ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D53(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D67(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D55(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D51(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  addss  0x8(%rsp),%xmm0
  addss  0xc(%rsp),%xmm0
  addss  0x10(%rsp),%xmm0
  addss  0x14(%rsp),%xmm0
  divss  D70(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D65(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D53
data_type: string
name: extract_symbolic_data
arguments: label: D67
data_type: string
name: extract_symbolic_data
arguments: label: D16
data_type: string
name: extract_symbolic_data
arguments: label: D55
data_type: string
name: extract_symbolic_data
arguments: label: D51
data_type: string
name: extract_symbolic_data
arguments: label: D38
data_type: string
name: extract_symbolic_data
arguments: label: D70
data_type: byte[4]
name: extract_symbolic_data
arguments: label: D65
data_type: string

tool: D53: What is value1 ? 

tool: D67: %f

tool: D16: What is value2 ? 

tool: D55: What is value3 ? 

tool: D51: What is value4 ? 

tool: D38: What is value5 ? 

tool: {"D70": [0x00,0x00,0xa0,0x40]}

tool: D65: Average = %f


assistant: ```c
int main() {
  float value1;
  float value2;
  float value3;
  float value4;
  float value5;
  float Average;
  printf("What is value1 ? ");
  scanf("%f", &value1);
  printf("What is value2 ? ");
  scanf("%f", &value2);
  printf("What is value3 ? ");
  scanf("%f", &value3);
  printf("What is value4 ? ");
  scanf("%f", &value4);
  printf("What is value5 ? ");
  scanf("%f", &value5);
  Average = value1;
  Average += value2;
  Average += value3;
  Average += value4;
  Average += value5;
  Average /= 5;
  printf("Average = %f\n", Average);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D32(%rip),%rdi
  lea    D2(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D53(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D43(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D64(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    0x14(%rsp),%rsi
  call   <scanf@plt>
  mov    $0x1,%eax
  lea    D16(%rip),%rdi
  movss  0x4(%rsp),%xmm0
  addss  0x8(%rsp),%xmm0
  addss  0xc(%rsp),%xmm0
  addss  0x10(%rsp),%xmm0
  addss  0x14(%rsp),%xmm0
  divss  D8(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D32
type: string
name: parse_data
arguments: tag: D2
type: string
name: parse_data
arguments: tag: D53
type: string
name: parse_data
arguments: tag: D43
type: string
name: parse_data
arguments: tag: D64
type: string
name: parse_data
arguments: tag: D18
type: string
name: parse_data
arguments: tag: D16
type: string
name: parse_data
arguments: tag: D8
type: byte[4]

tool: D32: What is value1 ? 

tool: D2: %f

tool: D53: What is value2 ? 

tool: D43: What is value3 ? 

tool: D64: What is value4 ? 

tool: D18: What is value5 ? 

tool: D16: Average = %f


tool: {"D8": [0x00,0x00,0xa0,0x40]}

assistant: ```c
int main() {
  float value1;
  float value2;
  float value3;
  float value4;
  float value5;
  float Average;
  printf("What is value1 ? ");
  scanf("%f", &value1);
  printf("What is value2 ? ");
  scanf("%f", &value2);
  printf("What is value3 ? ");
  scanf("%f", &value3);
  printf("What is value4 ? ");
  scanf("%f", &value4);
  printf("What is value5 ? ");
  scanf("%f", &value5);
  Average = value1;
  Average += value2;
  Average += value3;
  Average += value4;
  Average += value5;
  Average /= 5;
  printf("Average = %f\n", Average);
  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
  push   %rbp
  lea    D4(%rip),%rdi
  lea    D59(%rip),%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D76(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D48(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D44(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  lea    0x14(%rsp),%rsi
  call   <scanf@plt>
  mov    $0x1,%eax
  lea    D78(%rip),%rdi
  movss  0x4(%rsp),%xmm0
  addss  0x8(%rsp),%xmm0
  addss  0xc(%rsp),%xmm0
  addss  0x10(%rsp),%xmm0
  addss  0x14(%rsp),%xmm0
  divss  D26(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D4
datatype: string
name: retrieve_data
arguments: tag: D59
datatype: string
name: retrieve_data
arguments: tag: D76
datatype: string
name: retrieve_data
arguments: tag: D48
datatype: string
name: retrieve_data
arguments: tag: D44
datatype: string
name: retrieve_data
arguments: tag: D20
datatype: string
name: retrieve_data
arguments: tag: D78
datatype: string
name: retrieve_data
arguments: tag: D26
datatype: byte[4]

tool: D4: What is value1 ? 

