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
<main>:
  endbr64
  push   %rbx
  mov    D6(%rip),%rbx
  nopl   0x0(%rax)
L31:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %eax,(%rbx)
  je     L49
L68:
  cmp    $0x9,%eax
  je     L32
  cmp    $0x8,%eax
  je     L48
  cmp    $0x5c,%eax
  je     L7
  mov    %eax,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %eax,(%rbx)
  jne    L68
L49:
  xor    %eax,%eax
  pop    %rbx
  ret
L7:
  mov    $0x5c,%edi
  call   <putchar@plt>
  mov    $0x5c,%edi
  call   <putchar@plt>
  jmp    L31
  cs nopw 0x0(%rax,%rax,1)
L32:
  mov    $0x5c,%edi
  call   <putchar@plt>
  mov    $0x74,%edi
  call   <putchar@plt>
  jmp    L31
  cs nopw 0x0(%rax,%rax,1)
L48:
  mov    $0x5c,%edi
  call   <putchar@plt>
  mov    $0x62,%edi
  call   <putchar@plt>
  jmp    L31
```

assistant: ```c
main() {
  int c = 0;
  while ((c = getchar()) != EOF) {
    if (c == '\t') {
      putchar('\\');
      putchar('t');
    } else if (c == '\b') {
      putchar('\\');
      putchar('b');
    } else if (c == '\\') {
      putchar('\\');
      putchar('\\');
    } else {
      putchar(c);
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D5(%rip),%rbx
  nopl   0x0(%rax)
L35:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %eax,(%rbx)
  je     L61
L30:
  cmp    $0x9,%eax
  je     L29
  cmp    $0x8,%eax
  je     L46
  cmp    $0x5c,%eax
  je     L74
  mov    %eax,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %eax,(%rbx)
  jne    L30
L61:
  xor    %eax,%eax
  pop    %rbx
  ret
L74:
  mov    $0x5c,%edi
  call   <putchar@plt>
  mov    $0x5c,%edi
  call   <putchar@plt>
  jmp    L35
  cs nopw 0x0(%rax,%rax,1)
L29:
  mov    $0x5c,%edi
  call   <putchar@plt>
  mov    $0x74,%edi
  call   <putchar@plt>
  jmp    L35
  cs nopw 0x0(%rax,%rax,1)
L46:
  mov    $0x5c,%edi
  call   <putchar@plt>
  mov    $0x62,%edi
  call   <putchar@plt>
  jmp    L35
```

assistant: ```c
main() {
  int c = 0;
  while ((c = getchar()) != EOF) {
    if (c == '\t') {
      putchar('\\');
      putchar('t');
    } else if (c == '\b') {
      putchar('\\');
      putchar('b');
    } else if (c == '\\') {
      putchar('\\');
      putchar('\\');
    } else {
      putchar(c);
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<obterDivisor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x9,-0x4(%rbp)
  jg     L15
  mov    $0xa,%eax
  jmp    L5
L15:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,%edi
  call   <obterDivisor>
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
L5:
  leave
  ret
```

assistant: ```c
int obterDivisor(int n) {
  if (n < 10) {
    return 10;
  } else {
    return 10 * obterDivisor(n / 10);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<obterDivisor>:
  endbr64
  mov    $0xa,%eax
  cmp    $0x9,%edi
  jg     L9
  ret
L9:
  sub    $0x8,%rsp
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edi
  call   <obterDivisor>
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int obterDivisor(int n) {
  if (n < 10) {
    return 10;
  } else {
    return 10 * obterDivisor(n / 10);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<obterDivisor>:
  endbr64
  mov    $0x1,%edx
  mov    $0xcccccccd,%esi
  cmp    $0x9,%edi
  jle    L19
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %edi,%edi
  mov    %edx,%eax
  lea    (%rdx,%rdx,4),%edx
  mov    %rdi,%rcx
  imul   %rsi,%rdi
  add    %edx,%edx
  shr    $0x23,%rdi
  cmp    $0x63,%ecx
  jg     L11
  imul   $0x64,%eax,%eax
  ret
  nopl   (%rax)
L19:
  mov    $0xa,%eax
