ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<write_all>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  jle    L6
  mov    %edi,%ebp
  mov    %rsi,%r13
  mov    %rdx,%rbx
  xor    %r12d,%r12d
  jmp    L24
L14:
  add    %rax,%r12
  cmp    %r12,%rbx
  jle    L7
L24:
  mov    %rbx,%rdx
  lea    0x0(%r13,%r12,1),%rsi
  mov    %ebp,%edi
  sub    %r12,%rdx
  call   <write@plt>
  test   %rax,%rax
  jg     L14
L7:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  xor    %r12d,%r12d
  jmp    L7
```

assistant: ```c
ssize_t write_all(int socket, void *buffer, ssize_t size) {
  ssize_t i = 0, res = 0;
  while (i < size) {
    res = write(socket, buffer + i, size - i);
    if (res < 1)
      break;
    i += res;
  }
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<write_all>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  jle    L15
  mov    %edi,%ebp
  mov    %rsi,%r13
  mov    %rdx,%rbx
  xor    %r12d,%r12d
  jmp    L22
L36:
  add    %rax,%r12
  cmp    %r12,%rbx
  jle    L47
L22:
  mov    %rbx,%rdx
  lea    0x0(%r13,%r12,1),%rsi
  mov    %ebp,%edi
  sub    %r12,%rdx
  call   <write@plt>
  test   %rax,%rax
  jg     L36
L47:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  xor    %r12d,%r12d
  jmp    L47
```

assistant: ```c
ssize_t write_all(int socket, void *buffer, ssize_t size) {
  ssize_t i = 0, res = 0;
  while (i < size) {
    res = write(socket, buffer + i, size - i);
    if (res < 1)
      break;
    i += res;
  }
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<order_dbl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %ecx,-0x38(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L13
L78:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x14(%rbp)
  addl   $0x1,-0xc(%rbp)
L13:
  mov    -0x34(%rbp),%edx
  mov    -0x38(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x14(%rbp)
  jl     L78
  movl   $0x1,-0x14(%rbp)
  jmp    L56
L70:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L58
L50:
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  comisd %xmm1,%xmm0
  ja     L25
  jmp    L66
L25:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  subl   $0x1,-0x10(%rbp)
L58:
  cmpl   $0x0,-0x10(%rbp)
  jns    L50
L66:
  addl   $0x1,-0x14(%rbp)
L56:
  mov    -0x14(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L70
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void order_dbl(double unord_arr[], double ord_arr[], int start, int num_elems) {
  int i, j, k;
  double temp;
  for (i = start, k = 0; i < start + num_elems; i++, k++)
    ord_arr[k] = unord_arr[i];
  for (i = 1; i < num_elems; i++) {
    for (j = i - 1; j >= 0; j--)
      if (ord_arr[j + 1] < ord_arr[j]) {
        temp = ord_arr[j];
        ord_arr[j] = ord_arr[j + 1];
        ord_arr[j + 1] = temp;
      } else
        break;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<order_dbl>:
  endbr64
  lea    (%rdx,%rcx,1),%r8d
  cmp    %r8d,%edx
  jge    L76
  movslq %edx,%rax
  mov    %rax,%rdx
  neg    %rdx
  lea    (%rsi,%rdx,8),%rdx
L68:
  movsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %eax,%r8d
  jg     L68
L76:
  cmp    $0x1,%ecx
  jle    L56
  add    $0x8,%rsi
  lea    -0x1(%rcx),%r8d
  mov    $0x0,%ecx
  jmp    L79
L12:
  add    $0x1,%rcx
