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

```asm
<update_max_zdrop>:
  endbr64
  mov    %rdi,%rax
  mov    (%rcx),%rdi
  cmp    %rax,%rdi
  jle    L16
  mov    %esi,%r11d
  mov    %edx,%r10d
  sub    (%r8),%r11d
  sub    (%r9),%r10d
  push   %rbx
  mov    %r10d,%ecx
  mov    %r11d,%ebx
  sub    %r10d,%ebx
  sub    %r11d,%ecx
  cmp    %r10d,%r11d
  cmovg  %ebx,%ecx
  sub    %eax,%edi
  mov    0x10(%rsp),%eax
  imul   %ecx,%eax
  sub    %eax,%edi
  mov    0x18(%rsp),%rax
  cmp    %edi,(%rax)
  jge    L13
  mov    %edi,(%rax)
  mov    0x20(%rsp),%rbx
  add    $0x1,%esi
  add    $0x1,%edx
  mov    (%r8),%eax
  mov    %esi,0x4(%rbx)
  mov    %eax,(%rbx)
  mov    (%r9),%eax
  mov    %edx,0xc(%rbx)
  mov    %eax,0x8(%rbx)
L13:
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  mov    %rax,(%rcx)
  mov    %esi,(%r8)
  mov    %edx,(%r9)
  ret
```

assistant: ```c
void update_max_zdrop(int32_t score, int i, int j, int32_t *max, int *max_i, int *max_j, int e, int *max_zdrop, int pos[2][2]) {
  if (score < *max) {
    int li = i - *max_i;
    int lj = j - *max_j;
    int diff = li > lj ? li - lj : lj - li;
    int z = *max - score - diff * e;
    if (z > *max_zdrop) {
      *max_zdrop = z;
      pos[0][0] = *max_i, pos[0][1] = i + 1;
      pos[1][0] = *max_j, pos[1][1] = j + 1;
    }
  } else
    *max = score, *max_i = i, *max_j = j;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xfe70,-0x8(%rbp)
  movl   $0x7ef,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  or     %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  xor    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int a = 0xFE70, b = 0x07EF;
#pragma omp parallel
  {
#pragma omp atomic
    a &= b;
#pragma omp atomic
    a |= b;
#pragma omp atomic
    a ^= b;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int a = 0xFE70, b = 0x07EF;
#pragma omp parallel
  {
#pragma omp atomic
    a &= b;
#pragma omp atomic
    a |= b;
#pragma omp atomic
    a ^= b;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int a = 0xFE70, b = 0x07EF;
#pragma omp parallel
  {
#pragma omp atomic
    a &= b;
#pragma omp atomic
    a |= b;
#pragma omp atomic
    a ^= b;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<geraValoresVetor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L35
L10:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rdx,%rcx
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  imul   $0x64,%edx,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,(%rcx)
  addl   $0x1,-0x4(%rbp)
L35:
  cmpl   $0x9,-0x4(%rbp)
  jle    L10
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L16
L39:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L16:
  cmpl   $0x9,-0x4(%rbp)
  jle    L39
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
type_infered: string
name: extract_symbolic_data
arguments: data_label: D14
type_infered: string

tool: D5: Vetor gerado aleatoriamente: 

tool: D14: v[%d] = %d


assistant: ```c
void geraValoresVetor(int *v) {
  int i;
  srand(time(NULL));
  for (i = 0; i < 10; i++)
    v[i] = rand() % 100;
  printf("Vetor gerado aleatoriamente: \n");
  for (i = 0; i < 10; i++)
    printf("v[%d] = %d\n", i, v[i]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<geraValoresVetor>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    %rbp,%rbx
  lea    0x28(%rbp),%r12
L16:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,(%rbx)
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L16
  lea    D18(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%ebx
  lea    D13(%rip),%r12
L4:
  mov    0x0(%rbp,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0xa,%rbx
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D18
type_guessed: string
name: retrieve_data
arguments: tag: D13
type_guessed: string

tool: D18: Vetor gerado aleatoriamente: 

tool: D13: v[%d] = %d


assistant: ```c
void geraValoresVetor(int *v) {
  int i;
  srand(time(NULL));
  for (i = 0; i < 10; i++)
    v[i] = rand() % 100;
  printf("Vetor gerado aleatoriamente: \n");
  for (i = 0; i < 10; i++)
    printf("v[%d] = %d\n", i, v[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<geraValoresVetor>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  mov    %rbx,%r12
  lea    0x28(%rbx),%rbp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
L9:
  xor    %eax,%eax
  add    $0x4,%r12
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x25,%rdx
  sub    %ecx,%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%r12)
  cmp    %rbp,%r12
  jne    L9
  lea    D3(%rip),%rdi
  xor    %ebp,%ebp
  lea    D18(%rip),%r12
  call   <puts@plt>
  nopl   0x0(%rax)
L4:
  mov    (%rbx,%rbp,4),%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    $0xa,%rbp
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: tag: D3
datatype: string
name: unpack_data
arguments: tag: D18
datatype: string

