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
<transpose>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    -0x2c(%rbp),%eax
  imul   -0x30(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L3
L2:
  movq   $0x0,-0x10(%rbp)
  jmp    L20
L26:
  mov    -0x30(%rbp),%eax
  cltq
  imul   -0x18(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  imul   -0x10(%rbp),%rax
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  addq   $0x1,-0x10(%rbp)
L20:
  mov    -0x30(%rbp),%eax
  cltq
  cmp    %rax,-0x10(%rbp)
  jl     L26
  addq   $0x1,-0x18(%rbp)
L3:
  mov    -0x2c(%rbp),%eax
  cltq
  cmp    %rax,-0x18(%rbp)
  jl     L2
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
float *transpose(float *mat, int m, int n) {
  float *result = malloc(m * n * sizeof(float));
  for (long i = 0; i < m; ++i)
    for (long j = 0; j < n; ++j)
      result[j * m + i] = mat[i * n + j];
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<transpose>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %edx,%r12d
  mov    %esi,%edi
  imul   %edx,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  movslq %ebx,%r10
  test   %r10,%r10
  jle    L41
  mov    %r10,%rdi
  shl    $0x2,%rdi
  mov    $0x0,%r8d
  movslq %r12d,%r9
  jmp    L11
L46:
  mov    %r9,%rsi
  imul   %r8,%rsi
  lea    0x0(%rbp,%rsi,4),%rdx
  lea    (%rax,%r8,4),%rcx
  add    %r9,%rsi
  lea    0x0(%rbp,%rsi,4),%rsi
L10:
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rcx)
  add    $0x4,%rdx
  add    %rdi,%rcx
  cmp    %rsi,%rdx
  jne    L10
L57:
  add    $0x1,%r8
  cmp    %r10,%r8
  jge    L41
L11:
  test   %r9,%r9
  jg     L46
  jmp    L57
L41:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
float *transpose(float *mat, int m, int n) {
  float *result = malloc(m * n * sizeof(float));
  for (long i = 0; i < m; ++i)
    for (long j = 0; j < n; ++j)
      result[j * m + i] = mat[i * n + j];
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<transpose>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%edi
  imul   %edx,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  movslq %ebp,%r9
  test   %r9,%r9
  jle    L9
  movslq %r12d,%r10
  lea    0x0(,%r9,4),%rdi
  xor    %r8d,%r8d
  lea    0x0(,%r10,4),%r11
  neg    %r10
  lea    (%rbx,%r11,1),%rsi
  shl    $0x2,%r10
  cs nopw 0x0(%rax,%rax,1)
L36:
  lea    (%rsi,%r10,1),%rdx
  lea    (%rax,%r8,4),%rcx
  test   %r12d,%r12d
  jle    L32
  nopl   (%rax)
L11:
  movss  (%rdx),%xmm0
  add    $0x4,%rdx
  movss  %xmm0,(%rcx)
  add    %rdi,%rcx
  cmp    %rsi,%rdx
  jne    L11
L32:
  add    $0x1,%r8
  add    %r11,%rsi
  cmp    %r9,%r8
  jne    L36
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
float *transpose(float *mat, int m, int n) {
  float *result = malloc(m * n * sizeof(float));
  for (long i = 0; i < m; ++i)
    for (long j = 0; j < n; ++j)
      result[j * m + i] = mat[i * n + j];
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<transpose>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%edi
  imul   %edx,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  movslq %r12d,%r10
  test   %r10,%r10
  jle    L20
  movslq %ebp,%r9
  test   %r9,%r9
  jle    L20
  lea    0x0(,%r9,4),%r11
  neg    %r9
  lea    0x0(,%r10,4),%rdi
  xor    %r8d,%r8d
  lea    (%rbx,%r11,1),%rsi
  shl    $0x2,%r9
  nopl   0x0(%rax,%rax,1)
L2:
  lea    (%rsi,%r9,1),%rdx
  lea    (%rax,%r8,4),%rcx
  nopl   0x0(%rax,%rax,1)
L19:
  movss  (%rdx),%xmm0
  add    $0x4,%rdx
  movss  %xmm0,(%rcx)
  add    %rdi,%rcx
  cmp    %rdx,%rsi
  jne    L19
  add    $0x1,%r8
  add    %r11,%rsi
  cmp    %r10,%r8
  jne    L2
L20:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
float *transpose(float *mat, int m, int n) {
  float *result = malloc(m * n * sizeof(float));
  for (long i = 0; i < m; ++i)
    for (long j = 0; j < n; ++j)
      result[j * m + i] = mat[i * n + j];
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D28
type_infered: string
name: retrieve_data
arguments: tag: D4
type_infered: string
name: retrieve_data
arguments: tag: D9
type_infered: string

