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
<fill_with_random>:
  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
  lea    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x10(%rbp),%eax
  shl    $0xb,%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0xc(%rbp),%edx
  mov    -0xc(%rbp),%eax
  shl    $0x11,%eax
  add    %edx,%eax
  xor    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L3
L28:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0xffffffff80808081,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x7,%edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  shl    $0x8,%edx
  sub    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x18(%rbp)
L3:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x2c(%rbp)
  ja     L28
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
void fill_with_random(unsigned char *buf, unsigned int size) {
  int i;
  struct timeval tv;
  unsigned int seed;
  gettimeofday(&tv, NULL);
  seed = (tv.tv_sec + (tv.tv_sec << 11)) ^ (tv.tv_usec + (tv.tv_usec << 17));
  srand(seed);
  for (i = 0; i < size; i++)
    buf[i] = rand() % 0xff;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fill_with_random>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    (%rsp),%eax
  mov    %eax,%edi
  shl    $0xb,%edi
  add    %eax,%edi
  mov    0x4(%rsp),%edx
  mov    %edx,%eax
  shl    $0x11,%eax
  add    %edx,%eax
  xor    %eax,%edi
  call   <srand@plt>
  test   %ebp,%ebp
  je     L9
  mov    %r12,%rbx
  lea    -0x1(%rbp),%eax
  lea    0x1(%r12,%rax,1),%rbp
L22:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0xffffffff80808081,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x7,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,%ecx
  shl    $0x8,%ecx
  sub    %edx,%ecx
  sub    %ecx,%eax
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L22
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void fill_with_random(unsigned char *buf, unsigned int size) {
  int i;
  struct timeval tv;
  unsigned int seed;
  gettimeofday(&tv, NULL);
  seed = (tv.tv_sec + (tv.tv_sec << 11)) ^ (tv.tv_usec + (tv.tv_usec << 17));
  srand(seed);
  for (i = 0; i < size; i++)
    buf[i] = rand() % 0xff;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fill_with_random>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  xor    %esi,%esi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%eax
  mov    0x4(%rsp),%edx
  mov    %eax,%edi
  shl    $0xb,%edi
  add    %eax,%edi
  mov    %edx,%eax
  shl    $0x11,%eax
  add    %edx,%eax
  xor    %eax,%edi
  call   <srand@plt>
  test   %ebx,%ebx
  je     L19
  lea    -0x1(%rbx),%eax
  lea    0x1(%rbp,%rax,1),%rbx
  nopl   0x0(%rax)
L10:
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0xffffffff80808081,%rdx,%rdx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x7,%edx
  sub    %ecx,%edx
  mov    %edx,%ecx
  shl    $0x8,%ecx
  sub    %edx,%ecx
  sub    %ecx,%eax
  mov    %al,-0x1(%rbp)
  cmp    %rbx,%rbp
  jne    L10
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void fill_with_random(unsigned char *buf, unsigned int size) {
  int i;
  struct timeval tv;
  unsigned int seed;
  gettimeofday(&tv, NULL);
  seed = (tv.tv_sec + (tv.tv_sec << 11)) ^ (tv.tv_usec + (tv.tv_usec << 17));
  srand(seed);
  for (i = 0; i < size; i++)
    buf[i] = rand() % 0xff;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fill_with_random>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  xor    %esi,%esi
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  mov    (%rsp),%eax
  mov    0x4(%rsp),%edx
  mov    %eax,%edi
  shl    $0xb,%edi
  add    %eax,%edi
  mov    %edx,%eax
  shl    $0x11,%eax
  add    %edx,%eax
  xor    %eax,%edi
  call   <srand@plt>
  test   %ebx,%ebx
  je     L10
  lea    -0x1(%rbx),%eax
  lea    0x1(%rbp,%rax,1),%rbx
  nopl   0x0(%rax)
L29:
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0xffffffff80808081,%rdx,%rdx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x7,%edx
  sub    %ecx,%edx
  mov    %edx,%ecx
  shl    $0x8,%ecx
  sub    %edx,%ecx
  sub    %ecx,%eax
  mov    %al,-0x1(%rbp)
  cmp    %rbx,%rbp
  jne    L29
L10:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void fill_with_random(unsigned char *buf, unsigned int size) {
  int i;
  struct timeval tv;
  unsigned int seed;
  gettimeofday(&tv, NULL);
  seed = (tv.tv_sec + (tv.tv_sec << 11)) ^ (tv.tv_usec + (tv.tv_usec << 17));
  srand(seed);
  for (i = 0; i < size; i++)
    buf[i] = rand() % 0xff;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<scalar_product>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L0:
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    (%rax),%ecx
  mov    -0x20(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  mov    (%rax),%eax
  imul   %ecx,%eax
  add    %eax,-0x4(%rbp)
L7:
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  test   %eax,%eax
  jg     L0
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int scalar_product(int *v1, int *v2, int size) {
  int result = 0;
  while (size-- > 0)
    result += (*v1++) * (*v2++);
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<scalar_product>:
  endbr64
  test   %edx,%edx
  jle    L7
  mov    %edx,%edx
  mov    $0x0,%eax
  mov    $0x0,%r8d
L21:
  mov    (%rdi,%rax,4),%ecx
  imul   (%rsi,%rax,4),%ecx
  add    %ecx,%r8d
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L21
L13:
  mov    %r8d,%eax
  ret
L7:
  mov    $0x0,%r8d
  jmp    L13
```

assistant: ```c
int scalar_product(int *v1, int *v2, int size) {
  int result = 0;
  while (size-- > 0)
    result += (*v1++) * (*v2++);
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<scalar_product>:
  endbr64
  test   %edx,%edx
  jle    L12
  movslq %edx,%rdx
  xor    %eax,%eax
  xor    %r8d,%r8d
L18:
  mov    (%rdi,%rax,4),%ecx
  imul   (%rsi,%rax,4),%ecx
  add    $0x1,%rax
  add    %ecx,%r8d
  cmp    %rdx,%rax
  jne    L18
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int scalar_product(int *v1, int *v2, int size) {
  int result = 0;
  while (size-- > 0)
    result += (*v1++) * (*v2++);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<scalar_product>:
  endbr64
  mov    %rsi,%rcx
  lea    -0x1(%rdx),%r8d
  test   %edx,%edx
  jle    L19
  cmp    $0x2,%r8d
  jbe    L55
  mov    %edx,%esi
  xor    %eax,%eax
  pxor   %xmm3,%xmm3
  shr    $0x2,%esi
  shl    $0x4,%rsi
  nopl   0x0(%rax)
L33:
  movdqu (%rdi,%rax,1),%xmm0
  movdqu (%rcx,%rax,1),%xmm2
  movdqu (%rdi,%rax,1),%xmm1
  movdqu (%rcx,%rax,1),%xmm4
  add    $0x10,%rax
  psrlq  $0x20,%xmm0
  psrlq  $0x20,%xmm2
  pmuludq %xmm4,%xmm1
  pmuludq %xmm2,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  paddd  %xmm1,%xmm3
  cmp    %rsi,%rax
  jne    L33
  movdqa %xmm3,%xmm0
  mov    %edx,%r9d
  psrldq $0x8,%xmm0
  and    $0xfffffffc,%r9d
  paddd  %xmm0,%xmm3
  mov    %r9d,%esi
  sub    %r9d,%r8d
  movdqa %xmm3,%xmm0
  shl    $0x2,%rsi
  psrldq $0x4,%xmm0
  add    %rsi,%rdi
  add    %rsi,%rcx
  paddd  %xmm0,%xmm3
  movd   %xmm3,%eax
  cmp    %r9d,%edx
  je     L69
L20:
  mov    (%rdi),%edx
  imul   (%rcx),%edx
  add    %edx,%eax
  test   %r8d,%r8d
  jle    L36
  mov    0x4(%rdi),%edx
  imul   0x4(%rcx),%edx
  add    %edx,%eax
  cmp    $0x1,%r8d
  jle    L36
  mov    0x8(%rdi),%edx
  imul   0x8(%rcx),%edx
  add    %edx,%eax
  ret
L19:
  xor    %eax,%eax
L36:
  ret
  nopl   0x0(%rax,%rax,1)
L69:
  ret
L55:
  xor    %eax,%eax
  jmp    L20
```

