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
<read_long_int_from_file>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <assert@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <read_int32_from_file@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int read_long_int_from_file(FILE *f, long int *num) {
  assert(f);
  assert(num);
  return read_int32_from_file(f, (int32_t *)num);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_long_int_from_file>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <assert@plt>
  mov    %r12,%rdi
  call   <assert@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <read_int32_from_file@plt>
```

assistant: ```c
int read_long_int_from_file(FILE *f, long int *num) {
  assert(f);
  assert(num);
  return read_int32_from_file(f, (int32_t *)num);
}
```
user: What is the c source code of the assembly code below:

```asm
<read_long_int_from_file>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <assert@plt>
  mov    %r12,%rdi
  call   <assert@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1070 <read_int32_from_file@plt>
```

assistant: ```c
int read_long_int_from_file(FILE *f, long int *num) {
  assert(f);
  assert(num);
  return read_int32_from_file(f, (int32_t *)num);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x61616161,0x4(%rsp)
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  lea    D42(%rip),%r13
  mov    %r13,%rdi
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  movsbl 0x5(%rsp),%edx
  movsbl 0x4(%rsp),%esi
  movsbl 0x7(%rsp),%r8d
  lea    D6(%rip),%r12
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D23(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movb   $0x5a,0x4(%rsp)
  movb   $0x65,0x5(%rsp)
  movb   $0x64,0x6(%rsp)
  movb   $0x41,0x7(%rsp)
  mov    $0x4,%r8d
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  movsbl 0x5(%rsp),%edx
  movsbl 0x4(%rsp),%esi
  movsbl 0x7(%rsp),%r8d
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D24(%rip),%rsi
  lea    D37(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%r8d
  mov    $0x64,%ecx
  mov    $0x65,%edx
  mov    $0x5a,%esi
  lea    D32(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D42
type_guessed: string
name: extract_symbolic_data
arguments: tag: D6
type_guessed: string
name: extract_symbolic_data
arguments: tag: D23
type_guessed: string
name: extract_symbolic_data
arguments: tag: D24
type_guessed: string
name: extract_symbolic_data
arguments: tag: D37
type_guessed: string
name: extract_symbolic_data
arguments: tag: D32
type_guessed: string

tool: D42: numbers: %d %d %d %d


tool: D6: name each: %c %c %c %c


tool: D23: name: %s


tool: D24: Zed

tool: D37: another: %s


tool: D32: another each: %c %c %c %c


assistant: ```c
int main(int argc, char *argv[]) {
  int numbers[4] = {0};
  char name[4] = {'a', 'a', 'a', 'a'};
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  numbers[0] = 1;
  numbers[1] = 2;
  numbers[2] = 3;
  numbers[3] = 4;
  name[0] = 'Z';
  name[1] = 'e';
  name[2] = 'd';
  name[3] = 'A';
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  char *another = "Zed";
  printf("another: %s\n", another);
  printf("another each: %c %c %c %c\n", another[0], another[1], another[2], another[3]);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r14
  lea    D0(%rip),%r14
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  push   %r13
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  push   %r12
  lea    D44(%rip),%r13
  push   %rbp
  lea    D18(%rip),%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x61616161,0x4(%rsp)
  lea    0x4(%rsp),%r12
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  mov    %r13,%rdi
  xor    %eax,%eax
  movsbl 0x5(%rsp),%edx
  movsbl 0x7(%rsp),%r8d
  movsbl 0x4(%rsp),%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%r8d
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  movl   $0x4164655a,0x4(%rsp)
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  mov    %r13,%rdi
  xor    %eax,%eax
  movsbl 0x5(%rsp),%edx
  movsbl 0x7(%rsp),%r8d
  movsbl 0x4(%rsp),%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D28(%rip),%rsi
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %r8d,%r8d
  xor    %eax,%eax
  mov    $0x64,%ecx
  mov    $0x65,%edx
  mov    $0x5a,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D0
type_infered: string
name: disassemble_data
arguments: tag: D44
type_infered: string
name: disassemble_data
arguments: tag: D18
type_infered: string
name: disassemble_data
arguments: tag: D28
type_infered: string
name: disassemble_data
arguments: tag: D4
type_infered: string
name: disassemble_data
arguments: tag: D5
type_infered: string