tool: D28: Usage: tdbdump [options] <filename>



tool: D4:    -h          this help message


tool: D9:    -k keyname  dumps value of keyname


assistant: ```c
void usage(void) {
  printf("Usage: tdbdump [options] <filename>\n\n");
  printf("   -h          this help message\n");
  printf("   -k keyname  dumps value of keyname\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D25
type_guessed: string
name: disassemble_data
arguments: data_label: D7
type_guessed: string
name: disassemble_data
arguments: data_label: D11
type_guessed: string

tool: D25: Usage: tdbdump [options] <filename>



tool: D7:    -h          this help message


tool: D11:    -k keyname  dumps value of keyname


assistant: ```c
void usage(void) {
  printf("Usage: tdbdump [options] <filename>\n\n");
  printf("   -h          this help message\n");
  printf("   -k keyname  dumps value of keyname\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D8(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: label: D26
type_guessed: string
name: parse_data
arguments: label: D0
type_guessed: string
name: parse_data
arguments: label: D8
type_guessed: string

tool: D26: Usage: tdbdump [options] <filename>



tool: D0:    -h          this help message


tool: D8:    -k keyname  dumps value of keyname


assistant: ```c
void usage(void) {
  printf("Usage: tdbdump [options] <filename>\n\n");
  printf("   -h          this help message\n");
  printf("   -k keyname  dumps value of keyname\n");
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    D11(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D16
datatype: string
name: unpack_data
arguments: data_tag: D3
datatype: string
name: unpack_data
arguments: data_tag: D11
datatype: string

tool: D16: Usage: tdbdump [options] <filename>



tool: D3:    -h          this help message


tool: D11:    -k keyname  dumps value of keyname


assistant: ```c
void usage(void) {
  printf("Usage: tdbdump [options] <filename>\n\n");
  printf("   -h          this help message\n");
  printf("   -k keyname  dumps value of keyname\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<executeCommand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    $0x0,%r8d
  mov    $0x2000,%ecx
  mov    %rax,%rdx
  mov    $0x4,%esi
  mov    $0x21,%edi
  call   <interrupt@plt>
  nop
  leave
  ret
```

assistant: ```c
void executeCommand(char *args[]) {
  interrupt(0x21, 0x4, args[1], 0x2000, 0);
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<executeCommand>:
  endbr64
  sub    $0x8,%rsp
  mov    0x8(%rdi),%rdx
  mov    $0x0,%r8d
  mov    $0x2000,%ecx
  mov    $0x4,%esi
  mov    $0x21,%edi
  call   <interrupt@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void executeCommand(char *args[]) {
  interrupt(0x21, 0x4, args[1], 0x2000, 0);
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<executeCommand>:
  endbr64
  mov    0x8(%rdi),%rdx
  xor    %r8d,%r8d
  mov    $0x2000,%ecx
  mov    $0x4,%esi
  mov    $0x21,%edi
  jmp    1050 <interrupt@plt>
```