  add    $0x8,%rsi
  cmp    %r8,%rcx
  je     L56
L79:
  test   %ecx,%ecx
  js     L12
  lea    -0x8(%rsi),%rax
  lea    -0x10(%rsi),%rdx
  lea    0x0(,%rcx,8),%rdi
  sub    %rdi,%rdx
L89:
  movsd  0x8(%rax),%xmm1
  movsd  (%rax),%xmm0
  comisd %xmm1,%xmm0
  jbe    L12
  movsd  %xmm1,(%rax)
  movsd  %xmm0,0x8(%rax)
  sub    $0x8,%rax
  cmp    %rdx,%rax
  jne    L89
  jmp    L12
L56:
  ret
```

assistant: ```c
void order_dbl(double unord_arr[], double ord_arr[], int start, int num_elems) {
  int i, j, k;
  double temp;
  for (i = start, k = 0; i < start + num_elems; i++, k++)
    ord_arr[k] = unord_arr[i];
  for (i = 1; i < num_elems; i++) {
    for (j = i - 1; j >= 0; j--)
      if (ord_arr[j + 1] < ord_arr[j]) {
        temp = ord_arr[j];
        ord_arr[j] = ord_arr[j + 1];
        ord_arr[j + 1] = temp;
      } else
        break;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<order_dbl>:
  endbr64
  lea    (%rdx,%rcx,1),%r8d
  cmp    %r8d,%edx
  jge    L36
  movslq %edx,%rax
  mov    %rax,%rdx
  neg    %rdx
  lea    (%rsi,%rdx,8),%rdx
  nopw   0x0(%rax,%rax,1)
L32:
  movsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %eax,%r8d
  jg     L32
L36:
  cmp    $0x1,%ecx
  jle    L40
  lea    -0x1(%rcx),%r8d
  add    $0x8,%rsi
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L46:
  lea    -0x10(%rsi),%rdx
  lea    0x0(,%rcx,8),%rdi
  lea    -0x8(%rsi),%rax
  sub    %rdi,%rdx
  nopl   0x0(%rax,%rax,1)
L31:
  movsd  0x8(%rax),%xmm1
  movsd  (%rax),%xmm0
  comisd %xmm1,%xmm0
  jbe    L3
  movsd  %xmm1,(%rax)
  sub    $0x8,%rax
  movsd  %xmm0,0x10(%rax)
  cmp    %rdx,%rax
  jne    L31
L3:
  add    $0x1,%rcx
  add    $0x8,%rsi
  cmp    %r8,%rcx
  jne    L46
L40:
  ret
```

assistant: ```c
void order_dbl(double unord_arr[], double ord_arr[], int start, int num_elems) {
  int i, j, k;
  double temp;
  for (i = start, k = 0; i < start + num_elems; i++, k++)
    ord_arr[k] = unord_arr[i];
  for (i = 1; i < num_elems; i++) {
    for (j = i - 1; j >= 0; j--)
      if (ord_arr[j + 1] < ord_arr[j]) {
        temp = ord_arr[j];
        ord_arr[j] = ord_arr[j + 1];
        ord_arr[j + 1] = temp;
      } else
        break;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<order_dbl>:
  endbr64
  lea    (%rdx,%rcx,1),%r9d
  mov    %rsi,%rax
  mov    %ecx,%r8d
  mov    %edx,%esi
  cmp    %r9d,%edx
  jge    L30
  movslq %edx,%rcx
  lea    0x1(%rcx),%rdx
  lea    0x0(,%rdx,8),%r10
  lea    (%rdi,%r10,1),%r11
  cmp    %r11,%rax
  je     L79
  lea    -0x1(%r8),%r11d
  cmp    $0x2,%r11d
  jbe    L79
  mov    %r8d,%ecx
  lea    -0x8(%rdi,%r10,1),%r9
  xor    %edx,%edx
  shr    %ecx
  shl    $0x4,%rcx
  nop
L56:
  movupd (%r9,%rdx,1),%xmm4
  movups %xmm4,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %rcx,%rdx
  jne    L56
  mov    %r8d,%ecx
  and    $0xfffffffe,%ecx
  lea    (%rsi,%rcx,1),%edx
  cmp    %ecx,%r8d
  je     L30
  movslq %edx,%rdx
  movsd  (%rdi,%rdx,8),%xmm0
  movslq %ecx,%rdx
  movsd  %xmm0,(%rax,%rdx,8)
L30:
  cmp    $0x1,%r8d
  jle    L116
  lea    0x8(%rax),%rcx
  sub    $0x1,%r8d
  xor    %esi,%esi
  nopl   0x0(%rax)
L78:
  lea    -0x10(%rcx),%rdx
  lea    0x0(,%rsi,8),%rdi
  lea    -0x8(%rcx),%rax
  sub    %rdi,%rdx
  nopl   0x0(%rax,%rax,1)
L109:
  movupd (%rax),%xmm0