tool: D0: numbers: %d %d %d %d


tool: D44: name each: %c %c %c %c


tool: D18: name: %s


tool: D28: Zed

tool: D4: another: %s


tool: D5: another each: %c %c %c %c


assistant: ```c
int main(int argc, char *argv[]) {
  int numbers[4] = {0};
  char name[4] = {'a', 'a', 'a', 'a'};
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  numbers[0] = 1;
  numbers[1] = 2;
  numbers[2] = 3;
  numbers[3] = 4;
  name[0] = 'Z';
  name[1] = 'e';
  name[2] = 'd';
  name[3] = 'A';
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  char *another = "Zed";
  printf("another: %s\n", another);
  printf("another each: %c %c %c %c\n", another[0], another[1], another[2], another[3]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  lea    D8(%rip),%r14
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  push   %r13
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  push   %r12
  lea    D53(%rip),%r13
  push   %rbp
  lea    D2(%rip),%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x61616161,0x4(%rsp)
  lea    0x4(%rsp),%r12
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  mov    %r13,%rdi
  xor    %eax,%eax
  movsbl 0x5(%rsp),%edx
  movsbl 0x7(%rsp),%r8d
  movsbl 0x4(%rsp),%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%r8d
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  movl   $0x4164655a,0x4(%rsp)
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  mov    %r13,%rdi
  xor    %eax,%eax
  movsbl 0x5(%rsp),%edx
  movsbl 0x7(%rsp),%r8d
  movsbl 0x4(%rsp),%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D38(%rip),%rsi
  lea    D47(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %r8d,%r8d
  xor    %eax,%eax
  mov    $0x64,%ecx
  mov    $0x65,%edx
  mov    $0x5a,%esi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type: string
name: unpack_data
arguments: data_tag: D53
type: string
name: unpack_data
arguments: data_tag: D2
type: string
name: unpack_data
arguments: data_tag: D38
type: string
name: unpack_data
arguments: data_tag: D47
type: string
name: unpack_data
arguments: data_tag: D12
type: string

tool: D8: numbers: %d %d %d %d


tool: D53: name each: %c %c %c %c


tool: D2: name: %s


tool: D38: Zed

tool: D47: another: %s


tool: D12: another each: %c %c %c %c


assistant: ```c
int main(int argc, char *argv[]) {
  int numbers[4] = {0};
  char name[4] = {'a', 'a', 'a', 'a'};
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  numbers[0] = 1;
  numbers[1] = 2;
  numbers[2] = 3;
  numbers[3] = 4;
  name[0] = 'Z';
  name[1] = 'e';
  name[2] = 'd';
  name[3] = 'A';
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  char *another = "Zed";
  printf("another: %s\n", another);
  printf("another each: %c %c %c %c\n", another[0], another[1], another[2], another[3]);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<stridentcnt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
L25:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L14
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
L14:
  mov    -0x8(%rbp),%rax
  jmp    L17
L6:
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
  jmp    L25
L17:
  pop    %rbp
  ret
```

assistant: ```c
uint32_t stridentcnt(const char *a, const char *b) {
  uint16_t i = 0;
  for (;;) {
    if (*a != *b || *b == '\0')
      return i;
    i++;
    a++;
    b++;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<stridentcnt>:
  endbr64
  movzbl (%rsi),%edx
  cmp    %dl,(%rdi)
  jne    L3
  mov    $0x0,%eax
  test   %dl,%dl
  je     L3
L29:
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  cmp    %dl,(%rdi,%rax,1)
  jne    L35
  test   %dl,%dl
  jne    L29
L35:
  ret
L3:
  mov    $0x0,%eax
  ret
```

