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

```asm
<strip_copy>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L55
  movzbl (%rbx),%edx
  mov    %rax,%rcx
  test   %dl,%dl
  je     L36
  nopl   (%rax)
L44:
  add    $0x1,%rbx
  cmp    $0xd,%dl
  je     L68
  cmp    $0xa,%dl
  je     L68
  mov    %dl,(%rcx)
  movzbl (%rbx),%edx
  lea    0x1(%rcx),%rsi
  mov    %rsi,%rcx
  test   %dl,%dl
  jne    L44
L36:
  movb   $0x0,(%rcx)
L55:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L68:
  movzbl (%rbx),%edx
  test   %dl,%dl
  jne    L44
  movb   $0x0,(%rcx)
  jmp    L55
```

assistant: ```c
char *strip_copy(const char *s) {
  char *p = malloc(strlen(s) + 1);
  if (p) {
    char *p2 = p;
    while (*s != '\0') {
      if (*s != '\r' && *s != '\n') {
        *p2++ = *s++;
      } else {
        ++s;
      }
    }
    *p2 = '\0';
  }
  return p;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L5
L7:
  lea    D19(%rip),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <print_results@plt>
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L5:
  cmpl   $0x2,-0xc(%rbp)
  jle    L7
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D19
data_type: string
name: read_data
arguments: data_label: D5
data_type: string
name: read_data
arguments: data_label: D3
data_type: string

tool: D19: floor

tool: D5: # c::native::%s


tool: D3: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "floor");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x0,%ebx
  lea    D19(%rip),%r13
  lea    D3(%rip),%r12
  lea    D25(%rip),%rbp
L2:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L2
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D19
type: string
name: unpack_data
arguments: data_label: D3
type: string
name: unpack_data
arguments: data_label: D25
type: string

tool: D19: floor

tool: D3: # c::native::%s


tool: D25: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "floor");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
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   %r13
  xor    %edi,%edi
  lea    D11(%rip),%r13
  push   %r12
  lea    D3(%rip),%r12
  push   %rbp
  lea    D15(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
L4:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  xor    %eax,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L4
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: label: D11
type: string
name: read_data
arguments: label: D3
type: string
name: read_data
arguments: label: D15
type: string

tool: D11: floor

tool: D3: # c::native::%s


tool: D15: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "floor");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  xor    %edi,%edi
  lea    D17(%rip),%r13
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  lea    D21(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
L6:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  xor    %eax,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L6
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D17
type_infered: string
name: disassemble_data
arguments: data_label: D4
type_infered: string
name: disassemble_data
arguments: data_label: D21
type_infered: string

tool: D17: floor

tool: D4: # c::native::%s


tool: D21: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "floor");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<utimensat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  add    $0x8,%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x38(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x18(%rbp)
  lea    -0x20(%rbp),%rdx
  mov    -0x30(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <futimesat@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int utimensat(int dfd, const char *pathname, const struct timespec times[2], int flags) {
  struct timeval tvs[2];
  if (flags) {
  }
  tvs[0].tv_sec = times[0].tv_sec;
  tvs[0].tv_usec = times[0].tv_nsec / 1000;
  tvs[1].tv_sec = times[1].tv_sec;
  tvs[1].tv_usec = times[1].tv_nsec / 1000;
  return futimesat(dfd, pathname, tvs);
}
```
user: What is the c source code of the assembly code below:

```asm
<utimensat>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    0x4(%rdx),%eax
  mov    %eax,0x4(%rsp)
  mov    (%rdx),%ecx
  movslq %ecx,%rax
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,(%rsp)
  mov    0xc(%rdx),%eax
  mov    %eax,0xc(%rsp)
  mov    0x8(%rdx),%edx
  movslq %edx,%rax
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,0x8(%rsp)
  mov    %rsp,%rdx
  call   <futimesat@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int utimensat(int dfd, const char *pathname, const struct timespec times[2], int flags) {
  struct timeval tvs[2];
  if (flags) {
  }
  tvs[0].tv_sec = times[0].tv_sec;
  tvs[0].tv_usec = times[0].tv_nsec / 1000;
  tvs[1].tv_sec = times[1].tv_sec;
  tvs[1].tv_usec = times[1].tv_nsec / 1000;
  return futimesat(dfd, pathname, tvs);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<utimensat>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  mov    0x4(%rdx),%eax
  mov    %eax,0x4(%rsp)
  movslq (%rdx),%rax
  mov    %rax,%rcx
  imul   $0x10624dd3,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x26,%rax
  sub    %ecx,%eax
  mov    %eax,(%rsp)
  mov    0xc(%rdx),%eax
  mov    %eax,0xc(%rsp)
  movslq 0x8(%rdx),%rax
  mov    %rax,%rdx
  imul   $0x10624dd3,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x26,%rax
  sub    %edx,%eax
  mov    %rsp,%rdx
  mov    %eax,0x8(%rsp)
  call   <futimesat@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int utimensat(int dfd, const char *pathname, const struct timespec times[2], int flags) {
  struct timeval tvs[2];
  if (flags) {
  }
  tvs[0].tv_sec = times[0].tv_sec;
  tvs[0].tv_usec = times[0].tv_nsec / 1000;
  tvs[1].tv_sec = times[1].tv_sec;
  tvs[1].tv_usec = times[1].tv_nsec / 1000;
  return futimesat(dfd, pathname, tvs);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<utimensat>:
  endbr64
  sub    $0x28,%rsp
  movd   0xc(%rdx),%xmm2
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movslq (%rdx),%rax
  mov    %rax,%rcx
  imul   $0x10624dd3,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x26,%rax
  sub    %ecx,%eax
  movd   %eax,%xmm0
  movslq 0x8(%rdx),%rax
  mov    %rax,%rcx
  imul   $0x10624dd3,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x26,%rax
  sub    %ecx,%eax
  movd   %eax,%xmm1
  mov    0x4(%rdx),%eax
  mov    %rsp,%rdx
  punpckldq %xmm2,%xmm1
  movd   %eax,%xmm3
  punpckldq %xmm3,%xmm0
  punpcklqdq %xmm1,%xmm0
  movaps %xmm0,(%rsp)
  call   <futimesat@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int utimensat(int dfd, const char *pathname, const struct timespec times[2], int flags) {
  struct timeval tvs[2];
  if (flags) {
  }
  tvs[0].tv_sec = times[0].tv_sec;
  tvs[0].tv_usec = times[0].tv_nsec / 1000;
  tvs[1].tv_sec = times[1].tv_sec;
  tvs[1].tv_usec = times[1].tv_nsec / 1000;
  return futimesat(dfd, pathname, tvs);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x18,-0x40(%rbp)
  movl   $0x1e,-0x3c(%rbp)
  movl   $0xf,-0x38(%rbp)
  movl   $0x2d,-0x34(%rbp)
  movl   $0x22,-0x30(%rbp)
  movl   $0x7,-0x2c(%rbp)
  lea    -0x40(%rbp),%rcx
  lea    -0x20(%rbp),%rax
  mov    $0x18,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <CopiarMatriz@plt>
  movl   $0x0,-0x48(%rbp)
  jmp    L10
L29:
  movl   $0x0,-0x44(%rbp)
  jmp    L21
L13:
  mov    -0x44(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x48(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rcx,%rax
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x44(%rbp)
L21:
  cmpl   $0x2,-0x44(%rbp)
  jle    L13
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x48(%rbp)
L10:
  cmpl   $0x1,-0x48(%rbp)
  jle    L29
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L45
  call   <__stack_chk_fail@plt>
L45:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
type_guessed: string

tool: D2: %d  

assistant: ```c
main() {
  int m1[2][3] = {24, 30, 15, 45, 34, 7};
  int m2[2][3], f, c;
  CopiarMatriz(m2, m1, sizeof(m1));
  for (f = 0; f < 2; f++) {
    for (c = 0; c < 3; c++)
      printf("%d  ", m2[f][c]);
    printf("\n");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x18,(%rsp)
  movl   $0x1e,0x4(%rsp)
  movl   $0xf,0x8(%rsp)
  movl   $0x2d,0xc(%rsp)
  movl   $0x22,0x10(%rsp)
  movl   $0x7,0x14(%rsp)
  mov    %rsp,%rsi
  lea    0x20(%rsp),%rdi
  mov    $0x18,%edx
  call   <CopiarMatriz@plt>
  mov    0x20(%rsp),%esi
  lea    D5(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x24(%rsp),%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x2c(%rsp),%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x30(%rsp),%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x34(%rsp),%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_infered: string

