ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<printregs>:
  endbr64
  mov    %esi,%r8d
  mov    %edx,%esi
  push   %r15
  mov    %ecx,%r11d
  mov    %r8d,%edx
  mov    %r8d,%r15d
  push   %r14
  mov    %r8d,%r14d
  sar    $0x18,%edx
  sar    $0x10,%r15d
  push   %r13
  mov    %edi,%r13d
  movzbl %dl,%edx
  movzbl %r15b,%r15d
  sar    $0x8,%r14d
  push   %r12
  shl    $0x8,%rdx
  push   %rbp
  movzbl %r14b,%r14d
  sar    $0x18,%r13d
  or     %r15,%rdx
  push   %rbx
  mov    %edi,%r12d
  mov    %ecx,%ebx
  shl    $0x8,%rdx
  sar    $0x10,%r12d
  mov    %edi,%ebp
  mov    %esi,%r9d
  or     %r14,%rdx
  movzbl %r8b,%r14d
  sar    $0x10,%ebx
  mov    %esi,%r10d
  shl    $0x8,%rdx
  sub    $0x38,%rsp
  movzbl %bl,%ebx
  sar    $0x8,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rdx,%r8
  mov    %esi,%eax
  movzbl %r13b,%edx
  or     %r14,%r8
  sar    $0x8,%eax
  movzbl %dil,%edi
  movzbl %sil,%esi
  shl    $0x8,%r8
  mov    %eax,(%rsp)
  mov    %ecx,%eax
  sar    $0x8,%r11d
  mov    %r8,%r13
  sar    $0x18,%eax
  movzbl %r11b,%r11d
  movzbl %cl,%ecx
  or     %rdx,%r13
  movzbl %al,%eax
  movzbl %r12b,%edx
  sar    $0x10,%r9d
  mov    %r13,%r12
  shl    $0x8,%rax
  movzbl %r9b,%r9d
  movb   $0x0,0x20(%rsp)
  shl    $0x8,%r12
  or     %rbx,%rax
  sar    $0x18,%r10d
  or     %rdx,%r12
  shl    $0x8,%rax
  movzbl %bpl,%edx
  movzbl %r10b,%r10d
  mov    %r12,%rbp
  or     %r11,%rax
  shl    $0x8,%rbp
  shl    $0x8,%rax
  or     %rdx,%rbp
  movzbl (%rsp),%edx
  shl    $0x8,%rbp
  or     %rcx,%rax
  shl    $0x8,%rax
  or     %rdi,%rbp
  lea    D7(%rip),%rdi
  or     %r10,%rax
  mov    %rbp,(%rsp)
  shl    $0x8,%rax
  or     %r9,%rax
  shl    $0x8,%rax
  or     %rdx,%rax
  shl    $0x8,%rax
  or     %rsi,%rax
  lea    0x10(%rsp),%rsi
  mov    %rax,0x8(%rsp)
  movdqa (%rsp),%xmm0
  movaps %xmm0,0x10(%rsp)
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D7
type: string

