ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<Heap_HeapSort>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  sar    %r12
  mov    %rsi,%rbp
  push   %rbx
  lea    -0x1(%rdi),%rbx
  sub    $0x8,%rsp
  test   %r12,%r12
  jle    L10
  nopl   0x0(%rax,%rax,1)
L28:
  sub    $0x1,%r12
  mov    %r13,%rcx
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <Heap_Sift@plt>
  test   %r12,%r12
  jne    L28
L10:
  test   %rbx,%rbx
  jle    L12
  nopw   0x0(%rax,%rax,1)
L39:
  mov    0x0(%rbp,%rbx,8),%rdx
  mov    0x0(%rbp),%rax
  mov    %r13,%rcx
  mov    %r12,%rdi
  mov    %rdx,0x0(%rbp)
  mov    %rbp,%rdx
  mov    %rax,0x0(%rbp,%rbx,8)
  sub    $0x1,%rbx
  mov    %rbx,%rsi
  call   <Heap_Sift@plt>
  test   %rbx,%rbx
  jne    L39
L12:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void Heap_HeapSort(long long n, long long *a, long long a__len) {
  long long l, r, x;
  l = ((long long)(n) >> (1));
  r = n - 1;
  while (l > 0) {
    l -= 1;
    Heap_Sift(l, r, (void *)a, a__len);
  }
  while (r > 0) {
    x = a[0];
    a[0] = a[r];
    a[r] = x;
    r -= 1;
    Heap_Sift(l, r, (void *)a, a__len);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Heap_HeapSort>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  sar    %r12
  mov    %rsi,%rbp
  push   %rbx
  lea    -0x1(%rdi),%rbx
  sub    $0x8,%rsp
  test   %r12,%r12
  jle    L36
  nopl   0x0(%rax,%rax,1)
L9:
  sub    $0x1,%r12
  mov    %r13,%rcx
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <Heap_Sift@plt>
  test   %r12,%r12
  jne    L9
L36:
  test   %rbx,%rbx
  jle    L34
  nopw   0x0(%rax,%rax,1)
L6:
  mov    0x0(%rbp,%rbx,8),%rdx
  mov    0x0(%rbp),%rax
  mov    %r13,%rcx
  mov    %r12,%rdi
  mov    %rdx,0x0(%rbp)
  mov    %rbp,%rdx
  mov    %rax,0x0(%rbp,%rbx,8)
  sub    $0x1,%rbx
  mov    %rbx,%rsi
  call   <Heap_Sift@plt>
  test   %rbx,%rbx
  jne    L6
L34:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void Heap_HeapSort(long long n, long long *a, long long a__len) {
  long long l, r, x;
  l = ((long long)(n) >> (1));
  r = n - 1;
  while (l > 0) {
    l -= 1;
    Heap_Sift(l, r, (void *)a, a__len);
  }
  while (r > 0) {
    x = a[0];
    a[0] = a[r];
    a[r] = x;
    r -= 1;
    Heap_Sift(l, r, (void *)a, a__len);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Gammln>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  movss  %xmm0,-0x34(%rbp)
  pxor   %xmm0,%xmm0
  comiss -0x34(%rbp),%xmm0
  jb     L28
  movss  D55(%rip),%xmm0
  jmp    L4
L28:
  pxor   %xmm0,%xmm0
  cvtss2sd -0x34(%rbp),%xmm0
  movsd  D37(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm1
  movsd  D2(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  D37(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movl   $0xa,-0x24(%rbp)
  jmp    L35
L29:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  lea    D85(%rip),%rax
  movsd  (%rdx,%rax,1),%xmm0
  divsd  -0x20(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x20(%rbp),%xmm0
  movsd  D37(%rip),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  subl   $0x1,-0x24(%rbp)
L35:
  cmpl   $0x0,-0x24(%rbp)
  jns    L29
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <log@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x18(%rbp)
  movsd  -0x8(%rbp),%xmm1
  movsd  D23(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x10(%rbp),%xmm1
  movsd  D23(%rip),%xmm0
  addsd  %xmm0,%xmm1
  movsd  %xmm1,-0x40(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <log@plt>
  movsd  -0x40(%rbp),%xmm1
  mulsd  %xmm0,%xmm1
  movsd  D11(%rip),%xmm0
  addsd  %xmm1,%xmm0
  subsd  -0x8(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  cvtsd2ss -0x18(%rbp),%xmm0
L4:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D55
datatype: f32
name: extract_symbolic_data
arguments: label: D37
datatype: f64
name: extract_symbolic_data
arguments: label: D2
datatype: f64
name: extract_symbolic_data
arguments: label: D23
datatype: f64
name: extract_symbolic_data
arguments: label: D11
datatype: f64