tool: D5: %d  

assistant: ```c
main() {
  int m1[2][3] = {24, 30, 15, 45, 34, 7};
  int m2[2][3], f, c;
  CopiarMatriz(m2, m1, sizeof(m1));
  for (f = 0; f < 2; f++) {
    for (c = 0; c < 3; c++)
      printf("%d  ", m2[f][c]);
    printf("\n");
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x18,%edx
  lea    D4(%rip),%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x1e00000018,%rax
  mov    %rsp,%rsi
  lea    0x20(%rsp),%rdi
  mov    %rax,(%rsp)
  movabs $0x2d0000000f,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x700000022,%rax
  mov    %rax,0x10(%rsp)
  call   <CopiarMatriz@plt>
  mov    0x20(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x24(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x2c(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x30(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x34(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type_guessed: string

tool: D4: %d  

assistant: ```c
main() {
  int m1[2][3] = {24, 30, 15, 45, 34, 7};
  int m2[2][3], f, c;
  CopiarMatriz(m2, m1, sizeof(m1));
  for (f = 0; f < 2; f++) {
    for (c = 0; c < 3; c++)
      printf("%d  ", m2[f][c]);
    printf("\n");
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x18,%edx
  lea    D9(%rip),%rbp
  sub    $0x40,%rsp
  movdqa D14(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    D12(%rip),%rax
  mov    %rsp,%rsi
  lea    0x20(%rsp),%rdi
  movaps %xmm0,(%rsp)
  mov    %rax,0x10(%rsp)
  call   <CopiarMatriz@plt>
  mov    0x20(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x24(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x2c(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x30(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x34(%rsp),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type_infered: string
name: extract_symbolic_data
arguments: label: D14
type_infered: byte[16]
name: extract_symbolic_data
arguments: label: D12
type_infered: byte[8]

tool: D9: %d  

tool: {"D14": [0x18,0x00,0x00,0x00,0x1e,0x00,0x00,0x00,0x0f,0x00,0x00,0x00,0x2d,0x00,0x00,0x00]}

tool: {"D12": [0x22,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

assistant: ```c
main() {
  int m1[2][3] = {24, 30, 15, 45, 34, 7};
  int m2[2][3], f, c;
  CopiarMatriz(m2, m1, sizeof(m1));
  for (f = 0; f < 2; f++) {
    for (c = 0; c < 3; c++)
      printf("%d  ", m2[f][c]);
    printf("\n");
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_dV>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  movsd  (%rax),%xmm1
  mov    -0x30(%rbp),%rax
  movsd  (%rax),%xmm0
  addsd  %xmm0,%xmm1
  movsd  D2(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  movsd  (%rax),%xmm0
  mov    -0x30(%rbp),%rax
  movsd  (%rax),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <get_dp@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  add    $0x10,%rax
  movsd  (%rax),%xmm0
  mov    -0x30(%rbp),%rax
  add    $0x10,%rax
  movsd  (%rax),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mulsd  -0x18(%rbp),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  mulsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D2
type: f32[2]

tool: {"D2": [0.0,1.75]}

assistant: ```c
double get_dV(const double *xp, const double *xm) {
  double r = .5 * (xp[0] + xm[0]);
  double dr = xp[0] - xm[0];
  double dphi = get_dp(xp[1], xm[1]);
  double dz = xp[2] - xm[2];
  return (r * dr * dphi * dz);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_dV>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    (%rdi),%rbx
  mov    (%rsi),%r13
  movq   %rbx,%xmm2
  movq   %r13,%xmm3
  subsd  %xmm3,%xmm2
  movsd  %xmm2,0x8(%rsp)
  movsd  0x8(%rdi),%xmm0
  movsd  0x8(%rsi),%xmm1
  call   <get_dp@plt>
  movapd %xmm0,%xmm1
  movq   %rbx,%xmm0
  movq   %r13,%xmm4
  addsd  %xmm4,%xmm0
  mulsd  D3(%rip),%xmm0
  mulsd  0x8(%rsp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  0x10(%r12),%xmm1
  subsd  0x10(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: tag: D3
type: f32[2]

