptCountLevels>:
  endbr64
  mov    %esi,%eax
  mov    $0x1,%r8d
  cmp    %edi,%esi
  jle    L14
L4:
  lea    -0x1(%rdi,%rax,1),%eax
  add    $0x1,%r8d
  cltd
  idiv   %edi
  cmp    %eax,%edi
  jl     L4
L14:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int bptCountLevels(int maxBlockSize, int itemCount) {
  int levels = 1;
  while (itemCount > maxBlockSize) {
    itemCount = (itemCount + maxBlockSize - 1) / maxBlockSize;
    levels += 1;
  }
  return levels;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bptCountLevels>:
  endbr64
  mov    %esi,%eax
  mov    $0x1,%r8d
  cmp    %edi,%esi
  jle    L4
L1:
  lea    -0x1(%rdi,%rax,1),%eax
  add    $0x1,%r8d
  cltd
  idiv   %edi
  cmp    %eax,%edi
  jl     L1
L4:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int bptCountLevels(int maxBlockSize, int itemCount) {
  int levels = 1;
  while (itemCount > maxBlockSize) {
    itemCount = (itemCount + maxBlockSize - 1) / maxBlockSize;
    levels += 1;
  }
  return levels;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<vector_dist_squared_3d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rdx
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <vector_subtract_3d@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <vector_magnitude_squared_3d@plt>
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double vector_dist_squared_3d(const double v1[3], const double v2[3]) {
  double v[3];
  vector_subtract_3d(v1, v2, v);
  return vector_magnitude_squared_3d(v);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vector_dist_squared_3d>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdx
  call   <vector_subtract_3d@plt>
  mov    %rbx,%rdi
  call   <vector_magnitude_squared_3d@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double vector_dist_squared_3d(const double v1[3], const double v2[3]) {
  double v[3];
  vector_subtract_3d(v1, v2, v);
  return vector_magnitude_squared_3d(v);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<vector_dist_squared_3d>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdx
  call   <vector_subtract_3d@plt>
  mov    %rbp,%rdi
  call   <vector_magnitude_squared_3d@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x20,%rsp
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double vector_dist_squared_3d(const double v1[3], const double v2[3]) {
  double v[3];
  vector_subtract_3d(v1, v2, v);
  return vector_magnitude_squared_3d(v);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<vector_dist_squared_3d>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdx
  call   <vector_subtract_3d@plt>
  mov    %rbp,%rdi
  call   <vector_magnitude_squared_3d@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x20,%rsp
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double vector_dist_squared_3d(const double v1[3], const double v2[3]) {
  double v[3];
  vector_subtract_3d(v1, v2, v);
  return vector_magnitude_squared_3d(v);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<UTL_ExpandToPointerArray>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x24(%rbp)
  mov    -0x48(%rbp),%rax
  movl   $0x1,(%rax)
  movl   $0x0,-0x28(%rbp)
  jmp    L49
L26:
  movl   $0x0,-0x20(%rbp)
  jmp    L0
L56:
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x40(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L28
  mov    -0x48(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x48(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L4
L28:
  addl   $0x1,-0x20(%rbp)
L0:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L56
L4:
  addl   $0x1,-0x28(%rbp)
L49:
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L26
  mov    -0x48(%rbp),%rax
  mov    (%rax),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x48(%rbp),%rax
  mov    (%rax),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x40(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x18(%rbp)
  jmp    L76
L71:
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,(%rdx)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x18(%rbp)