  movapd %xmm0,%xmm3
  movapd %xmm0,%xmm2
  unpckhpd %xmm3,%xmm3
  shufpd $0x1,%xmm0,%xmm2
  comisd %xmm3,%xmm0
  jbe    L74
  movups %xmm2,(%rax)
  sub    $0x8,%rax
  cmp    %rax,%rdx
  jne    L109
L74:
  add    $0x1,%rsi
  add    $0x8,%rcx
  cmp    %r8,%rsi
  jne    L78
L116:
  ret
L79:
  mov    %rcx,%rsi
  neg    %rsi
  lea    (%rax,%rsi,8),%rsi
  jmp    L10
  nopl   0x0(%rax)
L33:
  add    $0x1,%rdx
L10:
  movsd  (%rdi,%rcx,8),%xmm0
  movsd  %xmm0,(%rsi,%rcx,8)
  mov    %rdx,%rcx
  cmp    %edx,%r9d
  jg     L33
  jmp    L30
```

assistant: ```c
void order_dbl(double unord_arr[], double ord_arr[], int start, int num_elems) {
  int i, j, k;
  double temp;
  for (i = start, k = 0; i < start + num_elems; i++, k++)
    ord_arr[k] = unord_arr[i];
  for (i = 1; i < num_elems; i++) {
    for (j = i - 1; j >= 0; j--)
      if (ord_arr[j + 1] < ord_arr[j]) {
        temp = ord_arr[j];
        ord_arr[j] = ord_arr[j + 1];
        ord_arr[j + 1] = temp;
      } else
        break;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<flush>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  movq   $0x0,-0x8(%rbp)
  jmp    L0
L15:
  mov    -0x28(%rbp),%rax
  sub    -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  add    %rax,%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0xffffffff,-0xc(%rbp)
  jne    L23
  mov    $0xffffffff,%eax
  jmp    L13
L23:
  mov    -0xc(%rbp),%eax
  cltq
  add    %rax,-0x8(%rbp)
L0:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L15
  mov    $0x0,%eax
L13:
  leave
  ret
```

assistant: ```c
int flush(int s, char *msg, size_t len) {
  msg[len] = 0;
  size_t sent = 0;
  while (sent < len) {
    int chars = write(s, msg + sent, len - sent);
    if (chars == -1) {
      return (-1);
    }
    sent += chars;
  }
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<flush>:
  endbr64
  movb   $0x0,(%rsi,%rdx,1)
  test   %rdx,%rdx
  je     L1
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    $0x0,%ebx
L10:
  mov    %rbp,%rdx
  sub    %rbx,%rdx
  lea    (%r12,%rbx,1),%rsi
  mov    %r13d,%edi
  call   <write@plt>
  cmp    $0xffffffff,%eax
  je     L18
  cltq
  add    %rax,%rbx
  cmp    %rbx,%rbp
  ja     L10
  mov    $0x0,%eax
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int flush(int s, char *msg, size_t len) {
  msg[len] = 0;
  size_t sent = 0;
  while (sent < len) {
    int chars = write(s, msg + sent, len - sent);
    if (chars == -1) {
      return (-1);
    }
    sent += chars;
  }
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<flush>:
  endbr64
  movb   $0x0,(%rsi,%rdx,1)
  test   %rdx,%rdx
  je     L37
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  jmp    L8
  nopl   (%rax)
L6:
  cltq
  add    %rax,%r13
  cmp    %r13,%rbx
  jbe    L34
L8:
  mov    %rbx,%rdx
  lea    0x0(%rbp,%r13,1),%rsi
  mov    %r12d,%edi
  sub    %r13,%rdx
  call   <write@plt>
  cmp    $0xffffffff,%eax
  jne    L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L34:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L37:
  xor    %eax,%eax
  ret
```

assistant: ```c
int flush(int s, char *msg, size_t len) {
  msg[len] = 0;
  size_t sent = 0;
  while (sent < len) {
    int chars = write(s, msg + sent, len - sent);
    if (chars == -1) {
