ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<xrop3_95_0_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x14(%rbp)
  mov    %al,-0x18(%rbp)
  movl   $0xff,-0x10(%rbp)
  movzbl -0x14(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x14(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0xc(%rbp)
  movzbl -0x18(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x18(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  and    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  mov    -0x8(%rbp),%eax
  not    %eax
  or     -0xc(%rbp),%eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  not    %eax
  and    -0x8(%rbp),%eax
  and    -0x10(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_95_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  stk1 = D & T;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<xrop3_95_0_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%esi
  mov    %esi,%edx
  shl    $0x8,%edx
  or     %esi,%edx
  mov    %edx,%eax
  not    %eax
  or     %edi,%eax
  not    %esi
  and    %esi,%eax
  not    %edi
  and    %edx,%edi
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_95_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  stk1 = D & T;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: What is the c source code of the assembly code below:

```asm
<xrop3_95_0_1>:
  endbr64
  movzbl %sil,%esi
  movzbl %dil,%eax
  mov    %esi,%edx
  mov    %eax,%edi
  shl    $0x8,%edx
  shl    $0x8,%edi
  or     %esi,%edx
  or     %eax,%edi
  not    %esi
  mov    %edx,%eax
  not    %eax
  or     %edi,%eax
  not    %edi
  and    %edx,%edi
  and    %esi,%eax
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_95_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  stk1 = D & T;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<xrop3_95_0_1>:
  endbr64
  movzbl %sil,%esi
  movzbl %dil,%eax
  mov    %esi,%edx
  mov    %eax,%edi
  shl    $0x8,%edx
  shl    $0x8,%edi
  or     %esi,%edx
  or     %eax,%edi
  not    %esi
  mov    %edx,%eax
  not    %eax
  or     %edi,%eax
  not    %edi
  and    %edx,%edi
  and    %esi,%eax
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_95_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  stk1 = D & T;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<load>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8,-0x30(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jne    L32
  mov    -0x20(%rbp),%eax
  mov    %eax,%edi
  call   <typeSize@plt>
  mov    %eax,%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <loadN@plt>
  jmp    L37
L32:
  cmpl   $0x1,-0x20(%rbp)
  jg     L68
  cmpl   $0x0,-0x20(%rbp)
  jne    L53
  cmpl   $0x4,-0x24(%rbp)
  je     L68
L53:
  mov    $0x1,%eax
  jmp    L6
L68:
  mov    $0x0,%eax
L6:
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x24(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <loadN@plt>
  mov    %eax,-0x4(%rbp)
  cmpq   $0x0,-0x30(%rbp)
  je     L43
  mov    -0x24(%rbp),%eax
  shl    $0x3,%eax
  mov    $0xffffffffffffffff,%rdx
  mov    %eax,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rax
  mov    %eax,%esi
  mov    -0x24(%rbp),%eax
  shl    $0x3,%eax
  sub    $0x1,%eax
  mov    -0x4(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  cltq
  imul   %eax,%esi
  mov    %esi,%edx
  mov    -0x4(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x4(%rbp)
L43:
  cmpl   $0x0,-0x20(%rbp)
  mov    -0x4(%rbp),%eax
L37:
  leave
  ret
```

