ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L1:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L1
  sub    $0xc58,%rsp
  mov    $0x2710,%esi
  xor    %ebx,%ebx
  xor    %r12d,%r12d
  lea    D50(%rip),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sieve@plt>
  jmp    L52
  nopl   0x0(%rax,%rax,1)
L33:
  add    $0x1,%rbx
  cmp    $0x2711,%rbx
  je     L32
L52:
  cmpl   $0x1,0x0(%rbp,%rbx,4)
  jne    L33
  pxor   %xmm0,%xmm0
  add    $0x1,%r12d
  mov    %r13,%rdi
  cvtsi2sd %ebx,%xmm0
  mov    %r12d,%esi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x2711,%rbx
  jne    L52
L32:
  movsd  D46(%rip),%xmm0
  mov    %r12d,%esi
  lea    D20(%rip),%rdi
  call   <printf@plt>
  mov    $0x2710,%esi
  mov    %rbp,%rdi
  call   <twin_primes@plt>
  lea    D44(%rip),%rdi
  mov    %eax,%r12d
  mov    D46(%rip),%rax
  mov    %r12d,%esi
  movq   %rax,%xmm0
  call   <printf@plt>
  mov    $0x2710,%esi
  mov    %rbp,%rdi
  call   <brun_constant@plt>
  mov    %r12d,%esi
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x9c58,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D50
data_type: string
name: retrieve_data
arguments: label: D46
data_type: byte[8]
name: retrieve_data
arguments: label: D20
data_type: string
name: retrieve_data
arguments: label: D44
data_type: string
name: retrieve_data
arguments: label: D23
data_type: string

tool: D50: Prime #%04i:  %4i


tool: {"D46": [0x00,0x00,0x00,0x00,0x00,0x88,0xc3,0x40]}

tool: D20: %i primes found between 0 and %i.


tool: D44: %i twin primes found between 0 and %i.


tool: D23: Brun's constant with %i twin primes is %1.12f.


assistant: ```c
int main(void) {
  int array[10001], count = 0, i;
  double bruns;
  sieve(array, 10001 - 1);
  for (i = 0; i < 10001; i++) {
    if (array[i] == 1) {
      count++;
      printf("Prime #%04i:  %4i\n", count, i);
    }
  }
  printf("%i primes found between 0 and %i.\n", count, 10001 - 1);
  count = twin_primes(array, 10001 - 1);
  printf("%i twin primes found between 0 and %i.\n", count, 10001 - 1);
  bruns = brun_constant(array, 10001 - 1);
  printf("Brun's constant with %i twin primes is %1.12f.\n", count, bruns);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0xc(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L1
L22:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <kiss_fft_next_fast_size@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  je     L9
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L9:
  addl   $0x1,-0x10(%rbp)
L1:
  cmpl   $0x3e8,-0x10(%rbp)
  jle    L22
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0xc(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L49
L8:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <kiss_fft_next_fast_size@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  je     L41
  mov    -0x8(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L41:
  addl   $0x1,-0x10(%rbp)
L49:
  cmpl   $0x3e8,-0x10(%rbp)
  jle    L8
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D5
type_infered: string

tool: D5: %d 

assistant: ```c
int main(void) {
  int n, m, m2, m1;
  m2 = 0;
  for (n = 1; n < 1001; n++) {
    m = kiss_fft_next_fast_size(n);
    if (m != m2) {
      printf("%d ", m);
      m2 = m;
    }
  }
  printf("\n");
  m2 = 0;
  for (n = 1; n < 1001; n++) {
    m = kiss_fft_next_fast_size(n);
    if (m != m2) {
      m1 = 2 * m;
      printf("%d ", m1);
      m2 = m;
    }
  }
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebx
  mov    $0x1,%ebp
  lea    D4(%rip),%r13
  jmp    L41
L75:
  add    $0x1,%ebp
  cmp    $0x3e9,%ebp
  je     L29
L41:
  mov    %ebx,%r12d
  mov    %ebp,%edi
  call   <kiss_fft_next_fast_size@plt>
  mov    %eax,%ebx
  cmp    %r12d,%eax
  je     L75
  mov    %eax,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L75
L29:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%ebx
  mov    $0x1,%ebp
  lea    D4(%rip),%r13
  jmp    L70
L64:
  add    $0x1,%ebp
  cmp    $0x3e9,%ebp
  je     L9
L70:
  mov    %ebx,%r12d
  mov    %ebp,%edi
  call   <kiss_fft_next_fast_size@plt>
  mov    %eax,%ebx
  cmp    %r12d,%eax
  je     L64
  lea    (%rax,%rax,1),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L64
L9:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: tag: D4
data_type: string