tool: D59: %f

tool: D76: What is value2 ? 

tool: D48: What is value3 ? 

tool: D44: What is value4 ? 

tool: D20: What is value5 ? 

tool: D78: Average = %f


tool: {"D26": [0x00,0x00,0xa0,0x40]}

assistant: ```c
int main() {
  float value1;
  float value2;
  float value3;
  float value4;
  float value5;
  float Average;
  printf("What is value1 ? ");
  scanf("%f", &value1);
  printf("What is value2 ? ");
  scanf("%f", &value2);
  printf("What is value3 ? ");
  scanf("%f", &value3);
  printf("What is value4 ? ");
  scanf("%f", &value4);
  printf("What is value5 ? ");
  scanf("%f", &value5);
  Average = value1;
  Average += value2;
  Average += value3;
  Average += value4;
  Average += value5;
  Average /= 5;
  printf("Average = %f\n", Average);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<wsrep_check_version_symbol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D14(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <dlsym@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L11
  mov    $0x0,%eax
  jmp    L1
L11:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    D12(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <wsrep_check_iface_version@plt>
L1:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D14
data_type: string
name: disassemble_data
arguments: tag: D12
data_type: string

tool: D14: wsrep_interface_version

tool: D12: 25

assistant: ```c
int wsrep_check_version_symbol(void *dlh) {
  char **dlversion = NULL;
  dlversion = (char **)dlsym(dlh, "wsrep_interface_version");
  if (dlversion == NULL)
    return 0;
  return wsrep_check_iface_version(*dlversion, "25");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<wsrep_check_version_symbol>:
  endbr64
  sub    $0x8,%rsp
  lea    D15(%rip),%rsi
  call   <dlsym@plt>
  mov    $0x0,%edx
  test   %rax,%rax
  jne    L5
L13:
  mov    %edx,%eax
  add    $0x8,%rsp
  ret
L5:
  lea    D5(%rip),%rsi
  mov    (%rax),%rdi
  call   <wsrep_check_iface_version@plt>
  mov    %eax,%edx
  jmp    L13
```

tool_calls: name: parse_data
arguments: data_label: D15
type_infered: string
name: parse_data
arguments: data_label: D5
type_infered: string

tool: D15: wsrep_interface_version

tool: D5: 25

assistant: ```c
int wsrep_check_version_symbol(void *dlh) {
  char **dlversion = NULL;
  dlversion = (char **)dlsym(dlh, "wsrep_interface_version");
  if (dlversion == NULL)
    return 0;
  return wsrep_check_iface_version(*dlversion, "25");
}
```
user: Translate the below assembly code into C programming language:

```asm
<wsrep_check_version_symbol>:
  endbr64
  sub    $0x8,%rsp
  lea    D16(%rip),%rsi
  call   <dlsym@plt>
  test   %rax,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L1:
  mov    (%rax),%rdi
  lea    D18(%rip),%rsi
  add    $0x8,%rsp
  jmp    1060 <wsrep_check_iface_version@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D16
type_guessed: string
name: retrieve_data
arguments: data_tag: D18
type_guessed: string

tool: D16: wsrep_interface_version

tool: D18: 25

assistant: ```c
int wsrep_check_version_symbol(void *dlh) {
  char **dlversion = NULL;
  dlversion = (char **)dlsym(dlh, "wsrep_interface_version");
  if (dlversion == NULL)
    return 0;
  return wsrep_check_iface_version(*dlversion, "25");
}
```
user: What is the c source code of the assembly code below?

```asm
<wsrep_check_version_symbol>:
  endbr64
  sub    $0x8,%rsp
  lea    D18(%rip),%rsi
  call   <dlsym@plt>
  test   %rax,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L0:
  mov    (%rax),%rdi
  lea    D12(%rip),%rsi
  add    $0x8,%rsp
  jmp    1060 <wsrep_check_iface_version@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D18
type_infered: string
name: retrieve_data
arguments: tag: D12
type_infered: string

tool: D18: wsrep_interface_version

tool: D12: 25

assistant: ```c
int wsrep_check_version_symbol(void *dlh) {
  char **dlversion = NULL;
  dlversion = (char **)dlsym(dlh, "wsrep_interface_version");
  if (dlversion == NULL)
    return 0;
  return wsrep_check_iface_version(*dlversion, "25");
}
```
user: What is the c source code of the assembly code below:

```asm
<p2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movb   $0x61,-0x1(%rbp)
  movsbl -0x1(%rbp),%eax
  mov    %eax,%edi
  call   <p1@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
int p2(int x, int y) {
  char c;
  c = 'a';
  p1(c);
  return x + y;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<p2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    $0x61,%edi
  call   <p1@plt>
  lea    (%rbx,%rbp,1),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int p2(int x, int y) {
  char c;
  c = 'a';
  p1(c);
  return x + y;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<p2>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  mov    $0x61,%edi
  sub    $0x8,%rsp
  call   <p1@plt>
  add    $0x8,%rsp
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int p2(int x, int y) {
  char c;
  c = 'a';
  p1(c);
  return x + y;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<p2>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  mov    $0x61,%edi
  sub    $0x8,%rsp
  call   <p1@plt>
  add    $0x8,%rsp
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int p2(int x, int y) {
  char c;
  c = 'a';
  p1(c);
  return x + y;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<LDA_absx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <c64_memAbsoluteAddrX@plt>
  mov    $0x0,%eax
  call   <LDA_@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void LDA_absx(void) {
  c64_memAbsoluteAddrX();
  LDA_();
}
```
user: Translate the below assembly code into C programming language:

```asm
<LDA_absx>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <c64_memAbsoluteAddrX@plt>
  mov    $0x0,%eax
  call   <LDA_@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void LDA_absx(void) {
  c64_memAbsoluteAddrX();
  LDA_();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<LDA_absx>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <c64_memAbsoluteAddrX@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <LDA_@plt>
```