assistant: ```c
int scalar_product(int *v1, int *v2, int size) {
  int result = 0;
  while (size-- > 0)
    result += (*v1++) * (*v2++);
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<frexp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  mov    %rdi,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  ucomisd -0x18(%rbp),%xmm0
  jp     L10
  pxor   %xmm0,%xmm0
  ucomisd -0x18(%rbp),%xmm0
  je     L0
L10:
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movzwl -0x8(%rbp),%eax
  add    %eax,%eax
  sar    $0x8,%ax
  movsbl %al,%eax
  lea    -0x80(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  movzwl -0x8(%rbp),%eax
  and    $0x807f,%ax
  or     $0x40,%ah
  mov    %ax,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  jmp    L6
L0:
  mov    -0x20(%rbp),%rax
  movl   $0x0,(%rax)
  pxor   %xmm0,%xmm0
L6:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double frexp(double value, int *eptr) {
  union {
    double v;
    struct {
      u_int u_mant1 : 7;
      u_int u_exp : 8;
      u_int u_sign : 1;
      u_int u_mant2 : 16;
      u_int u_mant3 : 32;
    } s;
  } u;
  if (value) {
    u.v = value;
    *eptr = u.s.u_exp - 128;
    u.s.u_exp = 128;
    return (u.v);
  } else {
    *eptr = 0;
    return (0.0);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<frexp>:
  endbr64
  ucomisd D8(%rip),%xmm0
  jp     L1
  jne    L1
  movl   $0x0,(%rdi)
  pxor   %xmm0,%xmm0
  ret
L1:
  movq   %xmm0,%rax
  lea    (%rax,%rax,1),%edx
  sar    $0x8,%dx
  movsbl %dl,%edx
  add    $0xffffff80,%edx
  mov    %edx,(%rdi)
  and    $0xffffffffffff807f,%rax
  or     $0x40,%ah
  movq   %rax,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D8
datatype: f64