tool: D4: %d 

assistant: ```c
int main(void) {
  int n, m, m2, m1;
  m2 = 0;
  for (n = 1; n < 1001; n++) {
    m = kiss_fft_next_fast_size(n);
    if (m != m2) {
      printf("%d ", m);
      m2 = m;
    }
  }
  printf("\n");
  m2 = 0;
  for (n = 1; n < 1001; n++) {
    m = kiss_fft_next_fast_size(n);
    if (m != m2) {
      m1 = 2 * m;
      printf("%d ", m1);
      m2 = m;
    }
  }
  printf("\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  lea    D3(%rip),%r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L39:
  mov    %ebp,%edi
  mov    %ebx,%r12d
  call   <kiss_fft_next_fast_size@plt>
  mov    %eax,%ebx
  cmp    %r12d,%eax
  je     L33
  mov    %eax,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L33:
  add    $0x1,%ebp
  cmp    $0x3e9,%ebp
  jne    L39
  mov    $0xa,%edi
  xor    %ebx,%ebx
  mov    $0x1,%ebp
  call   <putchar@plt>
  lea    D3(%rip),%r13
L26:
  mov    %ebp,%edi
  mov    %ebx,%r12d
  call   <kiss_fft_next_fast_size@plt>
  mov    %eax,%ebx
  cmp    %r12d,%eax
  je     L27
  lea    (%rax,%rax,1),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L27:
  add    $0x1,%ebp
  cmp    $0x3e9,%ebp
  jne    L26
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
data_type: string

tool: D3: %d 

assistant: ```c
int main(void) {
  int n, m, m2, m1;
  m2 = 0;
  for (n = 1; n < 1001; n++) {
    m = kiss_fft_next_fast_size(n);
    if (m != m2) {
      printf("%d ", m);
      m2 = m;
    }
  }
  printf("\n");
  m2 = 0;
  for (n = 1; n < 1001; n++) {
    m = kiss_fft_next_fast_size(n);
    if (m != m2) {
      m1 = 2 * m;
      printf("%d ", m1);
      m2 = m;
    }
  }
  printf("\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  lea    D14(%rip),%r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L23:
  mov    %ebp,%edi
  mov    %ebx,%r12d
  call   <kiss_fft_next_fast_size@plt>
  mov    %eax,%ebx
  cmp    %r12d,%eax
  je     L3
  mov    %eax,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L3:
  add    $0x1,%ebp
  cmp    $0x3e9,%ebp
  jne    L23
  mov    $0xa,%edi
  xor    %ebx,%ebx
  mov    $0x1,%ebp
  call   <putchar@plt>
  lea    D14(%rip),%r13
L30:
  mov    %ebp,%edi
  mov    %ebx,%r12d
  call   <kiss_fft_next_fast_size@plt>
  mov    %eax,%ebx
  cmp    %r12d,%eax
  je     L13
  lea    (%rax,%rax,1),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L13:
  add    $0x1,%ebp
  cmp    $0x3e9,%ebp
  jne    L30
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: label: D14
data_type: string