tool: D7: %s

assistant: ```c
void printregs(int eax, int ebx, int ecx, int edx) {
  int j;
  char string[17];
  string[16] = '\0';
  for (j = 0; j < 4; j++) {
    string[j] = eax >> (8 * j);
    string[j + 4] = ebx >> (8 * j);
    string[j + 8] = ecx >> (8 * j);
    string[j + 12] = edx >> (8 * j);
  }
  printf("%s", string);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<halfAdder>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  mov    %rcx,-0x18(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <gXor@plt>
  mov    -0x10(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <gAnd@plt>
  mov    -0x18(%rbp),%rdx
  mov    %eax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void halfAdder(const int a, const int b, int *sum, int *carry) {
  *sum = gXor(a, b);
  *carry = gAnd(a, b);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<halfAdder>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %rdx,%r13
  mov    %rcx,%r12
  call   <gXor@plt>
  mov    %eax,0x0(%r13)
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <gAnd@plt>
  mov    %eax,(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void halfAdder(const int a, const int b, int *sum, int *carry) {
  *sum = gXor(a, b);
  *carry = gAnd(a, b);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<halfAdder>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x8,%rsp
  call   <gXor@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  mov    %eax,0x0(%r13)
  call   <gAnd@plt>
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void halfAdder(const int a, const int b, int *sum, int *carry) {
  *sum = gXor(a, b);
  *carry = gAnd(a, b);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<halfAdder>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x8,%rsp
  call   <gXor@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  mov    %eax,0x0(%r13)
  call   <gAnd@plt>
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void halfAdder(const int a, const int b, int *sum, int *carry) {
  *sum = gXor(a, b);
  *carry = gAnd(a, b);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<popcount_8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    %eax
  and    $0x55,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  and    $0x55,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x2,%eax
  and    $0x33,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  and    $0x33,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x4,%eax
  and    $0xf,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  and    $0xf,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int popcount_8(unsigned int v) {
  v = ((v >> 1) & 0x55) + (v & 0x55);
  v = ((v >> 2) & 0x33) + (v & 0x33);
  v = ((v >> 4) & 0x0F) + (v & 0x0F);
  return v;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<popcount_8>:
  endbr64
  mov    %edi,%eax
  shr    %eax
  and    $0x55,%eax
  and    $0x55,%edi
  add    %eax,%edi
  mov    %edi,%edx
  shr    $0x2,%edx
  and    $0x33,%edx
  and    $0x33,%edi
  add    %edi,%edx
  mov    %edx,%eax
  shr    $0x4,%eax
  and    $0xf,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
int popcount_8(unsigned int v) {
  v = ((v >> 1) & 0x55) + (v & 0x55);
  v = ((v >> 2) & 0x33) + (v & 0x33);
  v = ((v >> 4) & 0x0F) + (v & 0x0F);
  return v;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<popcount_8>:
  endbr64
  mov    %edi,%eax
  and    $0x55,%edi
  shr    %eax
  and    $0x55,%eax
  add    %eax,%edi
  mov    %edi,%edx
  and    $0x33,%edi
  shr    $0x2,%edx
  and    $0x33,%edx
  add    %edi,%edx
  mov    %edx,%eax
  and    $0xf,%edx
  shr    $0x4,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
int popcount_8(unsigned int v) {
  v = ((v >> 1) & 0x55) + (v & 0x55);
  v = ((v >> 2) & 0x33) + (v & 0x33);
  v = ((v >> 4) & 0x0F) + (v & 0x0F);
  return v;
}
```
user: Translate the below assembly code into C programming language:

```asm
<popcount_8>:
  endbr64
  mov    %edi,%eax
  and    $0x55,%edi
  shr    %eax
  and    $0x55,%eax
  add    %eax,%edi
  mov    %edi,%edx
  and    $0x33,%edi
  shr    $0x2,%edx
  and    $0x33,%edx
  add    %edi,%edx
  mov    %edx,%eax
  and    $0xf,%edx
  shr    $0x4,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
int popcount_8(unsigned int v) {
  v = ((v >> 1) & 0x55) + (v & 0x55);
  v = ((v >> 2) & 0x33) + (v & 0x33);
  v = ((v >> 4) & 0x0F) + (v & 0x0F);
  return v;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<benchmark>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x28(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L18
L47:
  mov    $0x0,%eax
  call   <rand_double@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x20(%rbp)
  mov    $0x0,%eax
  call   <rand_double@plt>
  movsd  D45(%rip),%xmm1
  mulsd  %xmm0,%xmm1
  movsd  D21(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    $0x0,%eax
  call   <rand_double@plt>
  movsd  D45(%rip),%xmm1
  mulsd  %xmm0,%xmm1
  movsd  D21(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x20(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <incbet@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x30(%rbp)
  movsd  -0x30(%rbp),%xmm0
  ucomisd -0x30(%rbp),%xmm0
  jnp    L11
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L9
L11:
  addl   $0x1,-0x34(%rbp)
L18:
  cmpl   $0xf423f,-0x34(%rbp)
  jle    L47
L9:
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  subsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x30(%rbp),%xmm0
  ucomisd -0x30(%rbp),%xmm0
  jnp    L12
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L12:
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D45
type: f64
name: unpack_data
arguments: data_label: D21
type: f64
name: unpack_data
arguments: data_label: D43
type: string