tool: D8: 0

assistant: ```c
double frexp(double value, int *eptr) {
  union {
    double v;
    struct {
      u_int u_mant1 : 7;
      u_int u_exp : 8;
      u_int u_sign : 1;
      u_int u_mant2 : 16;
      u_int u_mant3 : 32;
    } s;
  } u;
  if (value) {
    u.v = value;
    *eptr = u.s.u_exp - 128;
    u.s.u_exp = 128;
    return (u.v);
  } else {
    *eptr = 0;
    return (0.0);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<frexp>:
  endbr64
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L10
  jne    L10
  movl   $0x0,(%rdi)
  pxor   %xmm0,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  movq   %xmm0,%rax
  lea    (%rax,%rax,1),%edx
  and    $0xffffffffffff807f,%rax
  sar    $0x8,%dx
  or     $0x40,%ah
  movsbl %dl,%edx
  movq   %rax,%xmm0
  add    $0xffffff80,%edx
  mov    %edx,(%rdi)
  ret
```

assistant: ```c
double frexp(double value, int *eptr) {
  union {
    double v;
    struct {
      u_int u_mant1 : 7;
      u_int u_exp : 8;
      u_int u_sign : 1;
      u_int u_mant2 : 16;
      u_int u_mant3 : 32;
    } s;
  } u;
  if (value) {
    u.v = value;
    *eptr = u.s.u_exp - 128;
    u.s.u_exp = 128;
    return (u.v);
  } else {
    *eptr = 0;
    return (0.0);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<frexp>:
  endbr64
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L18
  jne    L18
  movl   $0x0,(%rdi)
  pxor   %xmm0,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  movq   %xmm0,%rax
  lea    (%rax,%rax,1),%edx
  and    $0xffffffffffff807f,%rax
  sar    $0x8,%dx
  or     $0x40,%ah
  movsbl %dl,%edx
  movq   %rax,%xmm0
  add    $0xffffff80,%edx
  mov    %edx,(%rdi)
  ret
```