tool: D55: 999999

tool: D37: 1

tool: D2: 11

tool: D23: 0.5

tool: D11: 0.918938533

assistant: ```c
float Gammln(float x) {
  int i;
  double xx, tx;
  double tmp, value;
  static double cof[11] = {4.694580336184385e+04, -1.560605207784446e+05, 2.065049568014106e+05, -1.388934775095388e+05, 5.031796415085709e+04, -9.601592329182778e+03, 8.785855930895250e+02, -3.155153906098611e+01, 2.908143421162229e-01, -2.319827630494973e-04, 1.251639670050933e-10};
  if (x <= 0.0)
    return 999999.;
  xx = x - 1.0;
  tx = tmp = xx + 11.0;
  value = 1.0;
  for (i = 10; i >= 0; i--) {
    value += cof[i] / tmp;
    tmp -= 1.0;
  }
  value = log(value);
  tx += 0.5;
  value += 0.918938533 + (xx + 0.5) * log(tx) - tx;
  return (float)value;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<eqep_2_pinmux_setup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x6b,%esi
  mov    $0x8b0,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8b4,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8b8,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8bc,%edi
  call   <GpioPinMuxSetup@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void eqep_2_pinmux_setup(void) {
  GpioPinMuxSetup((0x08b0), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08b4), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08b8), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08bc), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<eqep_2_pinmux_setup>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6b,%esi
  mov    $0x8b0,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8b4,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8b8,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8bc,%edi
  call   <GpioPinMuxSetup@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void eqep_2_pinmux_setup(void) {
  GpioPinMuxSetup((0x08b0), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08b4), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08b8), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08bc), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<eqep_2_pinmux_setup>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6b,%esi
  mov    $0x8b0,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8b4,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8b8,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8bc,%edi
  add    $0x8,%rsp
  jmp    1050 <GpioPinMuxSetup@plt>
```

assistant: ```c
void eqep_2_pinmux_setup(void) {
  GpioPinMuxSetup((0x08b0), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08b4), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08b8), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08bc), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
}
```
user: Translate the below assembly code into C programming language:

```asm
<eqep_2_pinmux_setup>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6b,%esi
  mov    $0x8b0,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8b4,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8b8,%edi
  call   <GpioPinMuxSetup@plt>
  mov    $0x6b,%esi
  mov    $0x8bc,%edi
  add    $0x8,%rsp
  jmp    1050 <GpioPinMuxSetup@plt>
```

assistant: ```c
void eqep_2_pinmux_setup(void) {
  GpioPinMuxSetup((0x08b0), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08b4), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08b8), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
  GpioPinMuxSetup((0x08bc), (0x00000040 | 0x00000020 | 0x00000008 | (3)));
}
```
user: Translate the below assembly code into C programming language:

```asm
<abs_and_sign>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jle    L15
  mov    $0x0,%eax
  jmp    L5
L15:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  neg    %eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int abs_and_sign(int *x) {
  if (*x > 0)
    return 0;
  *x *= -1;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<abs_and_sign>:
  endbr64
  mov    (%rdi),%eax
  mov    $0x0,%edx
  test   %eax,%eax
  jg     L2
  neg    %eax
  mov    %eax,(%rdi)
  mov    $0x1,%edx
L2:
  mov    %edx,%eax
  ret
```

assistant: ```c
int abs_and_sign(int *x) {
  if (*x > 0)
    return 0;
  *x *= -1;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<abs_and_sign>:
  endbr64
  mov    (%rdi),%eax
  xor    %r8d,%r8d
  test   %eax,%eax
  jg     L4
  neg    %eax
  mov    $0x1,%r8d
  mov    %eax,(%rdi)
L4:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int abs_and_sign(int *x) {
  if (*x > 0)
    return 0;
  *x *= -1;
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<abs_and_sign>:
  endbr64
  mov    (%rdi),%eax
  xor    %r8d,%r8d
  test   %eax,%eax
  jg     L3
  neg    %eax
  mov    $0x1,%r8d
  mov    %eax,(%rdi)
L3:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int abs_and_sign(int *x) {
  if (*x > 0)
    return 0;
  *x *= -1;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L39
L15:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0x989680,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L14
L25:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D11(%rip),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L14:
  cmpl   $0x2,-0x14(%rbp)
  jle    L25
  addl   $0x1,-0x18(%rbp)
L39:
  cmpl   $0x6,-0x18(%rbp)
  jle    L15
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D11
type: string
name: parse_data
arguments: data_label: D24
type: string
name: parse_data
arguments: data_label: D27
type: string

tool: D11: dmskrsqrt

tool: D24: # c::%s:len=%d


tool: D27: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dmskrsqrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D29(%rip),%r15
  jmp    L17
L27:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L2
L17:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0x989680,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L15:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L15
  jmp    L27
L2:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: label: D29
type_infered: string
name: retrieve_data
arguments: label: D20
type_infered: string
name: retrieve_data
arguments: label: D11
type_infered: string

tool: D29: dmskrsqrt

tool: D20: # c::%s:len=%d


tool: D11: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dmskrsqrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D2(%rip),%r15
  push   %r14
  lea    D19(%rip),%r14
  push   %r13
  lea    D18(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L5:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0x989680,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L2:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L2
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L5
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
type_infered: string
name: read_data
arguments: data_tag: D19
type_infered: string
name: read_data
arguments: data_tag: D18
type_infered: string

tool: D2: dmskrsqrt

tool: D19: # c::%s:len=%d


tool: D18: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dmskrsqrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D26(%rip),%r15
  push   %r14
  lea    D21(%rip),%r14
  push   %r13
  lea    D18(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L10:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0x989680,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L3:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L3
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L10
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D26
type_infered: string
name: parse_data
arguments: label: D21
type_infered: string
name: parse_data
arguments: label: D18
type_infered: string

tool: D26: dmskrsqrt

tool: D21: # c::%s:len=%d


tool: D18: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 10000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dmskrsqrt", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strprbrk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x140,%rsp
  mov    %rdi,-0x138(%rbp)
  mov    %rsi,-0x140(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x138(%rbp),%rax
  mov    %rax,-0x120(%rbp)
  movq   $0x0,-0x118(%rbp)
  lea    -0x110(%rbp),%rax
  mov    $0x100,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L26
L11:
  movsbq -0x121(%rbp),%rax
  movb   $0x1,-0x110(%rbp,%rax,1)
L26:
  mov    -0x140(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x140(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x121(%rbp)
  cmpb   $0x0,-0x121(%rbp)
  jne    L11
  jmp    L38
L6:
  movsbq -0x121(%rbp),%rax
  movzbl -0x110(%rbp,%rax,1),%eax
  test   %al,%al
  jne    L38
  mov    -0x120(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x118(%rbp)
L38:
  mov    -0x120(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x120(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x121(%rbp)
  cmpb   $0x0,-0x121(%rbp)
  jne    L6
  mov    -0x118(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L50
  call   <__stack_chk_fail@plt>
L50:
  leave
  ret
```