assistant: ```c
uint64_t load(uint8_t *mem, uint64_t addr, uint64_t ty, uint64_t packing, uint8_t sign_extend) {
  if (packing == 0)
    return loadN(mem, addr, typeSize(ty));
  else {
    assert(ty < 2 && !(ty == 0 && packing == 4));
    uint64_t res = loadN(mem, addr, packing);
    if (sign_extend) {
      res = res | (0xffffffffffffffff << (8 * packing)) * (res >> (8 * packing - 1));
    }
    if (ty == 0)
      res = res & 0xffffffff;
    else
      res = res & 0xffffffffffffffff;
    return res;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<load>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r13d
  test   %ecx,%ecx
  je     L1
  mov    %ecx,%ebx
  mov    %r8,%r12
  mov    $0x0,%eax
  cmp    $0x1,%edx
  jle    L13
L19:
  mov    %eax,%edi
  call   <assert@plt>
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <loadN@plt>
  test   %r12,%r12
  je     L17
  shl    $0x3,%ebx
  lea    -0x1(%rbx),%ecx
  mov    %eax,%edx
  sar    %cl,%edx
  mov    $0xffffffffffffffff,%rsi
  mov    %ebx,%ecx
  shl    %cl,%rsi
  imul   %esi,%edx
  or     %edx,%eax
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  mov    %edx,%edi
  call   <typeSize@plt>
  mov    %eax,%edx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <loadN@plt>
  jmp    L17
L13:
  test   %edx,%edx
  setne  %al
  cmp    $0x4,%ecx
  setne  %dl
  or     %edx,%eax
  movzbl %al,%eax
  jmp    L19
```

assistant: ```c
uint64_t load(uint8_t *mem, uint64_t addr, uint64_t ty, uint64_t packing, uint8_t sign_extend) {
  if (packing == 0)
    return loadN(mem, addr, typeSize(ty));
  else {
    assert(ty < 2 && !(ty == 0 && packing == 4));
    uint64_t res = loadN(mem, addr, packing);
    if (sign_extend) {
      res = res | (0xffffffffffffffff << (8 * packing)) * (res >> (8 * packing - 1));
    }
    if (ty == 0)
      res = res & 0xffffffff;
    else
      res = res & 0xffffffffffffffff;
    return res;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<load>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %ecx,%ecx
  je     L35
  mov    %r8,%r13
  mov    %ecx,%ebx
  xor    %r8d,%r8d
  cmp    $0x1,%edx
  jle    L39
L34:
  mov    %r8d,%edi
  call   <assert@plt>
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <loadN@plt>
  test   %r13,%r13
  je     L13
  shl    $0x3,%ebx
  mov    %eax,%edx
  mov    $0xffffffffffffffff,%rsi
  lea    -0x1(%rbx),%ecx
  sar    %cl,%edx
  mov    %ebx,%ecx
  shl    %cl,%rsi
  imul   %esi,%edx
  or     %edx,%eax
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L39:
  test   %edx,%edx
  setne  %r8b
  cmp    $0x4,%ecx
  setne  %al
  or     %eax,%r8d
  movzbl %r8b,%r8d
  jmp    L34
  nopl   (%rax)
L35:
  mov    %edx,%edi
  call   <typeSize@plt>
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %rbp,%rdi
  pop    %rbx
  mov    %eax,%edx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1090 <loadN@plt>
```

assistant: ```c
uint64_t load(uint8_t *mem, uint64_t addr, uint64_t ty, uint64_t packing, uint8_t sign_extend) {
  if (packing == 0)
    return loadN(mem, addr, typeSize(ty));
  else {
    assert(ty < 2 && !(ty == 0 && packing == 4));
    uint64_t res = loadN(mem, addr, packing);
    if (sign_extend) {
      res = res | (0xffffffffffffffff << (8 * packing)) * (res >> (8 * packing - 1));
    }
    if (ty == 0)
      res = res & 0xffffffff;
    else
      res = res & 0xffffffffffffffff;
    return res;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<load>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %ecx,%ecx
  je     L16
  mov    %r8,%r13
  mov    %ecx,%ebx
  xor    %r8d,%r8d
  cmp    $0x1,%edx
  jle    L35
L29:
  mov    %r8d,%edi
  call   <assert@plt>
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <loadN@plt>
  test   %r13,%r13
  je     L8
  shl    $0x3,%ebx
  mov    %eax,%edx
  mov    $0xffffffffffffffff,%rsi
  lea    -0x1(%rbx),%ecx
  sar    %cl,%edx
  mov    %ebx,%ecx
  shl    %cl,%rsi
  imul   %esi,%edx
  or     %edx,%eax
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L35:
  test   %edx,%edx
  setne  %r8b
  cmp    $0x4,%ecx
  setne  %al
  or     %eax,%r8d
  movzbl %r8b,%r8d
  jmp    L29
  nopl   (%rax)
L16:
  mov    %edx,%edi
  call   <typeSize@plt>
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %rbp,%rdi
  pop    %rbx
  mov    %eax,%edx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1090 <loadN@plt>
```

assistant: ```c
uint64_t load(uint8_t *mem, uint64_t addr, uint64_t ty, uint64_t packing, uint8_t sign_extend) {
  if (packing == 0)
    return loadN(mem, addr, typeSize(ty));
  else {
    assert(ty < 2 && !(ty == 0 && packing == 4));
    uint64_t res = loadN(mem, addr, packing);
    if (sign_extend) {
      res = res | (0xffffffffffffffff << (8 * packing)) * (res >> (8 * packing - 1));
    }
    if (ty == 0)
      res = res & 0xffffffff;
    else
      res = res & 0xffffffffffffffff;
    return res;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<hash_bits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L7
L10:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  lea    0x0(,%rax,8),%edx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  sar    $0x1d,%eax
  mov    %edx,%ecx
  or     %eax,%ecx
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  xor    %eax,%ecx
  mov    %ecx,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x1,-0x8(%rbp)
L7:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L10
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void hash_bits(uint32_t *hash, const void *ptr, size_t len) {
  for (size_t n = 0; n < len; n++)
    *hash = ((*hash << 3) | (*hash >> 29)) ^ ((char *)ptr)[n];
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<hash_bits>:
  endbr64
  test   %rdx,%rdx
  je     L16
  mov    %rsi,%rcx
  add    %rdx,%rsi
L10:
  mov    (%rdi),%edx
  lea    0x0(,%rdx,8),%eax
  sar    $0x1d,%edx
  or     %edx,%eax
  movsbl (%rcx),%edx
  xor    %edx,%eax
  mov    %eax,(%rdi)
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  jne    L10
L16:
  ret
```

