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

```asm
<fraction_to_double>:
  endbr64
  push   %rbp
  mov    $0x2f,%esi
  mov    %rdi,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <strchr@plt>
  test   %rax,%rax
  je     L9
  lea    0x2(%rbp),%rsi
  lea    0xf(%rsp),%rbp
  mov    $0x5,%edx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <strtod@plt>
  pxor   %xmm0,%xmm0
  mov    %eax,%r8d
  mov    $0x1,%eax
  cltd
  idiv   %r8d
  cvtsi2sd %eax,%xmm0
L0:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x20,%rsp
  pop    %rbp
  ret
  nopl   (%rax)
L9:
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <strtod@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  jmp    L0
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double fraction_to_double(char *fraction) {
  double fraction_as_db;
  char denominator[9];
  char *ptr = strstr(fraction, "/");
  if (ptr) {
    strncpy(denominator, fraction + 2, 5);
    fraction_as_db = 1 / strtod(denominator, NULL);
  } else {
    fraction_as_db = strtod(fraction, NULL);
  }
  return fraction_as_db;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<script_arc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  sar    $0x8,%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <script_write_byte@plt>
  mov    -0x4(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <script_write_byte@plt>
  mov    -0x8(%rbp),%eax
  sar    $0x8,%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <script_write_byte@plt>
  mov    -0x8(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <script_write_byte@plt>
  nop
  leave
  ret
```

assistant: ```c
void script_arc(int speed, int radius) {
  script_write_byte(((speed & 0xFFFF) >> 8));
  script_write_byte((speed & 0xFF));
  script_write_byte(((radius & 0xFFFF) >> 8));
  script_write_byte((radius & 0xFF));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<script_arc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %edi,%eax
  movzbl %ah,%edi
  call   <script_write_byte@plt>
  movzbl %bpl,%edi
  call   <script_write_byte@plt>
  movzbl %bh,%edi
  call   <script_write_byte@plt>
  movzbl %bl,%edi
  call   <script_write_byte@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void script_arc(int speed, int radius) {
  script_write_byte(((speed & 0xFFFF) >> 8));
  script_write_byte((speed & 0xFF));
  script_write_byte(((radius & 0xFFFF) >> 8));
  script_write_byte((radius & 0xFF));
}
```
user: What is the c source code of the assembly code below?

```asm
<script_arc>:
  endbr64
  push   %rbp
  mov    %edi,%eax
  mov    %edi,%ebp
  push   %rbx
  movzbl %ah,%edi
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <script_write_byte@plt>
  movzbl %bpl,%edi
  call   <script_write_byte@plt>
  movzbl %bh,%edi
  call   <script_write_byte@plt>
  add    $0x8,%rsp
  movzbl %bl,%edi
  pop    %rbx
  pop    %rbp
  jmp    1050 <script_write_byte@plt>
```

assistant: ```c
void script_arc(int speed, int radius) {
  script_write_byte(((speed & 0xFFFF) >> 8));
  script_write_byte((speed & 0xFF));
  script_write_byte(((radius & 0xFFFF) >> 8));
  script_write_byte((radius & 0xFF));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<script_arc>:
  endbr64
  push   %rbp
  mov    %edi,%eax
  mov    %edi,%ebp
  push   %rbx
  movzbl %ah,%edi
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <script_write_byte@plt>
  movzbl %bpl,%edi
  call   <script_write_byte@plt>
  movzbl %bh,%edi
  call   <script_write_byte@plt>
  add    $0x8,%rsp
  movzbl %bl,%edi
  pop    %rbx
  pop    %rbp
  jmp    1050 <script_write_byte@plt>
```