assistant: ```c
char *(strprbrk)(const char *string, const char *stopset) {
  char set[256];
  char c;
  const char *s = string;
  const char *r = NULL;
  memset(set, 0, 256);
  while ((c = *stopset++))
    set[(size_t)c] = 1;
  while ((c = *s++))
    if (!set[(size_t)c])
      r = s - 1;
  return r;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strprbrk>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x100,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  lea    0x1(%rbp),%rdx
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L10
L34:
  movsbq %al,%rax
  movb   $0x1,(%rsp,%rax,1)
  add    $0x1,%rdx
  movzbl -0x1(%rdx),%eax
  test   %al,%al
  jne    L34
L10:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L59
  mov    %rbx,%rdx
  mov    $0x0,%ecx
L12:
  movsbq %al,%rax
  cmpb   $0x0,(%rsp,%rax,1)
  cmove  %rdx,%rcx
  add    $0x1,%rdx
  movzbl (%rdx),%eax
  test   %al,%al
  jne    L12
L31:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  mov    %rcx,%rax
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  ret
L59:
  mov    $0x0,%ecx
  jmp    L31
L54:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *(strprbrk)(const char *string, const char *stopset) {
  char set[256];
  char c;
  const char *s = string;
  const char *r = NULL;
  memset(set, 0, 256);
  while ((c = *stopset++))
    set[(size_t)c] = 1;
  while ((c = *s++))
    if (!set[(size_t)c])
      r = s - 1;
  return r;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strprbrk>:
  endbr64
  push   %rbp
  mov    $0x100,%edx
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memset@plt>
  movsbq 0x0(%rbp),%rax
  test   %al,%al
  je     L55
  lea    0x1(%rbp),%rdx
  nopw   0x0(%rax,%rax,1)
L14:
  add    $0x1,%rdx
  movb   $0x1,(%rsp,%rax,1)
  movsbq -0x1(%rdx),%rax
  test   %al,%al
  jne    L14
L55:
  movsbq (%rbx),%rax
  test   %al,%al
  je     L48
  mov    %rbx,%rdi
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L44:
  cmpb   $0x0,(%rsp,%rax,1)
  cmove  %rdi,%r8
  movsbq 0x1(%rdi),%rax
  add    $0x1,%rdi
  test   %al,%al
  jne    L44
L2:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x118,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L48:
  xor    %r8d,%r8d
  jmp    L2
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *(strprbrk)(const char *string, const char *stopset) {
  char set[256];
  char c;
  const char *s = string;
  const char *r = NULL;
  memset(set, 0, 256);
  while ((c = *stopset++))
    set[(size_t)c] = 1;
  while ((c = *s++))
    if (!set[(size_t)c])
      r = s - 1;
  return r;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strprbrk>:
  endbr64
  push   %rbp
  mov    $0x100,%edx
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memset@plt>
  movsbq 0x0(%rbp),%rax
  test   %al,%al
  je     L33
  lea    0x1(%rbp),%rdx
  nopw   0x0(%rax,%rax,1)
L54:
  add    $0x1,%rdx
  movb   $0x1,(%rsp,%rax,1)
  movsbq -0x1(%rdx),%rax
  test   %al,%al
  jne    L54
L33:
  movsbq (%rbx),%rax
  test   %al,%al
  je     L36
  mov    %rbx,%rdi
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L32:
  cmpb   $0x0,(%rsp,%rax,1)
  cmove  %rdi,%r8
  movsbq 0x1(%rdi),%rax
  add    $0x1,%rdi
  test   %al,%al
  jne    L32
L47:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x118,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L36:
  xor    %r8d,%r8d
  jmp    L47
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *(strprbrk)(const char *string, const char *stopset) {
  char set[256];
  char c;
  const char *s = string;
  const char *r = NULL;
  memset(set, 0, 256);
  while ((c = *stopset++))
    set[(size_t)c] = 1;
  while ((c = *s++))
    if (!set[(size_t)c])
      r = s - 1;
  return r;
}
```