assistant: ```c
void hash_bits(uint32_t *hash, const void *ptr, size_t len) {
  for (size_t n = 0; n < len; n++)
    *hash = ((*hash << 3) | (*hash >> 29)) ^ ((char *)ptr)[n];
}
```
user: What is the c source code of the assembly code below?

```asm
<hash_bits>:
  endbr64
  test   %rdx,%rdx
  je     L9
  mov    (%rdi),%eax
  add    %rsi,%rdx
  xchg   %ax,%ax
L1:
  lea    0x0(,%rax,8),%ecx
  sar    $0x1d,%eax
  add    $0x1,%rsi
  or     %eax,%ecx
  movsbl -0x1(%rsi),%eax
  xor    %ecx,%eax
  mov    %eax,(%rdi)
  cmp    %rsi,%rdx
  jne    L1
L9:
  ret
```

assistant: ```c
void hash_bits(uint32_t *hash, const void *ptr, size_t len) {
  for (size_t n = 0; n < len; n++)
    *hash = ((*hash << 3) | (*hash >> 29)) ^ ((char *)ptr)[n];
}
```
user: What is the c source code of the assembly code below:

```asm
<hash_bits>:
  endbr64
  test   %rdx,%rdx
  je     L1
  mov    (%rdi),%eax
  add    %rsi,%rdx
  xchg   %ax,%ax
L16:
  lea    0x0(,%rax,8),%ecx
  sar    $0x1d,%eax
  add    $0x1,%rsi
  or     %eax,%ecx
  movsbl -0x1(%rsi),%eax
  xor    %ecx,%eax
  mov    %eax,(%rdi)
  cmp    %rsi,%rdx
  jne    L16
L1:
  ret
```

assistant: ```c
void hash_bits(uint32_t *hash, const void *ptr, size_t len) {
  for (size_t n = 0; n < len; n++)
    *hash = ((*hash << 3) | (*hash >> 29)) ^ ((char *)ptr)[n];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L18
L9:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    %dl,(%rax)
L18:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L9
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *memset(void *s, int c, size_t count) {
  char *xs = s;
  while (count--) {
    *xs++ = c;
  }
  return s;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<memset>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L6
  add    %rdi,%rdx
  mov    %rdi,%rcx
L2:
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  cmp    %rcx,%rdx
  jne    L2
L6:
  ret
```

assistant: ```c
void *memset(void *s, int c, size_t count) {
  char *xs = s;
  while (count--) {
    *xs++ = c;
  }
  return s;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<memset>:
  endbr64
  test   %rdx,%rdx
  je     L7
  sub    $0x8,%rsp
  movsbl %sil,%esi
  call   <memset@plt>
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memset(void *s, int c, size_t count) {
  char *xs = s;
  while (count--) {
    *xs++ = c;
  }
  return s;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<memset>:
  endbr64
  test   %rdx,%rdx
  je     L1
  sub    $0x8,%rsp
  movsbl %sil,%esi
  call   <memset@plt>
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memset(void *s, int c, size_t count) {
  char *xs = s;
  while (count--) {
    *xs++ = c;
  }
  return s;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<substring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L9
L20:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L9:
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x8(%rbp)
  jl     L20
  cmpl   $0x0,-0x8(%rbp)
  je     L20
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void substring(char source[], int start, int count, char result[]) {
  int i;
  int j = 0;
  for (i = start; i < start + count || i == '\0'; ++i) {
    result[j] = source[i];
    ++j;
  }
  result[j] = '\0';
}
```
user: Translate the below assembly code into C programming language:

```asm
<substring>:
  endbr64
  add    %esi,%edx
  test   %esi,%esi
  je     L22
  cmp    %edx,%esi
  jge    L25
L22:
  mov    %esi,%r8d
  mov    $0x1,%eax
  movslq %esi,%rsi
  add    %rsi,%rdi
L9:
  movzbl -0x1(%rdi,%rax,1),%esi
  mov    %sil,-0x1(%rcx,%rax,1)
  mov    %eax,%esi
  add    $0x1,%rax
  add    $0x1,%r8d
  je     L9
  cmp    %edx,%r8d
  jl     L9
L47:
  movslq %esi,%rsi
  movb   $0x0,(%rcx,%rsi,1)
  ret
L25:
  mov    $0x0,%esi
  jmp    L47
```

