ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<getComplex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    D2(%rip),%eax
  lea    0x1(%rax),%edx
  mov    %edx,D7(%rip)
  test   %eax,%eax
  je     L8
  call   <abort@plt>
L8:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    %ecx,%ecx
  mov    %rax,%rsi
  movabs $0xffffffff00000000,%rax
  and    %rsi,%rax
  or     %rcx,%rax
  mov    %edx,%edx
  shl    $0x20,%rdx
  mov    %eax,%eax
  or     %rdx,%rax
  leave
  ret
```

assistant: ```c
_Complex int getComplex(_Complex int val) {
  static int count;
  if (count++)
    abort();
  return val;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<getComplex>:
  endbr64
  mov    %edi,%eax
  shr    $0x20,%rdi
  mov    D14(%rip),%edx
  lea    0x1(%rdx),%ecx
  mov    %ecx,D9(%rip)
  test   %edx,%edx
  jne    L1
  shl    $0x20,%rdi
  mov    %eax,%eax
  or     %rdi,%rax
  ret
L1:
  sub    $0x8,%rsp
  call   <abort@plt>
```

assistant: ```c
_Complex int getComplex(_Complex int val) {
  static int count;
  if (count++)
    abort();
  return val;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<getComplex>:
  endbr64
  mov    D2(%rip),%edx
  mov    %edi,%eax
  shr    $0x20,%rdi
  lea    0x1(%rdx),%ecx
  mov    %ecx,D5(%rip)
  test   %edx,%edx
  jne    1060 <getComplex.cold>
  shl    $0x20,%rdi
  or     %rdi,%rax
  ret
```

assistant: ```c
_Complex int getComplex(_Complex int val) {
  static int count;
  if (count++)
    abort();
  return val;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getComplex>:
  endbr64
  mov    D7(%rip),%edx
  mov    %edi,%eax
  shr    $0x20,%rdi
  lea    0x1(%rdx),%ecx
  mov    %ecx,D11(%rip)
  test   %edx,%edx
  jne    1060 <getComplex.cold>
  shl    $0x20,%rdi
  or     %rdi,%rax
  ret
```

assistant: ```c
_Complex int getComplex(_Complex int val) {
  static int count;
  if (count++)
    abort();
  return val;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fibSumEven>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movq   $0x2,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x2,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L13
L4:
  mov    -0x10(%rbp),%rax
  add    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L13:
  cmpq   $0x3d08ff,-0x10(%rbp)
  jle    L4
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int fibSumEven() {
  long fib3 = 2;
  long fib6 = 0;
  long result = 2;
  long summed = 0;
  while (result < 4000000) {
    summed += result;
    result = 4 * fib3 + fib6;
    fib6 = fib3;
    fib3 = result;
  }
  return summed;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fibSumEven>:
  endbr64
  mov    $0x0,%eax
  mov    $0x0,%esi
  mov    $0x2,%edx
  jmp    L12
L11:
  mov    %rcx,%rdx
L12:
  add    %rdx,%rax
  lea    (%rsi,%rdx,4),%rcx
  mov    %rdx,%rsi
  cmp    $0x3d08ff,%rcx
  jle    L11
  ret
```

assistant: ```c
int fibSumEven() {
  long fib3 = 2;
  long fib6 = 0;
  long result = 2;
  long summed = 0;
  while (result < 4000000) {
    summed += result;
    result = 4 * fib3 + fib6;
    fib6 = fib3;
    fib3 = result;
  }
  return summed;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fibSumEven>:
  endbr64
  xor    %eax,%eax
  xor    %esi,%esi
  mov    $0x2,%edx
  jmp    L15
  nop
L17:
  mov    %rcx,%rdx
L15:
  lea    (%rsi,%rdx,4),%rcx
  add    %rdx,%rax
  mov    %rdx,%rsi
  cmp    $0x3d08ff,%rcx
  jle    L17
  ret
```