assistant: ```c
double frexp(double value, int *eptr) {
  union {
    double v;
    struct {
      u_int u_mant1 : 7;
      u_int u_exp : 8;
      u_int u_sign : 1;
      u_int u_mant2 : 16;
      u_int u_mant3 : 32;
    } s;
  } u;
  if (value) {
    u.v = value;
    *eptr = u.s.u_exp - 128;
    u.s.u_exp = 128;
    return (u.v);
  } else {
    *eptr = 0;
    return (0.0);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x2,-0x4(%rbp)
  jmp    L28
L22:
  mov    $0x7b5,%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L21
  mov    $0x7b5,%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L10
L21:
  addl   $0x1,-0x4(%rbp)
L28:
  mov    $0x7b5,%edi
  call   <sqrt@plt>
  cmp    %eax,-0x4(%rbp)
  jl     L22
  mov    $0x7b5,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
L10:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type: string
name: unpack_data
arguments: data_tag: D15
type: string

tool: D0: NUM is  %d * %d and so it is not prime


tool: D15: %d IS prime


assistant: ```c
int main() {
  int i;
  for (i = 2; i < sqrt(1973); i++)
    if ((1973 % i == 0)) {
      printf("NUM is  %d * %d and so it is not prime\n", i, 1973 / i);
      return 0;
    }
  printf("%d IS prime\n", 1973);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2,%ebx
  jmp    L5
L10:
  add    $0x1,%ebx
L5:
  mov    $0x7b5,%edi
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jle    L23
  mov    $0x7b5,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  jne    L10
  mov    $0x7b5,%eax
  cltd
  idiv   %ebx
  mov    %eax,%edx
  mov    %ebx,%esi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L2
L23:
  mov    $0x7b5,%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L2:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D11
datatype: string
name: disassemble_data
arguments: data_tag: D4
datatype: string

tool: D11: NUM is  %d * %d and so it is not prime


tool: D4: %d IS prime


assistant: ```c
int main() {
  int i;
  for (i = 2; i < sqrt(1973); i++)
    if ((1973 % i == 0)) {
      printf("NUM is  %d * %d and so it is not prime\n", i, 1973 / i);
      return 0;
    }
  printf("%d IS prime\n", 1973);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x2,%ebp
  jmp    L0
  nopl   0x0(%rax)
L33:
  mov    $0x7b5,%eax
  cltd
  idiv   %ebp
  test   %edx,%edx
  je     L28
  add    $0x1,%ebp
L0:
  mov    $0x7b5,%edi
  call   <sqrt@plt>
  cmp    %ebp,%eax
  jg     L33
  mov    $0x7b5,%esi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L37:
  xor    %eax,%eax
  pop    %rbp
  ret
L28:
  mov    %eax,%edx
  mov    %ebp,%esi
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L37
```

tool_calls: name: read_data
arguments: tag: D11
type_infered: string
name: read_data
arguments: tag: D18
type_infered: string

tool: D11: %d IS prime


tool: D18: NUM is  %d * %d and so it is not prime


assistant: ```c
int main() {
  int i;
  for (i = 2; i < sqrt(1973); i++)
    if ((1973 % i == 0)) {
      printf("NUM is  %d * %d and so it is not prime\n", i, 1973 / i);
      return 0;
    }
  printf("%d IS prime\n", 1973);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x2,%ebp
  jmp    L30
  nopl   0x0(%rax)
L9:
  mov    $0x7b5,%eax
  cltd
  idiv   %ebp
  test   %edx,%edx
  je     L12
  add    $0x1,%ebp
L30:
  mov    $0x7b5,%edi
  call   <sqrt@plt>
  cmp    %ebp,%eax
  jg     L9
  mov    $0x7b5,%esi
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L26:
  xor    %eax,%eax
  pop    %rbp
  ret
L12:
  mov    %eax,%edx
  mov    %ebp,%esi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L26
```

tool_calls: name: unpack_data
arguments: label: D3
datatype: string
name: unpack_data
arguments: label: D15
datatype: string

