ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <getid@plt>
  movss  -0x20(%rbp),%xmm0
  movss  %xmm0,-0x24(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L13
L35:
  mov    -0x2c(%rbp),%eax
  cltq
  movss  -0x20(%rbp,%rax,4),%xmm0
  comiss -0x24(%rbp),%xmm0
  jbe    L39
  mov    -0x2c(%rbp),%eax
  cltq
  movss  -0x20(%rbp,%rax,4),%xmm0
  movss  %xmm0,-0x24(%rbp)
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x28(%rbp)
L39:
  addl   $0x1,-0x2c(%rbp)
L13:
  cmpl   $0x4,-0x2c(%rbp)
  jle    L35
  movss  -0x24(%rbp),%xmm0
  cvttss2si %xmm0,%eax
  mov    %eax,%edi
  call   <printid@plt>
  mov    -0x28(%rbp),%eax
  mov    %eax,%edi
  call   <printid@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

assistant: ```c
void main() {
  int i, k;
  float A[5], max;
  getid(A);
  max = A[0];
  for (i = 0; i < 5; i++) {
    if (A[i] > max) {
      max = A[i];
      k = i + 1;
    }
  }
  printid(max);
  printid(k);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <getid@plt>
  movss  0x4(%rsp),%xmm0
  maxss  (%rsp),%xmm0
  movss  0x8(%rsp),%xmm1
  mov    $0x3,%ebx
  comiss %xmm0,%xmm1
  jbe    L18
L65:
  movss  0xc(%rsp),%xmm0
  comiss %xmm1,%xmm0
  jbe    L52
  mov    $0x4,%ebx
L57:
  movss  0x10(%rsp),%xmm1
  comiss %xmm0,%xmm1
  jbe    L50
  mov    $0x5,%ebx
L40:
  cvttss2si %xmm1,%edi
  call   <printid@plt>
  mov    %ebx,%edi
  call   <printid@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x20,%rsp
  pop    %rbx
  ret
L18:
  mov    $0x2,%ebx
  movaps %xmm0,%xmm1
  jmp    L65
L52:
  movaps %xmm1,%xmm0
  jmp    L57
L50:
  movaps %xmm0,%xmm1
  jmp    L40
L33:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void main() {
  int i, k;
  float A[5], max;
  getid(A);
  max = A[0];
  for (i = 0; i < 5; i++) {
    if (A[i] > max) {
      max = A[i];
      k = i + 1;
    }
  }
  printid(max);
  printid(k);
}
```
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    $0x3,%ebp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <getid@plt>
  movss  0x4(%rsp),%xmm0
  maxss  (%rsp),%xmm0
  movss  0x8(%rsp),%xmm1
  comiss %xmm0,%xmm1
  ja     L46
  movaps %xmm0,%xmm1
  mov    $0x2,%ebp
L46:
  movss  0xc(%rsp),%xmm0
  comiss %xmm1,%xmm0
  jbe    L35
  mov    $0x4,%ebp
L14:
  movss  0x10(%rsp),%xmm1
  comiss %xmm0,%xmm1
  jbe    L42
  mov    $0x5,%ebp
L52:
  cvttss2si %xmm1,%edi
  call   <printid@plt>
  mov    %ebp,%edi
  call   <printid@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x20,%rsp
  pop    %rbp
  ret
L35:
  movaps %xmm1,%xmm0
  jmp    L14
L42:
  movaps %xmm0,%xmm1
  jmp    L52
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void main() {
  int i, k;
  float A[5], max;
  getid(A);
  max = A[0];
  for (i = 0; i < 5; i++) {
    if (A[i] > max) {
      max = A[i];
      k = i + 1;
    }
  }
  printid(max);
  printid(k);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x3,%ebp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <getid@plt>
  movss  0x4(%rsp),%xmm0
  maxss  (%rsp),%xmm0
  movss  0x8(%rsp),%xmm1
  comiss %xmm0,%xmm1
  ja     L18
  movaps %xmm0,%xmm1
  mov    $0x2,%ebp
L18:
  movss  0xc(%rsp),%xmm0
  comiss %xmm1,%xmm0
  jbe    L48
  mov    $0x4,%ebp
L49:
  movss  0x10(%rsp),%xmm1
  comiss %xmm0,%xmm1
  jbe    L41
  mov    $0x5,%ebp
L29:
  cvttss2si %xmm1,%edi
  call   <printid@plt>
  mov    %ebp,%edi
  call   <printid@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x20,%rsp
  pop    %rbp
  ret
L48:
  movaps %xmm1,%xmm0
  jmp    L49
L41:
  movaps %xmm0,%xmm1
  jmp    L29
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void main() {
  int i, k;
  float A[5], max;
  getid(A);