assistant: ```c
int fibSumEven() {
  long fib3 = 2;
  long fib6 = 0;
  long result = 2;
  long summed = 0;
  while (result < 4000000) {
    summed += result;
    result = 4 * fib3 + fib6;
    fib6 = fib3;
    fib3 = result;
  }
  return summed;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fibSumEven>:
  endbr64
  xor    %eax,%eax
  xor    %esi,%esi
  mov    $0x2,%edx
  jmp    L0
  nop
L10:
  mov    %rcx,%rdx
L0:
  lea    (%rsi,%rdx,4),%rcx
  add    %rdx,%rax
  mov    %rdx,%rsi
  cmp    $0x3d08ff,%rcx
  jle    L10
  ret
```

assistant: ```c
int fibSumEven() {
  long fib3 = 2;
  long fib6 = 0;
  long result = 2;
  long summed = 0;
  while (result < 4000000) {
    summed += result;
    result = 4 * fib3 + fib6;
    fib6 = fib3;
    fib3 = result;
  }
  return summed;
}
```
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
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x42,-0x22(%rbp)
  movb   $0x2b,-0x21(%rbp)
  lea    -0x21(%rbp),%rcx
  lea    -0x22(%rbp),%rax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <swap@plt>
  movzbl -0x21(%rbp),%eax
  movsbl %al,%edx
  movzbl -0x22(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0xa,-0x20(%rbp)
  movl   $0x6,-0x1c(%rbp)
  lea    -0x1c(%rbp),%rcx
  lea    -0x20(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <swap@plt>
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsd  D13(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  D26(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  lea    -0x10(%rbp),%rcx
  lea    -0x18(%rbp),%rax
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <swap@plt>
  movsd  -0x10(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type: string
name: unpack_data
arguments: data_tag: D45
type: string
name: unpack_data
arguments: data_tag: D13
type: f64
name: unpack_data
arguments: data_tag: D26
type: f64
name: unpack_data
arguments: data_tag: D9
type: string

tool: D6: %c %c


tool: D45: %d %d


tool: D13: 3.14

tool: D26: 1.23567

tool: D9: %.f2 %.f2


assistant: ```c
int main() {
  char letter = 'B', symbol = '+';
  swap(&letter, &symbol, sizeof(char));
  printf("%c %c\n", letter, symbol);
  int a = 10, b = 6;
  swap(&a, &b, sizeof(int));
  printf("%d %d\n", a, b);
  double d = 3.14, f = 1.23567;
  swap(&d, &f, sizeof(double));
  printf("%.f2 %.f2\n", d, f);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movb   $0x42,0xe(%rsp)
  movb   $0x2b,0xf(%rsp)
  lea    0xf(%rsp),%rsi
  lea    0xe(%rsp),%rdi
  mov    $0x1,%edx
  call   <swap@plt>
  movsbl 0xf(%rsp),%edx
  movsbl 0xe(%rsp),%esi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0xa,0x10(%rsp)
  movl   $0x6,0x14(%rsp)
  lea    0x14(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    $0x4,%edx
  mov    $0x0,%eax
  call   <swap@plt>
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%esi
  lea    D49(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    D9(%rip),%rax
  mov    %rax,0x18(%rsp)
  mov    D39(%rip),%rax
  mov    %rax,0x20(%rsp)
  lea    0x20(%rsp),%rsi
  lea    0x18(%rsp),%rdi
  mov    $0x8,%edx
  mov    $0x0,%eax
  call   <swap@plt>
  movsd  0x20(%rsp),%xmm1
  movsd  0x18(%rsp),%xmm0
  lea    D42(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x38,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D2
datatype: string
name: parse_data
arguments: tag: D49
datatype: string
name: parse_data
arguments: tag: D9
datatype: f64
name: parse_data
arguments: tag: D39
datatype: f64
name: parse_data
arguments: tag: D42
datatype: string

tool: D2: %c %c


tool: D49: %d %d


tool: D9: 3.14

tool: D39: 1.23567

tool: D42: %.f2 %.f2


assistant: ```c
int main() {
  char letter = 'B', symbol = '+';
  swap(&letter, &symbol, sizeof(char));
  printf("%c %c\n", letter, symbol);
  int a = 10, b = 6;
  swap(&a, &b, sizeof(int));
  printf("%d %d\n", a, b);
  double d = 3.14, f = 1.23567;
  swap(&d, &f, sizeof(double));
  printf("%.f2 %.f2\n", d, f);
  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    $0x38,%rsp
  mov    $0x1,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xf(%rsp),%rsi
  lea    0xe(%rsp),%rdi
  movb   $0x42,0xe(%rsp)
  movb   $0x2b,0xf(%rsp)
  call   <swap@plt>
  movsbl 0xf(%rsp),%edx
  movsbl 0xe(%rsp),%esi
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  xor    %eax,%eax
  mov    $0x4,%edx
  movl   $0xa,0x10(%rsp)
  movl   $0x6,0x14(%rsp)
  call   <swap@plt>
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%esi
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    0x18(%rsp),%rdi
  lea    0x20(%rsp),%rsi
  mov    D5(%rip),%rax
  mov    $0x8,%edx
  mov    %rax,0x18(%rsp)
  mov    D21(%rip),%rax
  mov    %rax,0x20(%rsp)
  xor    %eax,%eax
  call   <swap@plt>
  movsd  0x20(%rsp),%xmm1
  movsd  0x18(%rsp),%xmm0
  lea    D35(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D11
datatype: string
name: unpack_data
arguments: data_tag: D16
datatype: string
name: unpack_data
arguments: data_tag: D5
datatype: f64
name: unpack_data
arguments: data_tag: D21
datatype: f64
name: unpack_data
arguments: data_tag: D35
datatype: string