      return (-1);
    }
    sent += chars;
  }
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<flush>:
  endbr64
  movb   $0x0,(%rsi,%rdx,1)
  test   %rdx,%rdx
  je     L15
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  jmp    L38
  nopl   (%rax)
L9:
  cltq
  add    %rax,%r13
  cmp    %r13,%rbx
  jbe    L3
L38:
  mov    %rbx,%rdx
  lea    0x0(%rbp,%r13,1),%rsi
  mov    %r12d,%edi
  sub    %r13,%rdx
  call   <write@plt>
  cmp    $0xffffffff,%eax
  jne    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L3:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  xor    %eax,%eax
  ret
```

assistant: ```c
int flush(int s, char *msg, size_t len) {
  msg[len] = 0;
  size_t sent = 0;
  while (sent < len) {
    int chars = write(s, msg + sent, len - sent);
    if (chars == -1) {
      return (-1);
    }
    sent += chars;
  }
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<banbks>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %ecx,%eax
  mov    0x28(%rsp),%rcx
  lea    0x1(%rdx,%rax,1),%ebp
  test   %rsi,%rsi
  je     L4
  mov    %r9,%r13
  movslq %edx,%r10
  lea    0x4(%rcx),%r11
  lea    0x8(%r8),%r9
  mov    $0x1,%ebx
  jmp    L81
L26:
  cmp    %rsi,%r10
  adc    $0x0,%r10
  lea    0x1(%rbx),%r12
  cmp    %r12,%r10
  jb     L29
  mov    %r12,%rax
  mov    $0x4,%edx
L44:
  mov    (%r9),%r8
  movss  (%r8,%rdx,1),%xmm1
  mulss  (%r11),%xmm1
  movss  (%rcx,%rax,4),%xmm0
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rcx,%rax,4)
  add    $0x1,%rax
  add    $0x4,%rdx
  cmp    %rax,%r10
  jae    L44
L29:
  add    $0x4,%r11
  add    $0x8,%r9
  cmp    %rbx,%rsi
  je     L5
  mov    %r12,%rbx
L81:
  mov    0x0(%r13,%rbx,8),%rax
  cmp    %rbx,%rax
  je     L26
  movss  (%r11),%xmm0
  lea    (%rcx,%rax,4),%rax
  movss  (%rax),%xmm1
  movss  %xmm1,(%r11)
  movss  %xmm0,(%rax)
  jmp    L26
L5:
  lea    (%rcx,%rsi,4),%r10
  mov    $0x1,%ecx
  movslq %ebp,%rbp
L24:
  mov    %r10,%r8
  movss  (%r10),%xmm1
  cmp    $0x1,%rcx
  jbe    L56
  mov    (%rdi,%rsi,8),%r9
  mov    $0x2,%eax
L67:
  movss  (%r9,%rax,4),%xmm0
  mulss  -0x4(%r8,%rax,4),%xmm0
  subss  %xmm0,%xmm1
  mov    %rax,%rdx
  add    $0x1,%rax
  cmp    %rcx,%rdx
  jne    L67
L56:
  mov    (%rdi,%rsi,8),%rax
  divss  0x4(%rax),%xmm1
  movss  %xmm1,(%r8)
  cmp    %rbp,%rcx
  adc    $0x0,%rcx
  sub    $0x4,%r10
  sub    $0x1,%rsi
  jne    L24
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void banbks(float **a, unsigned long n, int m1, int m2, float **al, unsigned long indx[], float b[]) {
  unsigned long i, k, l;
  int mm;
  float dum;
  mm = m1 + m2 + 1;
  l = m1;
  for (k = 1; k <= n; k++) {
    i = indx[k];
    if (i != k) {
      dum = (b[k]);
      (b[k]) = (b[i]);
      (b[i]) = dum;
    }
    if (l < n)
      l++;
    for (i = k + 1; i <= l; i++)
      b[i] -= al[k][i - k] * b[k];
  }
  l = 1;
  for (i = n; i >= 1; i--) {
    dum = b[i];
    for (k = 2; k <= l; k++)
      dum -= a[i][k] * b[k + i - 1];
    b[i] = dum / a[i][1];
    if (l < mm)
      l++;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<banbks>:
  endbr64
  mov    %ecx,%r10d
  push   %r13
  lea    0x1(%rdx,%r10,1),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x28(%rsp),%rcx
  test   %rsi,%rsi
  je     L21
  mov    %r8,%r12