assistant: ```c
void executeCommand(char *args[]) {
  interrupt(0x21, 0x4, args[1], 0x2000, 0);
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<executeCommand>:
  endbr64
  mov    0x8(%rdi),%rdx
  xor    %r8d,%r8d
  mov    $0x2000,%ecx
  mov    $0x4,%esi
  mov    $0x21,%edi
  jmp    1050 <interrupt@plt>
```

assistant: ```c
void executeCommand(char *args[]) {
  interrupt(0x21, 0x4, args[1], 0x2000, 0);
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sortColors>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L13
L69:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  mov    -0x14(%rbp,%rdx,4),%edx
  add    $0x1,%edx
  cltq
  mov    %edx,-0x14(%rbp,%rax,4)
  addl   $0x1,-0x1c(%rbp)
L13:
  mov    -0x1c(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L69
  movl   $0x0,-0x1c(%rbp)
  jmp    L22
L14:
  movl   $0x0,-0x18(%rbp)
  jmp    L63
L8:
  mov    -0x28(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x18(%rbp)
L63:
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x14(%rbp,%rax,4),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L8
  addl   $0x1,-0x1c(%rbp)
L22:
  cmpl   $0x2,-0x1c(%rbp)
  jle    L14
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

assistant: ```c
void sortColors(int *nums, int numsSize) {
  int count[3] = {0};
  int i, j;
  for (i = 0; i < numsSize; i++) {
    count[nums[i]]++;
  }
  for (i = 0; i < 3; i++) {
    for (j = 0; j < count[i]; j++) {
      *(nums++) = i;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sortColors>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,0xc(%rsp)
  movl   $0x0,0x14(%rsp)
  test   %esi,%esi
  jle    L32
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rcx
L72:
  movslq (%rax),%rdx
  addl   $0x1,0xc(%rsp,%rdx,4)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L72
L32:
  mov    $0x0,%ecx
L40:
  mov    %ecx,%edx
  mov    0xc(%rsp,%rcx,4),%esi
  mov    %esi,%eax
  lea    (%rdi,%rax,4),%rax
  test   %esi,%esi
  jle    L65
L7:
  add    $0x4,%rdi
  mov    %edx,-0x4(%rdi)
  cmp    %rax,%rdi
  jne    L7
L6:
  add    $0x1,%rcx
  cmp    $0x3,%rcx
  je     L16
  mov    %rax,%rdi
  jmp    L40
L65:
  mov    %rdi,%rax
  jmp    L6
L16:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L78
  add    $0x28,%rsp
  ret
L78:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sortColors(int *nums, int numsSize) {
  int count[3] = {0};
  int i, j;
  for (i = 0; i < numsSize; i++) {
    count[nums[i]]++;
  }
  for (i = 0; i < 3; i++) {
    for (j = 0; j < count[i]; j++) {
      *(nums++) = i;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sortColors>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  movq   $0x0,0xc(%rsp)
  test   %esi,%esi
  jle    L68
  lea    -0x1(%rsi),%edx
  mov    %rdi,%rax
  lea    0x4(%rdi,%rdx,4),%rcx
  nopl   0x0(%rax)
L65:
  movslq (%rax),%rdx
  add    $0x4,%rax
  addl   $0x1,0xc(%rsp,%rdx,4)
  cmp    %rcx,%rax
  jne    L65
  mov    0xc(%rsp),%esi
L84:
  xor    %ecx,%ecx
  movslq %esi,%rax
  mov    %ecx,%edx
  lea    (%rdi,%rax,4),%rax
  test   %esi,%esi
  jle    L29
  nopl   0x0(%rax)
L5:
  add    $0x4,%rdi
  mov    %edx,-0x4(%rdi)
  cmp    %rax,%rdi
  jne    L5
  add    $0x1,%rcx
  cmp    $0x3,%rcx
  je     L82
L88:
  mov    0xc(%rsp,%rcx,4),%esi
  mov    %rax,%rdi
  mov    %ecx,%edx
  movslq %esi,%rax
  lea    (%rdi,%rax,4),%rax
  test   %esi,%esi
  jg     L5
L29:
  add    $0x1,%rcx
  mov    %rdi,%rax
  cmp    $0x3,%rcx
  jne    L88
  nop
L82:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L67
  add    $0x28,%rsp
  ret
L68:
  xor    %esi,%esi
  jmp    L84
L67:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sortColors(int *nums, int numsSize) {
  int count[3] = {0};
  int i, j;
  for (i = 0; i < numsSize; i++) {
    count[nums[i]]++;
  }
  for (i = 0; i < 3; i++) {
    for (j = 0; j < count[i]; j++) {
      *(nums++) = i;
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<sortColors>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  movq   $0x0,0xc(%rsp)