  ret
```

assistant: ```c
int obterDivisor(int n) {
  if (n < 10) {
    return 10;
  } else {
    return 10 * obterDivisor(n / 10);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<obterDivisor>:
  endbr64
  mov    $0x1,%edx
  mov    $0xcccccccd,%esi
  cmp    $0x9,%edi
  jle    L4
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %edi,%edi
  mov    %edx,%eax
  lea    (%rdx,%rdx,4),%edx
  mov    %rdi,%rcx
  imul   %rsi,%rdi
  add    %edx,%edx
  shr    $0x23,%rdi
  cmp    $0x63,%ecx
  jg     L0
  imul   $0x64,%eax,%eax
  ret
  nopl   (%rax)
L4:
  mov    $0xa,%eax
  ret
```

assistant: ```c
int obterDivisor(int n) {
  if (n < 10) {
    return 10;
  } else {
    return 10 * obterDivisor(n / 10);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<epp2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  movsd  %xmm0,-0x40(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L21
L51:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L21:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L51
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  movsd  -0x40(%rbp),%xmm0
  comisd -0x10(%rbp),%xmm0
  jb     L4
  movl   $0x0,-0x14(%rbp)
  jmp    L42
L46:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x14(%rbp)
L42:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L46
  jmp    L16
L4:
  movsd  -0x10(%rbp),%xmm0
  subsd  -0x40(%rbp),%xmm0
  divsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L35
L0:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mulsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x14(%rbp)
L35:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L0
L16:
  nop
  leave
  ret
```

assistant: ```c
void epp2(double *x, double *v, int n, double rho) {
  int i;
  double v2 = 0, ratio;
  for (i = 0; i < n; i++) {
    v2 += v[i] * v[i];
  }
  v2 = sqrt(v2);
  if (rho >= v2)
    for (i = 0; i < n; i++)
      x[i] = 0;
  else {
    ratio = (v2 - rho) / v2;
    for (i = 0; i < n; i++)
      x[i] = v[i] * ratio;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<epp2>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %edx,%ebx
  movq   %xmm0,%r13
  test   %edx,%edx
  jle    L37
  mov    %rsi,%rax
  lea    -0x1(%rdx),%edx
  lea    0x8(%rsi,%rdx,8),%rdx
  pxor   %xmm0,%xmm0
L104:
  movsd  (%rax),%xmm1
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L104
  pxor   %xmm1,%xmm1
  ucomisd %xmm0,%xmm1
  ja     L86
L59:
  sqrtsd %xmm0,%xmm0
  movq   %r13,%xmm2
  comisd %xmm0,%xmm2
  jae    L66
L11:
  movapd %xmm0,%xmm1
  movq   %r13,%xmm3
  subsd  %xmm3,%xmm1
  divsd  %xmm0,%xmm1
  test   %ebx,%ebx
  jle    L92
  mov    %ebx,%ebx
  mov    $0x0,%eax
L74:
  movapd %xmm1,%xmm0
  mulsd  (%r12,%rax,8),%xmm0
  movsd  %xmm0,0x0(%rbp,%rax,8)
  add    $0x1,%rax
  cmp    %rbx,%rax
  jne    L74
  jmp    L92
L37:
  pxor   %xmm0,%xmm0
  jmp    L59
L86:
  call   <sqrt@plt>
  movq   %r13,%xmm4
  comisd %xmm0,%xmm4
  jae    L84
  jmp    L11
L66:
  test   %ebx,%ebx
  jle    L92
L84:
  mov    %rbp,%rax
  lea    -0x1(%rbx),%edx
  lea    0x8(%rbp,%rdx,8),%rdx
L88:
  movq   $0x0,(%rax)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L88
L92:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void epp2(double *x, double *v, int n, double rho) {
  int i;
  double v2 = 0, ratio;
  for (i = 0; i < n; i++) {
    v2 += v[i] * v[i];
  }
  v2 = sqrt(v2);
  if (rho >= v2)
    for (i = 0; i < n; i++)
      x[i] = 0;
  else {
    ratio = (v2 - rho) / v2;
    for (i = 0; i < n; i++)
      x[i] = v[i] * ratio;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<epp2>:
  endbr64
  test   %edx,%edx
  jle    L68
  lea    -0x1(%rdx),%ecx
  pxor   %xmm3,%xmm3
  movapd %xmm0,%xmm2
  sub    $0x28,%rsp
  mov    %rsi,%rax
  lea    0x8(%rsi,%rcx,8),%rcx
  movapd %xmm3,%xmm0
  nopw   0x0(%rax,%rax,1)
L21:
  movsd  (%rax),%xmm1
  add    $0x8,%rax
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rcx,%rax
  jne    L21
  ucomisd %xmm0,%xmm3
  ja     L4
  sqrtsd %xmm0,%xmm0
  comisd %xmm0,%xmm2
  jae    L74
  movapd %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  divsd  %xmm0,%xmm1
L30:
  movslq %edx,%rdx
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L70:
  movsd  (%rsi,%rax,8),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdi,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L70
L11:
  add    $0x28,%rsp
  ret
  nopl   0x0(%rax)
L74:
  test   %edx,%edx
  jle    L11
L66:
  mov    %edx,%edx
  xor    %esi,%esi
  add    $0x28,%rsp
  shl    $0x3,%rdx
  jmp    1060 <memset@plt>
  nopl   (%rax)
L68:
  ret
L4:
  mov    %edx,0x14(%rsp)
  mov    %rdi,(%rsp)
  movsd  %xmm2,0x18(%rsp)
  mov    %rsi,0x8(%rsp)
  call   <sqrt@plt>
  movsd  0x18(%rsp),%xmm2
  mov    (%rsp),%rdi
  mov    0x14(%rsp),%edx
  comisd %xmm0,%xmm2
  jae    L66
  movapd %xmm0,%xmm1
  mov    0x8(%rsp),%rsi
  subsd  %xmm2,%xmm1
  divsd  %xmm0,%xmm1
  jmp    L30
```

assistant: ```c
void epp2(double *x, double *v, int n, double rho) {
  int i;
  double v2 = 0, ratio;
  for (i = 0; i < n; i++) {
    v2 += v[i] * v[i];
  }
  v2 = sqrt(v2);
  if (rho >= v2)
    for (i = 0; i < n; i++)
      x[i] = 0;
  else {
    ratio = (v2 - rho) / v2;
    for (i = 0; i < n; i++)
      x[i] = v[i] * ratio;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<epp2>:
  endbr64
  push   %rbp
  mov    %edx,%ebp
  sub    $0x20,%rsp
  test   %ebp,%ebp
  jle    L93
  movapd %xmm0,%xmm3
  cmp    $0x1,%ebp
  je     L50
  mov    %ebp,%edx
  pxor   %xmm4,%xmm4
  mov    %rsi,%rax
  shr    %edx
  movapd %xmm4,%xmm0
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rsi,%rdx,1),%rdx
  nopl   0x0(%rax,%rax,1)
L17:
  movupd (%rax),%xmm1
  add    $0x10,%rax
  mulpd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  unpckhpd %xmm1,%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rdx,%rax
  jne    L17
  mov    %ebp,%eax
  and    $0xfffffffe,%eax
  test   $0x1,%bpl
  je     L157
L142:
  cltq
  movsd  (%rsi,%rax,8),%xmm1
  mulsd  %xmm1,%xmm1
  addsd  %xmm1,%xmm0
L157:
  ucomisd %xmm0,%xmm4
  ja     L98
  sqrtsd %xmm0,%xmm0
  comisd %xmm0,%xmm3
  jae    L114
  movapd %xmm0,%xmm1
  subsd  %xmm3,%xmm1
  divsd  %xmm0,%xmm1
L25:
  lea    -0x1(%rbp),%eax
  cmp    $0x1,%eax
  jbe    L16
  lea    0x8(%rsi),%rax
  cmp    %rax,%rdi
  je     L16
  mov    %ebp,%edx
  movapd %xmm1,%xmm2
  xor    %eax,%eax
  shr    %edx
  unpcklpd %xmm2,%xmm2
  shl    $0x4,%rdx
  nopw   0x0(%rax,%rax,1)
L34:
  movupd (%rsi,%rax,1),%xmm0
  mulpd  %xmm2,%xmm0
  movups %xmm0,(%rdi,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L34
  mov    %ebp,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%ebp
  je     L93
  mulsd  (%rsi,%rax,8),%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
L93:
  add    $0x20,%rsp
  pop    %rbp
  ret
L114:
  test   %ebp,%ebp
  jle    L93
L155:
  mov    %ebp,%edx
  add    $0x20,%rsp
  xor    %esi,%esi
  shl    $0x3,%rdx
  pop    %rbp
  jmp    1060 <memset@plt>
L16:
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L52:
  movsd  (%rsi,%rax,8),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdi,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rbp