L76:
  cmpq   $0x0,-0x18(%rbp)
  jne    L71
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
char **UTL_ExpandToPointerArray(const char *inputText, const char *delimiters, int *numberOfEntries) {
  int idx;
  int memorySize = 0;
  int arrayIndex = 0;
  char **array;
  char *outputPtr;
  char *token;
  *numberOfEntries = 1;
  for (idx = 0; inputText[idx] != '\0'; idx++) {
    int j;
    for (j = 0; delimiters[j] != '\0'; j++) {
      if (inputText[idx] == delimiters[j]) {
        (*numberOfEntries)++;
        break;
      }
    }
  }
  memorySize = (sizeof(char *)) * (*numberOfEntries);
  memorySize += strlen(inputText) + 1;
  array = (char **)malloc((memorySize));
  outputPtr = ((char *)array) + ((sizeof(char *)) * (*numberOfEntries));
  strcpy(outputPtr, inputText);
  token = strtok(outputPtr, delimiters);
  while (token != NULL) {
    array[arrayIndex++] = token;
    token = strtok(NULL, delimiters);
  }
  return array;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<UTL_ExpandToPointerArray>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  movl   $0x1,(%rdx)
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L17
  lea    0x1(%rdi),%rsi
  jmp    L0
L50:
  addl   $0x1,0x0(%rbp)
L76:
  add    $0x1,%rsi
  movzbl -0x1(%rsi),%ecx
  test   %cl,%cl
  je     L17
L0:
  movzbl (%rbx),%eax
  lea    0x1(%rbx),%rdx
  test   %al,%al
  je     L76
L69:
  cmp    %cl,%al
  je     L50
  add    $0x1,%rdx
  movzbl -0x1(%rdx),%eax
  test   %al,%al
  jne    L69
  jmp    L76
L17:
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    0x0(%rbp),%edx
  lea    0x1(%rax,%rdx,8),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  movslq 0x0(%rbp),%rax
  lea    0x0(%r13,%rax,8),%rbp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L4
  mov    %r13,%rbp
L88:
  mov    %rax,0x0(%rbp)
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  add    $0x8,%rbp
  test   %rax,%rax
  jne    L88
L4:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **UTL_ExpandToPointerArray(const char *inputText, const char *delimiters, int *numberOfEntries) {
  int idx;
  int memorySize = 0;
  int arrayIndex = 0;
  char **array;
  char *outputPtr;
  char *token;
  *numberOfEntries = 1;
  for (idx = 0; inputText[idx] != '\0'; idx++) {
    int j;
    for (j = 0; delimiters[j] != '\0'; j++) {
      if (inputText[idx] == delimiters[j]) {
        (*numberOfEntries)++;
        break;
      }
    }
  }
  memorySize = (sizeof(char *)) * (*numberOfEntries);
  memorySize += strlen(inputText) + 1;
  array = (char **)malloc((memorySize));
  outputPtr = ((char *)array) + ((sizeof(char *)) * (*numberOfEntries));
  strcpy(outputPtr, inputText);
  token = strtok(outputPtr, delimiters);
  while (token != NULL) {
    array[arrayIndex++] = token;
    token = strtok(NULL, delimiters);
  }
  return array;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<UTL_ExpandToPointerArray>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    $0x9,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movl   $0x1,(%rdx)
  movzbl (%rdi),%esi
  test   %sil,%sil
  je     L5
  add    $0x1,%rdi
  mov    $0x1,%r8d
  nopl   (%rax)
L18:
  movzbl (%rbx),%eax
  lea    0x1(%rbx),%rcx
  test   %al,%al
  jne    L58
  jmp    L28
  nopl   (%rax)
L37:
  movzbl (%rcx),%eax
  add    $0x1,%rcx
  test   %al,%al
  je     L28
L58:
  cmp    %sil,%al
  jne    L37
  add    $0x1,%r8d
  mov    %r8d,0x0(%rbp)
L28:
  movzbl (%rdi),%esi
  add    $0x1,%rdi
  test   %sil,%sil
  jne    L18
  lea    0x1(,%r8,8),%r12d
L5:
  mov    %r13,%rdi
  call   <strlen@plt>
  lea    (%r12,%rax,1),%edi
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %rax,%r12
  movslq 0x0(%rbp),%rax
  lea    (%r12,%rax,8),%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  mov    %rbx,%rsi
  mov    %r12,%rbp
  call   <strtok@plt>
  test   %rax,%rax
  je     L64
  xchg   %ax,%ax
L14:
  mov    %rax,0x0(%rbp)
  mov    %rbx,%rsi
  xor    %edi,%edi
  add    $0x8,%rbp
  call   <strtok@plt>
  test   %rax,%rax
  jne    L14