tool: D3: %d IS prime


tool: D15: NUM is  %d * %d and so it is not prime


assistant: ```c
int main() {
  int i;
  for (i = 2; i < sqrt(1973); i++)
    if ((1973 % i == 0)) {
      printf("NUM is  %d * %d and so it is not prime\n", i, 1973 / i);
      return 0;
    }
  printf("%d IS prime\n", 1973);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<kernel_dgemv_t_1_lib4>:
  endbr64
  push   %rbp
  push   %rbx
  mov    0x18(%rsp),%r11d
  test   %edi,%edi
  jle    L42
  mov    %rcx,%rax
  cmp    $0x3,%edi
  jle    L27
  shl    $0x2,%edx
  movslq %edx,%rcx
  shl    $0x3,%rcx
  lea    -0x4(%rdi),%ebx
  shr    $0x2,%ebx
  lea    0x1(%rbx),%ebp
  mov    %ebp,%ebp
  mov    %rbp,%r10
  shl    $0x5,%r10
  add    %rax,%r10
  mov    %rsi,%rdx
  pxor   %xmm0,%xmm0
L45:
  movsd  (%rdx),%xmm1
  mulsd  (%rax),%xmm1
  addsd  %xmm0,%xmm1
  movsd  0x8(%rdx),%xmm0
  mulsd  0x8(%rax),%xmm0
  addsd  %xmm1,%xmm0
  movsd  0x10(%rdx),%xmm1
  mulsd  0x10(%rax),%xmm1
  addsd  %xmm0,%xmm1
  movsd  0x18(%rdx),%xmm0
  mulsd  0x18(%rax),%xmm0
  addsd  %xmm1,%xmm0
  add    %rcx,%rdx
  add    $0x20,%rax
  cmp    %r10,%rax
  jne    L45
  imul   %rbp,%rcx
  add    %rcx,%rsi
  lea    0x4(,%rbx,4),%ecx
  cmp    %edi,%ecx
  jge    L49
L77:
  mov    $0x0,%eax
L48:
  movsd  (%rsi,%rax,8),%xmm1
  mulsd  (%r10,%rax,8),%xmm1
  addsd  %xmm1,%xmm0
  add    $0x1,%rax
  lea    (%rcx,%rax,1),%edx
  cmp    %edx,%edi
  jg     L48
L49:
  test   %r11d,%r11d
  jne    L66
  movsd  %xmm0,(%r8)
L42:
  pop    %rbx
  pop    %rbp
  ret
L27:
  mov    %rcx,%r10
  pxor   %xmm0,%xmm0
  mov    $0x0,%ecx
  jmp    L77
L66:
  cmp    $0x1,%r11d
  je     L86
  movsd  (%r8),%xmm1
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%r9)
  jmp    L42
L86:
  addsd  (%r8),%xmm0
  movsd  %xmm0,(%r9)
  jmp    L42
```

assistant: ```c
void kernel_dgemv_t_1_lib4(int kmax, double *A, int sda, double *x, double *y, double *z, int alg) {
  if (kmax <= 0)
    return;
  const int lda = 4;
  const int bs = 4;
  int k, ka = kmax;
  double x_0, x_1, x_2, x_3, y_0 = 0;
  k = 0;
  for (; k < ka - bs + 1; k += bs) {
    x_0 = x[0];
    x_1 = x[1];
    x_2 = x[2];
    x_3 = x[3];
    y_0 += A[0 + lda * 0] * x_0;
    y_0 += A[1 + lda * 0] * x_1;
    y_0 += A[2 + lda * 0] * x_2;
    y_0 += A[3 + lda * 0] * x_3;
    A += sda * bs;
    x += 4;
  }
  for (; k < ka; k++) {
    x_0 = x[0];
    y_0 += A[0 + lda * 0] * x_0;
    A += 1;
    x += 1;
  }
  if (alg == 0) {
    y[0] = y_0;
  } else if (alg == 1) {
    z[0] = y[0] + y_0;
  } else {
    z[0] = y[0] - y_0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<kernel_dgemv_t_1_lib4>:
  endbr64
  push   %rbp
  push   %rbx
  mov    0x18(%rsp),%r11d
  test   %edi,%edi
  jle    L70
  mov    %rcx,%rax
  cmp    $0x3,%edi
  jle    L75
  shl    $0x2,%edx
  pxor   %xmm0,%xmm0
  movslq %edx,%rcx
  lea    -0x4(%rdi),%edx
  shr    $0x2,%edx
  shl    $0x3,%rcx
  lea    0x1(%rdx),%ebp
  mov    %rsi,%rdx
  mov    %rbp,%r10
  mov    %rbp,%rbx
  shl    $0x5,%r10
  add    %rax,%r10
  cs nopw 0x0(%rax,%rax,1)
L5:
  movsd  (%rdx),%xmm1
  mulsd  (%rax),%xmm1
  add    $0x20,%rax
  addsd  %xmm0,%xmm1
  movsd  0x8(%rdx),%xmm0
  mulsd  -0x18(%rax),%xmm0
  addsd  %xmm1,%xmm0
  movsd  0x10(%rdx),%xmm1
  mulsd  -0x10(%rax),%xmm1
  addsd  %xmm0,%xmm1
  movsd  0x18(%rdx),%xmm0
  mulsd  -0x8(%rax),%xmm0
  add    %rcx,%rdx
  addsd  %xmm1,%xmm0
  cmp    %rax,%r10
  jne    L5
  imul   %rbp,%rcx
  lea    0x0(,%rbx,4),%edx
  add    %rcx,%rsi
  cmp    %edx,%edi
  jle    L64
L14:
  xor    %eax,%eax
L29:
  movsd  (%rsi,%rax,8),%xmm1
  mulsd  (%r10,%rax,8),%xmm1
  add    $0x1,%rax
  lea    (%rdx,%rax,1),%ecx
  addsd  %xmm1,%xmm0
  cmp    %ecx,%edi
  jg     L29
L64:
  test   %r11d,%r11d
  je     L51
  movsd  (%r8),%xmm1
  cmp    $0x1,%r11d
  je     L39
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%r9)
L70:
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L51:
  pop    %rbx
  movsd  %xmm0,(%r8)
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L39:
  addsd  %xmm1,%xmm0
  pop    %rbx
  pop    %rbp
  movsd  %xmm0,(%r9)
  ret
  nopl   0x0(%rax)
L75:
  mov    %rcx,%r10
  pxor   %xmm0,%xmm0
  xor    %edx,%edx
  jmp    L14
```