  mov    %r9,%rbp
  movslq %edx,%rdx
  mov    $0xfffffffffffffffc,%rbx
  lea    0x4(%rcx),%r10
  mov    $0x1,%r8d
  nopw   0x0(%rax,%rax,1)
L65:
  mov    0x0(%rbp,%r8,8),%rax
  cmp    %r8,%rax
  je     L6
  lea    (%rcx,%rax,4),%rax
  movss  (%r10),%xmm0
  movss  (%rax),%xmm1
  movss  %xmm1,(%r10)
  movss  %xmm0,(%rax)
L6:
  cmp    %rsi,%rdx
  lea    0x1(%r8),%r9
  adc    $0x0,%rdx
  cmp    %r9,%rdx
  jb     L8
  mov    -0x8(%r12,%r9,8),%r11
  mov    %r9,%rax
  add    %rbx,%r11
  nopl   0x0(%rax,%rax,1)
L81:
  movss  (%r11,%rax,4),%xmm1
  mulss  (%r10),%xmm1
  movss  (%rcx,%rax,4),%xmm0
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jae    L81
L8:
  sub    $0x4,%rbx
  add    $0x4,%r10
  cmp    %r8,%rsi
  je     L84
  mov    %r9,%r8
  jmp    L65
  nopl   0x0(%rax)
L84:
  lea    (%rcx,%rsi,4),%r8
  movslq %r13d,%r13
  mov    $0x1,%ecx
  nopl   0x0(%rax)
L77:
  movss  (%r8),%xmm1
  mov    (%rdi,%rsi,8),%r9
  mov    $0x2,%eax
  cmp    $0x1,%rcx
  jbe    L7
  nopl   0x0(%rax)
L83:
  movss  (%r9,%rax,4),%xmm0
  mulss  -0x4(%r8,%rax,4),%xmm0
  mov    %rax,%rdx
  add    $0x1,%rax
  subss  %xmm0,%xmm1
  cmp    %rcx,%rdx
  jne    L83
L7:
  cmp    %r13,%rcx
  divss  0x4(%r9),%xmm1
  movss  %xmm1,(%r8)
  adc    $0x0,%rcx
  sub    $0x4,%r8
  sub    $0x1,%rsi
  jne    L77
L21:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void banbks(float **a, unsigned long n, int m1, int m2, float **al, unsigned long indx[], float b[]) {
  unsigned long i, k, l;
  int mm;
  float dum;
  mm = m1 + m2 + 1;
  l = m1;
  for (k = 1; k <= n; k++) {
    i = indx[k];
    if (i != k) {
      dum = (b[k]);
      (b[k]) = (b[i]);
      (b[i]) = dum;
    }
    if (l < n)
      l++;
    for (i = k + 1; i <= l; i++)
      b[i] -= al[k][i - k] * b[k];
  }
  l = 1;
  for (i = n; i >= 1; i--) {
    dum = b[i];
    for (k = 2; k <= l; k++)
      dum -= a[i][k] * b[k + i - 1];
    b[i] = dum / a[i][1];
    if (l < mm)
      l++;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<max1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x8(%rbp)
  movsd  %xmm1,-0x10(%rbp)
  movsd  %xmm2,-0x18(%rbp)
  movsd  -0x10(%rbp),%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L8
  movsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
L8:
  movsd  -0x18(%rbp),%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L19
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
L19:
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double max1(double a, double b, double c) {
  if (a < b)
    a = b;
  if (a < c)
    a = c;
  return a;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<max1>:
  endbr64
  maxsd  %xmm0,%xmm1
  maxsd  %xmm1,%xmm2
  movapd %xmm2,%xmm0
  ret
```

assistant: ```c
double max1(double a, double b, double c) {
  if (a < b)
    a = b;
  if (a < c)
    a = c;
  return a;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<max1>:
  endbr64
  maxsd  %xmm0,%xmm1
  maxsd  %xmm1,%xmm2
  movapd %xmm2,%xmm0
  ret
```

assistant: ```c
double max1(double a, double b, double c) {
  if (a < b)
    a = b;
  if (a < c)
    a = c;
  return a;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<max1>:
  endbr64
  maxsd  %xmm0,%xmm1
  maxsd  %xmm1,%xmm2
  movapd %xmm2,%xmm0
  ret
```

assistant: ```c