tool: D3: Vetor gerado aleatoriamente: 

tool: D18: v[%d] = %d


assistant: ```c
void geraValoresVetor(int *v) {
  int i;
  srand(time(NULL));
  for (i = 0; i < 10; i++)
    v[i] = rand() % 100;
  printf("Vetor gerado aleatoriamente: \n");
  for (i = 0; i < 10; i++)
    printf("v[%d] = %d\n", i, v[i]);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<geraValoresVetor>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  mov    %rbx,%r12
  lea    0x28(%rbx),%rbp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
L2:
  xor    %eax,%eax
  add    $0x4,%r12
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x25,%rdx
  sub    %ecx,%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%r12)
  cmp    %rbp,%r12
  jne    L2
  lea    D15(%rip),%rdi
  xor    %ebp,%ebp
  lea    D13(%rip),%r12
  call   <puts@plt>
  nopl   0x0(%rax)
L19:
  mov    (%rbx,%rbp,4),%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    $0xa,%rbp
  jne    L19
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D15
type_guessed: string
name: retrieve_data
arguments: data_label: D13
type_guessed: string

tool: D15: Vetor gerado aleatoriamente: 

tool: D13: v[%d] = %d


assistant: ```c
void geraValoresVetor(int *v) {
  int i;
  srand(time(NULL));
  for (i = 0; i < 10; i++)
    v[i] = rand() % 100;
  printf("Vetor gerado aleatoriamente: \n");
  for (i = 0; i < 10; i++)
    printf("v[%d] = %d\n", i, v[i]);
}
```
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    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  int result = 0;
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  int result = 0;
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int result = 0;
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int result = 0;
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<gen_cvin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  shl    $0x1c,%eax
  and    $0x30000000,%eax
  mov    %eax,-0x4(%rbp)
  orl    $0xc0000000,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int gen_cvin(int cvin) {
  int r = ((cvin & 2) ? (1 << 29) : 0) | ((cvin & 1) ? (1 << 28) : 0);
  r |= (1 << 31) | (1 << 30);
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<gen_cvin>:
  endbr64
  mov    %edi,%eax
  shl    $0x1c,%eax
  and    $0x30000000,%eax
  or     $0xc0000000,%eax
  ret
```

assistant: ```c
int gen_cvin(int cvin) {
  int r = ((cvin & 2) ? (1 << 29) : 0) | ((cvin & 1) ? (1 << 28) : 0);
  r |= (1 << 31) | (1 << 30);
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<gen_cvin>:
  endbr64
  mov    %edi,%eax
  shl    $0x1c,%eax
  and    $0x30000000,%eax
  or     $0xc0000000,%eax
  ret
```

assistant: ```c
int gen_cvin(int cvin) {
  int r = ((cvin & 2) ? (1 << 29) : 0) | ((cvin & 1) ? (1 << 28) : 0);
  r |= (1 << 31) | (1 << 30);
  return r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gen_cvin>:
  endbr64
  mov    %edi,%eax
  shl    $0x1c,%eax
  and    $0x30000000,%eax
  or     $0xc0000000,%eax
  ret
```

assistant: ```c
int gen_cvin(int cvin) {
  int r = ((cvin & 2) ? (1 << 29) : 0) | ((cvin & 1) ? (1 << 28) : 0);
  r |= (1 << 31) | (1 << 30);
  return r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movsd  -0x18(%rbp),%xmm1
  movsd  D19(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x10(%rbp),%xmm2
  movsd  D47(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movsd  D11(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D28
type_infered: string
name: retrieve_data
arguments: label: D19
type_infered: f64
name: retrieve_data
arguments: label: D47
type_infered: f64
name: retrieve_data
arguments: label: D11
type_infered: byte[8]
name: retrieve_data
arguments: label: D39
type_infered: string