  max = A[0];
  for (i = 0; i < 5; i++) {
    if (A[i] > max) {
      max = A[i];
      k = i + 1;
    }
  }
  printid(max);
  printid(k);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<union2>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  movzbl (%rdi),%esi
  test   %sil,%sil
  je     L162
  mov    %rdi,%r13
  cmpb   $0x0,(%r12)
  je     L162
  mov    %rdi,%r14
  mov    $0x1,%ebx
  mov    $0x0,%ebp
  jmp    L89
L60:
  cltq
  lea    0x0(%r13,%rax,1),%r14
  movzbl (%r14),%esi
  add    $0x1,%ebx
  test   %sil,%sil
  je     L9
L89:
  lea    -0x1(%rbx),%edx
  movsbl %sil,%esi
  mov    %r13,%rdi
  call   <ft_check_char@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  test   %edx,%edx
  jne    L60
  movslq %ebp,%rax
  movzbl (%r14),%edx
  mov    %dl,(%rsp,%rax,1)
  mov    %ebx,%eax
  add    $0x1,%ebp
  jmp    L60
L9:
  movzbl (%r12),%esi
  test   %sil,%sil
  je     L151
  mov    %r12,%r13
  mov    $0x1,%ebx
  jmp    L29
L45:
  cltq
  lea    (%r12,%rax,1),%r13
  movzbl 0x0(%r13),%esi
  add    $0x1,%ebx
  test   %sil,%sil
  je     L151
L29:
  lea    -0x1(%rbx),%edx
  movsbl %sil,%esi
  mov    %r12,%rdi
  call   <ft_check_char@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  test   %edx,%edx
  jne    L45
  movslq %ebp,%rax
  movzbl 0x0(%r13),%edx
  mov    %dl,(%rsp,%rax,1)
  mov    %ebx,%eax
  add    $0x1,%ebp
  jmp    L45
L151:
  movslq %ebp,%rbp
  movb   $0x0,(%rsp,%rbp,1)
  movzbl (%rsp),%esi
  test   %sil,%sil
  je     L162
  mov    $0x0,%ebx
  mov    %rsp,%rbp
  jmp    L139
L21:
  cltq
  movzbl (%rsp,%rax,1),%esi
  add    $0x1,%rbx
  test   %sil,%sil
  je     L162
L139:
  mov    %ebx,%edx
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <ft_check_char@plt>
  mov    %eax,%edx
  lea    0x1(%rbx),%eax
  test   %edx,%edx
  jne    L21
  movsbl (%rbx,%rbp,1),%edi
  call   <ft_putchar@plt>
  lea    0x1(%rbx),%eax
  jmp    L21
L162:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L62
  mov    $0x0,%eax
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L62:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int union2(char *s1, char *s2) {
  int i;
  char cup[127];
  int j;
  i = 0;
  j = 0;
  if (*s1 == '\0' || *s2 == '\0')
    return (0);
  while (s1[i] != '\0')
    if (!(ft_check_char(s1, s1[i], i))) {
      cup[j] = s1[i];
      i++;
      j++;
    } else
      i++;
  i = 0;
  while (s2[i] != '\0')
    if (!(ft_check_char(s2, s2[i], i))) {
      cup[j] = s2[i];
      i++;
      j++;
    } else
      i++;
  cup[j] = '\0';
  i = 0;
  while (cup[i] != '\0')
    if (!(ft_check_char(cup, cup[i], i))) {
      ft_putchar(cup[i]);
      i++;
    } else
      i++;
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<union2>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  movsbl (%rdi),%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  test   %sil,%sil
  je     L120
  cmpb   $0x0,(%r12)
  je     L120
  mov    %rdi,%r13
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  nopl   (%rax)
L127:
  mov    %ebx,%edx
  mov    %r13,%rdi
  call   <ft_check_char@plt>
  test   %eax,%eax
  jne    L78
  movzbl 0x0(%r13,%rbx,1),%edx
  movslq %ebp,%rax
  add    $0x1,%ebp
  mov    %dl,(%rsp,%rax,1)
L78:
  add    $0x1,%rbx
  movsbl 0x0(%r13,%rbx,1),%esi
  test   %sil,%sil
  jne    L127
  movsbl (%r12),%esi
  test   %sil,%sil
  je     L116
  xor    %ebx,%ebx
L27:
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <ft_check_char@plt>
  test   %eax,%eax
  jne    L9
  movzbl (%r12,%rbx,1),%edx
  movslq %ebp,%rax
  add    $0x1,%ebp
  mov    %dl,(%rsp,%rax,1)
L9:
  add    $0x1,%rbx
  movsbl (%r12,%rbx,1),%esi