  test   %esi,%esi
  jle    L9
  lea    -0x1(%rsi),%edx
  mov    %rdi,%rax
  lea    0x4(%rdi,%rdx,4),%rcx
  nopl   0x0(%rax,%rax,1)
L63:
  movslq (%rax),%rdx
  add    $0x4,%rax
  addl   $0x1,0xc(%rsp,%rdx,4)
  cmp    %rcx,%rax
  jne    L63
L9:
  xor    %esi,%esi
  lea    0xc(%rsp),%r9
L21:
  mov    (%r9,%rsi,4),%edx
  mov    %esi,%r8d
  test   %edx,%edx
  jle    L57
  lea    -0x1(%rdx),%eax
  cmp    $0x2,%eax
  jbe    L70
  mov    %edx,%ecx
  movd   %r8d,%xmm1
  mov    %rdi,%rax
  shr    $0x2,%ecx
  pshufd $0x0,%xmm1,%xmm0
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nopl   0x0(%rax)
L18:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rax,%rcx
  jne    L18
  mov    %edx,%eax
  and    $0xfffffffc,%eax
  mov    %eax,%ecx
  lea    (%rdi,%rcx,4),%rcx
  cmp    %eax,%edx
  je     L35
L92:
  lea    0x1(%rax),%r10d
  mov    %r8d,(%rcx)
  cmp    %edx,%r10d
  jge    L35
  add    $0x2,%eax
  mov    %r8d,0x4(%rcx)
  cmp    %eax,%edx
  jle    L35
  mov    %r8d,0x8(%rcx)
L35:
  lea    (%rdi,%rdx,4),%rdi
L57:
  add    $0x1,%rsi
  cmp    $0x3,%rsi
  jne    L21
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x28,%rsp
  ret
L70:
  mov    %rdi,%rcx
  xor    %eax,%eax
  jmp    L92
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sortColors(int *nums, int numsSize) {
  int count[3] = {0};
  int i, j;
  for (i = 0; i < numsSize; i++) {
    count[nums[i]]++;
  }
  for (i = 0; i < 3; i++) {
    for (j = 0; j < count[i]; j++) {
      *(nums++) = i;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<str_to_longlong_u>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x8,-0x1c(%rbp)
  jmp    L30
L48:
  mov    -0x18(%rbp),%rax
  shl    $0x8,%rax
  mov    %rax,%rbx
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  or     %rbx,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L30
  addq   $0x1,-0x28(%rbp)
L30:
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  test   %eax,%eax
  jne    L48
  jmp    L35
L3:
  mov    -0x18(%rbp),%rax
  shl    $0x8,%rax
  mov    %rax,%rbx
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  or     %rbx,%rax
  mov    %rax,-0x18(%rbp)
  addq   $0x1,-0x28(%rbp)
L35:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
unsigned long long str_to_longlong_u(unsigned char *s) {
  unsigned long long temp = 0;
  int loop = 8;
  while (loop--) {
    temp = (temp << 8) | (toupper(*s));
    if (*s)
      s++;
  }
  while (*s) {
    temp = (temp << 8) | (toupper(*s));
    s++;
  }
  return (temp);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<str_to_longlong_u>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0x8,%r12d
  mov    $0x0,%ebx
L17:
  shl    $0x8,%rbx
  movzbl 0x0(%rbp),%edi
  call   <toupper@plt>
  or     %rax,%rbx
  cmpb   $0x1,0x0(%rbp)
  sbb    $0xffffffffffffffff,%rbp
  sub    $0x1,%r12d
  jne    L17
  movzbl 0x0(%rbp),%edi
  test   %dil,%dil
  je     L9
L12:
  shl    $0x8,%rbx
  movzbl %dil,%edi
  call   <toupper@plt>
  or     %rax,%rbx
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%edi
  test   %dil,%dil
  jne    L12
L9:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned long long str_to_longlong_u(unsigned char *s) {
  unsigned long long temp = 0;
  int loop = 8;
  while (loop--) {
    temp = (temp << 8) | (toupper(*s));
    if (*s)
      s++;
  }
  while (*s) {
    temp = (temp << 8) | (toupper(*s));
    s++;
  }
  return (temp);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<str_to_longlong_u>:
  endbr64
  push   %r12
  mov    $0x8,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movzbl (%rdi),%edi
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L10:
  call   <toupper@plt>
  movzbl 0x0(%rbp),%edi
  shl    $0x8,%rbx
  or     %rax,%rbx
  test   %dil,%dil
  je     L14
  movzbl 0x1(%rbp),%edi
  add    $0x1,%rbp
L14:
  sub    $0x1,%r12d
  jne    L10
  test   %dil,%dil
  je     L15
  nopl   0x0(%rax,%rax,1)
L2:
  call   <toupper@plt>
  movzbl 0x1(%rbp),%edi
  shl    $0x8,%rbx
  add    $0x1,%rbp
  or     %rax,%rbx
  test   %dil,%dil
