ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<EMACTimestampPPSCommandModeSet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  nop
L7:
  mov    -0x4(%rbp),%eax
  add    $0x72c,%eax
  cltq
  mov    (%rax),%eax
  and    $0xf,%eax
  test   %eax,%eax
  jne    L7
  mov    -0x4(%rbp),%eax
  add    $0x72c,%eax
  cltq
  mov    %rax,%rdx
  mov    -0x8(%rbp),%eax
  or     $0x10,%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void EMACTimestampPPSCommandModeSet(uint32_t ui32Base, uint32_t ui32Config) {
  ;

  ;
  while ((*((volatile uint32_t *)(ui32Base + 0x0000072C))) & 0x0000000F) {
  }
  (*((volatile uint32_t *)(ui32Base + 0x0000072C))) = (0x00000010 | ui32Config);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<EMACTimestampPPSCommandModeSet>:
  endbr64
  add    $0x72c,%edi
  movslq %edi,%rdi
L8:
  mov    (%rdi),%eax
  test   $0xf,%al
  jne    L8
  or     $0x10,%esi
  mov    %esi,(%rdi)
  ret
```

assistant: ```c
void EMACTimestampPPSCommandModeSet(uint32_t ui32Base, uint32_t ui32Config) {
  ;

  ;
  while ((*((volatile uint32_t *)(ui32Base + 0x0000072C))) & 0x0000000F) {
  }
  (*((volatile uint32_t *)(ui32Base + 0x0000072C))) = (0x00000010 | ui32Config);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<EMACTimestampPPSCommandModeSet>:
  endbr64
  add    $0x72c,%edi
  movslq %edi,%rdi
  nopl   (%rax)
L0:
  mov    (%rdi),%eax
  test   $0xf,%al
  jne    L0
  or     $0x10,%esi
  mov    %esi,(%rdi)
  ret
```

assistant: ```c
void EMACTimestampPPSCommandModeSet(uint32_t ui32Base, uint32_t ui32Config) {
  ;

  ;
  while ((*((volatile uint32_t *)(ui32Base + 0x0000072C))) & 0x0000000F) {
  }
  (*((volatile uint32_t *)(ui32Base + 0x0000072C))) = (0x00000010 | ui32Config);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<EMACTimestampPPSCommandModeSet>:
  endbr64
  add    $0x72c,%edi
  movslq %edi,%rdi
  nopl   (%rax)
L8:
  mov    (%rdi),%eax
  test   $0xf,%al
  jne    L8
  or     $0x10,%esi
  mov    %esi,(%rdi)
  ret
```

assistant: ```c
void EMACTimestampPPSCommandModeSet(uint32_t ui32Base, uint32_t ui32Config) {
  ;

  ;
  while ((*((volatile uint32_t *)(ui32Base + 0x0000072C))) & 0x0000000F) {
  }
  (*((volatile uint32_t *)(ui32Base + 0x0000072C))) = (0x00000010 | ui32Config);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ti_medprice_start>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int ti_medprice_start(double const *options) {
  (void)options;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

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

assistant: ```c
int ti_medprice_start(double const *options) {
  (void)options;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

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

assistant: ```c
int ti_medprice_start(double const *options) {
  (void)options;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

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

assistant: ```c
int ti_medprice_start(double const *options) {
  (void)options;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getSumOfDigits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L13:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  add    %edx,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L18:
  cmpl   $0x0,-0x14(%rbp)
  jne    L13
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int getSumOfDigits(int num) {
  int sum = 0;
  while (num != 0) {
    sum += num % 10;
    num /= 10;
  }
  return sum;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getSumOfDigits>:
  endbr64
  test   %edi,%edi
  je     L3
  mov    $0x0,%ecx
L28:
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  add    %edi,%ecx
  mov    %eax,%edi
  test   %eax,%eax
  jne    L28
L12:
  mov    %ecx,%eax
  ret
L3:
  mov    %edi,%ecx
  jmp    L12
```