  jne    L52
  jmp    L93
L50:
  pxor   %xmm4,%xmm4
  xor    %eax,%eax
  movapd %xmm4,%xmm0
  jmp    L142
L98:
  mov    %rdi,0x8(%rsp)
  movsd  %xmm3,0x18(%rsp)
  mov    %rsi,0x10(%rsp)
  call   <sqrt@plt>
  movsd  0x18(%rsp),%xmm3
  mov    0x8(%rsp),%rdi
  comisd %xmm0,%xmm3
  jae    L155
  movapd %xmm0,%xmm1
  mov    0x10(%rsp),%rsi
  subsd  %xmm3,%xmm1
  divsd  %xmm0,%xmm1
  jmp    L25
```

assistant: ```c
void epp2(double *x, double *v, int n, double rho) {
  int i;
  double v2 = 0, ratio;
  for (i = 0; i < n; i++) {
    v2 += v[i] * v[i];
  }
  v2 = sqrt(v2);
  if (rho >= v2)
    for (i = 0; i < n; i++)
      x[i] = 0;
  else {
    ratio = (v2 - rho) / v2;
    for (i = 0; i < n; i++)
      x[i] = v[i] * ratio;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<_ackermann>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L17
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  jmp    L29
L17:
  cmpq   $0x0,-0x10(%rbp)
  jne    L18
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <_ackermann>
  jmp    L29
L18:
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <_ackermann>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <_ackermann>
L29:
  leave
  ret
```

assistant: ```c
long _ackermann(long EV_m, long EV_n) {
  if ((EV_m == 0)) {
    return (EV_n + 1);
  }
  if ((EV_n == 0)) {
    return _ackermann((EV_m - 1), 1);
  } else {
    return _ackermann((EV_m - 1), _ackermann(EV_m, (EV_n - 1)));
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_ackermann>:
  endbr64
  test   %rdi,%rdi
  je     L18
  push   %rbx
  mov    %rdi,%rbx
  test   %rsi,%rsi
  jne    L7
  lea    -0x1(%rdi),%rdi
  mov    $0x1,%esi
  call   <_ackermann>
L4:
  pop    %rbx
  ret
L18:
  lea    0x1(%rsi),%rax
  ret
L7:
  sub    $0x1,%rsi
  call   <_ackermann>
  mov    %rax,%rsi
  lea    -0x1(%rbx),%rdi
  call   <_ackermann>
  jmp    L4
```

assistant: ```c
long _ackermann(long EV_m, long EV_n) {
  if ((EV_m == 0)) {
    return (EV_n + 1);
  }
  if ((EV_n == 0)) {
    return _ackermann((EV_m - 1), 1);
  } else {
    return _ackermann((EV_m - 1), _ackermann(EV_m, (EV_n - 1)));
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<TEST_HOST_REG_L>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <AND_HOST_REG_L@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int TEST_HOST_REG_L(int dst_reg, int src_reg) {
  AND_HOST_REG_L(dst_reg, src_reg);
  return dst_reg;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<TEST_HOST_REG_L>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <AND_HOST_REG_L@plt>
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int TEST_HOST_REG_L(int dst_reg, int src_reg) {
  AND_HOST_REG_L(dst_reg, src_reg);
  return dst_reg;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<TEST_HOST_REG_L>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  call   <AND_HOST_REG_L@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int TEST_HOST_REG_L(int dst_reg, int src_reg) {
  AND_HOST_REG_L(dst_reg, src_reg);
  return dst_reg;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<TEST_HOST_REG_L>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  call   <AND_HOST_REG_L@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int TEST_HOST_REG_L(int dst_reg, int src_reg) {
  AND_HOST_REG_L(dst_reg, src_reg);
  return dst_reg;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<foo3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L15
L19:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x3,-0x4(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L19