double max1(double a, double b, double c) {
  if (a < b)
    a = b;
  if (a < c)
    a = c;
  return a;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L50
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L29
L50:
  movl   $0xa,-0x4(%rbp)
L29:
  mov    $0x0,%edi
  call   <srand@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L28
L4:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0xffffffff8fb823ef,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x72,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  add    $0xe,%eax
  mov    %al,-0x9(%rbp)
  movsbl -0x9(%rbp),%eax
  cmp    $0x5f,%eax
  je     L99
  cmp    $0x5f,%eax
  jg     L45
  cmp    $0x3c,%eax
  je     L99
  cmp    $0x3c,%eax
  jg     L45
  test   %eax,%eax
  je     L99
  test   %eax,%eax
  js     L45
  sub    $0x13,%eax
  cmp    $0xc,%eax
  ja     L45
L99:
  subl   $0x1,-0x8(%rbp)
  jmp    L48
L45:
  movsbl -0x9(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
L48:
  addl   $0x1,-0x8(%rbp)
L28:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L4
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D1
data_type: string

tool: D1: 


assistant: ```c
int main(int argc, char **argv) {
  int i, j, len;
  char c;
  if (argv[1])
    len = atoi(argv[1]);
  else
    len = 10;
  srand((int)(time)(NULL));
  for (i = 0; i < len; i++) {
    c = rand() % (0x80 - 0x0E) + 0x0E;
    switch (c) {
    case 0x3C:
    case 0x5F:
    case 0x13:
    case 0x14:
    case 0x15:
    case 0x16:
    case 0x17:
    case 0x18:
    case 0x19:
    case 0x1A:
    case 0x1B:
    case 0x1C:
    case 0x1D:
    case 0x1E:
    case 0x1F:
    case 0x00:
      i--;
      break;
    default:
      putchar(c);
    }
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdi
  test   %rdi,%rdi
  je     L98
  call   <atoi@plt>
  mov    %eax,%ebp
  mov    $0x0,%edi
  call   <srand@plt>
  test   %ebp,%ebp
  jg     L44
L12:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L62:
  test   %dil,%dil
  je     L63
  sub    $0x5,%eax
  cmp    $0xc,%al
  ja     L59
L63:
  sub    $0x1,%ebx
L1:
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jge    L12
L20:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0xffffffff8fb823ef,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x72,%edx,%edx
  sub    %edx,%eax
  lea    0xe(%rax),%edi
  cmp    $0x2e,%al
  je     L63
  cmp    $0x3c,%dil
  jbe    L62
  cmp    $0x5f,%dil
  je     L63
L59:
  movsbl %dil,%edi
  call   <putchar@plt>
  jmp    L1
L98:
  mov    $0x0,%edi
  call   <srand@plt>
  mov    $0xa,%ebp
L44:
  mov    $0x0,%ebx
  jmp    L20
```

tool_calls: name: disassemble_data
arguments: label: D8
type_guessed: string

tool: D8: 


assistant: ```c
int main(int argc, char **argv) {
  int i, j, len;
  char c;
  if (argv[1])
    len = atoi(argv[1]);
  else
    len = 10;
  srand((int)(time)(NULL));
  for (i = 0; i < len; i++) {
    c = rand() % (0x80 - 0x0E) + 0x0E;
    switch (c) {
    case 0x3C:
    case 0x5F:
    case 0x13:
    case 0x14:
    case 0x15:
    case 0x16:
    case 0x17:
    case 0x18:
    case 0x19:
    case 0x1A:
    case 0x1B:
    case 0x1C:
    case 0x1D:
    case 0x1E:
    case 0x1F:
    case 0x00:
      i--;
      break;
    default:
      putchar(c);
    }
  }
  printf("\n");
}
```