  test   %sil,%sil
  jne    L27
L116:
  movslq %ebp,%rbp
  movb   $0x0,(%rsp,%rbp,1)
  movsbl (%rsp),%esi
  test   %sil,%sil
  je     L120
  xor    %ebx,%ebx
  mov    %rsp,%rbp
  jmp    L109
  nopl   0x0(%rax)
L18:
  add    $0x1,%rbx
  movsbl 0x0(%rbp,%rbx,1),%esi
  test   %sil,%sil
  je     L120
L109:
  mov    %ebx,%edx
  mov    %rbp,%rdi
  call   <ft_check_char@plt>
  test   %eax,%eax
  jne    L18
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  call   <ft_putchar@plt>
  movsbl 0x0(%rbp,%rbx,1),%esi
  test   %sil,%sil
  jne    L109
  nopl   0x0(%rax)
L120:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L81
  add    $0x98,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L81:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int union2(char *s1, char *s2) {
  int i;
  char cup[127];
  int j;
  i = 0;
  j = 0;
  if (*s1 == '\0' || *s2 == '\0')
    return (0);
  while (s1[i] != '\0')
    if (!(ft_check_char(s1, s1[i], i))) {
      cup[j] = s1[i];
      i++;
      j++;
    } else
      i++;
  i = 0;
  while (s2[i] != '\0')
    if (!(ft_check_char(s2, s2[i], i))) {
      cup[j] = s2[i];
      i++;
      j++;
    } else
      i++;
  cup[j] = '\0';
  i = 0;
  while (cup[i] != '\0')
    if (!(ft_check_char(cup, cup[i], i))) {
      ft_putchar(cup[i]);
      i++;
    } else
      i++;
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<union2>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  movsbl (%rdi),%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  test   %sil,%sil
  je     L135
  cmpb   $0x0,(%r12)
  je     L135
  mov    %rdi,%r13
  xor    %ebx,%ebx
  xor    %ebp,%ebp
  nopl   (%rax)
L124:
  mov    %ebx,%edx
  mov    %r13,%rdi
  call   <ft_check_char@plt>
  test   %eax,%eax
  jne    L98
  movzbl 0x0(%r13,%rbx,1),%edx
  movslq %ebp,%rax
  add    $0x1,%rbx
  add    $0x1,%ebp
  mov    %dl,(%rsp,%rax,1)
  movsbl 0x0(%r13,%rbx,1),%esi
  test   %sil,%sil
  jne    L124
L80:
  movsbl (%r12),%esi
  test   %sil,%sil
  je     L119
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L104:
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <ft_check_char@plt>
  test   %eax,%eax
  jne    L156
  movzbl (%r12,%rbx,1),%edx
  add    $0x1,%rbx
  movslq %ebp,%rax
  add    $0x1,%ebp
  movsbl (%r12,%rbx,1),%esi
  mov    %dl,(%rsp,%rax,1)
  test   %sil,%sil
  jne    L104
L119:
  movslq %ebp,%rbp
  movb   $0x0,(%rsp,%rbp,1)
  movsbl (%rsp),%esi
  test   %sil,%sil
  je     L135
  xor    %ebx,%ebx
  mov    %rsp,%rbp
  nopw   0x0(%rax,%rax,1)
L85:
  mov    %ebx,%edx
  mov    %rbp,%rdi
  call   <ft_check_char@plt>
  test   %eax,%eax
  je     L116
  add    $0x1,%rbx
  movsbl 0x0(%rbp,%rbx,1),%esi
  test   %sil,%sil
  jne    L85
L135:
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L87
  add    $0x98,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L156:
  add    $0x1,%rbx
  movsbl (%r12,%rbx,1),%esi
  test   %sil,%sil
  jne    L104
  jmp    L119
  nopl   0x0(%rax)
L98:
  add    $0x1,%rbx
  movsbl 0x0(%r13,%rbx,1),%esi
  test   %sil,%sil
  jne    L124
  jmp    L80
L116:
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  call   <ft_putchar@plt>
  movsbl 0x0(%rbp,%rbx,1),%esi
  test   %sil,%sil
  jne    L85
  jmp    L135
L87:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int union2(char *s1, char *s2) {
  int i;
  char cup[127];
  int j;
  i = 0;
  j = 0;
  if (*s1 == '\0' || *s2 == '\0')
    return (0);
  while (s1[i] != '\0')
    if (!(ft_check_char(s1, s1[i], i))) {
      cup[j] = s1[i];
      i++;
      j++;
    } else
      i++;
  i = 0;
  while (s2[i] != '\0')
    if (!(ft_check_char(s2, s2[i], i))) {