tool: D28: %lf %lf

tool: D19: 3.5

tool: D47: 7.5

tool: {"D11": [0x00,0x00,0x00,0x00,0x00,0x00,0x26,0x40]}

tool: D39: MEDIA = %.5lf


assistant: ```c
int main() {
  double a, b;
  scanf("%lf %lf", &a, &b);
  printf("MEDIA = %.5lf\n", (3.5 * a + 7.5 * b) / 11);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  movsd  D25(%rip),%xmm0
  mulsd  0x8(%rsp),%xmm0
  movsd  D19(%rip),%xmm1
  mulsd  0x10(%rsp),%xmm1
  addsd  %xmm1,%xmm0
  divsd  D11(%rip),%xmm0
  lea    D35(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D18
datatype: string
name: extract_symbolic_data
arguments: data_label: D25
datatype: f64
name: extract_symbolic_data
arguments: data_label: D19
datatype: f64
name: extract_symbolic_data
arguments: data_label: D11
datatype: byte[8]
name: extract_symbolic_data
arguments: data_label: D35
datatype: string

tool: D18: %lf %lf

tool: D25: 3.5

tool: D19: 7.5

tool: {"D11": [0x00,0x00,0x00,0x00,0x00,0x00,0x26,0x40]}

tool: D35: MEDIA = %.5lf


assistant: ```c
int main() {
  double a, b;
  scanf("%lf %lf", &a, &b);
  printf("MEDIA = %.5lf\n", (3.5 * a + 7.5 * b) / 11);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D45(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  movsd  D43(%rip),%xmm0
  movsd  D15(%rip),%xmm1
  lea    D44(%rip),%rdi
  mulsd  0x8(%rsp),%xmm0
  mulsd  0x10(%rsp),%xmm1
  addsd  %xmm1,%xmm0
  divsd  D40(%rip),%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D45
datatype: string
name: unpack_data
arguments: label: D43
datatype: f64
name: unpack_data
arguments: label: D15
datatype: f64
name: unpack_data
arguments: label: D44
datatype: string
name: unpack_data
arguments: label: D40
datatype: byte[8]

tool: D45: %lf %lf

tool: D43: 3.5

tool: D15: 7.5

tool: D44: MEDIA = %.5lf


tool: {"D40": [0x00,0x00,0x00,0x00,0x00,0x00,0x26,0x40]}

assistant: ```c
int main() {
  double a, b;
  scanf("%lf %lf", &a, &b);
  printf("MEDIA = %.5lf\n", (3.5 * a + 7.5 * b) / 11);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D37(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  movsd  D5(%rip),%xmm0
  movsd  D14(%rip),%xmm1
  lea    D34(%rip),%rdi
  mulsd  0x8(%rsp),%xmm0
  mulsd  0x10(%rsp),%xmm1
  addsd  %xmm1,%xmm0
  divsd  D44(%rip),%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D37
data_type: string
name: read_data
arguments: tag: D5
data_type: f64
name: read_data
arguments: tag: D14
data_type: f64
name: read_data
arguments: tag: D34
data_type: string
name: read_data
arguments: tag: D44
data_type: byte[8]

