ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L13
  lea    0x4(%rsp),%rdx
  mov    0x8(%rsi),%rdi
  lea    D3(%rip),%rsi
  call   <sscanf@plt>
  movss  0x4(%rsp),%xmm0
  call   <my_sqrt@plt>
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L13:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L4
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
data_type: string
name: extract_symbolic_data
arguments: tag: D10
data_type: string

tool: D3: %f

tool: D10: test_mysqrt <num>


assistant: ```c
int main(int argc, char *argv[]) {
  float x;
  if (argc < (1 + 1)) {
    printf("test_mysqrt <num>\n");
    return 0;
  }
  sscanf(argv[1], "%f", &x);
  my_sqrt(x);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L7
  mov    0x8(%rsi),%rdi
  lea    0x4(%rsp),%rdx
  lea    D0(%rip),%rsi
  call   <sscanf@plt>
  movss  0x4(%rsp),%xmm0
  call   <my_sqrt@plt>
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L9
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D0
type: string
name: unpack_data
arguments: data_label: D6
type: string

tool: D0: %f

tool: D6: test_mysqrt <num>


assistant: ```c
int main(int argc, char *argv[]) {
  float x;
  if (argc < (1 + 1)) {
    printf("test_mysqrt <num>\n");
    return 0;
  }
  sscanf(argv[1], "%f", &x);
  my_sqrt(x);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L21
  mov    0x8(%rsi),%rdi
  lea    0x4(%rsp),%rdx
  lea    D2(%rip),%rsi
  call   <sscanf@plt>
  movss  0x4(%rsp),%xmm0
  call   <my_sqrt@plt>
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L21:
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L9
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type_guessed: string
name: disassemble_data
arguments: data_label: D14
type_guessed: string

tool: D2: %f

tool: D14: test_mysqrt <num>


assistant: ```c
int main(int argc, char *argv[]) {
  float x;
  if (argc < (1 + 1)) {
    printf("test_mysqrt <num>\n");
    return 0;
  }
  sscanf(argv[1], "%f", &x);
  my_sqrt(x);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<processLineWithScan>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rsi
  lea    -0x18(%rbp),%rcx
  lea    -0x1c(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rsi,%r8
  lea    D11(%rip),%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  movl   $0x1,-0x10(%rbp)
  jmp    L5
L25:
  movl   $0x1,-0xc(%rbp)
  mov    -0x1c(%rbp),%ecx
  mov    -0x10(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  jne    L60
  movl   $0x0,-0xc(%rbp)
  mov    $0x46,%edi
  call   <putchar@plt>
L60:
  mov    -0x18(%rbp),%ecx
  mov    -0x10(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  jne    L66
  movl   $0x0,-0xc(%rbp)
  mov    $0x42,%edi
  call   <putchar@plt>
L66:
  cmpl   $0x0,-0xc(%rbp)
  je     L43
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L43:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  je     L14
  mov    $0x20,%edi
  call   <putchar@plt>
L14:
  addl   $0x1,-0x10(%rbp)
L5:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L25
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D11
type: string
name: retrieve_data
arguments: data_tag: D1
type: string

tool: D11: %d %d %d

tool: D1: %d

assistant: ```c
void processLineWithScan(char line[]) {
  int x, y, count;
  sscanf(line, "%d %d %d", &x, &y, &count);
  for (int i = 1; i <= count; i++) {
    int printNum = 1;
    if ((i % x) == 0) {
      printNum = 0;
      printf("F");
    }
    if ((i % y) == 0) {
      printNum = 0;
      printf("B");
    }
    if (printNum) {
      printf("%d", i);
    }
    if (i != count) {
      printf(" ");
    }
  }
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<processLineWithScan>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  lea    D19(%rip),%rsi
  call   <sscanf@plt>
  cmpl   $0x0,0x14(%rsp)
  jle    L3
  mov    $0x1,%ebx
  lea    D16(%rip),%rbp
  jmp    L23
L7:
  mov    $0x46,%edi
  call   <putchar@plt>
  mov    %ebx,%eax
  cltd
  idivl  0x10(%rsp)
  test   %edx,%edx
  je     L26
L0:
  cmp    %ebx,0x14(%rsp)
  jne    L50
L36:
  add    $0x1,%ebx
  cmp    %ebx,0x14(%rsp)
  jl     L3
L23:
  mov    %ebx,%eax
  cltd
  idivl  0xc(%rsp)
  test   %edx,%edx
  je     L7
  mov    %ebx,%eax
  cltd
  idivl  0x10(%rsp)
  test   %edx,%edx
  jne    L106
L26:
  mov    $0x42,%edi
  call   <putchar@plt>
  jmp    L0
L106:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L0
L50:
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L36
L3:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D19
type: string
name: read_data
arguments: data_label: D16
type: string