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void foo3(int j, int n, int *a) {
  int i;
#pragma omp target teams
#pragma omp distribute parallel for shared(a) firstprivate(n) private(i) firstprivate(j)
  for (i = j + 1; i < n; i += 3)
    a[i] = i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo3>:
  endbr64
  add    $0x1,%edi
  cmp    %esi,%edi
  jge    L5
  movslq %edi,%rdi
L14:
  mov    %edi,(%rdx,%rdi,4)
  add    $0x3,%rdi
  cmp    %edi,%esi
  jg     L14
L5:
  ret
```

assistant: ```c
void foo3(int j, int n, int *a) {
  int i;
#pragma omp target teams
#pragma omp distribute parallel for shared(a) firstprivate(n) private(i) firstprivate(j)
  for (i = j + 1; i < n; i += 3)
    a[i] = i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo3>:
  endbr64
  add    $0x1,%edi
  cmp    %esi,%edi
  jge    L9
  movslq %edi,%rdi
  xchg   %ax,%ax
L12:
  mov    %edi,(%rdx,%rdi,4)
  add    $0x3,%rdi
  cmp    %edi,%esi
  jg     L12
L9:
  ret
```

assistant: ```c
void foo3(int j, int n, int *a) {
  int i;
#pragma omp target teams
#pragma omp distribute parallel for shared(a) firstprivate(n) private(i) firstprivate(j)
  for (i = j + 1; i < n; i += 3)
    a[i] = i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<foo3>:
  endbr64
  add    $0x1,%edi
  cmp    %esi,%edi
  jge    L11
  movslq %edi,%rdi
  xchg   %ax,%ax
L17:
  mov    %edi,(%rdx,%rdi,4)
  add    $0x3,%rdi
  cmp    %edi,%esi
  jg     L17
L11:
  ret
```

assistant: ```c
void foo3(int j, int n, int *a) {
  int i;
#pragma omp target teams
#pragma omp distribute parallel for shared(a) firstprivate(n) private(i) firstprivate(j)
  for (i = j + 1; i < n; i += 3)
    a[i] = i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<binderfs_test_unprivileged>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <change_to_userns@plt>
  mov    $0x0,%eax
  call   <__do_binderfs_test@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void binderfs_test_unprivileged() {
  change_to_userns();
  __do_binderfs_test();
}
```
user: Translate the below assembly code into C programming language:

```asm
<binderfs_test_unprivileged>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <change_to_userns@plt>
  mov    $0x0,%eax
  call   <__do_binderfs_test@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void binderfs_test_unprivileged() {
  change_to_userns();
  __do_binderfs_test();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<binderfs_test_unprivileged>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <change_to_userns@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <__do_binderfs_test@plt>
```

assistant: ```c
void binderfs_test_unprivileged() {
  change_to_userns();
  __do_binderfs_test();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<binderfs_test_unprivileged>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <change_to_userns@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <__do_binderfs_test@plt>
```

assistant: ```c
void binderfs_test_unprivileged() {
  change_to_userns();
  __do_binderfs_test();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<getIndex>:
  endbr64
  mov    %edx,%ecx
  shr    %cl,%edi
  cmp    $0x14,%esi
  ja     L0
  mov    %esi,%esi
  lea    D12(%rip),%rdx
  movslq (%rdx,%rsi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    %edi,%eax
  and    $0x1,%eax
  ret
  mov    %edi,%eax
  and    $0x3,%eax
  ret
  mov    %edi,%eax
  and    $0x7,%eax
  ret
  mov    %edi,%eax
  and    $0xf,%eax
  ret