assistant: ```c
int getSumOfDigits(int num) {
  int sum = 0;
  while (num != 0) {
    sum += num % 10;
    num /= 10;
  }
  return sum;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<getSumOfDigits>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L16
  nopl   0x0(%rax,%rax,1)
L4:
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  add    %edi,%r8d
  mov    %eax,%edi
  test   %eax,%eax
  jne    L4
L16:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int getSumOfDigits(int num) {
  int sum = 0;
  while (num != 0) {
    sum += num % 10;
    num /= 10;
  }
  return sum;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getSumOfDigits>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L10
  nopl   0x0(%rax,%rax,1)
L12:
  movslq %edi,%rax
  mov    %edi,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  add    %edi,%r8d
  mov    %eax,%edi
  test   %eax,%eax
  jne    L12
L10:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int getSumOfDigits(int num) {
  int sum = 0;
  while (num != 0) {
    sum += num % 10;
    num /= 10;
  }
  return sum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x24(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  sub    -0x18(%rbp),%rax
  sar    $0x2,%rax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  neg    %edx
  cmovns %edx,%eax
  cmp    $0x1,%eax
  je     L17
  mov    $0x0,%eax
  call   <abort@plt>
L17:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
main() {
  int a, b;
  int *pa, *pb;
  pa = &a;
  pb = &b;
  if (({
        int i = (pb - pa);
        i < 0 ? -i : i;
      }) != 1)
    abort();
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
main() {
  int a, b;
  int *pa, *pb;
  pa = &a;
  pb = &b;
  if (({
        int i = (pb - pa);
        i < 0 ? -i : i;
      }) != 1)
    abort();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
main() {
  int a, b;
  int *pa, *pb;
  pa = &a;
  pb = &b;
  if (({
        int i = (pb - pa);
        i < 0 ? -i : i;
      }) != 1)
    abort();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
main() {
  int a, b;
  int *pa, *pb;
  pa = &a;
  pb = &b;
  if (({
        int i = (pb - pa);
        i < 0 ? -i : i;
      }) != 1)
    abort();
}
```
user: Translate the below assembly code into C programming language:

```asm
<factorial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jg     L18
  cmpl   $0x0,-0x4(%rbp)
  jns    L7
  mov    $0xffffffff,%eax
  jmp    L2
L7:
  mov    $0x1,%eax
  jmp    L2
L18:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <factorial>
  imul   -0x4(%rbp),%eax
L2:
  leave
  ret
```

assistant: ```c
int factorial(int n) {
  if (n <= 1) {
    if (n < 0) {
      return -1;
    }
    return 1;
  }
  return factorial(n - 1) * n;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<factorial>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0x1,%edi
  jg     L5
  mov    %edi,%eax
  sar    $0x1f,%eax
  or     $0x1,%eax
L0:
  pop    %rbx
  ret
L5:
  lea    -0x1(%rdi),%edi
  call   <factorial>
  imul   %ebx,%eax
  jmp    L0
```

assistant: ```c
int factorial(int n) {
  if (n <= 1) {
    if (n < 0) {
      return -1;
    }
    return 1;
  }
  return factorial(n - 1) * n;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<factorial>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jle    L15
  xchg   %ax,%ax
L11:
  mov    %edi,%edx
  sub    $0x1,%edi
  imul   %edx,%eax
  cmp    $0x1,%edi
  jne    L11
  ret
  xchg   %ax,%ax
L15:
  mov    %edi,%eax
  sar    $0x1f,%eax
  or     $0x1,%eax
  ret
```

assistant: ```c
int factorial(int n) {
  if (n <= 1) {
    if (n < 0) {
      return -1;
    }
    return 1;
  }
  return factorial(n - 1) * n;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<factorial>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jle    L5
  xchg   %ax,%ax
L18:
  mov    %edi,%edx
  sub    $0x1,%edi
  imul   %edx,%eax
  cmp    $0x1,%edi
  jne    L18
  ret
  xchg   %ax,%ax
L5:
  mov    %edi,%eax
  sar    $0x1f,%eax
  or     $0x1,%eax
  ret
```

assistant: ```c
int factorial(int n) {
  if (n <= 1) {
    if (n < 0) {
      return -1;
    }
    return 1;
  }
  return factorial(n - 1) * n;
}
```
user: What is the c source code of the assembly code below:

```asm
<vec2_multiply>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  %xmm1,%xmm0
  mov    -0x8(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mulss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void vec2_multiply(float *dst, float *b) {
  dst[0] = dst[0] * b[0];
  dst[1] = dst[1] * b[1];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<vec2_multiply>:
  endbr64
  movss  (%rdi),%xmm0
  mulss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  movss  0x4(%rdi),%xmm0
  mulss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdi)
  ret
```

assistant: ```c
void vec2_multiply(float *dst, float *b) {
  dst[0] = dst[0] * b[0];
  dst[1] = dst[1] * b[1];
}
```
user: What is the c source code of the assembly code below:

```asm
<vec2_multiply>:
  endbr64
  movss  (%rdi),%xmm0
  mulss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  movss  0x4(%rdi),%xmm0
  mulss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdi)
  ret
```

assistant: ```c
void vec2_multiply(float *dst, float *b) {
  dst[0] = dst[0] * b[0];
  dst[1] = dst[1] * b[1];
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<vec2_multiply>:
  endbr64
  movss  (%rdi),%xmm0
  mulss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  movss  0x4(%rdi),%xmm0
  mulss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdi)
  ret
```

assistant: ```c
void vec2_multiply(float *dst, float *b) {
  dst[0] = dst[0] * b[0];
  dst[1] = dst[1] * b[1];
}
```
user: What is the c source code of the assembly code below?

```asm
<getopt_long_only>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    -0x28(%rbp),%rdi
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x4(%rbp),%eax
  mov    $0x1,%r9d
  mov    %rdi,%r8
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <_getopt_internal@plt>
  leave
  ret
```

assistant: ```c
int getopt_long_only(argc, argv, options, long_options, opt_index)
int argc;
char *const *argv;
const char *options;
const struct gg_option *long_options;
int *opt_index;
{ return _getopt_internal(argc, argv, options, long_options, opt_index, 1); }
```
user: What would the C source look like for the assembly code provided here?

```asm
<getopt_long_only>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%r9d
  mov    $0x0,%eax
  call   <_getopt_internal@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int getopt_long_only(argc, argv, options, long_options, opt_index)
int argc;
char *const *argv;
const char *options;
const struct gg_option *long_options;
int *opt_index;
{ return _getopt_internal(argc, argv, options, long_options, opt_index, 1); }
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getopt_long_only>:
  endbr64
  mov    $0x1,%r9d
  xor    %eax,%eax
  jmp    1050 <_getopt_internal@plt>
```

assistant: ```c
int getopt_long_only(argc, argv, options, long_options, opt_index)
int argc;
char *const *argv;
const char *options;
const struct gg_option *long_options;
int *opt_index;
{ return _getopt_internal(argc, argv, options, long_options, opt_index, 1); }
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<getopt_long_only>:
  endbr64
  mov    $0x1,%r9d
  xor    %eax,%eax
  jmp    1050 <_getopt_internal@plt>
```

assistant: ```c
int getopt_long_only(argc, argv, options, long_options, opt_index)
int argc;
char *const *argv;
const char *options;
const struct gg_option *long_options;
int *opt_index;
{ return _getopt_internal(argc, argv, options, long_options, opt_index, 1); }
```
user: What is the c source code of the assembly code below:

```asm
<matrix_multiply4x4>:
  endbr64
  mov    %rsi,%rax
  add    $0x10,%rsi
L7:
  movss  (%rax),%xmm0
  movss  0x10(%rax),%xmm4
  movss  0x20(%rax),%xmm3
  movss  0x30(%rax),%xmm2
  movaps %xmm0,%xmm1
  mulss  (%rdx),%xmm1
  movaps %xmm4,%xmm5
  mulss  0x4(%rdx),%xmm5
  addss  %xmm5,%xmm1
  movaps %xmm3,%xmm5
  mulss  0x8(%rdx),%xmm5
  addss  %xmm5,%xmm1
  movaps %xmm2,%xmm5
  mulss  0xc(%rdx),%xmm5
  addss  %xmm5,%xmm1
  movss  %xmm1,(%rdi)
  movaps %xmm0,%xmm1
  mulss  0x10(%rdx),%xmm1
  movaps %xmm4,%xmm5
  mulss  0x14(%rdx),%xmm5
  addss  %xmm5,%xmm1
  movaps %xmm3,%xmm5
  mulss  0x18(%rdx),%xmm5
  addss  %xmm5,%xmm1
  movaps %xmm2,%xmm5
  mulss  0x1c(%rdx),%xmm5
  addss  %xmm5,%xmm1
  movss  %xmm1,0x10(%rdi)
  movaps %xmm0,%xmm1
  mulss  0x20(%rdx),%xmm1
  movaps %xmm4,%xmm5
  mulss  0x24(%rdx),%xmm5
  addss  %xmm5,%xmm1
  movaps %xmm3,%xmm5
  mulss  0x28(%rdx),%xmm5
  addss  %xmm5,%xmm1
  movaps %xmm2,%xmm5
  mulss  0x2c(%rdx),%xmm5
  addss  %xmm5,%xmm1
  movss  %xmm1,0x20(%rdi)
  mulss  0x30(%rdx),%xmm0
  mulss  0x34(%rdx),%xmm4
  addss  %xmm4,%xmm0
  mulss  0x38(%rdx),%xmm3
  addss  %xmm3,%xmm0
  mulss  0x3c(%rdx),%xmm2
  addss  %xmm2,%xmm0
  movss  %xmm0,0x30(%rdi)
  add    $0x4,%rax
  add    $0x4,%rdi
  cmp    %rsi,%rax
  jne    L7
  ret
```

assistant: ```c
void matrix_multiply4x4(float *result, const float *a, const float *b) {
  int i;
  for (i = 0; i < 4; i++) {
    const float ai0 = a[(0 << 2) + i], ai1 = a[(1 << 2) + i], ai2 = a[(2 << 2) + i], ai3 = a[(3 << 2) + i];
    result[(0 << 2) + i] = ai0 * b[(0 << 2) + 0] + ai1 * b[(0 << 2) + 1] + ai2 * b[(0 << 2) + 2] + ai3 * b[(0 << 2) + 3];
    result[(1 << 2) + i] = ai0 * b[(1 << 2) + 0] + ai1 * b[(1 << 2) + 1] + ai2 * b[(1 << 2) + 2] + ai3 * b[(1 << 2) + 3];
    result[(2 << 2) + i] = ai0 * b[(2 << 2) + 0] + ai1 * b[(2 << 2) + 1] + ai2 * b[(2 << 2) + 2] + ai3 * b[(2 << 2) + 3];
    result[(3 << 2) + i] = ai0 * b[(3 << 2) + 0] + ai1 * b[(3 << 2) + 1] + ai2 * b[(3 << 2) + 2] + ai3 * b[(3 << 2) + 3];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<matrix_multiply4x4>:
  endbr64
  lea    0x10(%rsi),%rax
L8:
  movss  (%rsi),%xmm0
  movss  (%rdx),%xmm1
  add    $0x4,%rsi
  add    $0x4,%rdi
  movss  0xc(%rsi),%xmm4
  movss  0x4(%rdx),%xmm5
  mulss  %xmm0,%xmm1
  movss  0x1c(%rsi),%xmm3
  movss  0x2c(%rsi),%xmm2
  mulss  %xmm4,%xmm5
  addss  %xmm5,%xmm1
  movss  0x8(%rdx),%xmm5
  mulss  %xmm3,%xmm5
  addss  %xmm5,%xmm1
  movss  0xc(%rdx),%xmm5
  mulss  %xmm2,%xmm5
  addss  %xmm5,%xmm1
  movss  %xmm1,-0x4(%rdi)
  movss  0x10(%rdx),%xmm1
  movss  0x14(%rdx),%xmm5
  mulss  %xmm0,%xmm1
  mulss  %xmm4,%xmm5
  addss  %xmm5,%xmm1
  movss  0x18(%rdx),%xmm5
  mulss  %xmm3,%xmm5
  addss  %xmm5,%xmm1
  movss  0x1c(%rdx),%xmm5
  mulss  %xmm2,%xmm5
  addss  %xmm5,%xmm1
  movss  %xmm1,0xc(%rdi)
  movss  0x20(%rdx),%xmm1
  movss  0x24(%rdx),%xmm5
  mulss  %xmm0,%xmm1
  mulss  %xmm4,%xmm5
  addss  %xmm5,%xmm1
  movss  0x28(%rdx),%xmm5
  mulss  %xmm3,%xmm5
  addss  %xmm5,%xmm1
  movss  0x2c(%rdx),%xmm5
  mulss  %xmm2,%xmm5
  addss  %xmm5,%xmm1
  movss  %xmm1,0x1c(%rdi)
  mulss  0x30(%rdx),%xmm0
  mulss  0x34(%rdx),%xmm4
  mulss  0x38(%rdx),%xmm3
  mulss  0x3c(%rdx),%xmm2
  addss  %xmm4,%xmm0
  addss  %xmm3,%xmm0
  addss  %xmm2,%xmm0
  movss  %xmm0,0x2c(%rdi)
  cmp    %rsi,%rax
  jne    L8
  ret
```