tool: D14: %d 

assistant: ```c
int main(void) {
  int n, m, m2, m1;
  m2 = 0;
  for (n = 1; n < 1001; n++) {
    m = kiss_fft_next_fast_size(n);
    if (m != m2) {
      printf("%d ", m);
      m2 = m;
    }
  }
  printf("\n");
  m2 = 0;
  for (n = 1; n < 1001; n++) {
    m = kiss_fft_next_fast_size(n);
    if (m != m2) {
      m1 = 2 * m;
      printf("%d ", m1);
      m2 = m;
    }
  }
  printf("\n");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<string_length>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L12:
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L12
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int string_length(char *s) {
  int c = 0;
  while (*(s + c))
    c++;
  return c;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<string_length>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L8
  mov    $0x1,%eax
L3:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L3
L4:
  mov    %edx,%eax
  ret
L8:
  mov    $0x0,%edx
  jmp    L4
```

assistant: ```c
int string_length(char *s) {
  int c = 0;
  while (*(s + c))
    c++;
  return c;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<string_length>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L3
  mov    $0x1,%eax
  xchg   %ax,%ax
L1:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L1
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int string_length(char *s) {
  int c = 0;
  while (*(s + c))
    c++;
  return c;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<string_length>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L1
  mov    $0x1,%eax
  xchg   %ax,%ax
L7:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L7
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int string_length(char *s) {
  int c = 0;
  while (*(s + c))
    c++;
  return c;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rollhash_arg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L10
L8:
  mov    -0x4(%rbp),%eax
  shl    $0x8,%eax
  movslq %eax,%rdx
  imul   $0x40400041,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x15,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  imul   $0x7f807f,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
L10:
  subl   $0x1,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L8
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t rollhash_arg(uint32_t leng) {
  uint32_t arg = 1;
  while (--leng)
    arg = arg * 256 % 8355967;
  return arg;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<rollhash_arg>:
  endbr64
  sub    $0x1,%edi
  je     L12
  mov    $0x1,%eax
L8:
  shl    $0x8,%eax
  mov    %eax,%edx
  cltq
  imul   $0x40400041,%rax,%rax
  sar    $0x35,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x7f807f,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  sub    $0x1,%edi
  jne    L8
  ret
L12:
  mov    $0x1,%eax
  ret
```

assistant: ```c
uint32_t rollhash_arg(uint32_t leng) {
  uint32_t arg = 1;
  while (--leng)
    arg = arg * 256 % 8355967;
  return arg;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rollhash_arg>:
  endbr64
  mov    $0x1,%eax
  sub    $0x1,%edi
  je     L8
  xchg   %ax,%ax
L10:
  shl    $0x8,%eax
  mov    %eax,%edx
  cltq
  imul   $0x40400041,%rax,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sar    $0x35,%rax
  sub    %ecx,%eax
  imul   $0x7f807f,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  sub    $0x1,%edi
  jne    L10
  ret
  nopl   0x0(%rax)
L8:
  ret
```

assistant: ```c
uint32_t rollhash_arg(uint32_t leng) {
  uint32_t arg = 1;
  while (--leng)
    arg = arg * 256 % 8355967;
  return arg;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rollhash_arg>:
  endbr64
  mov    $0x1,%eax
  sub    $0x1,%edi
  je     L9
  xchg   %ax,%ax
L16:
  shl    $0x8,%eax
  mov    %eax,%edx
  cltq
  imul   $0x40400041,%rax,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sar    $0x35,%rax
  sub    %ecx,%eax
  imul   $0x7f807f,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  sub    $0x1,%edi
  jne    L16
  ret
  nopl   0x0(%rax)
L9:
  ret
```

assistant: ```c
uint32_t rollhash_arg(uint32_t leng) {
  uint32_t arg = 1;
  while (--leng)
    arg = arg * 256 % 8355967;
  return arg;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<adjust>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L21
L13:
  addq   $0x1,-0x8(%rbp)
L21:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L4
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x3a,%al
  jne    L13
L4:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L7
  addq   $0x1,-0x8(%rbp)
L7:
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void adjust(char **s) {
  char *t;
  t = *s;
  while (*t && (*t != ':')) {
    t++;
  }
  if (*t) {
    t++;
  }
  *s = t;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<adjust>:
  endbr64
  mov    (%rdi),%rax
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L31
  cmp    $0x3a,%dl
  je     L31
L29:
  add    $0x1,%rax
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L31
  cmp    $0x3a,%dl
  jne    L29
L31:
  cmp    $0x1,%dl
  sbb    $0xffffffffffffffff,%rax
  mov    %rax,(%rdi)
  ret
```

assistant: ```c
void adjust(char **s) {
  char *t;
  t = *s;
  while (*t && (*t != ':')) {
    t++;
  }
  if (*t) {
    t++;
  }
  *s = t;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<adjust>:
  endbr64
  mov    (%rdi),%rax
  movzbl (%rax),%edx
  test   %dl,%dl
  jne    L23
  jmp    L36
L6:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  test   %dl,%dl
  je     L36
L23:
  cmp    $0x3a,%dl
  jne    L6
L36:
  cmp    $0x1,%dl
  sbb    $0xffffffffffffffff,%rax
  mov    %rax,(%rdi)
  ret
```