assistant: ```c
void kernel_dgemv_t_1_lib4(int kmax, double *A, int sda, double *x, double *y, double *z, int alg) {
  if (kmax <= 0)
    return;
  const int lda = 4;
  const int bs = 4;
  int k, ka = kmax;
  double x_0, x_1, x_2, x_3, y_0 = 0;
  k = 0;
  for (; k < ka - bs + 1; k += bs) {
    x_0 = x[0];
    x_1 = x[1];
    x_2 = x[2];
    x_3 = x[3];
    y_0 += A[0 + lda * 0] * x_0;
    y_0 += A[1 + lda * 0] * x_1;
    y_0 += A[2 + lda * 0] * x_2;
    y_0 += A[3 + lda * 0] * x_3;
    A += sda * bs;
    x += 4;
  }
  for (; k < ka; k++) {
    x_0 = x[0];
    y_0 += A[0 + lda * 0] * x_0;
    A += 1;
    x += 1;
  }
  if (alg == 0) {
    y[0] = y_0;
  } else if (alg == 1) {
    z[0] = y[0] + y_0;
  } else {
    z[0] = y[0] - y_0;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<unjulian>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x5c,-0xc(%rbp)
  jmp    L43
L26:
  mov    -0xc(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L16
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L16
  mov    $0x1,%eax
  jmp    L2
L16:
  mov    $0x0,%eax
L2:
  add    $0x16d,%eax
  cltq
  add    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L43:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x18(%rbp),%rcx
  movabs $0x20c49ba5e353f7cf,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x7,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  cmp    %rax,%rsi
  jl     L26
  mov    -0x18(%rbp),%rcx
  movabs $0x20c49ba5e353f7cf,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x7,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  imul   $0x3e8,%rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  sub    $0x1,%rax
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long unjulian(long date) {
  int i;
  long res = 0;
  for (i = 92001 / 1000; i < date / 1000; i++)
    res += 365 + ((!(i % 4) && (i % 100)) ? 1 : 0);
  res += date % 1000 - 1;
  return (res);
}
```