assistant: ```c
void matrix_multiply4x4(float *result, const float *a, const float *b) {
  int i;
  for (i = 0; i < 4; i++) {
    const float ai0 = a[(0 << 2) + i], ai1 = a[(1 << 2) + i], ai2 = a[(2 << 2) + i], ai3 = a[(3 << 2) + i];
    result[(0 << 2) + i] = ai0 * b[(0 << 2) + 0] + ai1 * b[(0 << 2) + 1] + ai2 * b[(0 << 2) + 2] + ai3 * b[(0 << 2) + 3];
    result[(1 << 2) + i] = ai0 * b[(1 << 2) + 0] + ai1 * b[(1 << 2) + 1] + ai2 * b[(1 << 2) + 2] + ai3 * b[(1 << 2) + 3];
    result[(2 << 2) + i] = ai0 * b[(2 << 2) + 0] + ai1 * b[(2 << 2) + 1] + ai2 * b[(2 << 2) + 2] + ai3 * b[(2 << 2) + 3];
    result[(3 << 2) + i] = ai0 * b[(3 << 2) + 0] + ai1 * b[(3 << 2) + 1] + ai2 * b[(3 << 2) + 2] + ai3 * b[(3 << 2) + 3];
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<CALC_CRC1a>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %esi,-0x18(%rbp)
  mov    %ax,-0x14(%rbp)
  movl   $0x8,-0x4(%rbp)
  mov    -0x18(%rbp),%eax
  shl    $0x8,%eax
  mov    %ax,-0x6(%rbp)
  jmp    L26
L10:
  movzwl -0x14(%rbp),%eax
  add    %eax,%eax
  mov    %eax,%edx
  movzwl -0x14(%rbp),%eax
  xor    -0x6(%rbp),%ax
  test   %ax,%ax
  jns    L34
  mov    $0x1021,%eax
  jmp    L28
L34:
  mov    $0x0,%eax
L28:
  xor    %edx,%eax
  mov    %ax,-0x14(%rbp)
  shlw   -0x6(%rbp)
L26:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jne    L10
  movzwl -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned short CALC_CRC1a(unsigned short crc, uint8_t byte) {
  int i = 8;
  unsigned short b = byte << 8;
  while (i--) {
    crc = (crc << 1) ^ (((crc ^ b) & 0x8000) ? 0x1021 : 0);
    b <<= 1;
  }
  return crc;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<CALC_CRC1a>:
  endbr64
  mov    %edi,%eax
  shl    $0x8,%esi
  mov    $0x8,%edx
L7:
  lea    (%rax,%rax,1),%ecx
  xor    %esi,%eax
  sar    $0xf,%ax
  and    $0x1021,%ax
  xor    %ecx,%eax
  add    %esi,%esi
  sub    $0x1,%edx
  jne    L7
  ret
```

assistant: ```c
unsigned short CALC_CRC1a(unsigned short crc, uint8_t byte) {
  int i = 8;
  unsigned short b = byte << 8;
  while (i--) {
    crc = (crc << 1) ^ (((crc ^ b) & 0x8000) ? 0x1021 : 0);
    b <<= 1;
  }
  return crc;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<CALC_CRC1a>:
  endbr64
  mov    %edi,%eax
  shl    $0x8,%esi
  mov    $0x8,%ecx
  xchg   %ax,%ax
L7:
  movzwl %ax,%edx
  add    %edx,%edx
  mov    %edx,%edi
  xor    $0x1021,%di
  xor    %si,%ax
  mov    %edi,%eax
  cmovns %edx,%eax
  add    %esi,%esi
  sub    $0x1,%ecx
  jne    L7
  ret
```