tool: D19: %d %d %d

tool: D16: %d

assistant: ```c
void processLineWithScan(char line[]) {
  int x, y, count;
  sscanf(line, "%d %d %d", &x, &y, &count);
  for (int i = 1; i <= count; i++) {
    int printNum = 1;
    if ((i % x) == 0) {
      printNum = 0;
      printf("F");
    }
    if ((i % y) == 0) {
      printNum = 0;
      printf("B");
    }
    if (printNum) {
      printf("%d", i);
    }
    if (i != count) {
      printf(" ");
    }
  }
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<processLineWithScan>:
  endbr64
  push   %r12
  lea    D18(%rip),%rsi
  mov    $0x1,%r12d
  push   %rbx
  lea    D0(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  call   <sscanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jg     L29
  jmp    L9
  nopl   (%rax)
L40:
  mov    %r12d,%eax
  cltd
  idivl  0x10(%rsp)
  test   %edx,%edx
  je     L61
  mov    %r12d,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L44:
  mov    %r12d,%eax
  cmp    %r12d,0x14(%rsp)
  je     L62
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    0x14(%rsp),%eax
L62:
  add    $0x1,%r12d
  cmp    %eax,%r12d
  jg     L9
L29:
  mov    %r12d,%eax
  cltd
  idivl  0xc(%rsp)
  test   %edx,%edx
  jne    L40
  mov    $0x46,%edi
  call   <putchar@plt>
  mov    %r12d,%eax
  cltd
  idivl  0x10(%rsp)
  test   %edx,%edx
  jne    L44
L61:
  mov    $0x42,%edi
  call   <putchar@plt>
  jmp    L44
  nopl   0x0(%rax,%rax,1)
L9:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  add    $0x28,%rsp
  pop    %rbx
  pop    %r12
  ret
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D18
type_guessed: string
name: read_data
arguments: label: D0
type_guessed: string

tool: D18: %d %d %d

tool: D0: %d

assistant: ```c
void processLineWithScan(char line[]) {
  int x, y, count;
  sscanf(line, "%d %d %d", &x, &y, &count);
  for (int i = 1; i <= count; i++) {
    int printNum = 1;
    if ((i % x) == 0) {
      printNum = 0;
      printf("F");
    }
    if ((i % y) == 0) {
      printNum = 0;
      printf("B");
    }
    if (printNum) {
      printf("%d", i);
    }
    if (i != count) {
      printf(" ");
    }
  }
  printf("\n");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<processLineWithScan>:
  endbr64
  push   %r12
  lea    D17(%rip),%rsi
  mov    $0x1,%r12d
  push   %rbx
  lea    D0(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%r8
  call   <sscanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L26
L11:
  mov    %r12d,%eax
  cltd
  idivl  0xc(%rsp)
  test   %edx,%edx
  je     L8
  mov    %r12d,%eax
  cltd
  idivl  0x10(%rsp)
  test   %edx,%edx
  je     L73
  mov    %r12d,%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L67:
  cmp    0x14(%rsp),%r12d
  je     L26
  mov    $0x20,%edi
  add    $0x1,%r12d
  call   <putchar@plt>
  cmp    %r12d,0x14(%rsp)
  jge    L11
L26:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x28,%rsp
  pop    %rbx
  pop    %r12
  ret
  xchg   %ax,%ax
L8:
  mov    $0x46,%edi
  call   <putchar@plt>
  mov    %r12d,%eax
  cltd
  idivl  0x10(%rsp)
  test   %edx,%edx
  jne    L67
L73:
  mov    $0x42,%edi
  call   <putchar@plt>
  jmp    L67
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D17
type_infered: string
name: unpack_data
arguments: data_label: D0
type_infered: string