assistant: ```c
void script_arc(int speed, int radius) {
  script_write_byte(((speed & 0xFFFF) >> 8));
  script_write_byte((speed & 0xFF));
  script_write_byte(((radius & 0xFFFF) >> 8));
  script_write_byte((radius & 0xFF));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<count_unique_benchmarks_in_sorted_list>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L12
L7:
  addl   $0x1,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <get_next_different_bench@plt>
  mov    %rax,-0x8(%rbp)
L12:
  cmpq   $0x0,-0x8(%rbp)
  jne    L7
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int count_unique_benchmarks_in_sorted_list(struct pmm_benchmark *first) {
  int count;
  struct pmm_benchmark *b;
  count = 0;
  b = first;
  while (b != NULL) {
    count++;
    b = get_next_different_bench(b);
  }
  return count;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<count_unique_benchmarks_in_sorted_list>:
  endbr64
  push   %rbx
  test   %rdi,%rdi
  je     L6
  mov    $0x0,%ebx
L16:
  add    $0x1,%ebx
  call   <get_next_different_bench@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L16
L10:
  mov    %ebx,%eax
  pop    %rbx
  ret
L6:
  mov    $0x0,%ebx
  jmp    L10
```

assistant: ```c
int count_unique_benchmarks_in_sorted_list(struct pmm_benchmark *first) {
  int count;
  struct pmm_benchmark *b;
  count = 0;
  b = first;
  while (b != NULL) {
    count++;
    b = get_next_different_bench(b);
  }
  return count;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<count_unique_benchmarks_in_sorted_list>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  test   %rdi,%rdi
  je     L4
  xchg   %ax,%ax
L13:
  call   <get_next_different_bench@plt>
  add    $0x1,%r12d
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L13
L4:
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int count_unique_benchmarks_in_sorted_list(struct pmm_benchmark *first) {
  int count;
  struct pmm_benchmark *b;
  count = 0;
  b = first;
  while (b != NULL) {
    count++;
    b = get_next_different_bench(b);
  }
  return count;
}
```
user: What is the c source code of the assembly code below:

```asm
<count_unique_benchmarks_in_sorted_list>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  test   %rdi,%rdi
  je     L19
  xchg   %ax,%ax
L11:
  call   <get_next_different_bench@plt>
  add    $0x1,%r12d
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L11
L19:
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int count_unique_benchmarks_in_sorted_list(struct pmm_benchmark *first) {
  int count;
  struct pmm_benchmark *b;
  count = 0;
  b = first;
  while (b != NULL) {
    count++;
    b = get_next_different_bench(b);
  }
  return count;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L1
L8:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0xc(%rbp),%edx
  add    %edx,%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rdx,%rcx
  mov    %eax,%edx
  lea    D3(%rip),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <sprintf@plt>
  addl   $0x1,-0xc(%rbp)
L1:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L8
  mov    -0x1c(%rbp),%eax
  add    %eax,%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D3
datatype: string

tool: D3: %.2x

assistant: ```c
char *hex(char *s, int len) {
  int i;
  char *buf = (char *)calloc((len * 2) + 1, sizeof(char));
  for (i = 0; i < len; i++)
    sprintf(&(buf[i * 2]), "%.2x", s[i] & 0xff);
  buf[len * 2] = 0;
  return buf;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<hex>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%r12d
  lea    (%rsi,%rsi,1),%r15d
  lea    0x1(%r15),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%r14
  test   %r12d,%r12d
  jle    L14
  mov    %r13,%rbx
  mov    %rax,%rbp
  lea    -0x1(%r12),%eax
  lea    0x1(%r13,%rax,1),%r13
  lea    D7(%rip),%r12
L8:
  movsbl (%rbx),%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  add    $0x1,%rbx
  add    $0x2,%rbp
  cmp    %r13,%rbx
  jne    L8
L14:
  movslq %r15d,%r15
  movb   $0x0,(%r14,%r15,1)
  mov    %r14,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

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

tool: D7: %.2x

assistant: ```c
char *hex(char *s, int len) {
  int i;
  char *buf = (char *)calloc((len * 2) + 1, sizeof(char));
  for (i = 0; i < len; i++)
    sprintf(&(buf[i * 2]), "%.2x", s[i] & 0xff);
  buf[len * 2] = 0;
  return buf;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<hex>:
  endbr64
  push   %r15
  push   %r14
  lea    (%rsi,%rsi,1),%r14d
  push   %r13
  push   %r12
  mov    %esi,%r12d
  mov    $0x1,%esi
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x1(%r14),%edi
  sub    $0x8,%rsp
  call   <calloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L3
  mov    %rax,%rbp
  lea    -0x1(%r12),%eax
  lea    D4(%rip),%r12
  lea    0x1(%rbx,%rax,1),%r15
  cs nopw 0x0(%rax,%rax,1)
L15:
  movsbl (%rbx),%edx
  mov    %rbp,%rdi
  mov    %r12,%rsi
  add    $0x1,%rbx
  add    $0x2,%rbp
  call   <sprintf@plt>
  cmp    %r15,%rbx
  jne    L15
L3:
  movslq %r14d,%r14
  mov    %r13,%rax
  movb   $0x0,0x0(%r13,%r14,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D4
datatype: string

tool: D4: %.2x

assistant: ```c
char *hex(char *s, int len) {
  int i;
  char *buf = (char *)calloc((len * 2) + 1, sizeof(char));
  for (i = 0; i < len; i++)
    sprintf(&(buf[i * 2]), "%.2x", s[i] & 0xff);
  buf[len * 2] = 0;
  return buf;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<hex>:
  endbr64
  push   %r15
  push   %r14
  lea    (%rsi,%rsi,1),%r14d
  push   %r13
  push   %r12
  mov    %esi,%r12d
  mov    $0x1,%esi
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x1(%r14),%edi
  sub    $0x8,%rsp
  call   <calloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L2
  mov    %rax,%rbp
  lea    -0x1(%r12),%eax
  lea    D6(%rip),%r12
  lea    0x1(%rbx,%rax,1),%r15
  cs nopw 0x0(%rax,%rax,1)
L17:
  movsbl (%rbx),%edx
  mov    %rbp,%rdi
  mov    %r12,%rsi
  add    $0x1,%rbx
  add    $0x2,%rbp
  call   <sprintf@plt>
  cmp    %r15,%rbx
  jne    L17
L2:
  movslq %r14d,%r14
  mov    %r13,%rax
  movb   $0x0,0x0(%r13,%r14,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

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

tool: D6: %.2x

assistant: ```c
char *hex(char *s, int len) {
  int i;
  char *buf = (char *)calloc((len * 2) + 1, sizeof(char));
  for (i = 0; i < len; i++)
    sprintf(&(buf[i * 2]), "%.2x", s[i] & 0xff);
  buf[len * 2] = 0;
  return buf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L13
L7:
  addl   $0x1,-0x8(%rbp)
L13:
  cmpl   $0xf423f,-0x8(%rbp)
  jle    L7
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D7
datatype: string

tool: D7: --Processor %d says hello!--


assistant: ```c
int main(int argc, char *argv[]) {
  int pid = 0;
  int i;
  pid = atoi(argv[2]);
  for (i = 0; i < 1000000; i++) {
  }
  printf("--Processor %d says hello!--\n", pid);
  return i;
}
```
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    0x10(%rsi),%rdi
  call   <atoi@plt>
  mov    %eax,%esi
  mov    $0xf4240,%edx
L3:
  sub    $0x1,%edx
  jne    L3
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0xf4240,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type: string