assistant: ```c
void adjust(char **s) {
  char *t;
  t = *s;
  while (*t && (*t != ':')) {
    t++;
  }
  if (*t) {
    t++;
  }
  *s = t;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<adjust>:
  endbr64
  mov    (%rdi),%rax
  movzbl (%rax),%edx
  test   %dl,%dl
  jne    L18
  jmp    L26
L20:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  test   %dl,%dl
  je     L26
L18:
  cmp    $0x3a,%dl
  jne    L20
L26:
  cmp    $0x1,%dl
  sbb    $0xffffffffffffffff,%rax
  mov    %rax,(%rdi)
  ret
```

assistant: ```c
void adjust(char **s) {
  char *t;
  t = *s;
  while (*t && (*t != ':')) {
    t++;
  }
  if (*t) {
    t++;
  }
  *s = t;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<recursion>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  jne    L29
  mov    -0x4(%rbp),%eax
  jmp    L11
L29:
  mov    -0x4(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L34
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <recursion>
  jmp    L11
L34:
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L44
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edi
  call   <recursion>
  jmp    L11
L44:
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <exit@plt>
L11:
  leave
  ret
```

assistant: ```c
int recursion(int n) {
  if (n == 4) {
    return n;
  } else if (n % 2 == 1) {
    return recursion(3 * n + 1);
  } else if (n % 2 == 0) {
    return recursion(n / 2);
  }
  exit(EXIT_FAILURE);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<recursion>:
  endbr64
  mov    %edi,%eax
  cmp    $0x4,%edi
  je     L2
  sub    $0x8,%rsp
  mov    %edi,%edx
  shr    $0x1f,%edx
  lea    (%rdi,%rdx,1),%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  je     L3
  test   $0x1,%dil
  je     L27
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
L17:
  add    $0x8,%rsp
  ret
L3:
  lea    0x1(%rdi,%rdi,2),%edi
  call   <recursion>
  jmp    L17
L27:
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %eax,%edi
  sar    %edi
  call   <recursion>
  jmp    L17
L2:
  ret
```

assistant: ```c
int recursion(int n) {
  if (n == 4) {
    return n;
  } else if (n % 2 == 1) {
    return recursion(3 * n + 1);
  } else if (n % 2 == 0) {
    return recursion(n / 2);
  }
  exit(EXIT_FAILURE);
}
```
user: Translate the below assembly code into C programming language:

```asm
<recursion>:
  endbr64
L22:
  cmp    $0x4,%edi
  je     L28
L38:
  mov    %edi,%edx
  shr    $0x1f,%edx
  lea    (%rdi,%rdx,1),%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  je     L24
  test   $0x1,%dil
  jne    L47
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %eax,%edi
  sar    %edi
  cmp    $0x4,%edi
  jne    L38
L28:
  mov    $0x4,%eax
  ret
  nopl   (%rax)
L24:
  lea    0x1(%rdi,%rdi,2),%edi
  jmp    L22
  xchg   %ax,%ax
L47:
  mov    D9(%rip),%rax
  mov    (%rax),%edi
  jmp    1050 <exit@plt>
```