assistant: ```c
void LDA_absx(void) {
  c64_memAbsoluteAddrX();
  LDA_();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<LDA_absx>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <c64_memAbsoluteAddrX@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <LDA_@plt>
```

assistant: ```c
void LDA_absx(void) {
  c64_memAbsoluteAddrX();
  LDA_();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<revComplement>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L21
L15:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x19(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x18(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x28(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <comp@plt>
  mov    %al,(%rbx)
  movsbl -0x19(%rbp),%eax
  mov    -0x14(%rbp),%edx
  sub    -0x18(%rbp),%edx
  movslq %edx,%rdx
  lea    -0x1(%rdx),%rcx
  mov    -0x28(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <comp@plt>
  mov    %al,(%rbx)
  addl   $0x1,-0x18(%rbp)
L21:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x18(%rbp)
  jl     L15
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L22
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x14(%rbp),%edx
  mov    %edx,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%edx
  sar    %edx
  movslq %edx,%rcx
  mov    -0x28(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <comp@plt>
  mov    %al,(%rbx)
L22:
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void revComplement(char *a) {
  int length = strlen(a);
  char lft;
  int i;
  for (i = 0; i < length / 2; i++) {
    lft = a[i];
    a[i] = comp(a[length - i - 1]);
    a[length - i - 1] = comp(lft);
  }
  if (length % 2)
    a[length / 2] = comp(a[length / 2]);
}
```
user: What is the c source code of the assembly code below?

```asm
<revComplement>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  call   <strlen@plt>
  mov    %eax,%r15d
  mov    %eax,%r14d
  shr    $0x1f,%r14d
  add    %eax,%r14d
  sar    %r14d
  cmp    $0x1,%eax
  jle    L29
  cltq
  lea    -0x1(%r13,%rax,1),%rbp
  mov    $0x0,%ebx
L37:
  movzbl 0x0(%r13,%rbx,1),%r12d
  movsbl 0x0(%rbp),%edi
  call   <comp@plt>
  mov    %al,0x0(%r13,%rbx,1)
  movsbl %r12b,%edi
  call   <comp@plt>
  mov    %al,0x0(%rbp)
  add    $0x1,%rbx
  sub    $0x1,%rbp
  cmp    %ebx,%r14d
  jg     L37
L29:
  test   $0x1,%r15b
  jne    L39
L22:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L39:
  movslq %r14d,%r14
  add    %r14,%r13
  movsbl 0x0(%r13),%edi
  call   <comp@plt>
  mov    %al,0x0(%r13)
  jmp    L22
```