tool: D11: %c %c


tool: D16: %d %d


tool: D5: 3.14

tool: D21: 1.23567

tool: D35: %.f2 %.f2


assistant: ```c
int main() {
  char letter = 'B', symbol = '+';
  swap(&letter, &symbol, sizeof(char));
  printf("%c %c\n", letter, symbol);
  int a = 10, b = 6;
  swap(&a, &b, sizeof(int));
  printf("%d %d\n", a, b);
  double d = 3.14, f = 1.23567;
  swap(&d, &f, sizeof(double));
  printf("%.f2 %.f2\n", d, f);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    $0x1,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xf(%rsp),%rsi
  lea    0xe(%rsp),%rdi
  movb   $0x42,0xe(%rsp)
  movb   $0x2b,0xf(%rsp)
  call   <swap@plt>
  movsbl 0xf(%rsp),%edx
  movsbl 0xe(%rsp),%esi
  xor    %eax,%eax
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  xor    %eax,%eax
  mov    $0x4,%edx
  movl   $0xa,0x10(%rsp)
  movl   $0x6,0x14(%rsp)
  call   <swap@plt>
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%esi
  xor    %eax,%eax
  lea    D42(%rip),%rdi
  call   <printf@plt>
  lea    0x18(%rsp),%rdi
  lea    0x20(%rsp),%rsi
  mov    D4(%rip),%rax
  mov    $0x8,%edx
  mov    %rax,0x18(%rsp)
  mov    D43(%rip),%rax
  mov    %rax,0x20(%rsp)
  xor    %eax,%eax
  call   <swap@plt>
  movsd  0x20(%rsp),%xmm1
  movsd  0x18(%rsp),%xmm0
  lea    D27(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D15
datatype: string
name: disassemble_data
arguments: label: D42
datatype: string
name: disassemble_data
arguments: label: D4
datatype: f64
name: disassemble_data
arguments: label: D43
datatype: f64
name: disassemble_data
arguments: label: D27
datatype: string

tool: D15: %c %c


tool: D42: %d %d


tool: D4: 3.14

tool: D43: 1.23567

tool: D27: %.f2 %.f2


assistant: ```c
int main() {
  char letter = 'B', symbol = '+';
  swap(&letter, &symbol, sizeof(char));
  printf("%c %c\n", letter, symbol);
  int a = 10, b = 6;
  swap(&a, &b, sizeof(int));
  printf("%d %d\n", a, b);
  double d = 3.14, f = 1.23567;
  swap(&d, &f, sizeof(double));
  printf("%.f2 %.f2\n", d, f);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<power>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movq   $0x1,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L7:
  mov    -0x14(%rbp),%eax
  cltq
  mov    -0x8(%rbp),%rdx
  imul   %rdx,%rax
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L3:
  mov    -0xc(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L7
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long power(int x, int y) {
  int i;
  long result = 1;
  for (i = 0; i < y; i++) {
    result *= x;
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<power>:
  endbr64
  test   %esi,%esi
  jle    L21
  mov    $0x1,%edx
  mov    $0x0,%eax
  movslq %edi,%rdi
L4:
  imul   %rdi,%rdx
  add    $0x1,%eax
  cmp    %eax,%esi
  jne    L4
L14:
  mov    %rdx,%rax
  ret
L21:
  mov    $0x1,%edx
  jmp    L14
```

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