tool: D45: 1000

tool: D21: 0.1

tool: D43: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double a;
  double b;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = rand_double();
    a = (1000.0 * rand_double()) + 0.1;
    b = (1000.0 * rand_double()) + 0.1;
    y = incbet(a, b, x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<benchmark>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%r13
  mov    $0xf4240,%r12d
L1:
  mov    $0x0,%eax
  call   <rand_double@plt>
  movq   %xmm0,%rbp
  mov    $0x0,%eax
  call   <rand_double@plt>
  movq   %xmm0,%rbx
  mov    $0x0,%eax
  call   <rand_double@plt>
  movsd  D48(%rip),%xmm1
  mulsd  %xmm0,%xmm1
  movq   %rbx,%xmm0
  mulsd  D48(%rip),%xmm0
  addsd  D8(%rip),%xmm0
  movq   %rbp,%xmm2
  addsd  D8(%rip),%xmm1
  call   <incbet@plt>
  ucomisd %xmm0,%xmm0
  jp     L4
  sub    $0x1,%r12d
  jne    L1
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r13,%xmm4
  subsd  %xmm4,%xmm0
  movq   %xmm0,%rbx
  jmp    L13
L4:
  lea    D47(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r13,%xmm3
  subsd  %xmm3,%xmm0
  movq   %xmm0,%rbx
  mov    %rbp,%rdi
  call   <printf@plt>
L13:
  movq   %rbx,%xmm0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D48
datatype: f64
name: disassemble_data
arguments: tag: D8
datatype: f64
name: disassemble_data
arguments: tag: D47
datatype: string

tool: D48: 1000

tool: D8: 0.1

tool: D47: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double a;
  double b;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = rand_double();
    a = (1000.0 * rand_double()) + 0.1;
    b = (1000.0 * rand_double()) + 0.1;
    y = incbet(a, b, x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<benchmark>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  mov    $0xf4240,%ebx
  sub    $0x28,%rsp
  call   <tic@plt>
  movsd  %xmm0,0x18(%rsp)
  jmp    L18
  xchg   %ax,%ax
L14:
  sub    $0x1,%ebx
  je     L1
L18:
  xor    %eax,%eax
  call   <rand_double@plt>
  xor    %eax,%eax
  movsd  %xmm0,0x10(%rsp)
  call   <rand_double@plt>
  xor    %eax,%eax
  movsd  %xmm0,0x8(%rsp)
  call   <rand_double@plt>
  movsd  0x8(%rsp),%xmm3
  mulsd  D5(%rip),%xmm3
  movsd  D5(%rip),%xmm1
  movsd  0x10(%rsp),%xmm2
  mulsd  %xmm0,%xmm1
  movsd  D49(%rip),%xmm0
  addsd  D49(%rip),%xmm1
  addsd  %xmm3,%xmm0
  call   <incbet@plt>
  ucomisd %xmm0,%xmm0
  jnp    L14
  lea    D45(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x18(%rsp),%xmm0
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x18(%rsp),%xmm0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
data_type: f64
name: retrieve_data
arguments: data_tag: D49
data_type: f64
name: retrieve_data
arguments: data_tag: D45
data_type: string

tool: D5: 1000

tool: D49: 0.1

tool: D45: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double a;
  double b;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = rand_double();
    a = (1000.0 * rand_double()) + 0.1;
    b = (1000.0 * rand_double()) + 0.1;
    y = incbet(a, b, x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<benchmark>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  mov    $0xf4240,%ebx
  sub    $0x28,%rsp
  call   <tic@plt>
  movsd  %xmm0,0x18(%rsp)
  jmp    L2
  xchg   %ax,%ax
L24:
  sub    $0x1,%ebx
  je     L12
L2:
  xor    %eax,%eax
  call   <rand_double@plt>
  xor    %eax,%eax
  movsd  %xmm0,0x10(%rsp)
  call   <rand_double@plt>
  xor    %eax,%eax
  movsd  %xmm0,0x8(%rsp)
  call   <rand_double@plt>
  movsd  0x8(%rsp),%xmm3
  mulsd  D24(%rip),%xmm3
  movsd  D24(%rip),%xmm1
  movsd  0x10(%rsp),%xmm2
  mulsd  %xmm0,%xmm1
  movsd  D19(%rip),%xmm0
  addsd  D19(%rip),%xmm1
  addsd  %xmm3,%xmm0
  call   <incbet@plt>
  ucomisd %xmm0,%xmm0
  jnp    L24
  lea    D45(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x18(%rsp),%xmm0
  mov    %rbp,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L12:
  xor    %eax,%eax
  call   <tic@plt>
  subsd  0x18(%rsp),%xmm0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D24
datatype: f64
name: unpack_data
arguments: tag: D19
datatype: f64
name: unpack_data
arguments: tag: D45
datatype: string