tool: D17: %d %d %d

tool: D0: %d

assistant: ```c
void processLineWithScan(char line[]) {
  int x, y, count;
  sscanf(line, "%d %d %d", &x, &y, &count);
  for (int i = 1; i <= count; i++) {
    int printNum = 1;
    if ((i % x) == 0) {
      printNum = 0;
      printf("F");
    }
    if ((i % y) == 0) {
      printNum = 0;
      printf("B");
    }
    if (printNum) {
      printf("%d", i);
    }
    if (i != count) {
      printf(" ");
    }
  }
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x14(%rbp)
  movl   $0x2,-0x10(%rbp)
  movl   $0x3,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x10(%rbp),%eax
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0x3,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  movl   $0xffffffff,-0x14(%rbp)
  movl   $0xfffffffe,-0x10(%rbp)
  movl   $0xfffffffd,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x10(%rbp),%eax
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0x3,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x14(%rbp),%eax
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x10(%rbp),%eax
  cmp    $0x2,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0x3,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int main() {
  int b[3] = {1, 2, 3};
#pragma omp task firstprivate(b)
  {
    assert(b[0] == 1);
    assert(b[1] == 2);
    assert(b[2] == 3);
    b[0] = -1;
    b[1] = -2;
    b[2] = -3;
  }
  assert(b[0] == 1);
  assert(b[1] == 2);
  assert(b[2] == 3);
#pragma omp taskwait
  assert(b[0] == 1);
  assert(b[1] == 2);
  assert(b[2] == 3);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int b[3] = {1, 2, 3};
#pragma omp task firstprivate(b)
  {
    assert(b[0] == 1);
    assert(b[1] == 2);
    assert(b[2] == 3);
    b[0] = -1;
    b[1] = -2;
    b[2] = -3;
  }
  assert(b[0] == 1);
  assert(b[1] == 2);
  assert(b[2] == 3);
#pragma omp taskwait
  assert(b[0] == 1);
  assert(b[1] == 2);
  assert(b[2] == 3);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int b[3] = {1, 2, 3};
#pragma omp task firstprivate(b)
  {
    assert(b[0] == 1);
    assert(b[1] == 2);
    assert(b[2] == 3);
    b[0] = -1;
    b[1] = -2;
    b[2] = -3;
  }
  assert(b[0] == 1);
  assert(b[1] == 2);
  assert(b[2] == 3);
#pragma omp taskwait
  assert(b[0] == 1);
  assert(b[1] == 2);
  assert(b[2] == 3);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x1,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int b[3] = {1, 2, 3};
#pragma omp task firstprivate(b)
  {
    assert(b[0] == 1);
    assert(b[1] == 2);
    assert(b[2] == 3);
    b[0] = -1;
    b[1] = -2;
    b[2] = -3;
  }
  assert(b[0] == 1);
  assert(b[1] == 2);
  assert(b[2] == 3);
#pragma omp taskwait
  assert(b[0] == 1);
  assert(b[1] == 2);
  assert(b[2] == 3);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<trim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L66
  mov    $0x0,%eax
  jmp    L46
L66:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L24
L50:
  addq   $0x1,-0x10(%rbp)
L24:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L59
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L50
  jmp    L59
L49:
  subq   $0x1,-0x8(%rbp)
L59:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jb     L79
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L49
L79:
  mov    -0x8(%rbp),%rax
  sub    -0x10(%rbp),%rax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x30(%rbp),%rax
L46:
  leave
  ret
```

assistant: ```c
char *trim(char *strBuf, char *strRst) {
  int iLen;
  char *strFirst = strBuf;
  char *strLast;
  if (NULL == strBuf) {
    return NULL;
  }
  strLast = strFirst + strlen(strFirst) - 1;
  while ('\0' != *strFirst && isspace(*strFirst)) {
    strFirst++;
  }
  while (strLast >= strFirst && isspace(*strLast)) {
    strLast--;
  }
  iLen = strLast - strFirst + 1;
  memmove(strRst, strFirst, iLen);
  strRst[iLen] = '\0';
  return strRst;
}
```
user: Translate the below assembly code into C programming language:

```asm
<trim>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L7
  mov    %rsi,%r12
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rbp
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L50
L45:
  movsbl %dil,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L50
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L45
L50:
  cmp    %rbx,%rbp
  jb     L1
L38:
  movsbl 0x0(%rbp),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L1
  sub    $0x1,%rbp
  cmp    %rbx,%rbp
  jae    L38
L1:
  sub    %rbx,%rbp
  add    $0x1,%ebp
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <memmove@plt>
  movslq %ebp,%rbp
  movb   $0x0,(%r12,%rbp,1)
  mov    %r12,%rax
L19:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  mov    %rdi,%rax
  jmp    L19
```

