ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x340,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x4,-0x338(%rbp)
  movl   $0x4,-0x334(%rbp)
  movl   $0x0,-0x340(%rbp)
  jmp    L118
L4:
  movl   $0x0,-0x33c(%rbp)
  jmp    L127
L111:
  lea    -0x330(%rbp),%rcx
  mov    -0x33c(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x340(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rsi,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x33c(%rbp)
L127:
  mov    -0x33c(%rbp),%eax
  cmp    -0x334(%rbp),%eax
  jl     L111
  addl   $0x1,-0x340(%rbp)
L118:
  mov    -0x340(%rbp),%eax
  cmp    -0x338(%rbp),%eax
  jl     L4
  movl   $0x0,-0x340(%rbp)
  jmp    L80
L93:
  movl   $0x0,-0x33c(%rbp)
  jmp    L103
L88:
  mov    -0x33c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x340(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  mov    -0x330(%rbp,%rax,4),%edx
  mov    -0x340(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x33c(%rbp),%eax
  movslq %eax,%rcx
  mov    %rcx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  add    %rax,%rax
  add    %rsi,%rax
  mov    %edx,-0x1a0(%rbp,%rax,4)
  addl   $0x1,-0x33c(%rbp)
L103:
  mov    -0x33c(%rbp),%eax
  cmp    -0x334(%rbp),%eax
  jl     L88
  addl   $0x1,-0x340(%rbp)
L80:
  mov    -0x340(%rbp),%eax
  cmp    -0x338(%rbp),%eax
  jl     L93
  movl   $0x0,-0x340(%rbp)
  jmp    L25
L42:
  movl   $0x0,-0x33c(%rbp)
  jmp    L82
L123:
  mov    -0x33c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x340(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rcx,%rax
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x33c(%rbp)
L82:
  mov    -0x33c(%rbp),%eax
  cmp    -0x338(%rbp),%eax
  jl     L123
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x340(%rbp)
L25:
  mov    -0x340(%rbp),%eax
  cmp    -0x334(%rbp),%eax
  jl     L42
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
type: string
name: retrieve_data
arguments: label: D16
type: string

tool: D3: %d

tool: D16: %d	

assistant: ```c
int main() {
  int a = 4, c = 4, i, j, m[10][10], t[10][10];
  for (i = 0; i < a; i++) {
    for (j = 0; j < c; j++)
      scanf("%d", &m[i][j]);
  }
  for (i = 0; i < a; i++) {
    for (j = 0; j < c; j++)
      t[j][i] = m[i][j];
  }
  for (i = 0; i < c; i++) {
    for (j = 0; j < a; j++)
      printf("%d\t", t[i][j]);
    printf("\n");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x330,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x328(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x10(%rsp),%rbp
  mov    $0x0,%r14d
  lea    D16(%rip),%r13
L8:
  lea    -0x10(%rbp),%rbx
L22:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L22
  add    $0xa,%r14
  add    $0x28,%rbp
  cmp    $0x28,%r14
  jne    L8
  lea    0x190(%rsp),%rbp
  lea    0xa0(%r12),%rcx
  mov    %rbp,%rax
L58:
  mov    (%r12),%edx
  mov    %edx,(%rax)
  mov    0x4(%r12),%edx
  mov    %edx,0x28(%rax)
  mov    0x8(%r12),%edx
  mov    %edx,0x50(%rax)
  mov    0xc(%r12),%edx
  mov    %edx,0x78(%rax)
  add    $0x28,%r12
  add    $0x4,%rax
  cmp    %rcx,%r12
  jne    L58
  lea    0xa0(%rbp),%r13
  lea    D0(%rip),%r12
L42:
  mov    $0x0,%ebx
L44:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x4,%rbx
  jne    L44
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x28,%rbp
  cmp    %r13,%rbp
  jne    L42
  mov    0x328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    $0x0,%eax
  add    $0x330,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D16
data_type: string
name: disassemble_data
arguments: data_label: D0
data_type: string