L64:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **UTL_ExpandToPointerArray(const char *inputText, const char *delimiters, int *numberOfEntries) {
  int idx;
  int memorySize = 0;
  int arrayIndex = 0;
  char **array;
  char *outputPtr;
  char *token;
  *numberOfEntries = 1;
  for (idx = 0; inputText[idx] != '\0'; idx++) {
    int j;
    for (j = 0; delimiters[j] != '\0'; j++) {
      if (inputText[idx] == delimiters[j]) {
        (*numberOfEntries)++;
        break;
      }
    }
  }
  memorySize = (sizeof(char *)) * (*numberOfEntries);
  memorySize += strlen(inputText) + 1;
  array = (char **)malloc((memorySize));
  outputPtr = ((char *)array) + ((sizeof(char *)) * (*numberOfEntries));
  strcpy(outputPtr, inputText);
  token = strtok(outputPtr, delimiters);
  while (token != NULL) {
    array[arrayIndex++] = token;
    token = strtok(NULL, delimiters);
  }
  return array;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<UTL_ExpandToPointerArray>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    $0x9,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movl   $0x1,(%rdx)
  movzbl (%rdi),%esi
  test   %sil,%sil
  je     L9
  movzbl (%rbx),%eax
  mov    $0x1,%edi
  mov    $0x1,%r8d
  test   %al,%al
  je     L42
  nopl   (%rax)
L43:
  lea    0x1(%rbx),%rcx
  jmp    L49
  cs nopw 0x0(%rax,%rax,1)
L16:
  movzbl (%rcx),%eax
  add    $0x1,%rcx
  test   %al,%al
  je     L52
L49:
  cmp    %sil,%al
  jne    L16
  add    $0x1,%r8d
  mov    %r8d,0x0(%r13)
L52:
  movzbl 0x0(%rbp,%rdi,1),%esi
  test   %sil,%sil
  je     L42
  movzbl (%rbx),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L43
L42:
  lea    0x1(,%r8,8),%r12d
L9:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    (%r12,%rax,1),%edi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%r12
  movslq 0x0(%r13),%rax
  mov    %r12,%rbp
  lea    (%r12,%rax,8),%r13
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L5
  nop
L66:
  mov    %rax,0x0(%rbp)
  mov    %rbx,%rsi
  xor    %edi,%edi
  add    $0x8,%rbp
  call   <strtok@plt>
  test   %rax,%rax
  jne    L66
L5:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **UTL_ExpandToPointerArray(const char *inputText, const char *delimiters, int *numberOfEntries) {
  int idx;
  int memorySize = 0;
  int arrayIndex = 0;
  char **array;
  char *outputPtr;
  char *token;
  *numberOfEntries = 1;
  for (idx = 0; inputText[idx] != '\0'; idx++) {
    int j;
    for (j = 0; delimiters[j] != '\0'; j++) {
      if (inputText[idx] == delimiters[j]) {
        (*numberOfEntries)++;
        break;
      }
    }
  }
  memorySize = (sizeof(char *)) * (*numberOfEntries);
  memorySize += strlen(inputText) + 1;
  array = (char **)malloc((memorySize));
  outputPtr = ((char *)array) + ((sizeof(char *)) * (*numberOfEntries));
  strcpy(outputPtr, inputText);
  token = strtok(outputPtr, delimiters);
  while (token != NULL) {
    array[arrayIndex++] = token;
    token = strtok(NULL, delimiters);
  }
  return array;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<asu_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  imul   -0x14(%rbp),%eax
  lea    (%rdx,%rax,1),%esi
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  mov    $0x20,%ecx
  mov    -0x14(%rbp),%edx
  sar    %cl,%edx
  imul   %edx,%eax
  add    %esi,%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x20,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t asu_hash(const uint64_t x, const uint64_t *k) {
  uint64_t t = k[0] + k[1] * (uint32_t)x + k[2] * (uint32_t)(x >> 32);
  return (uint32_t)(t >> 32);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<asu_hash>:
  endbr64
  mov    %edi,%eax
  imul   0x4(%rsi),%eax
  add    (%rsi),%eax
  mov    $0x20,%ecx
  sar    %cl,%edi
  imul   0x8(%rsi),%edi
  add    %edi,%eax
  sar    %cl,%eax
  ret
```

assistant: ```c
uint32_t asu_hash(const uint64_t x, const uint64_t *k) {
  uint64_t t = k[0] + k[1] * (uint32_t)x + k[2] * (uint32_t)(x >> 32);
  return (uint32_t)(t >> 32);
}
```
user: What is the c source code of the assembly code below?