assistant: ```c
int recursion(int n) {
  if (n == 4) {
    return n;
  } else if (n % 2 == 1) {
    return recursion(3 * n + 1);
  } else if (n % 2 == 0) {
    return recursion(n / 2);
  }
  exit(EXIT_FAILURE);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<recursion>:
  endbr64
L17:
  cmp    $0x4,%edi
  je     L36
L22:
  mov    %edi,%edx
  shr    $0x1f,%edx
  lea    (%rdi,%rdx,1),%eax
  and    $0x1,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  je     L32
  test   $0x1,%dil
  jne    L28
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %eax,%edi
  sar    %edi
  cmp    $0x4,%edi
  jne    L22
L36:
  mov    $0x4,%eax
  ret
  nopl   (%rax)
L32:
  lea    0x1(%rdi,%rdi,2),%edi
  jmp    L17
  xchg   %ax,%ax
L28:
  mov    D3(%rip),%rax
  mov    (%rax),%edi
  jmp    1050 <exit@plt>
```

assistant: ```c
int recursion(int n) {
  if (n == 4) {
    return n;
  } else if (n % 2 == 1) {
    return recursion(3 * n + 1);
  } else if (n % 2 == 0) {
    return recursion(n / 2);
  }
  exit(EXIT_FAILURE);
}
```
user: What is the c source code of the assembly code below:

```asm
<do_user>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L7
  mov    D11(%rip),%rax
  mov    (%rax),%eax
  jmp    L15
L7:
  mov    D19(%rip),%rax
  mov    (%rax),%eax
L15:
  pop    %rbp
  ret
```

assistant: ```c
int do_user(unsigned int userid, struct stat attr) {
  if (userid == attr.st_uid) {
    return EXIT_SUCCESS;
  }
  return EXIT_FAILURE;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<do_user>:
  endbr64
  cmp    %edi,%esi
  je     L7
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  ret
L7:
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
int do_user(unsigned int userid, struct stat attr) {
  if (userid == attr.st_uid) {
    return EXIT_SUCCESS;
  }
  return EXIT_FAILURE;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<do_user>:
  endbr64
  cmp    %edi,%esi
  je     L8
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  mov    D11(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
int do_user(unsigned int userid, struct stat attr) {
  if (userid == attr.st_uid) {
    return EXIT_SUCCESS;
  }
  return EXIT_FAILURE;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<do_user>:
  endbr64
  cmp    %edi,%esi
  je     L9
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
int do_user(unsigned int userid, struct stat attr) {
  if (userid == attr.st_uid) {
    return EXIT_SUCCESS;
  }
  return EXIT_FAILURE;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<xil_fpo_lessequal_flt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  movss  %xmm1,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movss  -0x18(%rbp),%xmm0
  mov    -0x14(%rbp),%edx
  lea    -0xc(%rbp),%rax
  movaps %xmm0,%xmm1
  movd   %edx,%xmm0
  mov    %rax,%rdi
  call   <xip_fpo_lessequal_flt@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int xil_fpo_lessequal_flt(float a, float b) {
  int res_int;
  xip_fpo_lessequal_flt(&res_int, a, b);
  return res_int;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<xil_fpo_lessequal_flt>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <xip_fpo_lessequal_flt@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int xil_fpo_lessequal_flt(float a, float b) {
  int res_int;
  xip_fpo_lessequal_flt(&res_int, a, b);
  return res_int;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<xil_fpo_lessequal_flt>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <xip_fpo_lessequal_flt@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int xil_fpo_lessequal_flt(float a, float b) {
  int res_int;
  xip_fpo_lessequal_flt(&res_int, a, b);
  return res_int;
}
```
user: Translate the below assembly code into C programming language:

```asm
<xil_fpo_lessequal_flt>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <xip_fpo_lessequal_flt@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int xil_fpo_lessequal_flt(float a, float b) {
  int res_int;
  xip_fpo_lessequal_flt(&res_int, a, b);
  return res_int;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<irc_ischan>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <irc_ischanchar@plt>
  test   %eax,%eax
  jne    L45
  mov    $0x0,%eax
  jmp    L23
L42:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <irc_ischanchar@plt>
  test   %eax,%eax
  jne    L45
  mov    $0x0,%eax
  jmp    L23
L45:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L42
  mov    $0x1,%eax
L23:
  leave
  ret
```

assistant: ```c
int irc_ischan(const char *str) {
  if (!irc_ischanchar(*str++, 1))
    return 0;
  while (*str) {
    if (!irc_ischanchar(*str++, 0))
      return 0;
  }
  return 1;
}
```