tool: D37: %lf %lf

tool: D5: 3.5

tool: D14: 7.5

tool: D34: MEDIA = %.5lf


tool: {"D44": [0x00,0x00,0x00,0x00,0x00,0x00,0x26,0x40]}

assistant: ```c
int main() {
  double a, b;
  scanf("%lf %lf", &a, &b);
  printf("MEDIA = %.5lf\n", (3.5 * a + 7.5 * b) / 11);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<mystrsep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,%eax
  mov    %al,-0x2c(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x20(%rbp)
  cmpl   $0x0,-0x20(%rbp)
  jle    L5
  movzbl -0x2c(%rbp),%eax
  movzbl %al,%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <memchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L31
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  sub    %rdx,%rax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L5
  mov    -0x1c(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L5
L31:
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L5
  mov    -0x20(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
L5:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *mystrsep(char **stringp, const char delim) {
  char *rv = NULL;
  char *mp = *stringp;
  int n = strlen(mp);
  if (n > 0) {
    char *dp = (char *)memchr(mp, (int)((unsigned char)delim), n);
    if (dp) {
      int nc;
      *stringp = dp + 1;
      nc = (int)((unsigned long)dp - (unsigned long)mp);
      rv = (char *)malloc(nc + 1);
      if (rv) {
        memcpy(rv, mp, nc);
        *(rv + nc) = '\0';
      }
    } else {
      rv = (char *)malloc(n + 1);
      if (rv) {
        memcpy(rv, mp, n);
        *(rv + n) = '\0';
        *stringp = mp + n;
      }
    }
  }
  return rv;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mystrsep>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %esi,%r12d
  mov    (%rdi),%r14
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    $0x0,%ebp
  test   %eax,%eax
  jle    L33
  mov    %eax,%ebx
  movzbl %r12b,%esi
  mov    %eax,%edx
  mov    %r14,%rdi
  call   <memchr@plt>
  test   %rax,%rax
  je     L20
  lea    0x1(%rax),%rdx
  mov    %rdx,0x0(%r13)
  sub    %r14,%rax
  mov    %rax,%rbx
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L48
L33:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L48:
  mov    %ebx,%edx
  mov    %r14,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rax
  movb   $0x0,0x0(%rbp,%rax,1)
  jmp    L33
L20:
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L33
  mov    %ebx,%edx
  mov    %r14,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rbx
  movb   $0x0,0x0(%rbp,%rbx,1)
  add    %rbx,%r14
  mov    %r14,0x0(%r13)
  jmp    L33
```

assistant: ```c
char *mystrsep(char **stringp, const char delim) {
  char *rv = NULL;
  char *mp = *stringp;
  int n = strlen(mp);
  if (n > 0) {
    char *dp = (char *)memchr(mp, (int)((unsigned char)delim), n);
    if (dp) {
      int nc;
      *stringp = dp + 1;
      nc = (int)((unsigned long)dp - (unsigned long)mp);
      rv = (char *)malloc(nc + 1);
      if (rv) {
        memcpy(rv, mp, nc);
        *(rv + nc) = '\0';
      }
    } else {
      rv = (char *)malloc(n + 1);
      if (rv) {
        memcpy(rv, mp, n);
        *(rv + n) = '\0';
        *stringp = mp + n;
      }
    }
  }
  return rv;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<mystrsep>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    (%rdi),%r13
  mov    %r13,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  test   %ebx,%ebx
  jle    L36
  movzbl %r12b,%esi
  mov    %ebx,%edx
  mov    %r13,%rdi
  call   <memchr@plt>
  test   %rax,%rax
  je     L13
  lea    0x1(%rax),%rdx
  sub    %r13,%rax
  mov    %rdx,0x0(%rbp)
  lea    0x1(%rax),%edi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L10
L36:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L13:
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L36
  mov    %ebx,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  add    %rbx,%r13
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  mov    %r13,0x0(%rbp)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L10:
  mov    %ebx,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rax
  movb   $0x0,(%r12,%rax,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *mystrsep(char **stringp, const char delim) {
  char *rv = NULL;
  char *mp = *stringp;
  int n = strlen(mp);
  if (n > 0) {
    char *dp = (char *)memchr(mp, (int)((unsigned char)delim), n);
    if (dp) {
      int nc;
      *stringp = dp + 1;
      nc = (int)((unsigned long)dp - (unsigned long)mp);
      rv = (char *)malloc(nc + 1);
      if (rv) {
        memcpy(rv, mp, nc);
        *(rv + nc) = '\0';
      }
    } else {
      rv = (char *)malloc(n + 1);
      if (rv) {
        memcpy(rv, mp, n);
        *(rv + n) = '\0';
        *stringp = mp + n;
      }
    }
  }
  return rv;
}
```
user: What is the c source code of the assembly code below?

```asm
<mystrsep>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    (%rdi),%r13
  mov    %r13,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  test   %ebx,%ebx
  jle    L39
  movzbl %r12b,%esi
  mov    %ebx,%edx
  mov    %r13,%rdi
  call   <memchr@plt>
  test   %rax,%rax
  je     L16
  lea    0x1(%rax),%rdx
  sub    %r13,%rax
  mov    %rdx,0x0(%rbp)
  lea    0x1(%rax),%edi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L21
L39:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L16:
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L39
  mov    %ebx,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  add    %rbx,%r13
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  mov    %r13,0x0(%rbp)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L21:
  mov    %ebx,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rax
  movb   $0x0,(%r12,%rax,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *mystrsep(char **stringp, const char delim) {
  char *rv = NULL;
  char *mp = *stringp;
  int n = strlen(mp);
  if (n > 0) {
    char *dp = (char *)memchr(mp, (int)((unsigned char)delim), n);
    if (dp) {
      int nc;
      *stringp = dp + 1;
      nc = (int)((unsigned long)dp - (unsigned long)mp);
      rv = (char *)malloc(nc + 1);
      if (rv) {
        memcpy(rv, mp, nc);
        *(rv + nc) = '\0';
      }
    } else {
      rv = (char *)malloc(n + 1);
      if (rv) {
        memcpy(rv, mp, n);
        *(rv + n) = '\0';
        *stringp = mp + n;
      }
    }
  }
  return rv;
}
```