assistant: ```c
char *trim(char *strBuf, char *strRst) {
  int iLen;
  char *strFirst = strBuf;
  char *strLast;
  if (NULL == strBuf) {
    return NULL;
  }
  strLast = strFirst + strlen(strFirst) - 1;
  while ('\0' != *strFirst && isspace(*strFirst)) {
    strFirst++;
  }
  while (strLast >= strFirst && isspace(*strLast)) {
    strLast--;
  }
  iLen = strLast - strFirst + 1;
  memmove(strRst, strFirst, iLen);
  strRst[iLen] = '\0';
  return strRst;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<trim>:
  endbr64
  test   %rdi,%rdi
  je     L21
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  movsbl 0x0(%rbp),%edi
  cltq
  lea    -0x1(%rbp,%rax,1),%rbx
  test   %dil,%dil
  jne    L26
  jmp    L78
  xchg   %ax,%ax
L3:
  movsbl 0x1(%rbp),%edi
  add    $0x1,%rbp
  test   %dil,%dil
  je     L78
L26:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L3
  cmp    %rbp,%rbx
  jb     L64
  nopl   0x0(%rax)
L70:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L64
  sub    $0x1,%rbx
L78:
  cmp    %rbp,%rbx
  jae    L70
L64:
  sub    %rbp,%rbx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  mov    %ebx,%edx
  movslq %ebx,%rbx
  call   <memmove@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L21:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *trim(char *strBuf, char *strRst) {
  int iLen;
  char *strFirst = strBuf;
  char *strLast;
  if (NULL == strBuf) {
    return NULL;
  }
  strLast = strFirst + strlen(strFirst) - 1;
  while ('\0' != *strFirst && isspace(*strFirst)) {
    strFirst++;
  }
  while (strLast >= strFirst && isspace(*strLast)) {
    strLast--;
  }
  iLen = strLast - strFirst + 1;
  memmove(strRst, strFirst, iLen);
  strRst[iLen] = '\0';
  return strRst;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<trim>:
  endbr64
  test   %rdi,%rdi
  je     L60
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  movsbl 0x0(%rbp),%edi
  cltq
  lea    -0x1(%rbp,%rax,1),%rbx
  test   %dil,%dil
  jne    L35
  jmp    L56
  xchg   %ax,%ax
L65:
  movsbl 0x1(%rbp),%edi
  add    $0x1,%rbp
  test   %dil,%dil
  je     L56
L35:
  call   <isspace@plt>
  test   %rax,%rax
  jne    L65
  cmp    %rbp,%rbx
  jb     L0
  nopl   0x0(%rax)
L19:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L0
  sub    $0x1,%rbx
L56:
  cmp    %rbp,%rbx
  jae    L19
L0:
  sub    %rbp,%rbx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  mov    %ebx,%edx
  movslq %ebx,%rbx
  call   <memmove@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L60:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *trim(char *strBuf, char *strRst) {
  int iLen;
  char *strFirst = strBuf;
  char *strLast;
  if (NULL == strBuf) {
    return NULL;
  }
  strLast = strFirst + strlen(strFirst) - 1;
  while ('\0' != *strFirst && isspace(*strFirst)) {
    strFirst++;
  }
  while (strLast >= strFirst && isspace(*strLast)) {
    strLast--;
  }
  iLen = strLast - strFirst + 1;
  memmove(strRst, strFirst, iLen);
  strRst[iLen] = '\0';
  return strRst;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_match>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0x24(%rbp)
  mov    -0x50(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x48(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x50(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movb   $0x0,(%rax)
  lea    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x10,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  cmpq   $0x0,-0x10(%rbp)
  je     L32
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cltq
  cmp    %rax,-0x10(%rbp)
  jne    L32
  movl   $0x0,-0x24(%rbp)
  jmp    L39
L32:
  movl   $0xffffffff,-0x24(%rbp)
L39:
  mov    -0x24(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
int get_match(const char *cell_ref, const char *cell_rel, size_t start, size_t len) {
  int ret = -1;
  size_t size_ref, size_rel;
  char *remaining_str;
  unsigned long val_integer;
  char *tmp_cell_rel;
  ;
  tmp_cell_rel = malloc(len + 1);
  strncpy(tmp_cell_rel, &cell_ref[start], len);
  tmp_cell_rel[2] = '\0';
  val_integer = strtoul(tmp_cell_rel, &remaining_str, 16);
  free(tmp_cell_rel);
  ;
  if (val_integer != 0 && val_integer == (strlen(cell_rel) / 2)) {
    ;
    ret = 0;
  } else
    ret = -1;
  return ret;
}
```