assistant: ```c
unsigned short CALC_CRC1a(unsigned short crc, uint8_t byte) {
  int i = 8;
  unsigned short b = byte << 8;
  while (i--) {
    crc = (crc << 1) ^ (((crc ^ b) & 0x8000) ? 0x1021 : 0);
    b <<= 1;
  }
  return crc;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<CALC_CRC1a>:
  endbr64
  movzwl %di,%ecx
  mov    %esi,%eax
  mov    %esi,%edx
  add    %ecx,%ecx
  shl    $0x8,%eax
  mov    %ecx,%esi
  xor    $0x1021,%si
  xor    %di,%ax
  cmovs  %esi,%ecx
  mov    %edx,%esi
  shl    $0x9,%esi
  movzwl %cx,%eax
  add    %eax,%eax
  mov    %eax,%edi
  xor    $0x1021,%di
  xor    %cx,%si
  mov    %edx,%esi
  cmovs  %edi,%eax
  shl    $0xa,%esi
  movzwl %ax,%ecx
  add    %ecx,%ecx
  mov    %ecx,%edi
  xor    $0x1021,%di
  xor    %ax,%si
  mov    %edx,%esi
  cmovs  %edi,%ecx
  shl    $0xb,%esi
  movzwl %cx,%eax
  add    %eax,%eax
  mov    %eax,%edi
  xor    $0x1021,%di
  xor    %cx,%si
  mov    %edx,%esi
  cmovs  %edi,%eax
  shl    $0xc,%esi
  movzwl %ax,%ecx
  add    %ecx,%ecx
  mov    %ecx,%edi
  xor    $0x1021,%di
  xor    %ax,%si
  mov    %edx,%esi
  cmovs  %edi,%ecx
  shl    $0xd,%esi
  movzwl %cx,%eax
  add    %eax,%eax
  mov    %eax,%edi
  xor    $0x1021,%di
  xor    %cx,%si
  mov    %edx,%esi
  cmovs  %edi,%eax
  shl    $0xe,%esi
  movzwl %ax,%ecx
  add    %ecx,%ecx
  mov    %ecx,%edi
  xor    $0x1021,%di
  xor    %ax,%si
  cmovs  %edi,%ecx
  shl    $0xf,%edx
  movzwl %cx,%eax
  add    %eax,%eax
  mov    %eax,%esi
  xor    $0x1021,%si
  xor    %cx,%dx
  cmovs  %esi,%eax
  ret
```

assistant: ```c
unsigned short CALC_CRC1a(unsigned short crc, uint8_t byte) {
  int i = 8;
  unsigned short b = byte << 8;
  while (i--) {
    crc = (crc << 1) ^ (((crc ^ b) & 0x8000) ? 0x1021 : 0);
    b <<= 1;
  }
  return crc;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ilog2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L8:
  addl   $0x1,-0x4(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rax
  cmp    %rax,-0x18(%rbp)
  ja     L8
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int ilog2(unsigned long v) {
  int l = 0;
  while ((1UL << l) < v)
    l++;
  return l;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ilog2>:
  endbr64
  cmp    $0x1,%rdi
  jbe    L16
  mov    $0x0,%ecx
  mov    $0x1,%edx
L22:
  add    $0x1,%ecx
  mov    %rdx,%rax
  shl    %cl,%rax
  cmp    %rdi,%rax
  jb     L22
L25:
  mov    %ecx,%eax
  ret
L16:
  mov    $0x0,%ecx
  jmp    L25
```

assistant: ```c
int ilog2(unsigned long v) {
  int l = 0;
  while ((1UL << l) < v)
    l++;
  return l;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ilog2>:
  endbr64
  xor    %ecx,%ecx
  cmp    $0x1,%rdi
  jbe    L14
  mov    $0x1,%edx
  nopl   0x0(%rax)
L13:
  add    $0x1,%ecx
  mov    %rdx,%rax
  shl    %cl,%rax
  cmp    %rdi,%rax
  jb     L13
L14:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int ilog2(unsigned long v) {
  int l = 0;
  while ((1UL << l) < v)
    l++;
  return l;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ilog2>:
  endbr64
  xor    %ecx,%ecx
  cmp    $0x1,%rdi
  jbe    L16
  mov    $0x1,%edx
  nopl   0x0(%rax)
L2:
  add    $0x1,%ecx
  mov    %rdx,%rax
  shl    %cl,%rax
  cmp    %rdi,%rax
  jb     L2
L16:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int ilog2(unsigned long v) {
  int l = 0;
  while ((1UL << l) < v)
    l++;
  return l;
}
```