tool: D16: %d

tool: D0: %d	

assistant: ```c
int main() {
  int a = 4, c = 4, i, j, m[10][10], t[10][10];
  for (i = 0; i < a; i++) {
    for (j = 0; j < c; j++)
      scanf("%d", &m[i][j]);
  }
  for (i = 0; i < a; i++) {
    for (j = 0; j < c; j++)
      t[j][i] = m[i][j];
  }
  for (i = 0; i < c; i++) {
    for (j = 0; j < a; j++)
      printf("%d\t", t[i][j]);
    printf("\n");
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x330,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x328(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x10(%rsp),%rbp
L2:
  lea    -0x10(%rbp),%rbx
L0:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbp,%rbx
  jne    L0
  add    $0xa,%r14
  lea    0x28(%rbx),%rbp
  cmp    $0x28,%r14
  jne    L2
  lea    0x190(%rsp),%rbp
  lea    0xa0(%r12),%rcx
  mov    %rbp,%rax
L30:
  mov    (%r12),%edx
  add    $0x28,%r12
  add    $0x4,%rax
  mov    %edx,-0x4(%rax)
  mov    -0x24(%r12),%edx
  mov    %edx,0x24(%rax)
  mov    -0x20(%r12),%edx
  mov    %edx,0x4c(%rax)
  mov    -0x1c(%r12),%edx
  mov    %edx,0x74(%rax)
  cmp    %rcx,%r12
  jne    L30
  lea    0xa0(%rbp),%r13
  lea    D5(%rip),%r12
L40:
  xor    %ebx,%ebx
L18:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x4,%rbx
  jne    L18
  mov    $0xa,%edi
  add    $0x28,%rbp
  call   <putchar@plt>
  cmp    %rbp,%r13
  jne    L40
  mov    0x328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0x330,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D0
type: string
name: parse_data
arguments: data_tag: D5
type: string

tool: D0: %d

tool: D5: %d	

assistant: ```c
int main() {
  int a = 4, c = 4, i, j, m[10][10], t[10][10];
  for (i = 0; i < a; i++) {
    for (j = 0; j < c; j++)
      scanf("%d", &m[i][j]);
  }
  for (i = 0; i < a; i++) {
    for (j = 0; j < c; j++)
      t[j][i] = m[i][j];
  }
  for (i = 0; i < c; i++) {
    for (j = 0; j < a; j++)
      printf("%d\t", t[i][j]);
    printf("\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x338,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x328(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
L16:
  lea    -0x10(%rbp),%rbx
L15:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbx,%rbp
  jne    L15
  add    $0xa,%r13
  add    $0x28,%rbp
  cmp    $0x28,%r13
  jne    L16
  movd   0x78(%rsp),%xmm2
  movd   0x28(%rsp),%xmm3
  lea    0x190(%rsp),%rbp
  lea    0x230(%rsp),%r13
  movd   0x50(%rsp),%xmm1
  movd   (%rsp),%xmm0
  lea    D2(%rip),%r12
  movd   0x7c(%rsp),%xmm4
  movd   0x2c(%rsp),%xmm5
  punpckldq %xmm2,%xmm1
  punpckldq %xmm3,%xmm0
  movd   0x30(%rsp),%xmm7
  movd   0x80(%rsp),%xmm6
  punpcklqdq %xmm1,%xmm0
  movd   0x54(%rsp),%xmm1
  movd   0x84(%rsp),%xmm2
  movaps %xmm0,0x190(%rsp)
  movd   0x4(%rsp),%xmm0
  movd   0x34(%rsp),%xmm3
  punpckldq %xmm4,%xmm1
  punpckldq %xmm5,%xmm0
  punpcklqdq %xmm1,%xmm0
  movd   0x58(%rsp),%xmm1
  movups %xmm0,0x1b8(%rsp)
  movd   0x8(%rsp),%xmm0
  punpckldq %xmm6,%xmm1
  punpckldq %xmm7,%xmm0
  punpcklqdq %xmm1,%xmm0
  movd   0x5c(%rsp),%xmm1
  movaps %xmm0,0x1e0(%rsp)
  movd   0xc(%rsp),%xmm0
  punpckldq %xmm2,%xmm1
  punpckldq %xmm3,%xmm0
  punpcklqdq %xmm1,%xmm0
  movups %xmm0,0x208(%rsp)
L36:
  xor    %ebx,%ebx
L43:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x4,%rbx
  jne    L43
  mov    $0xa,%edi
  add    $0x28,%rbp
  call   <putchar@plt>
  cmp    %rbp,%r13
  jne    L36
  mov    0x328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x338,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D6
data_type: string
name: unpack_data
arguments: data_label: D2
data_type: string