      cup[j] = s2[i];
      i++;
      j++;
    } else
      i++;
  cup[j] = '\0';
  i = 0;
  while (cup[i] != '\0')
    if (!(ft_check_char(cup, cup[i], i))) {
      ft_putchar(cup[i]);
      i++;
    } else
      i++;
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ypit_itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x2c(%rbp),%eax
  cmp    $0xffff,%eax
  ja     L9
  movl   $0x1,-0x18(%rbp)
  mov    -0x30(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L12
L4:
  mov    -0x2c(%rbp),%eax
  shr    %eax
  imul   $0x6667,%eax,%eax
  shr    $0x11,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x2c(%rbp),%eax
  lea    0x30(%rax),%edx
  mov    -0x10(%rbp),%eax
  imul   $0xfffffff6,%eax,%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0xc(%rbp)
  je     L7
  mov    -0x18(%rbp),%eax
  or     %eax,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
L7:
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x2c(%rbp)
  shll   -0x18(%rbp)
  subq   $0x1,-0x8(%rbp)
L12:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jae    L4
L9:
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int ypit_itoa(char *p, int val, int ndigits) {
  int n, ch, bit, flags = 0;
  char *wp;
  if ((unsigned)val <= 65535) {
    bit = 1;
    wp = p + ndigits - 1;
    while (wp >= p) {
      n = ((unsigned)val >> 1) * 26215 >> 17;
      ch = '0' + val - n * 10;
      if (*wp != ch) {
        flags |= bit;
        *wp = ch;
      }
      val = n;
      bit <<= 1;
      wp--;
    }
  }
  return flags;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ypit_itoa>:
  endbr64
  mov    $0x0,%r9d
  cmp    $0xffff,%esi
  ja     L36
  movslq %edx,%rdx
  lea    -0x1(%rdi,%rdx,1),%rax
  cmp    %rax,%rdi
  ja     L57
  sub    $0x1,%rdi
  mov    $0x1,%edx
  jmp    L9
L18:
  add    %edx,%edx
  sub    $0x1,%rax
  cmp    %rax,%rdi
  je     L36
L9:
  mov    %esi,%r8d
  shr    %esi
  imul   $0x6667,%esi,%esi
  shr    $0x11,%esi
  imul   $0xfffffff6,%esi,%ecx
  lea    0x30(%r8,%rcx,1),%ecx
  movsbl (%rax),%r8d
  cmp    %ecx,%r8d
  je     L18
  or     %edx,%r9d
  mov    %cl,(%rax)
  jmp    L18
L57:
  mov    $0x0,%r9d
L36:
  mov    %r9d,%eax
  ret
```

assistant: ```c
int ypit_itoa(char *p, int val, int ndigits) {
  int n, ch, bit, flags = 0;
  char *wp;
  if ((unsigned)val <= 65535) {
    bit = 1;
    wp = p + ndigits - 1;
    while (wp >= p) {
      n = ((unsigned)val >> 1) * 26215 >> 17;
      ch = '0' + val - n * 10;
      if (*wp != ch) {
        flags |= bit;
        *wp = ch;
      }
      val = n;
      bit <<= 1;
      wp--;
    }
  }
  return flags;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ypit_itoa>:
  endbr64
  xor    %r9d,%r9d
  cmp    $0xffff,%esi
  ja     L9
  movslq %edx,%rdx
  lea    -0x1(%rdi,%rdx,1),%rax
  cmp    %rax,%rdi
  ja     L9
  sub    $0x1,%rdi
  mov    $0x1,%ecx
  nopl   (%rax)
L16:
  mov    %esi,%edx
  shr    %esi
  imul   $0x6667,%esi,%esi
  shr    $0x11,%esi
  imul   $0xfffffff6,%esi,%r8d
  lea    0x30(%rdx,%r8,1),%edx
  movsbl (%rax),%r8d
  cmp    %edx,%r8d
  je     L24
  mov    %dl,(%rax)
  or     %ecx,%r9d
L24:
  sub    $0x1,%rax
  add    %ecx,%ecx
  cmp    %rdi,%rax
  jne    L16
L9:
  mov    %r9d,%eax
  ret
```

assistant: ```c
int ypit_itoa(char *p, int val, int ndigits) {
  int n, ch, bit, flags = 0;
  char *wp;
  if ((unsigned)val <= 65535) {
    bit = 1;
    wp = p + ndigits - 1;
    while (wp >= p) {
      n = ((unsigned)val >> 1) * 26215 >> 17;
      ch = '0' + val - n * 10;
      if (*wp != ch) {
        flags |= bit;
        *wp = ch;
      }
      val = n;
      bit <<= 1;
      wp--;
    }