```asm
<asu_hash>:
  endbr64
  mov    0x4(%rsi),%eax
  mov    $0x20,%ecx
  imul   %edi,%eax
  sar    %cl,%edi
  add    (%rsi),%eax
  imul   0x8(%rsi),%edi
  add    %edi,%eax
  sar    %cl,%eax
  ret
```

assistant: ```c
uint32_t asu_hash(const uint64_t x, const uint64_t *k) {
  uint64_t t = k[0] + k[1] * (uint32_t)x + k[2] * (uint32_t)(x >> 32);
  return (uint32_t)(t >> 32);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<asu_hash>:
  endbr64
  mov    0x4(%rsi),%eax
  mov    $0x20,%ecx
  imul   %edi,%eax
  sar    %cl,%edi
  add    (%rsi),%eax
  imul   0x8(%rsi),%edi
  add    %edi,%eax
  sar    %cl,%eax
  ret
```

assistant: ```c
uint32_t asu_hash(const uint64_t x, const uint64_t *k) {
  uint64_t t = k[0] + k[1] * (uint32_t)x + k[2] * (uint32_t)(x >> 32);
  return (uint32_t)(t >> 32);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<calc_checksum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0x11(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L27
L42:
  mov    -0x24(%rbp),%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  cltq
  mov    %dl,-0xc(%rbp,%rax,1)
  shrl   $0x8,-0x24(%rbp)
  addl   $0x1,-0x10(%rbp)
L27:
  mov    -0x10(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L42
  movl   $0x0,-0x10(%rbp)
  jmp    L25
L3:
  mov    -0x10(%rbp),%eax
  cltq
  movzbl -0xc(%rbp,%rax,1),%eax
  add    %al,-0x11(%rbp)
  addl   $0x1,-0x10(%rbp)
L25:
  mov    -0x10(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L3
  movzbl -0x11(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
unsigned char calc_checksum(unsigned int value, int nbytes) {
  int i;
  unsigned char sum = 0;
  unsigned char bytes[4];
  for (i = 0; i < nbytes; i++) {
    bytes[i] = value & 0x000000ff;
    value >>= 8;
  }
  for (i = 0; i < nbytes; i++) {
    sum += bytes[i];
  }
  return sum;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<calc_checksum>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L17
  lea    0x4(%rsp),%rax
  lea    -0x1(%rsi),%edx
  lea    0x5(%rsp,%rdx,1),%rcx
  mov    %rax,%rdx
L28:
  mov    %dil,(%rdx)
  shr    $0x8,%edi
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L28
  mov    $0x0,%edx
L31:
  add    (%rax),%dl
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L31
L5:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  mov    %edx,%eax
  add    $0x18,%rsp
  ret
L17:
  mov    $0x0,%edx
  jmp    L5
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char calc_checksum(unsigned int value, int nbytes) {
  int i;
  unsigned char sum = 0;
  unsigned char bytes[4];
  for (i = 0; i < nbytes; i++) {
    bytes[i] = value & 0x000000ff;
    value >>= 8;
  }
  for (i = 0; i < nbytes; i++) {
    sum += bytes[i];
  }
  return sum;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<calc_checksum>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L41
  lea    -0x1(%rsi),%edx
  lea    0x4(%rsp),%rax
  lea    0x5(%rsp,%rdx,1),%rcx
  mov    %rax,%rdx
L15:
  mov    %dil,(%rdx)
  add    $0x1,%rdx
  shr    $0x8,%edi
  cmp    %rcx,%rdx
  jne    L15
  xor    %r8d,%r8d
L1:
  add    (%rax),%r8b
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L1
L18:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    %r8d,%eax
  add    $0x18,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L41:
  xor    %r8d,%r8d
  jmp    L18
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char calc_checksum(unsigned int value, int nbytes) {
  int i;
  unsigned char sum = 0;
  unsigned char bytes[4];
  for (i = 0; i < nbytes; i++) {
    bytes[i] = value & 0x000000ff;