tool: D6: %d

tool: D2: %d	

assistant: ```c
int main() {
  int a = 4, c = 4, i, j, m[10][10], t[10][10];
  for (i = 0; i < a; i++) {
    for (j = 0; j < c; j++)
      scanf("%d", &m[i][j]);
  }
  for (i = 0; i < a; i++) {
    for (j = 0; j < c; j++)
      t[j][i] = m[i][j];
  }
  for (i = 0; i < c; i++) {
    for (j = 0; j < a; j++)
      printf("%d\t", t[i][j]);
    printf("\n");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<exact_square>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L12
L13:
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  cmp    %eax,-0x14(%rbp)
  jne    L32
  mov    $0x1,%eax
  jmp    L16
L32:
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L13
  mov    $0x0,%eax
L16:
  pop    %rbp
  ret
```

assistant: ```c
int exact_square(int num) {
  int i;
  for (i = 1; i <= num; i++) {
    if (i * i == num)
      return 1;
  }
  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
<exact_square>:
  endbr64
  test   %edi,%edi
  jle    L6
  cmp    $0x1,%edi
  je     L4
  mov    $0x2,%edx
L33:
  mov    %edx,%eax
  imul   %edx,%eax
  cmp    %edi,%eax
  je     L17
  add    $0x1,%edx
  cmp    %edx,%edi
  jge    L33
  mov    $0x0,%eax
  ret
L6:
  mov    $0x0,%eax
  ret
L4:
  mov    %edi,%eax
  ret
L17:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int exact_square(int num) {
  int i;
  for (i = 1; i <= num; i++) {
    if (i * i == num)
      return 1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<exact_square>:
  endbr64
  test   %edi,%edi
  jle    L23
  cmp    $0x1,%edi
  je     L29
  mov    $0x2,%eax
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %eax,%edx
  imul   %eax,%edx
  cmp    %edi,%edx
  je     L29
  add    $0x1,%eax
  cmp    %eax,%edi
  jge    L1
L23:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L29:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int exact_square(int num) {
  int i;
  for (i = 1; i <= num; i++) {
    if (i * i == num)
      return 1;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<exact_square>:
  endbr64
  test   %edi,%edi
  jle    L38
  cmp    $0x1,%edi
  je     L22
  mov    $0x2,%eax
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %eax,%edx
  imul   %eax,%edx
  cmp    %edi,%edx
  je     L22
  add    $0x1,%eax
  cmp    %eax,%edi
  jge    L0
L38:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L22:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int exact_square(int num) {
  int i;
  for (i = 1; i <= num; i++) {
    if (i * i == num)
      return 1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <doit@plt>
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
data_type: string

tool: D8: hello


assistant: ```c
int main() {
  doit();
  printf("hello\n");
  exit(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <doit@plt>
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
type_infered: string

tool: D6: hello


assistant: ```c
int main() {
  doit();
  printf("hello\n");
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <doit@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D9
data_type: string

tool: D9: hello


assistant: ```c
int main() {
  doit();
  printf("hello\n");
  exit(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
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <doit@plt>
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D7
type_infered: string

tool: D7: hello


assistant: ```c
int main() {
  doit();
  printf("hello\n");
  exit(0);
}
```
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
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
#pragma omp taskwait noflush
  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
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
#pragma omp taskwait noflush
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

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

assistant: ```c
int main(int argc, char *argv[]) {
#pragma omp taskwait noflush
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