assistant: ```c
uint32_t stridentcnt(const char *a, const char *b) {
  uint16_t i = 0;
  for (;;) {
    if (*a != *b || *b == '\0')
      return i;
    i++;
    a++;
    b++;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<stridentcnt>:
  endbr64
  movzbl (%rsi),%edx
  cmp    %dl,(%rdi)
  jne    L36
  xor    %eax,%eax
  test   %dl,%dl
  jne    L47
  jmp    L36
  nopl   0x0(%rax,%rax,1)
L7:
  test   %dl,%dl
  je     L21
L47:
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  cmp    %dl,(%rdi,%rax,1)
  je     L7
L21:
  ret
  nopw   0x0(%rax,%rax,1)
L36:
  xor    %eax,%eax
  ret
```

assistant: ```c
uint32_t stridentcnt(const char *a, const char *b) {
  uint16_t i = 0;
  for (;;) {
    if (*a != *b || *b == '\0')
      return i;
    i++;
    a++;
    b++;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<stridentcnt>:
  endbr64
  movzbl (%rsi),%edx
  cmp    %dl,(%rdi)
  jne    L21
  xor    %eax,%eax
  test   %dl,%dl
  jne    L30
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L35:
  test   %dl,%dl
  je     L18
L30:
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  cmp    %dl,(%rdi,%rax,1)
  je     L35
L18:
  ret
  nopw   0x0(%rax,%rax,1)
L21:
  xor    %eax,%eax
  ret
```

assistant: ```c
uint32_t stridentcnt(const char *a, const char *b) {
  uint16_t i = 0;
  for (;;) {
    if (*a != *b || *b == '\0')
      return i;
    i++;
    a++;
    b++;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<poly_sub2>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rax
  mov    (%rdx),%r14d
  mov    0x4(%rdx),%ebx
  mov    %ebx,-0x4(%rsp)
  mov    (%rcx),%esi
  mov    0x4(%rcx),%r15d
  cmp    %esi,%r14d
  mov    %esi,%edx
  cmovge %r14d,%edx
  mov    %edx,(%r9)
  cmp    %r15d,%ebx
  cmovl  %r15d,%ebx
  mov    %ebx,0x4(%r9)
  test   %ebx,%ebx
  js     L111
  mov    %rdi,%rbp
  mov    %r8,%rbx
  mov    %rax,%r8
  mov    $0x0,%edi
  movq   D2(%rip),%xmm1
  jmp    L79
L93:
  movsd  (%r11,%rax,8),%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
L1:
  add    $0x1,%rax
  cmp    %eax,(%r9)
  jl     L10
L51:
  cmp    %eax,%esi
  jl     L93
  test   %r10b,%r10b
  jne    L93
  cmp    %r14d,%eax
  jg     L91
  test   %r13b,%r13b
  jne    L91
  movsd  (%r11,%rax,8),%xmm0
  subsd  (%r12,%rax,8),%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
  jmp    L1
L91:
  movsd  (%r12,%rax,8),%xmm0
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
  jmp    L1
L10:
  add    $0x1,%edi
  add    $0x28,%rbp
  add    $0x28,%rbx
  add    $0x28,%r8
  cmp    %edi,0x4(%r9)
  jl     L111
L79:
  cmp    %edi,-0x4(%rsp)
  setl   %r13b
  cmp    %edi,%r15d
  setl   %r10b
  cmpl   $0x0,(%r9)
  js     L10
  mov    %rbp,%r11
  mov    %r8,%r12
  mov    %rbx,%rcx
  mov    $0x0,%eax
  jmp    L51
L111:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
data_type: byte[8]