tool: D3: --Processor %d says hello!--


assistant: ```c
int main(int argc, char *argv[]) {
  int pid = 0;
  int i;
  pid = atoi(argv[2]);
  for (i = 0; i < 1000000; i++) {
  }
  printf("--Processor %d says hello!--\n", pid);
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    0x10(%rsi),%rdi
  call   <atoi@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xf4240,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D6
data_type: string

tool: D6: --Processor %d says hello!--


assistant: ```c
int main(int argc, char *argv[]) {
  int pid = 0;
  int i;
  pid = atoi(argv[2]);
  for (i = 0; i < 1000000; i++) {
  }
  printf("--Processor %d says hello!--\n", pid);
  return i;
}
```
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    0x10(%rsi),%rdi
  call   <atoi@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xf4240,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D0
type: string

tool: D0: --Processor %d says hello!--


assistant: ```c
int main(int argc, char *argv[]) {
  int pid = 0;
  int i;
  pid = atoi(argv[2]);
  for (i = 0; i < 1000000; i++) {
  }
  printf("--Processor %d says hello!--\n", pid);
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<multiply91>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L4
L1:
  movl   $0x0,-0x3c(%rbp)
  jmp    L26
L31:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L26:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L31
  addl   $0x1,-0x40(%rbp)
L4:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L1
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply91(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<multiply91>:
  endbr64
  test   %edi,%edi
  jle    L13
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L18:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L21:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L21
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L18
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  ret
```