assistant: ```c
void revComplement(char *a) {
  int length = strlen(a);
  char lft;
  int i;
  for (i = 0; i < length / 2; i++) {
    lft = a[i];
    a[i] = comp(a[length - i - 1]);
    a[length - i - 1] = comp(lft);
  }
  if (length % 2)
    a[length / 2] = comp(a[length / 2]);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<revComplement>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%r14d
  mov    %eax,%r15d
  shr    $0x1f,%r14d
  add    %eax,%r14d
  sar    %r14d
  cmp    $0x1,%eax
  jle    L4
  cltq
  xor    %ebx,%ebx
  lea    -0x1(%r13,%rax,1),%rbp
  nopl   0x0(%rax,%rax,1)
L13:
  movsbl 0x0(%rbp),%edi
  movzbl 0x0(%r13,%rbx,1),%r12d
  sub    $0x1,%rbp
  call   <comp@plt>
  movsbl %r12b,%edi
  mov    %al,0x0(%r13,%rbx,1)
  add    $0x1,%rbx
  call   <comp@plt>
  mov    %al,0x1(%rbp)
  cmp    %ebx,%r14d
  jg     L13
L4:
  and    $0x1,%r15d
  jne    L1
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  movslq %r14d,%r14
  add    %r14,%r13
  movsbl 0x0(%r13),%edi
  call   <comp@plt>
  mov    %al,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void revComplement(char *a) {
  int length = strlen(a);
  char lft;
  int i;
  for (i = 0; i < length / 2; i++) {
    lft = a[i];
    a[i] = comp(a[length - i - 1]);
    a[length - i - 1] = comp(lft);
  }
  if (length % 2)
    a[length / 2] = comp(a[length / 2]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<revComplement>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%r14d
  mov    %eax,%r15d
  shr    $0x1f,%r14d
  add    %eax,%r14d
  sar    %r14d
  cmp    $0x1,%eax
  jle    L26
  cltq
  xor    %ebx,%ebx
  lea    -0x1(%r13,%rax,1),%rbp
  nopl   0x0(%rax,%rax,1)
L28:
  movsbl 0x0(%rbp),%edi
  movzbl 0x0(%r13,%rbx,1),%r12d
  sub    $0x1,%rbp
  call   <comp@plt>
  movsbl %r12b,%edi
  mov    %al,0x0(%r13,%rbx,1)
  add    $0x1,%rbx
  call   <comp@plt>
  mov    %al,0x1(%rbp)
  cmp    %ebx,%r14d
  jg     L28
L26:
  and    $0x1,%r15d
  jne    L8
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  movslq %r14d,%r14
  add    %r14,%r13
  movsbl 0x0(%r13),%edi
  call   <comp@plt>
  mov    %al,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void revComplement(char *a) {
  int length = strlen(a);
  char lft;
  int i;
  for (i = 0; i < length / 2; i++) {
    lft = a[i];
    a[i] = comp(a[length - i - 1]);
    a[length - i - 1] = comp(lft);
  }
  if (length % 2)
    a[length / 2] = comp(a[length / 2]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<foo4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rsi
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rcx
  mov    $0x0,%ebx
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%ebx
  mov    $0x0,%edx
  div    %rbx
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L32:
  cmp    %rdx,%rsp
  je     L25
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L32
L25:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L24
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L24:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movb   $0x1,(%rax)
  mov    $0x0,%eax
  mov    %rsi,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int foo4(int n) {
  char arr[n];
  arr[0] = 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<foo4>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int foo4(int n) {
  char arr[n];
  arr[0] = 1;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<foo4>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int foo4(int n) {
  char arr[n];
  arr[0] = 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<foo4>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int foo4(int n) {
  char arr[n];
  arr[0] = 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
<autohelperowl_attackpat185>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0x0,%esi
  mov    $0xffffffff,%edi
  call   <offset@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0xffffffff,%esi
  mov    $0x0,%edi
  call   <offset@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <offset@plt>
  mov    %eax,-0x4(%rbp)
  mov    $0x3,%eax
  sub    -0x24(%rbp),%eax
  mov    -0xc(%rbp),%r8d
  mov    -0x8(%rbp),%edi
  mov    -0xc(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  sub    $0x8,%rsp
  mov    -0x4(%rbp),%esi
  push   %rsi
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <play_break_through_n@plt>
  add    $0x10,%rsp
  cmp    $0x3,%eax
  sete   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int autohelperowl_attackpat185(struct pattern *patt, int transformation, int move, int color, int action) {
  int a, b, c;
  (void)patt;
  (void)color;
  (void)action;
  a = offset(-1, 0, move, transformation);
  b = offset(0, -1, move, transformation);
  c = offset(0, 1, move, transformation);
  return play_break_through_n((1 + 2 - (color)), 2, move, a, b, a, c) == 3;
}
```