tool: D24: 1000

tool: D19: 0.1

tool: D45: should not return NaN


assistant: ```c
double benchmark() {
  double elapsed;
  double x;
  double a;
  double b;
  double y;
  double t;
  int i;
  t = tic();
  for (i = 0; i < 1000000; i++) {
    x = rand_double();
    a = (1000.0 * rand_double()) + 0.1;
    b = (1000.0 * rand_double()) + 0.1;
    y = incbet(a, b, x);
    if (y != y) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y != y) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memcpy>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%rax
  test   $0x7,%dil
  je     L106
  lea    -0x1(%rdx),%rsi
  mov    %rdi,%rcx
  test   %rdx,%rdx
  je     L115
L135:
  add    $0x1,%rax
  add    $0x1,%rcx
  movzbl -0x1(%rax),%edx
  mov    %dl,-0x1(%rcx)
  test   $0x7,%cl
  je     L101
  sub    $0x1,%rsi
  cmp    $0xffffffffffffffff,%rsi
  jne    L135
  jmp    L115
L106:
  mov    %rdi,%rcx
  mov    %rdx,%rsi
L101:
  cmp    $0x7,%rsi
  ja     L85
L8:
  mov    $0x0,%edx
  test   %rsi,%rsi
  je     L115
L12:
  movzbl (%rax,%rdx,1),%edi
  mov    %dil,(%rcx,%rdx,1)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L12
L115:
  mov    %r8,%rax
  ret
L85:
  cmp    $0x1f,%rsi
  jbe    L59
  lea    -0x20(%rsi),%rdx
  and    $0xffffffffffffffe0,%rdx
  lea    0x20(%rdx),%r10
  lea    (%rax,%r10,1),%r9
  mov    %rcx,%rdx
L25:
  mov    (%rax),%rdi
  mov    %rdi,(%rdx)
  mov    0x8(%rax),%rdi
  mov    %rdi,0x8(%rdx)
  mov    0x10(%rax),%rdi
  mov    %rdi,0x10(%rdx)
  add    $0x20,%rax
  add    $0x20,%rdx
  mov    -0x8(%rax),%rdi
  mov    %rdi,-0x8(%rdx)
  cmp    %rax,%r9
  jne    L25
  add    %r10,%rcx
  mov    %rsi,%r10
  and    $0x1f,%r10d
  and    $0x18,%esi
  mov    %r10,%rsi
  je     L8
L82:
  lea    -0x8(%r10),%rdi
  shr    $0x3,%rdi
  lea    0x1(%rdi),%r9
  mov    $0x0,%edx
L23:
  mov    (%rax,%rdx,8),%rsi
  mov    %rsi,(%rcx,%rdx,8)
  mov    %rdx,%rsi
  add    $0x1,%rdx
  cmp    %rdi,%rsi
  jne    L23
  lea    0x0(,%r9,8),%rdx
  add    %rdx,%rax
  add    %rdx,%rcx
  mov    %r10,%rsi
  and    $0x7,%esi
  jmp    L8
L59:
  mov    %rsi,%r10
  jmp    L82
```

assistant: ```c
void *memcpy(void *restrict dest, const void *restrict src, size_t count) {
  const char *s = (const char *)src;
  const unsigned long *ns;
  char *d = (char *)dest;
  unsigned long *nd;
  while ((uintptr_t)d & (sizeof(unsigned long) - 1)) {
    if (count--) {
      *d++ = *s++;
    } else {
      return dest;
    }
  }
  if (count >= sizeof(unsigned long)) {
    nd = (unsigned long *)d;