  }
  return flags;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ypit_itoa>:
  endbr64
  xor    %r9d,%r9d
  cmp    $0xffff,%esi
  ja     L11
  movslq %edx,%rdx
  lea    -0x1(%rdi,%rdx,1),%rax
  cmp    %rax,%rdi
  ja     L11
  sub    $0x1,%rdi
  mov    $0x1,%ecx
  nopl   (%rax)
L17:
  mov    %esi,%edx
  shr    %esi
  imul   $0x6667,%esi,%esi
  shr    $0x11,%esi
  imul   $0xfffffff6,%esi,%r8d
  lea    0x30(%rdx,%r8,1),%edx
  movsbl (%rax),%r8d
  cmp    %edx,%r8d
  je     L15
  mov    %dl,(%rax)
  or     %ecx,%r9d
L15:
  sub    $0x1,%rax
  add    %ecx,%ecx
  cmp    %rdi,%rax
  jne    L17
L11:
  mov    %r9d,%eax
  ret
```

assistant: ```c
int ypit_itoa(char *p, int val, int ndigits) {
  int n, ch, bit, flags = 0;
  char *wp;
  if ((unsigned)val <= 65535) {
    bit = 1;
    wp = p + ndigits - 1;
    while (wp >= p) {
      n = ((unsigned)val >> 1) * 26215 >> 17;
      ch = '0' + val - n * 10;
      if (*wp != ch) {
        flags |= bit;
        *wp = ch;
      }
      val = n;
      bit <<= 1;
      wp--;
    }
  }
  return flags;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<get_mem4Dint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    $0x8,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L9
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <no_mem_exit@plt>
L9:
  movl   $0x0,-0x4(%rbp)
  jmp    L28
L8:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    (%rdx,%rax,1),%rdi
  mov    -0x28(%rbp),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  call   <get_mem3Dint@plt>
  addl   $0x1,-0x4(%rbp)
L28:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L8
  mov    -0x1c(%rbp),%eax
  imul   -0x20(%rbp),%eax
  imul   -0x24(%rbp),%eax
  imul   -0x28(%rbp),%eax
  cltq
  shl    $0x2,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
datatype: string

tool: D5: get_mem4Dint: array4D

assistant: ```c
int get_mem4Dint(int *****array4D, int idx, int frames, int rows, int columns) {
  int j;
  if (((*array4D) = (int ****)calloc(idx, sizeof(int **))) == NULL)
    no_mem_exit("get_mem4Dint: array4D");
  for (j = 0; j < idx; j++)
    get_mem3Dint((*array4D) + j, frames, rows, columns);
  return idx * frames * rows * columns * sizeof(int);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_mem4Dint>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r14
  mov    %esi,%edi
  mov    %esi,0xc(%rsp)
  mov    %edx,%r13d
  mov    %ecx,%r12d
  mov    %r8d,%ebp
  mov    $0x8,%esi
  call   <calloc@plt>
  mov    %rax,(%r14)
  test   %rax,%rax
  je     L12
L4:
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L24
  mov    %eax,%r15d
  shl    $0x3,%r15
  mov    $0x0,%ebx
L7:
  mov    %rbx,%rdi
  add    (%r14),%rdi
  mov    %ebp,%ecx
  mov    %r12d,%edx
  mov    %r13d,%esi
  call   <get_mem3Dint@plt>
  add    $0x8,%rbx
  cmp    %r15,%rbx
  jne    L7
L24:
  mov    0xc(%rsp),%eax
  imul   %r13d,%eax
  imul   %r12d,%eax
  imul   %ebp,%eax
  shl    $0x2,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L12:
  lea    D5(%rip),%rdi
  call   <no_mem_exit@plt>
  jmp    L4
```

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

tool: D5: get_mem4Dint: array4D

assistant: ```c
int get_mem4Dint(int *****array4D, int idx, int frames, int rows, int columns) {
  int j;
  if (((*array4D) = (int ****)calloc(idx, sizeof(int **))) == NULL)
    no_mem_exit("get_mem4Dint: array4D");
  for (j = 0; j < idx; j++)
    get_mem3Dint((*array4D) + j, frames, rows, columns);
  return idx * frames * rows * columns * sizeof(int);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_mem4Dint>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  mov    %esi,%edi