  jne    L2
L15:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned long long str_to_longlong_u(unsigned char *s) {
  unsigned long long temp = 0;
  int loop = 8;
  while (loop--) {
    temp = (temp << 8) | (toupper(*s));
    if (*s)
      s++;
  }
  while (*s) {
    temp = (temp << 8) | (toupper(*s));
    s++;
  }
  return (temp);
}
```
user: Translate the below assembly code into C programming language:

```asm
<str_to_longlong_u>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    $0x8,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  movzbl (%rdi),%edi
  nopl   0x0(%rax)
L23:
  call   <toupper@plt>
  movzbl 0x0(%r13),%edi
  shl    $0x8,%rbx
  or     %rax,%rbx
  test   %dil,%dil
  je     L5
  movzbl 0x1(%r13),%edi
  lea    0x1(%r13),%rbp
  sub    $0x1,%r12d
  je     L17
  mov    %rbp,%r13
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L18:
  call   <toupper@plt>
  shl    $0x8,%rbx
  movzbl 0x1(%rbp),%edi
  add    $0x1,%rbp
  or     %rax,%rbx
L17:
  test   %dil,%dil
  jne    L18
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  sub    $0x1,%r12d
  jne    L23
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
unsigned long long str_to_longlong_u(unsigned char *s) {
  unsigned long long temp = 0;
  int loop = 8;
  while (loop--) {
    temp = (temp << 8) | (toupper(*s));
    if (*s)
      s++;
  }
  while (*s) {
    temp = (temp << 8) | (toupper(*s));
    s++;
  }
  return (temp);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bitcount>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x55555555,-0x14(%rbp)
  movl   $0x33333333,-0x10(%rbp)
  movl   $0x1010101,-0xc(%rbp)
  movl   $0xf0f0f0f,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  sar    %eax
  and    -0x14(%rbp),%eax
  sub    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  and    -0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  sar    $0x2,%eax
  and    -0x10(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  sar    $0x4,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  and    -0x8(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    $0x38,%edx
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int bitcount(uint16_t word) {
  uint64_t x = word;
  const uint64_t m1 = 0x5555555555555555ULL;
  const uint64_t m2 = 0x3333333333333333ULL;
  const uint64_t h01 = 0x0101010101010101ULL;
  const uint64_t m4 = 0x0f0f0f0f0f0f0f0fULL;
  x -= (x >> 1) & m1;
  x = (x & m2) + ((x >> 2) & m2);
  x = (x + (x >> 4)) & m4;
  int retval = (x * h01) >> 56;
  return retval;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<bitcount>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int bitcount(uint16_t word) {
  uint64_t x = word;
  const uint64_t m1 = 0x5555555555555555ULL;
  const uint64_t m2 = 0x3333333333333333ULL;
  const uint64_t h01 = 0x0101010101010101ULL;
  const uint64_t m4 = 0x0f0f0f0f0f0f0f0fULL;
  x -= (x >> 1) & m1;
  x = (x & m2) + ((x >> 2) & m2);
  x = (x + (x >> 4)) & m4;
  int retval = (x * h01) >> 56;
  return retval;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bitcount>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int bitcount(uint16_t word) {
  uint64_t x = word;
  const uint64_t m1 = 0x5555555555555555ULL;
  const uint64_t m2 = 0x3333333333333333ULL;
  const uint64_t h01 = 0x0101010101010101ULL;
  const uint64_t m4 = 0x0f0f0f0f0f0f0f0fULL;
  x -= (x >> 1) & m1;
  x = (x & m2) + ((x >> 2) & m2);
  x = (x + (x >> 4)) & m4;
  int retval = (x * h01) >> 56;
  return retval;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitcount>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int bitcount(uint16_t word) {
  uint64_t x = word;
  const uint64_t m1 = 0x5555555555555555ULL;
  const uint64_t m2 = 0x3333333333333333ULL;
  const uint64_t h01 = 0x0101010101010101ULL;
  const uint64_t m4 = 0x0f0f0f0f0f0f0f0fULL;
  x -= (x >> 1) & m1;
  x = (x & m2) + ((x >> 2) & m2);
  x = (x + (x >> 4)) & m4;
  int retval = (x * h01) >> 56;
  return retval;
}
```