tool: {"D3": [0.0,1.75]}

assistant: ```c
double get_dV(const double *xp, const double *xm) {
  double r = .5 * (xp[0] + xm[0]);
  double dr = xp[0] - xm[0];
  double dphi = get_dp(xp[1], xm[1]);
  double dz = xp[2] - xm[2];
  return (r * dr * dphi * dz);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_dV>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  movsd  (%rdi),%xmm2
  movsd  (%rsi),%xmm3
  movsd  0x8(%rdi),%xmm0
  movsd  0x8(%rsi),%xmm1
  movapd %xmm2,%xmm4
  movsd  %xmm2,0x18(%rsp)
  subsd  %xmm3,%xmm4
  movsd  %xmm3,0x10(%rsp)
  movsd  %xmm4,0x8(%rsp)
  call   <get_dp@plt>
  movsd  0x10(%rsp),%xmm3
  movsd  0x18(%rsp),%xmm2
  addsd  %xmm3,%xmm2
  mulsd  D7(%rip),%xmm2
  mulsd  0x8(%rsp),%xmm2
  mulsd  %xmm0,%xmm2
  movsd  0x10(%rbp),%xmm0
  subsd  0x10(%rbx),%xmm0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  mulsd  %xmm2,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: label: D7
type: f32[2]

tool: {"D7": [0.0,1.75]}

assistant: ```c
double get_dV(const double *xp, const double *xm) {
  double r = .5 * (xp[0] + xm[0]);
  double dr = xp[0] - xm[0];
  double dphi = get_dp(xp[1], xm[1]);
  double dz = xp[2] - xm[2];
  return (r * dr * dphi * dz);
}
```
user: What is the c source code of the assembly code below:

```asm
<get_dV>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  movsd  (%rdi),%xmm2
  movsd  (%rsi),%xmm3
  movsd  0x8(%rdi),%xmm0
  movsd  0x8(%rsi),%xmm1
  movapd %xmm2,%xmm4
  movsd  %xmm2,0x18(%rsp)
  subsd  %xmm3,%xmm4
  movsd  %xmm3,0x10(%rsp)
  movsd  %xmm4,0x8(%rsp)
  call   <get_dp@plt>
  movsd  0x10(%rsp),%xmm3
  movsd  0x18(%rsp),%xmm2
  addsd  %xmm3,%xmm2
  mulsd  D3(%rip),%xmm2
  mulsd  0x8(%rsp),%xmm2
  mulsd  %xmm0,%xmm2
  movsd  0x10(%rbp),%xmm0
  subsd  0x10(%rbx),%xmm0
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  mulsd  %xmm2,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D3
type: f32[2]

tool: {"D3": [0.0,1.75]}

assistant: ```c
double get_dV(const double *xp, const double *xm) {
  double r = .5 * (xp[0] + xm[0]);
  double dr = xp[0] - xm[0];
  double dphi = get_dp(xp[1], xm[1]);
  double dz = xp[2] - xm[2];
  return (r * dr * dphi * dz);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_wild>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L20
L28:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2a,%al
  je     L3
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x3f,%al
  je     L3
  addl   $0x1,-0x4(%rbp)
L20:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
L3:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  setne  %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int is_wild(char *s) {
  int i;
  for (i = 0; s[i] != '\0'; i++)
    if (s[i] == '*' || s[i] == '?')
      break;
  return (s[i] != '\0');
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<is_wild>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L31
  add    $0x1,%rdi
L38:
  cmp    $0x2a,%al
  je     L31
  cmp    $0x3f,%al
  je     L31
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  jne    L38
L31:
  test   %al,%al
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int is_wild(char *s) {
  int i;
  for (i = 0; s[i] != '\0'; i++)
    if (s[i] == '*' || s[i] == '?')
      break;
  return (s[i] != '\0');
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<is_wild>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L13
  add    $0x1,%rdi
  nop
L2:
  cmp    $0x2a,%al
  je     L7
  cmp    $0x3f,%al
  je     L7
  movzbl (%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L2
L13:
  xor    %eax,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int is_wild(char *s) {
  int i;
  for (i = 0; s[i] != '\0'; i++)
    if (s[i] == '*' || s[i] == '?')
      break;
  return (s[i] != '\0');
}
```