    ns = (const unsigned long *)s;
    while (count >= (sizeof(unsigned long) * 4)) {
      *nd++ = *ns++;
      *nd++ = *ns++;
      *nd++ = *ns++;
      *nd++ = *ns++;
      count -= sizeof(unsigned long) * 4;
    }
    while (count >= sizeof(unsigned long)) {
      *nd++ = *ns++;
      count -= sizeof(unsigned long);
    }
    d = (char *)nd;
    s = (const char *)ns;
  }
  while (count--)
    *d++ = *s++;
  return dest;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<memcpy>:
  endbr64
  mov    %rdi,%r8
  and    $0x7,%edi
  je     L79
  lea    -0x1(%rdx),%rax
  mov    %r8,%rdi
  test   %rdx,%rdx
  jne    L83
  jmp    L140
  nopl   0x0(%rax)
L94:
  sub    $0x1,%rax
  jb     L140
L83:
  movsb  %ds:(%rsi),%es:(%rdi)
  test   $0x7,%dil
  jne    L94
L44:
  push   %rbp
  push   %rbx
  cmp    $0x7,%rax
  jbe    L117
  mov    %rax,%rbx
  cmp    $0x1f,%rax
  jbe    L141
  lea    -0x20(%rax),%rdx
  and    $0xffffffffffffffe0,%rdx
  lea    0x20(%rdx),%r10
  mov    %rdi,%rdx
  lea    (%rsi,%r10,1),%r9
  nopl   0x0(%rax)
L69:
  mov    (%rsi),%rcx
  add    $0x20,%rsi
  add    $0x20,%rdx
  mov    %rcx,-0x20(%rdx)
  mov    -0x18(%rsi),%rcx
  mov    %rcx,-0x18(%rdx)
  mov    -0x10(%rsi),%rcx
  mov    %rcx,-0x10(%rdx)
  mov    -0x8(%rsi),%rcx
  mov    %rcx,-0x8(%rdx)
  cmp    %rsi,%r9
  jne    L69
  mov    %rax,%rbx
  add    %r10,%rdi
  and    $0x1f,%ebx
  and    $0x18,%eax
  mov    %rbx,%rax
  je     L117
L141:
  lea    -0x8(%rbx),%rax
  lea    0x8(%rdi),%r9
  mov    %rsi,%rbp
  shr    $0x3,%rax
  and    $0xfffffffffffffff8,%r9
  lea    0x8(,%rax,8),%r11
  mov    (%rsi),%rax
  mov    %rax,(%rdi)
  mov    %r11d,%eax
  mov    -0x8(%rsi,%rax,1),%rdx
  mov    %rdx,-0x8(%rdi,%rax,1)
  mov    %rdi,%rax
  sub    %r9,%rax
  sub    %rax,%rbp
  add    %r11d,%eax
  and    $0xfffffff8,%eax
  cmp    $0x8,%eax
  jb     L115
  and    $0xfffffff8,%eax
  xor    %edx,%edx
L16:
  mov    %edx,%ecx
  add    $0x8,%edx
  mov    0x0(%rbp,%rcx,1),%r10
  mov    %r10,(%r9,%rcx,1)
  cmp    %eax,%edx
  jb     L16
L115:
  mov    %rbx,%rax
  add    %r11,%rsi
  add    %r11,%rdi
  and    $0x7,%eax
L117:
  test   %rax,%rax
  je     L58
  mov    %eax,%r9d
  test   %eax,%eax
  je     L58
  xor    %eax,%eax
L90:
  mov    %eax,%edx
  add    $0x1,%eax
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rdi,%rdx,1)
  cmp    %r9d,%eax
  jb     L90
L58:
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
L79:
  mov    %r8,%rdi
  mov    %rdx,%rax
  jmp    L44
L140:
  mov    %r8,%rax
  ret
```