assistant: ```c
void substring(char source[], int start, int count, char result[]) {
  int i;
  int j = 0;
  for (i = start; i < start + count || i == '\0'; ++i) {
    result[j] = source[i];
    ++j;
  }
  result[j] = '\0';
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<substring>:
  endbr64
  add    %esi,%edx
  test   %esi,%esi
  je     L32
  cmp    %edx,%esi
  jge    L39
L32:
  movslq %esi,%r8
  mov    $0x1,%eax
  add    %rdi,%r8
  nopl   0x0(%rax)
L37:
  movzbl -0x1(%r8,%rax,1),%edi
  mov    %dil,-0x1(%rcx,%rax,1)
  movslq %eax,%rdi
  add    $0x1,%rax
  add    $0x1,%esi
  je     L37
  cmp    %edx,%esi
  jl     L37
  add    %rdi,%rcx
L39:
  movb   $0x0,(%rcx)
  ret
```

assistant: ```c
void substring(char source[], int start, int count, char result[]) {
  int i;
  int j = 0;
  for (i = start; i < start + count || i == '\0'; ++i) {
    result[j] = source[i];
    ++j;
  }
  result[j] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<substring>:
  endbr64
  add    %esi,%edx
  test   %esi,%esi
  je     L10
  cmp    %edx,%esi
  jge    L30
L10:
  movslq %esi,%r8
  mov    $0x1,%eax
  add    %rdi,%r8
  nopl   0x0(%rax)
L15:
  movzbl -0x1(%r8,%rax,1),%edi
  mov    %dil,-0x1(%rcx,%rax,1)
  movslq %eax,%rdi
  add    $0x1,%rax
  add    $0x1,%esi
  je     L15
  cmp    %edx,%esi
  jl     L15
  add    %rdi,%rcx
L30:
  movb   $0x0,(%rcx)
  ret
```

assistant: ```c
void substring(char source[], int start, int count, char result[]) {
  int i;
  int j = 0;
  for (i = start; i < start + count || i == '\0'; ++i) {
    result[j] = source[i];
    ++j;
  }
  result[j] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<PJWHash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movl   $0x20,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  shr    $0x2,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  shr    $0x3,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0xc(%rbp),%eax
  mov    $0xffffffff,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L13:
  mov    -0xc(%rbp),%eax
  mov    -0x18(%rbp),%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%ecx
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %ecx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  and    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L11
  mov    -0x10(%rbp),%eax
  mov    -0x4(%rbp),%edx
  mov    %eax,%ecx
  shr    %cl,%edx
  mov    %edx,%eax
  xor    -0x18(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  not    %eax
  and    %edx,%eax
  mov    %eax,-0x18(%rbp)
L11:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L13
  mov    -0x18(%rbp),%eax
  and    $0x7fffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int PJWHash(const char *str) {
  unsigned int BitsInUnignedInt = (unsigned int)(sizeof(unsigned int) * 8);
  unsigned int ThreeQuarters = (unsigned int)((BitsInUnignedInt * 3) / 4);
  unsigned int OneEighth = (unsigned int)(BitsInUnignedInt / 8);
  unsigned int HighBits = (unsigned int)(0xFFFFFFFF) << (BitsInUnignedInt - OneEighth);
  unsigned int hash = 0;
  unsigned int test = 0;
  while (*str) {
    hash = (hash << OneEighth) + (*str++);
    if ((test = hash & HighBits) != 0) {
      hash = ((hash ^ (test >> ThreeQuarters)) & (~HighBits));
    }
  }
  return (hash & 0x7FFFFFFF);
}
```
user: Translate the below assembly code into C programming language:

```asm
<PJWHash>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L19
  mov    $0x0,%eax
  jmp    L25
L10:
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L11
L25:
  shl    $0x4,%eax
  add    $0x1,%rdi
  movsbl %dl,%edx
  add    %edx,%eax
  mov    %eax,%edx
  and    $0xf0000000,%edx
  je     L10
  shr    $0x18,%edx
  xor    %edx,%eax
  and    $0xfffffff,%eax
  jmp    L10
L19:
  mov    $0x0,%eax
L11:
  and    $0x7fffffff,%eax
  ret
```

assistant: ```c
unsigned int PJWHash(const char *str) {
  unsigned int BitsInUnignedInt = (unsigned int)(sizeof(unsigned int) * 8);
  unsigned int ThreeQuarters = (unsigned int)((BitsInUnignedInt * 3) / 4);
  unsigned int OneEighth = (unsigned int)(BitsInUnignedInt / 8);
  unsigned int HighBits = (unsigned int)(0xFFFFFFFF) << (BitsInUnignedInt - OneEighth);
  unsigned int hash = 0;
  unsigned int test = 0;
  while (*str) {
    hash = (hash << OneEighth) + (*str++);
    if ((test = hash & HighBits) != 0) {
      hash = ((hash ^ (test >> ThreeQuarters)) & (~HighBits));
    }
  }
  return (hash & 0x7FFFFFFF);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<PJWHash>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L27
  nopl   (%rax)
L29:
  shl    $0x4,%eax
  add    $0x1,%rdi
  add    %edx,%eax
  mov    %eax,%edx
  and    $0xf0000000,%edx
  je     L1
  shr    $0x18,%edx
  xor    %edx,%eax
  and    $0xfffffff,%eax
L1:
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L29
  ret
  nopl   (%rax)
L27:
  ret
```

assistant: ```c
unsigned int PJWHash(const char *str) {
  unsigned int BitsInUnignedInt = (unsigned int)(sizeof(unsigned int) * 8);
  unsigned int ThreeQuarters = (unsigned int)((BitsInUnignedInt * 3) / 4);
  unsigned int OneEighth = (unsigned int)(BitsInUnignedInt / 8);
  unsigned int HighBits = (unsigned int)(0xFFFFFFFF) << (BitsInUnignedInt - OneEighth);
  unsigned int hash = 0;
  unsigned int test = 0;
  while (*str) {
    hash = (hash << OneEighth) + (*str++);
    if ((test = hash & HighBits) != 0) {
      hash = ((hash ^ (test >> ThreeQuarters)) & (~HighBits));
    }
  }
  return (hash & 0x7FFFFFFF);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<PJWHash>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L9
  nopl   (%rax)
L28:
  shl    $0x4,%eax
  add    $0x1,%rdi
  add    %edx,%eax
  mov    %eax,%edx
  and    $0xf0000000,%edx
  je     L26
  shr    $0x18,%edx
  xor    %edx,%eax
  and    $0xfffffff,%eax
L26:
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L28
  ret
  nopl   (%rax)
L9:
  ret
```

assistant: ```c
unsigned int PJWHash(const char *str) {
  unsigned int BitsInUnignedInt = (unsigned int)(sizeof(unsigned int) * 8);
  unsigned int ThreeQuarters = (unsigned int)((BitsInUnignedInt * 3) / 4);
  unsigned int OneEighth = (unsigned int)(BitsInUnignedInt / 8);
  unsigned int HighBits = (unsigned int)(0xFFFFFFFF) << (BitsInUnignedInt - OneEighth);
  unsigned int hash = 0;
  unsigned int test = 0;
  while (*str) {
    hash = (hash << OneEighth) + (*str++);
    if ((test = hash & HighBits) != 0) {
      hash = ((hash ^ (test >> ThreeQuarters)) & (~HighBits));
    }
  }
  return (hash & 0x7FFFFFFF);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x54,%edi
  call   <g@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  g(84);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x54,%edi
  call   <g@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  g(84);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x54,%edi
  call   <g@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  g(84);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x54,%edi
  call   <g@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  g(84);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<callee>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x2,%eax
  pop    %rbp
  ret
```

assistant: ```c
int callee(int x) {
  int y = x * x;
  return (y - 2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<callee>:
  endbr64
  imul   %edi,%edi
  lea    -0x2(%rdi),%eax
  ret
```

assistant: ```c
int callee(int x) {
  int y = x * x;
  return (y - 2);
}
```