assistant: ```c
void multiply91(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<multiply91>:
  endbr64
  test   %edi,%edi
  jle    L19
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L21:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L21
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L3
  pop    %rbx
  ret
L19:
  ret
```

assistant: ```c
void multiply91(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<multiply91>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L31
  mov    %rbp,%r11
  mov    %esi,%ebx
  lea    0x4(%rdx),%r10
  mov    %r8,%rax
  mov    %r11d,%edx
  mov    %r11d,%esi
  lea    0x4(%rcx),%r9
  shl    $0x2,%rbp
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  movd   %ebx,%xmm4
  xor    %r12d,%r12d
  lea    0x1(%rsi),%ecx
  lea    0x2(%rsi),%edi
  mov    %esi,-0x24(%rsp)
  movslq %esi,%r13
  lea    -0x1(%rdx),%r8d
  mov    %r11d,%esi
  lea    -0x1(%r11),%edx
  mov    %ecx,-0x20(%rsp)
  pshufd $0x0,%xmm4,%xmm2
  mov    %edi,-0x1c(%rsp)
  add    $0x1,%r8
  movslq %ecx,%r14
  mov    %edx,-0x28(%rsp)
  movdqa %xmm2,%xmm3
  shl    $0x4,%r8
  movslq %edi,%r15
  mov    %rsi,-0x10(%rsp)
  psrlq  $0x20,%xmm3
  nopl   (%rax)
L66:
  mov    %rax,%rdx
  lea    -0x4(%r10),%rcx
  lea    -0x4(%r9),%rsi
  sub    %r10,%rdx
  cmp    $0x8,%rdx
  mov    %rax,%rdx
  seta   %dil
  sub    %r9,%rdx
  cmp    $0x8,%rdx
  seta   %dl
  test   %dl,%dil
  je     L2
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L2
  nopl   0x0(%rax)
L61:
  movdqu (%rcx,%rdx,1),%xmm0
  movdqu (%rcx,%rdx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqu (%rsi,%rdx,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %r8,%rdx
  jne    L61
  cmp    -0x24(%rsp),%r11d
  je     L25
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L25
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L25
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L25:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L66
L31:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L2:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L33:
  mov    (%rcx,%rdx,4),%edi
  imul   %ebx,%edi
  add    (%rsi,%rdx,4),%edi
  mov    %edi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r8
  jne    L33
  mov    -0x18(%rsp),%r8
  jmp    L25
```

assistant: ```c
void multiply91(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_vscprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <va_copy@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%ecx
  mov    %rax,%rdx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <vsnprintf@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int _vscprintf(const char *format, va_list pargs) {
  int retval;
  va_list argcopy;
  va_copy(argcopy, pargs);
  retval = vsnprintf(NULL, 0, format, argcopy);
  va_end(argcopy);
  return retval;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_vscprintf>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_copy@plt>
  mov    %ebp,%ecx
  mov    %rbx,%rdx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <vsnprintf@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int _vscprintf(const char *format, va_list pargs) {
  int retval;
  va_list argcopy;
  va_copy(argcopy, pargs);
  retval = vsnprintf(NULL, 0, format, argcopy);
  va_end(argcopy);
  return retval;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<_vscprintf>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_copy@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int _vscprintf(const char *format, va_list pargs) {
  int retval;
  va_list argcopy;
  va_copy(argcopy, pargs);
  retval = vsnprintf(NULL, 0, format, argcopy);
  va_end(argcopy);
  return retval;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_vscprintf>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_copy@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int _vscprintf(const char *format, va_list pargs) {
  int retval;
  va_list argcopy;
  va_copy(argcopy, pargs);
  retval = vsnprintf(NULL, 0, format, argcopy);
  va_end(argcopy);
  return retval;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edi
  call   <test_shellcode@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  test_shellcode(0);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

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

assistant: ```c
int main() {
  test_shellcode(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
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <test_shellcode@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  test_shellcode(0);
  return 0;
}
```