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %r8d,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %esi,0xc(%rsp)
  mov    $0x8,%esi
  call   <calloc@plt>
  mov    %rax,(%r14)
  test   %rax,%rax
  je     L27
L21:
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L8
  movslq %eax,%r15
  xor    %ebx,%ebx
  shl    $0x3,%r15
  nopw   0x0(%rax,%rax,1)
L39:
  mov    (%r14),%rdi
  mov    %ebp,%ecx
  mov    %r12d,%edx
  mov    %r13d,%esi
  add    %rbx,%rdi
  add    $0x8,%rbx
  call   <get_mem3Dint@plt>
  cmp    %r15,%rbx
  jne    L39
L8:
  mov    0xc(%rsp),%eax
  add    $0x18,%rsp
  pop    %rbx
  imul   %r13d,%eax
  imul   %r12d,%eax
  imul   %ebp,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  shl    $0x2,%eax
  ret
  nopl   (%rax)
L27:
  lea    D1(%rip),%rdi
  call   <no_mem_exit@plt>
  jmp    L21
```

tool_calls: name: read_data
arguments: data_tag: D1
type: string

tool: D1: get_mem4Dint: array4D

assistant: ```c
int get_mem4Dint(int *****array4D, int idx, int frames, int rows, int columns) {
  int j;
  if (((*array4D) = (int ****)calloc(idx, sizeof(int **))) == NULL)
    no_mem_exit("get_mem4Dint: array4D");
  for (j = 0; j < idx; j++)
    get_mem3Dint((*array4D) + j, frames, rows, columns);
  return idx * frames * rows * columns * sizeof(int);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_mem4Dint>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  mov    %esi,%edi
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %r8d,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %esi,0xc(%rsp)
  mov    $0x8,%esi
  call   <calloc@plt>
  mov    %rax,(%r14)
  test   %rax,%rax
  je     L13
L2:
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L22
  movslq %eax,%r15
  xor    %ebx,%ebx
  shl    $0x3,%r15
  nopw   0x0(%rax,%rax,1)
L7:
  mov    (%r14),%rdi
  mov    %ebp,%ecx
  mov    %r12d,%edx
  mov    %r13d,%esi
  add    %rbx,%rdi
  add    $0x8,%rbx
  call   <get_mem3Dint@plt>
  cmp    %rbx,%r15
  jne    L7
L22:
  mov    0xc(%rsp),%eax
  add    $0x18,%rsp
  pop    %rbx
  imul   %r13d,%eax
  imul   %r12d,%eax
  imul   %ebp,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  shl    $0x2,%eax
  ret
  nopl   (%rax)
L13:
  lea    D4(%rip),%rdi
  call   <no_mem_exit@plt>
  jmp    L2
```

tool_calls: name: parse_data
arguments: data_tag: D4
datatype: string

tool: D4: get_mem4Dint: array4D

assistant: ```c
int get_mem4Dint(int *****array4D, int idx, int frames, int rows, int columns) {
  int j;
  if (((*array4D) = (int ****)calloc(idx, sizeof(int **))) == NULL)
    no_mem_exit("get_mem4Dint: array4D");
  for (j = 0; j < idx; j++)
    get_mem3Dint((*array4D) + j, frames, rows, columns);
  return idx * frames * rows * columns * sizeof(int);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<f2m_rand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L9
L18:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,(%rbx)
  addl   $0x1,-0x14(%rbp)
L9:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jl     L18
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%esi
  mov    -0x28(%rbp),%eax
  cltd
  shr    $0x1b,%edx
  add    %edx,%eax
  and    $0x1f,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  mov    $0x20,%eax
  sub    %edx,%eax
  mov    $0xffffffff,%edx
  mov    %eax,%ecx
  shr    %cl,%edx
  mov    %edx,%eax
  and    %eax,%esi
  mov    %esi,%edx
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void f2m_rand(uint32_t t, uint32_t m, uint32_t *A) {
  uint32_t i;
  for (i = 0; i < t - 1; i++)
    A[i] = rand();
  A[t - 1] = rand() & (0xFFFFFFFF >> (32 - m % 32));
}
```