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
void poly_sub2(double u1[5][5], double u2[5][5], int p1[2], int p2[2], double u3[5][5], int p3[2]) {
  int i, j, p11, p12, p21, p22;
  int i1_ok, i2_ok;
  p11 = p1[0];
  p12 = p1[1];
  p21 = p2[0];
  p22 = p2[1];
  p3[0] = ((p11) > (p21)) ? (p11) : (p21);
  p3[1] = ((p12) > (p22)) ? (p12) : (p22);
  for (i = 0; i <= p3[1]; i++) {
    i1_ok = (i > p12);
    i2_ok = (i > p22);
    for (j = 0; j <= p3[0]; j++) {
      if (i2_ok || (j > p21))
        u3[i][j] = u1[i][j];
      else if (i1_ok || (j > p11))
        u3[i][j] = -u2[i][j];
      else
        u3[i][j] = u1[i][j] - u2[i][j];
    }
  }
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<poly_sub2>:
  endbr64
  push   %r15
  mov    %rsi,%rax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rdx),%ebx
  mov    (%rcx),%esi
  mov    0x4(%rdx),%r12d
  mov    0x4(%rcx),%r13d
  cmp    %esi,%ebx
  mov    %esi,%r14d
  cmovge %ebx,%r14d
  mov    %r13d,%ebp
  cmp    %r13d,%r12d
  cmovge %r12d,%ebp
  mov    %r14d,(%r9)
  mov    %ebp,0x4(%r9)
  test   %ebp,%ebp
  js     L66
  mov    %rdi,%rdx
  movq   D4(%rip),%xmm2
  mov    %rax,%r9
  xor    %edi,%edi
  add    $0x1,%ebp
  movslq %r14d,%rcx
  nopl   (%rax)
L42:
  cmp    %edi,%r12d
  setl   %r11b
  cmp    %edi,%r13d
  setl   %r10b
  xor    %eax,%eax
  test   %r14d,%r14d
  jns    L41
  jmp    L3
  nop
L79:
  movsd  (%r9,%rax,8),%xmm1
  cmp    %eax,%ebx
  jl     L39
  test   %r11b,%r11b
  jne    L39
  movsd  (%rdx,%rax,8),%xmm0
  subsd  %xmm1,%xmm0
L64:
  movsd  %xmm0,(%r8,%rax,8)
  lea    0x1(%rax),%r15
  cmp    %rax,%rcx
  je     L3
L63:
  mov    %r15,%rax
L41:
  cmp    %eax,%esi
  jl     L65
  test   %r10b,%r10b
  je     L79
L65:
  movsd  (%rdx,%rax,8),%xmm0
  lea    0x1(%rax),%r15
  movsd  %xmm0,(%r8,%rax,8)
  cmp    %rax,%rcx
  jne    L63
L3:
  add    $0x1,%edi
  add    $0x28,%r8
  add    $0x28,%rdx
  add    $0x28,%r9
  cmp    %ebp,%edi
  jne    L42
L66:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L39:
  xorpd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  jmp    L64
```

tool_calls: name: extract_symbolic_data
arguments: label: D4
type_infered: byte[8]

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
void poly_sub2(double u1[5][5], double u2[5][5], int p1[2], int p2[2], double u3[5][5], int p3[2]) {
  int i, j, p11, p12, p21, p22;
  int i1_ok, i2_ok;
  p11 = p1[0];
  p12 = p1[1];
  p21 = p2[0];
  p22 = p2[1];
  p3[0] = ((p11) > (p21)) ? (p11) : (p21);
  p3[1] = ((p12) > (p22)) ? (p12) : (p22);
  for (i = 0; i <= p3[1]; i++) {
    i1_ok = (i > p12);
    i2_ok = (i > p22);
    for (j = 0; j <= p3[0]; j++) {
      if (i2_ok || (j > p21))
        u3[i][j] = u1[i][j];
      else if (i1_ok || (j > p11))
        u3[i][j] = -u2[i][j];
      else
        u3[i][j] = u1[i][j] - u2[i][j];
    }
  }
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<kissf_fft_next_fast_size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
L6:
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L9
L70:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L70
  jmp    L25
L16:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
L25:
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  add    %ecx,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L16
  jmp    L30
L46:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
L30:
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    %eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x2,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L46
  cmpl   $0x1,-0x4(%rbp)
  jle    L50
  addl   $0x1,-0x14(%rbp)
  jmp    L6
L50:
  nop
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int kissf_fft_next_fast_size(int n) {
  while (1) {
    int m = n;
    while ((m % 2) == 0)
      m /= 2;
    while ((m % 3) == 0)
      m /= 3;
    while ((m % 5) == 0)
      m /= 5;
    if (m <= 1)
      break;
    n++;
  }
  return n;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<kissf_fft_next_fast_size>:
  endbr64
  mov    %edi,%esi
L23:
  test   $0x1,%sil
  jne    L26
  mov    %esi,%eax
L41:
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  mov    %edx,%eax
  test   $0x1,%dl
  je     L41
L38:
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,2),%edx
  cmp    %edx,%eax
  jne    L69
L50:
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  sub    %eax,%edx
  mov    %edx,%eax
  movslq %edx,%rcx
  imul   $0x55555556,%rcx,%rcx
  shr    $0x20,%rcx
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%ecx
  lea    (%rcx,%rcx,2),%ecx
  cmp    %ecx,%edx
  je     L50
L69:
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x21,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  cmp    %edx,%eax
  jne    L58
L42:
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x21,%rdx
  sar    $0x1f,%eax
  sub    %eax,%edx
  mov    %edx,%eax
  movslq %edx,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x21,%rcx
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%ecx
  lea    (%rcx,%rcx,4),%ecx
  cmp    %ecx,%edx
  je     L42
L58:
  cmp    $0x1,%eax
  jle    L40
  add    $0x1,%esi
  jmp    L23
L26:
  mov    %esi,%eax
  jmp    L38
L40:
  mov    %esi,%eax
  ret
```