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

assistant: ```c
int main(int argc, char *argv[]) {
#pragma omp taskwait noflush
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L14
L19:
  mov    -0x8(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L14:
  cmpl   $0xa,-0x4(%rbp)
  jle    L19
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D9
type_guessed: string

tool: D9: %d


assistant: ```c
int main() {
  int fact = 1, c;
  for (c = 1; c <= 10; c += 1) {
    fact = fact * c;
  }
  printf("%d\n", fact);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%eax
  mov    $0x1,%esi
L8:
  imul   %eax,%esi
  add    $0x1,%eax
  cmp    $0xb,%eax
  jne    L8
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D7
type_guessed: string

tool: D7: %d


assistant: ```c
int main() {
  int fact = 1, c;
  for (c = 1; c <= 10; c += 1) {
    fact = fact * c;
  }
  printf("%d\n", fact);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%eax
  mov    $0x1,%esi
  nopw   0x0(%rax,%rax,1)
L0:
  imul   %eax,%esi
  add    $0x1,%eax
  cmp    $0xb,%eax
  jne    L0
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D4
datatype: string

tool: D4: %d


assistant: ```c
int main() {
  int fact = 1, c;
  for (c = 1; c <= 10; c += 1) {
    fact = fact * c;
  }
  printf("%d\n", fact);
  return 0;
}
```
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    $0x375f00,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D0
type_infered: string

tool: D0: %d


assistant: ```c
int main() {
  int fact = 1, c;
  for (c = 1; c <= 10; c += 1) {
    fact = fact * c;
  }
  printf("%d\n", fact);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<lt_split_name_value>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L4
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L20
L4:
  mov    $0x1,%eax
  jmp    L6
L20:
  mov    -0x28(%rbp),%rax
  mov    $0x3d,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L35
  mov    $0x1,%eax
  jmp    L6
L35:
  addq   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <xstrdup@plt>
  cltq
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <xmalloc@plt>
  cltq
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x28(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  sub    $0x1,%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    $0x0,%eax
L6:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int lt_split_name_value(const char *arg, char **name, char **value) {
  const char *p;
  int len;
  if (!arg || !*arg)
    return 1;
  p = strchr(arg, (int)'=');
  if (!p)
    return 1;
  *value = xstrdup(++p);
  len = strlen(arg) - strlen(*value);
  *name = ((char *)xmalloc((len) * sizeof(char)));
  strncpy(*name, arg, len - 1);
  (*name)[len - 1] = '\0';
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<lt_split_name_value>:
  endbr64
  test   %rdi,%rdi
  je     L10
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    $0x1,%edx
  cmpb   $0x0,(%rdi)
  je     L15
  mov    $0x3d,%esi
  call   <strchr@plt>
  mov    $0x1,%edx
  test   %rax,%rax
  je     L15
  lea    0x1(%rax),%rdi
  mov    $0x0,%eax
  call   <xstrdup@plt>
  movslq %eax,%r13
  mov    %r13,0x0(%rbp)
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rax,%rbp
  mov    %r13,%rdi
  call   <strlen@plt>
  sub    %eax,%ebp
  movslq %ebp,%r13
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <xmalloc@plt>
  movslq %eax,%rdi
  mov    %rdi,(%r12)
  lea    -0x1(%rbp),%edx
  movslq %edx,%rdx
  mov    %rbx,%rsi
  call   <strncpy@plt>
  mov    (%r12),%rax
  movb   $0x0,-0x1(%rax,%r13,1)
  mov    $0x0,%edx
L15:
  mov    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  mov    $0x1,%edx
  mov    %edx,%eax
  ret
```