assistant: ```c
void *memcpy(void *restrict dest, const void *restrict src, size_t count) {
  const char *s = (const char *)src;
  const unsigned long *ns;
  char *d = (char *)dest;
  unsigned long *nd;
  while ((uintptr_t)d & (sizeof(unsigned long) - 1)) {
    if (count--) {
      *d++ = *s++;
    } else {
      return dest;
    }
  }
  if (count >= sizeof(unsigned long)) {
    nd = (unsigned long *)d;
    ns = (const unsigned long *)s;
    while (count >= (sizeof(unsigned long) * 4)) {
      *nd++ = *ns++;
      *nd++ = *ns++;
      *nd++ = *ns++;
      *nd++ = *ns++;
      count -= sizeof(unsigned long) * 4;
    }
    while (count >= sizeof(unsigned long)) {
      *nd++ = *ns++;
      count -= sizeof(unsigned long);
    }
    d = (char *)nd;
    s = (const char *)ns;
  }
  while (count--)
    *d++ = *s++;
  return dest;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memcpy>:
  endbr64
  mov    %rdi,%r8
  and    $0x7,%edi
  je     L74
  lea    -0x1(%rdx),%rax
  mov    %r8,%rdi
  test   %rdx,%rdx
  jne    L128
  jmp    L18
  nopl   0x0(%rax)
L132:
  sub    $0x1,%rax
  jb     L18
L128:
  movsb  %ds:(%rsi),%es:(%rdi)
  test   $0x7,%dil
  jne    L132
L7:
  push   %rbp
  push   %rbx
  cmp    $0x7,%rax
  jbe    L41
  mov    %rax,%rbx
  cmp    $0x1f,%rax
  jbe    L49
  lea    -0x20(%rax),%r9
  xor    %edx,%edx
  xor    %ecx,%ecx
  shr    $0x5,%r9
  add    $0x1,%r9
  xchg   %ax,%ax
L70:
  movdqu (%rsi,%rdx,1),%xmm0
  movdqu 0x10(%rsi,%rdx,1),%xmm1
  add    $0x1,%rcx
  movups %xmm0,(%rdi,%rdx,1)
  movups %xmm1,0x10(%rdi,%rdx,1)
  add    $0x20,%rdx
  cmp    %rcx,%r9
  ja     L70
  shl    $0x5,%r9
  mov    %rax,%rbx
  and    $0x1f,%ebx
  add    %r9,%rsi
  add    %r9,%rdi
  and    $0x18,%eax
  mov    %rbx,%rax
  je     L41
L49:
  lea    -0x8(%rbx),%rax
  lea    0x8(%rdi),%r9
  mov    %rsi,%rbp
  shr    $0x3,%rax
  and    $0xfffffffffffffff8,%r9
  lea    0x8(,%rax,8),%r11
  mov    (%rsi),%rax
  mov    %rax,(%rdi)
  mov    %r11d,%eax
  mov    -0x8(%rsi,%rax,1),%rdx
  mov    %rdx,-0x8(%rdi,%rax,1)
  mov    %rdi,%rax
  sub    %r9,%rax
  sub    %rax,%rbp
  add    %r11d,%eax
  and    $0xfffffff8,%eax
  cmp    $0x8,%eax
  jb     L15
  and    $0xfffffff8,%eax
  xor    %edx,%edx
L25:
  mov    %edx,%ecx
  add    $0x8,%edx
  mov    0x0(%rbp,%rcx,1),%r10
  mov    %r10,(%r9,%rcx,1)
  cmp    %eax,%edx
  jb     L25
L15:
  mov    %rbx,%rax
  add    %r11,%rsi
  add    %r11,%rdi
  and    $0x7,%eax
L41:
  test   %rax,%rax
  je     L92
  mov    %eax,%r9d
  test   %eax,%eax
  je     L92
  xor    %eax,%eax
L50:
  mov    %eax,%edx
  add    $0x1,%eax
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rdi,%rdx,1)
  cmp    %r9d,%eax
  jb     L50
L92:
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
L74:
  mov    %r8,%rdi
  mov    %rdx,%rax
  jmp    L7
L18:
  mov    %r8,%rax
  ret
```

assistant: ```c
void *memcpy(void *restrict dest, const void *restrict src, size_t count) {
  const char *s = (const char *)src;
  const unsigned long *ns;
  char *d = (char *)dest;
  unsigned long *nd;
  while ((uintptr_t)d & (sizeof(unsigned long) - 1)) {
    if (count--) {
      *d++ = *s++;
    } else {
      return dest;
    }
  }
  if (count >= sizeof(unsigned long)) {
    nd = (unsigned long *)d;
    ns = (const unsigned long *)s;
    while (count >= (sizeof(unsigned long) * 4)) {
      *nd++ = *ns++;
      *nd++ = *ns++;
      *nd++ = *ns++;
      *nd++ = *ns++;
      count -= sizeof(unsigned long) * 4;
    }
    while (count >= sizeof(unsigned long)) {
      *nd++ = *ns++;
      count -= sizeof(unsigned long);
    }
    d = (char *)nd;
    s = (const char *)ns;
  }
  while (count--)
    *d++ = *s++;
  return dest;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<snprint_sd32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L1
  mov    -0x8(%rbp),%rax
  movb   $0x2d,(%rax)
  addq   $0x1,-0x8(%rbp)
  subq   $0x1,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  neg    %eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <snprint_d32@plt>
  add    $0x1,%rax
  jmp    L3
L1:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <snprint_d32@plt>
L3:
  leave
  ret
```

assistant: ```c
uint8_t snprint_sd32(char *buffer, size_t buf_size, int32_t num) {
  if (num < 0) {
    buffer[0] = '-';
    buffer++;
    buf_size--;
    return (snprint_d32(buffer, buf_size, -num) + 1);
  } else {
    return (snprint_d32(buffer, buf_size, num));
  }
}
```