  mov    %edi,%eax
  and    $0x1f,%eax
  ret
  mov    %edi,%eax
  and    $0x3f,%eax
  ret
  mov    %edi,%eax
  and    $0x7f,%eax
  ret
  movzbl %dil,%eax
  ret
  mov    %edi,%eax
  and    $0x1ff,%eax
  ret
  mov    %edi,%eax
  and    $0x3ff,%eax
  ret
  mov    %edi,%eax
  and    $0x7ff,%eax
  ret
  mov    %edi,%eax
  and    $0xfff,%eax
  ret
  mov    %edi,%eax
  and    $0x1fff,%eax
  ret
  mov    %edi,%eax
  and    $0x3fff,%eax
  ret
  mov    %edi,%eax
  and    $0x7fff,%eax
  ret
  movzwl %di,%eax
  ret
  mov    %edi,%eax
  and    $0x1ffff,%eax
  ret
  mov    %edi,%eax
  and    $0x3ffff,%eax
  ret
  mov    %edi,%eax
  and    $0x7ffff,%eax
  ret
  mov    %edi,%eax
  and    $0xfffff,%eax
  ret
L0:
  sub    $0x8,%rsp
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
  mov    $0x0,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D18
data_type: string

tool: D18: *** ERROR: Index should not be greater than 20 (Index size = %d). Returning 0.


assistant: ```c
unsigned int getIndex(unsigned int address, const int indexSize, const int offsetSize) {
  unsigned int tempAddress = address >> offsetSize;
  switch (indexSize) {
  case 0:
    return 0;
  case 1:
    return (tempAddress & 0x1);
  case 2:
    return (tempAddress & 0x3);
  case 3:
    return (tempAddress & 0x7);
  case 4:
    return (tempAddress & 0xF);
  case 5:
    return (tempAddress & 0x1F);
  case 6:
    return (tempAddress & 0x3F);
  case 7:
    return (tempAddress & 0x7F);
  case 8:
    return (tempAddress & 0xFF);
  case 9:
    return (tempAddress & 0x1FF);
  case 10:
    return (tempAddress & 0x3FF);
  case 11:
    return (tempAddress & 0x7FF);
  case 12:
    return (tempAddress & 0xFFF);
  case 13:
    return (tempAddress & 0x1FFF);
  case 14:
    return (tempAddress & 0x3FFF);
  case 15:
    return (tempAddress & 0x7FFF);
  case 16:
    return (tempAddress & 0xFFFF);
  case 17:
    return (tempAddress & 0x1FFFF);
  case 18:
    return (tempAddress & 0x3FFFF);
  case 19:
    return (tempAddress & 0x7FFFF);
  case 20:
    return (tempAddress & 0xFFFFF);
  default:
    printf("*** ERROR: Index should not be greater than 20 (Index size = %d). Returning 0.\n", indexSize);
    return 0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cofactorMatrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x2c(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <allocateDoubleMatrix@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L9
L24:
  movl   $0x0,-0x20(%rbp)
  jmp    L28
L8:
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  mov    $0xffffffff,%edi
  call   <pow@plt>
  movsd  %xmm0,-0x38(%rbp)
  mov    -0x2c(%rbp),%eax
  lea    -0x1(%rax),%ebx
  mov    -0x1c(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x2c(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <minorMatrix@plt>
  mov    %ebx,%esi
  mov    %eax,%edi
  call   <matrixDeterminant@plt>
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mulsd  -0x38(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x20(%rbp)
L28:
  mov    -0x20(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L8
  addl   $0x1,-0x1c(%rbp)
L9:
  mov    -0x1c(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L24
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
double **cofactorMatrix(double **A, int n) {
  double **B;
  int i, j;
  B = allocateDoubleMatrix(n, n);
  for (j = 0; j < n; j++) {
    for (i = 0; i < n; i++) {
      B[j][i] = pow(-1, i + j) * matrixDeterminant(minorMatrix(A, n, i, j), n - 1);
    }
  }
  return B;
}
```