```asm
<power>:
  endbr64
  test   %esi,%esi
  jle    L11
  movslq %edi,%rdi
  mov    $0x1,%r8d
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L7:
  add    $0x1,%eax
  imul   %rdi,%r8
  cmp    %eax,%esi
  jne    L7
  mov    %r8,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  mov    $0x1,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
long power(int x, int y) {
  int i;
  long result = 1;
  for (i = 0; i < y; i++) {
    result *= x;
  }
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<power>:
  endbr64
  test   %esi,%esi
  jle    L7
  movslq %edi,%rdi
  mov    $0x1,%r8d
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L11:
  add    $0x1,%eax
  imul   %rdi,%r8
  cmp    %eax,%esi
  jne    L11
  mov    %r8,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  mov    $0x1,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
long power(int x, int y) {
  int i;
  long result = 1;
  for (i = 0; i < y; i++) {
    result *= x;
  }
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<SaveTokenString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,D12(%rip)
  mov    D12(%rip),%rax
  test   %rax,%rax
  jne    L12
  mov    $0x0,%eax
  jmp    L19
L12:
  mov    D12(%rip),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%rcx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  lea    D130(%rip),%rax
  mov    %rcx,(%rdx,%rax,1)
  mov    D12(%rip),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  lea    D130(%rip),%rax
  mov    (%rcx,%rax,1),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L21
L5:
  addl   $0x1,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,D116(%rip)
  mov    D116(%rip),%rax
  test   %rax,%rax
  je     L37
  mov    D116(%rip),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%rcx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  lea    D130(%rip),%rax
  mov    %rcx,(%rdx,%rax,1)
  mov    D116(%rip),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  lea    D130(%rip),%rax
  mov    (%rcx,%rax,1),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L21
L37:
  mov    D116(%rip),%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  lea    D130(%rip),%rdx
  mov    %rax,(%rcx,%rdx,1)
L21:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  lea    D130(%rip),%rax
  mov    (%rdx,%rax,1),%rax
  test   %rax,%rax
  jne    L5
  lea    D130(%rip),%rax
L19:
  leave
  ret
```

assistant: ```c
char **SaveTokenString(char *instring, char *delimeters) {
  int NDEX;
  static char *token[500], *TOKEN;
  NDEX = 0;
  TOKEN = strtok(instring, delimeters);
  if (TOKEN == NULL)
    return NULL;
  token[NDEX] = (char *)malloc(sizeof(char) * (strlen(TOKEN) + 1));
  strcpy(token[NDEX], TOKEN);
  while (token[NDEX] != NULL) {
    NDEX++;
    TOKEN = strtok(NULL, delimeters);
    if (TOKEN != NULL) {
      token[NDEX] = (char *)malloc(sizeof(char) * (strlen(TOKEN) + 1));
      strcpy(token[NDEX], TOKEN);
    } else
      token[NDEX] = TOKEN;
  }
  return token;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<SaveTokenString>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r12
  call   <strtok@plt>
  mov    %rax,D45(%rip)
  test   %rax,%rax
  je     L58
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,D48(%rip)
  mov    D45(%rip),%rsi
  call   <strcpy@plt>
  lea    D48(%rip),%rax
  cmpq   $0x0,D48(%rip)
  je     L58
  mov    $0x1,%ebx
  mov    %rax,%r14
  lea    -0x8(%rax),%r13
  jmp    L7
L55:
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,(%r14,%rbx,8)
  mov    D45(%rip),%rsi
  call   <strcpy@plt>
  add    $0x1,%rbx
  cmpq   $0x0,0x0(%r13,%rbx,8)
  je     L13
L7:
  mov    %r12,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rdi
  mov    %rax,D89(%rip)
  test   %rax,%rax
  jne    L55
  lea    D48(%rip),%rax
  movslq %ebx,%rbp
  movq   $0x0,(%rax,%rbp,8)
L58:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  lea    D48(%rip),%rax
  jmp    L58
```

assistant: ```c
char **SaveTokenString(char *instring, char *delimeters) {
  int NDEX;
  static char *token[500], *TOKEN;
  NDEX = 0;
  TOKEN = strtok(instring, delimeters);
  if (TOKEN == NULL)
    return NULL;
  token[NDEX] = (char *)malloc(sizeof(char) * (strlen(TOKEN) + 1));
  strcpy(token[NDEX], TOKEN);
  while (token[NDEX] != NULL) {
    NDEX++;
    TOKEN = strtok(NULL, delimeters);
    if (TOKEN != NULL) {
      token[NDEX] = (char *)malloc(sizeof(char) * (strlen(TOKEN) + 1));
      strcpy(token[NDEX], TOKEN);
    } else
      token[NDEX] = TOKEN;
  }
  return token;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<SaveTokenString>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  call   <strtok@plt>
  mov    %rax,D75(%rip)
  test   %rax,%rax
  je     L24
  mov    %rax,%rdi
  lea    D12(%rip),%r13
  mov    $0x1,%ebx
  call   <strlen@plt>
  lea    -0x8(%r13),%r14
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    D75(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,D43(%rip)
  call   <strcpy@plt>
  cmpq   $0x0,D43(%rip)
  jne    L14
  jmp    L49
  cs nopw 0x0(%rax,%rax,1)
L16:
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    D75(%rip),%rsi
  mov    %rax,0x0(%r13,%rbx,8)
  mov    %rax,%rdi
  add    $0x1,%rbx
  call   <strcpy@plt>
  cmpq   $0x0,(%r14,%rbx,8)
  je     L49
L14:
  xor    %edi,%edi
  mov    %r12,%rsi
  movslq %ebx,%rbp
  call   <strtok@plt>
  mov    %rax,D43(%rip)
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L16
  movq   $0x0,0x0(%r13,%rbp,8)
L49:
  lea    D43(%rip),%rax
L24:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char **SaveTokenString(char *instring, char *delimeters) {
  int NDEX;
  static char *token[500], *TOKEN;
  NDEX = 0;
  TOKEN = strtok(instring, delimeters);
  if (TOKEN == NULL)
    return NULL;
  token[NDEX] = (char *)malloc(sizeof(char) * (strlen(TOKEN) + 1));
  strcpy(token[NDEX], TOKEN);
  while (token[NDEX] != NULL) {
    NDEX++;
    TOKEN = strtok(NULL, delimeters);
    if (TOKEN != NULL) {
      token[NDEX] = (char *)malloc(sizeof(char) * (strlen(TOKEN) + 1));
      strcpy(token[NDEX], TOKEN);
    } else
      token[NDEX] = TOKEN;
  }
  return token;
}
```
user: What is the c source code of the assembly code below:

```asm
<SaveTokenString>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  call   <strtok@plt>
  mov    %rax,D78(%rip)
  test   %rax,%rax
  je     L38
  mov    %rax,%rdi
  lea    D22(%rip),%r13
  mov    $0x1,%ebx
  call   <strlen@plt>
  lea    -0x8(%r13),%r14
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    D78(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,D21(%rip)
  call   <strcpy@plt>
  cmpq   $0x0,D21(%rip)
  jne    L10
  jmp    L11
  cs nopw 0x0(%rax,%rax,1)
L20:
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    D78(%rip),%rsi
  mov    %rax,0x0(%r13,%rbx,8)
  mov    %rax,%rdi
  add    $0x1,%rbx
  call   <strcpy@plt>
  cmpq   $0x0,(%r14,%rbx,8)
  je     L11
L10:
  xor    %edi,%edi
  mov    %r12,%rsi
  movslq %ebx,%rbp
  call   <strtok@plt>
  mov    %rax,D21(%rip)
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L20
  movq   $0x0,0x0(%r13,%rbp,8)
L11:
  lea    D21(%rip),%rax
L38:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char **SaveTokenString(char *instring, char *delimeters) {
  int NDEX;
  static char *token[500], *TOKEN;
  NDEX = 0;
  TOKEN = strtok(instring, delimeters);
  if (TOKEN == NULL)
    return NULL;
  token[NDEX] = (char *)malloc(sizeof(char) * (strlen(TOKEN) + 1));
  strcpy(token[NDEX], TOKEN);
  while (token[NDEX] != NULL) {
    NDEX++;
    TOKEN = strtok(NULL, delimeters);
    if (TOKEN != NULL) {
      token[NDEX] = (char *)malloc(sizeof(char) * (strlen(TOKEN) + 1));
      strcpy(token[NDEX], TOKEN);
    } else
      token[NDEX] = TOKEN;
  }
  return token;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x20(%rbp)
  movl   $0x7,-0x1c(%rbp)
  movl   $0x7,-0x18(%rbp)
  movl   $0x8,-0x14(%rbp)
  movl   $0x8,-0x10(%rbp)
  movl   $0xa,-0xc(%rbp)
  lea    -0x2c(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x8,%edx
  mov    $0x6,%esi
  mov    %rax,%rdi
  call   <searchRange@plt>
  mov    %rax,-0x28(%rbp)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int main(void) {
  int arr[] = {
      5, 7, 7, 8, 8, 10,
  };
  int *a, n;
  a = searchRange(arr, sizeof(arr) / sizeof(arr[0]), 8, &n);
  return (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x5,0x10(%rsp)
  movl   $0x7,0x14(%rsp)
  movl   $0x7,0x18(%rsp)
  movl   $0x8,0x1c(%rsp)
  movl   $0x8,0x20(%rsp)
  movl   $0xa,0x24(%rsp)
  lea    0xc(%rsp),%rcx
  lea    0x10(%rsp),%rdi
  mov    $0x8,%edx
  mov    $0x6,%esi
  call   <searchRange@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x38,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int arr[] = {
      5, 7, 7, 8, 8, 10,
  };
  int *a, n;
  a = searchRange(arr, sizeof(arr) / sizeof(arr[0]), 8, &n);
  return (0);
}
```