assistant: ```c
int kissf_fft_next_fast_size(int n) {
  while (1) {
    int m = n;
    while ((m % 2) == 0)
      m /= 2;
    while ((m % 3) == 0)
      m /= 3;
    while ((m % 5) == 0)
      m /= 5;
    if (m <= 1)
      break;
    n++;
  }
  return n;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<kissf_fft_next_fast_size>:
  endbr64
  mov    %edi,%r8d
  nopw   0x0(%rax,%rax,1)
L49:
  mov    %r8d,%eax
  test   $0x1,%r8b
  jne    L77
  nopl   0x0(%rax)
L10:
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  mov    %edx,%eax
  sar    %eax
  and    $0x2,%edx
  je     L10
  jmp    L77
  nopw   0x0(%rax,%rax,1)
L12:
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x55555556,%rdx,%rdx
  mov    %eax,%ecx
  shr    $0x20,%rdx
  mov    %edx,%eax
  sub    %ecx,%eax
L77:
  imul   $0xaaaaaaab,%eax,%edx
  add    $0x2aaaaaaa,%edx
  cmp    $0x55555554,%edx
  jbe    L12
  jmp    L67
  nopl   (%rax)
L37:
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x66666667,%rdx,%rdx
  mov    %eax,%ecx
  sar    $0x21,%rdx
  mov    %edx,%eax
  sub    %ecx,%eax
L67:
  imul   $0xcccccccd,%eax,%edx
  add    $0x19999999,%edx
  cmp    $0x33333332,%edx
  jbe    L37
  cmp    $0x1,%eax
  jle    L53
  add    $0x1,%r8d
  jmp    L49
  nopl   0x0(%rax)
L53:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int kissf_fft_next_fast_size(int n) {
  while (1) {
    int m = n;
    while ((m % 2) == 0)
      m /= 2;
    while ((m % 3) == 0)
      m /= 3;
    while ((m % 5) == 0)
      m /= 5;
    if (m <= 1)
      break;
    n++;
  }
  return n;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<kissf_fft_next_fast_size>:
  endbr64
  mov    %edi,%r8d
  nopw   0x0(%rax,%rax,1)
L23:
  mov    %r8d,%eax
  test   $0x1,%r8b
  jne    L42
  nopl   0x0(%rax)
L6:
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  mov    %edx,%eax
  sar    %eax
  and    $0x2,%edx
  je     L6
  jmp    L42
  nopw   0x0(%rax,%rax,1)
L0:
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x55555556,%rdx,%rdx
  mov    %eax,%ecx
  shr    $0x20,%rdx
  mov    %edx,%eax
  sub    %ecx,%eax
L42:
  imul   $0xaaaaaaab,%eax,%edx
  add    $0x2aaaaaaa,%edx
  cmp    $0x55555554,%edx
  jbe    L0
  jmp    L72
  nopl   (%rax)
L15:
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x66666667,%rdx,%rdx
  mov    %eax,%ecx
  sar    $0x21,%rdx
  mov    %edx,%eax
  sub    %ecx,%eax
L72:
  imul   $0xcccccccd,%eax,%edx
  add    $0x19999999,%edx
  cmp    $0x33333332,%edx
  jbe    L15
  cmp    $0x1,%eax
  jle    L41
  add    $0x1,%r8d
  jmp    L23
  nopl   0x0(%rax)
L41:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int kissf_fft_next_fast_size(int n) {
  while (1) {
    int m = n;
    while ((m % 2) == 0)
      m /= 2;
    while ((m % 3) == 0)
      m /= 3;
    while ((m % 5) == 0)
      m /= 5;
    if (m <= 1)
      break;
    n++;
  }
  return n;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<GiveInstructions>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D12
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D19
type_guessed: string

tool: D12: This program displays a calendar for a full


tool: D19: year. The year must not be before 1900.


assistant: ```c
void GiveInstructions(void) {
  printf("This program displays a calendar for a full\n");
  printf("year. The year must not be before 1900.\n");
}
```