assistant: ```c
int lt_split_name_value(const char *arg, char **name, char **value) {
  const char *p;
  int len;
  if (!arg || !*arg)
    return 1;
  p = strchr(arg, (int)'=');
  if (!p)
    return 1;
  *value = xstrdup(++p);
  len = strlen(arg) - strlen(*value);
  *name = ((char *)xmalloc((len) * sizeof(char)));
  strncpy(*name, arg, len - 1);
  (*name)[len - 1] = '\0';
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<lt_split_name_value>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L6
  cmpb   $0x0,(%rdi)
  mov    %rdi,%rbp
  mov    $0x1,%r13d
  je     L2
  mov    %rsi,%r12
  mov    $0x3d,%esi
  mov    %rdx,%rbx
  call   <strchr@plt>
  test   %rax,%rax
  je     L2
  lea    0x1(%rax),%rdi
  xor    %eax,%eax
  call   <xstrdup@plt>
  mov    %rbp,%rdi
  movslq %eax,%r13
  mov    %r13,(%rbx)
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  sub    %eax,%ebx
  xor    %eax,%eax
  movslq %ebx,%r13
  mov    %r13,%rdi
  call   <xmalloc@plt>
  lea    -0x1(%rbx),%edx
  mov    %rbp,%rsi
  movslq %eax,%rdi
  movslq %edx,%rdx
  mov    %rdi,(%r12)
  call   <strncpy@plt>
  mov    (%r12),%rax
  movb   $0x0,-0x1(%rax,%r13,1)
  xor    %r13d,%r13d
L2:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L6:
  add    $0x8,%rsp
  mov    $0x1,%r13d
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int lt_split_name_value(const char *arg, char **name, char **value) {
  const char *p;
  int len;
  if (!arg || !*arg)
    return 1;
  p = strchr(arg, (int)'=');
  if (!p)
    return 1;
  *value = xstrdup(++p);
  len = strlen(arg) - strlen(*value);
  *name = ((char *)xmalloc((len) * sizeof(char)));
  strncpy(*name, arg, len - 1);
  (*name)[len - 1] = '\0';
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<lt_split_name_value>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L13
  cmpb   $0x0,(%rdi)
  mov    %rdi,%rbp
  mov    $0x1,%r13d
  je     L18
  mov    %rsi,%r12
  mov    $0x3d,%esi
  mov    %rdx,%rbx
  call   <strchr@plt>
  test   %rax,%rax
  je     L18
  lea    0x1(%rax),%rdi
  xor    %eax,%eax
  call   <xstrdup@plt>
  mov    %rbp,%rdi
  movslq %eax,%r13
  mov    %r13,(%rbx)
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%rbx
  call   <strlen@plt>
  sub    %eax,%ebx
  xor    %eax,%eax
  movslq %ebx,%r13
  mov    %r13,%rdi
  call   <xmalloc@plt>
  lea    -0x1(%rbx),%edx
  mov    %rbp,%rsi
  movslq %eax,%rdi
  movslq %edx,%rdx
  mov    %rdi,(%r12)
  call   <strncpy@plt>
  mov    (%r12),%rax
  movb   $0x0,-0x1(%rax,%r13,1)
  xor    %r13d,%r13d
L18:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L13:
  add    $0x8,%rsp
  mov    $0x1,%r13d
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int lt_split_name_value(const char *arg, char **name, char **value) {
  const char *p;
  int len;
  if (!arg || !*arg)
    return 1;
  p = strchr(arg, (int)'=');
  if (!p)
    return 1;
  *value = xstrdup(++p);
  len = strlen(arg) - strlen(*value);
  *name = ((char *)xmalloc((len) * sizeof(char)));
  strncpy(*name, arg, len - 1);
  (*name)[len - 1] = '\0';
  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
<quickSortTest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    $0x0,%eax
  call   <getSystemTime@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quickSort@plt>
  mov    $0x0,%eax
  call   <getSystemTime@plt>
  sub    -0x8(%rbp),%rax
  mov    -0x20(%rbp),%edx
  lea    0x1(%rdx),%ecx
  mov    %rax,%rdx
  mov    %ecx,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
data_type: string

tool: D7: quickSort count %d use time: %ld 


assistant: ```c
void quickSortTest(int *arr, int start, int end) {
  long startTime = getSystemTime();
  quickSort(arr, start, end);
  printf("quickSort count %d use time: %ld \n", (end + 1), (getSystemTime() - startTime));
}
```