    value >>= 8;
  }
  for (i = 0; i < nbytes; i++) {
    sum += bytes[i];
  }
  return sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<calc_checksum>:
  endbr64
  test   %esi,%esi
  jle    L15
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%ecx
  mov    %edi,%eax
  shr    $0x8,%ebx
  cmp    $0x1,%esi
  je     L34
  xor    %edx,%edx
  mov    %edi,%eax
  mov    %bl,%dh
  shr    $0x10,%eax
  cmp    $0x2,%esi
  je     L41
  movzbl %al,%eax
  and    $0xff00ffff,%edx
  shl    $0x10,%eax
  or     %eax,%edx
  mov    %edi,%eax
  shr    $0x18,%eax
  cmp    $0x3,%esi
  je     L41
  shl    $0x18,%eax
  and    $0xffffff,%edx
  or     %eax,%edx
L41:
  movzbl %dh,%eax
  add    %edi,%eax
  cmp    $0x2,%ecx
  je     L34
  mov    %edx,%esi
  shr    $0x18,%edx
  shr    $0x10,%esi
  add    %esi,%eax
  add    %eax,%edx
  cmp    $0x3,%ecx
  cmovne %edx,%eax
L34:
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned char calc_checksum(unsigned int value, int nbytes) {
  int i;
  unsigned char sum = 0;
  unsigned char bytes[4];
  for (i = 0; i < nbytes; i++) {
    bytes[i] = value & 0x000000ff;
    value >>= 8;
  }
  for (i = 0; i < nbytes; i++) {
    sum += bytes[i];
  }
  return sum;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<scmp3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L47
L30:
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
  addl   $0x1,-0x4(%rbp)
  subl   $0x1,-0x2c(%rbp)
L47:
  cmpl   $0x0,-0x2c(%rbp)
  jle    L23
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L30
L23:
  mov    -0x28(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  cmpl   $0x0,-0x2c(%rbp)
  jle    L49
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  jmp    L11
L49:
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
L11:
  pop    %rbp
  ret
```

assistant: ```c
int scmp3(unsigned char *p, unsigned char *q, int *l, int maxl) {
  int i;
  i = 0;
  while (maxl > 0 && *p == *q) {
    p++;
    q++;
    i++;
    maxl--;
  }
  *l = i;
  if (maxl > 0)
    return *p - *q;
  return q - p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<scmp3>:
  endbr64
  mov    $0x0,%r8d
  test   %ecx,%ecx
  jle    L19
L28:
  mov    %ecx,%eax
  sub    %r8d,%eax
  movzbl (%rsi),%r9d
  cmp    %r9b,(%rdi)
  jne    L43
  add    $0x1,%rdi
  add    $0x1,%rsi
  add    $0x1,%r8d
  cmp    %r8d,%ecx
  jne    L28
L15:
  mov    %ecx,(%rdx)
L8:
  mov    %esi,%eax
  sub    %edi,%eax
  ret
L19:
  mov    $0x0,%ecx
  jmp    L15
L43:
  mov    %r8d,(%rdx)
  test   %eax,%eax
  jle    L8
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int scmp3(unsigned char *p, unsigned char *q, int *l, int maxl) {
  int i;
  i = 0;
  while (maxl > 0 && *p == *q) {
    p++;
    q++;
    i++;
    maxl--;
  }
  *l = i;
  if (maxl > 0)
    return *p - *q;
  return q - p;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<scmp3>:
  endbr64
  xor    %eax,%eax
  test   %ecx,%ecx
  jg     L15
  jmp    L29
  nopl   0x0(%rax)
L6:
  add    $0x1,%eax
  add    $0x1,%rdi
  add    $0x1,%rsi
  cmp    %eax,%ecx
  je     L8
L15:
  movzbl (%rsi),%r8d
  cmp    %r8b,(%rdi)
  je     L6
  mov    %eax,(%rdx)
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  sub    %edx,%eax
  ret
L29:
  xor    %ecx,%ecx
L8:
  mov    %esi,%eax
  mov    %ecx,(%rdx)
  sub    %edi,%eax
  ret
```

assistant: ```c
int scmp3(unsigned char *p, unsigned char *q, int *l, int maxl) {
  int i;
  i = 0;
  while (maxl > 0 && *p == *q) {
    p++;
    q++;
    i++;
    maxl--;
  }
  *l = i;
  if (maxl > 0)
    return *p - *q;
  return q - p;
}
```