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

```asm
<c_srev>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  jle    L66
  sar    %edi
  lea    -0x1(%rax),%r8d
  cmp    $0x1,%edx
  je     L107
  test   %edx,%edx
  js     L30
  imul   %edx,%r8d
  xor    %eax,%eax
L64:
  test   %edi,%edi
  je     L66
  movslq %edx,%rdx
  cltq
  movslq %r8d,%r8
  lea    0x0(,%rdx,4),%rcx
  lea    (%rsi,%rax,4),%rdx
  lea    (%rsi,%r8,4),%rax
  xor    %esi,%esi
  nopl   0x0(%rax)
L95:
  movss  (%rdx),%xmm0
  movss  (%rax),%xmm1
  add    $0x1,%esi
  movss  %xmm1,(%rdx)
  add    %rcx,%rdx
  movss  %xmm0,(%rax)
  sub    %rcx,%rax
  cmp    %esi,%edi
  jne    L95
L66:
  ret
  cs nopw 0x0(%rax,%rax,1)
L107:
  mov    $0xaaaaaaab,%edx
  mov    %edi,%ecx
  imul   %rdx,%rcx
  shr    $0x21,%rcx
  lea    (%rcx,%rcx,2),%edx
  mov    %edi,%ecx
  sub    %edx,%ecx
  je     L46
  movslq %r8d,%rdx
  movss  (%rsi),%xmm0
  shl    $0x2,%rdx
  lea    (%rsi,%rdx,1),%r9
  movss  (%r9),%xmm1
  movss  %xmm1,(%rsi)
  movss  %xmm0,(%r9)
  cmp    $0x2,%ecx
  jne    L101
  lea    -0x4(%rsi,%rdx,1),%rdx
  movss  0x4(%rsi),%xmm0
  movss  (%rdx),%xmm1
  movss  %xmm1,0x4(%rsi)
  movss  %xmm0,(%rdx)
L101:
  sub    %ecx,%r8d
L46:
  cmp    $0x5,%eax
  jle    L66
  cmp    %ecx,%edi
  jle    L66
  movslq %ecx,%rax
  movslq %r8d,%r8
  lea    (%rsi,%rax,4),%rdx
  lea    (%rsi,%r8,4),%rax
  nopl   0x0(%rax,%rax,1)
L28:
  movss  (%rdx),%xmm0
  movss  (%rax),%xmm1
  add    $0x3,%ecx
  add    $0xc,%rdx
  sub    $0xc,%rax
  movss  %xmm1,-0xc(%rdx)
  movss  0x8(%rax),%xmm1
  movss  %xmm0,0xc(%rax)
  movss  -0x8(%rdx),%xmm0
  movss  %xmm1,-0x8(%rdx)
  movss  0x4(%rax),%xmm1
  movss  %xmm0,0x8(%rax)
  movss  -0x4(%rdx),%xmm0
  movss  %xmm1,-0x4(%rdx)
  movss  %xmm0,0x4(%rax)
  cmp    %ecx,%edi
  jg     L28
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  mov    $0x1,%ecx
  xor    %r8d,%r8d
  sub    %eax,%ecx
  mov    %ecx,%eax
  imul   %edx,%eax
  jmp    L64
```

assistant: ```c
void c_srev(const int64_t N, float *X, const int64_t stride) {
  int64_t ix;
  int64_t iy;
  int64_t m;
  int64_t n;
  int64_t i;
  float tmp;
  if (N <= 0) {
    return;
  }
  n = N / 2;
  if (stride == 1) {
    m = n % 3;
    iy = N - 1;
    if (m > 0) {
      for (ix = 0; ix < m; ix++) {
        tmp = X[ix];
        X[ix] = X[iy];
        X[iy] = tmp;
        iy -= 1;
      }
    }
    if (n < 3) {
      return;
    }
    for (ix = m; ix < n; ix += 3) {
      tmp = X[ix];
      X[ix] = X[iy];
      X[iy] = tmp;
      tmp = X[ix + 1];
      X[ix + 1] = X[iy - 1];
      X[iy - 1] = tmp;
      tmp = X[ix + 2];
      X[ix + 2] = X[iy - 2];
      X[iy - 2] = tmp;
      iy -= 3;
    }
    return;
  }
  if (stride < 0) {
    ix = (1 - N) * stride;
  } else {
    ix = 0;
  }
  iy = ix + ((N - 1) * stride);
  for (i = 0; i < n; i++) {
    tmp = X[ix];
    X[ix] = X[iy];
    X[iy] = tmp;
    ix += stride;
    iy -= stride;
  }
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strspn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L76
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  jmp    L72
L35:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L36
  addq   $0x1,-0x10(%rbp)
  jmp    L13
L36:
  addq   $0x1,-0x20(%rbp)
L59:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L35
L13:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
  mov    -0x10(%rbp),%rax
  jmp    L72
L6:
  addq   $0x1,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L76:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L59
  mov    -0x10(%rbp),%rax
L72:
  leave
  ret
```

assistant: ```c
size_t strspn(const char *s, const char *accept) {
  size_t num = 0;
  const char *start = (const char *)accept;
  if (!accept)
    return strlen(s);
  while (*s) {
    while (*accept) {
      if (*s == *accept) {
        num++;
        break;
      } else
        accept++;
    }
    if (!*accept)
      return num;
    else {
      s++;
      accept = start;
    }
  }
  return num;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strspn>:
  endbr64
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L46
  movzbl (%rsi),%r9d
  mov    $0x0,%r8d
  jmp    L20
L35:
  add    $0x1,%r8
  test   %al,%al
  je     L40
  movzbl (%rdi,%r8,1),%ecx
  test   %cl,%cl
  je     L40
L20:
  test   %r9b,%r9b
  je     L40
  mov    %rsi,%rdx
  mov    %r9d,%eax
L56:
  cmp    %cl,%al
  je     L35
  add    $0x1,%rdx
  movzbl (%rdx),%eax
  test   %al,%al
  jne    L56
L40:
  mov    %r8,%rax
  ret
L46:
  mov    $0x0,%r8d
  jmp    L40
```

assistant: ```c
size_t strspn(const char *s, const char *accept) {
  size_t num = 0;
  const char *start = (const char *)accept;
  if (!accept)
    return strlen(s);
  while (*s) {
    while (*accept) {
      if (*s == *accept) {
        num++;
        break;
      } else
        accept++;
    }
    if (!*accept)
      return num;
    else {
      s++;
      accept = start;
    }
  }
  return num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strspn>:
  endbr64
  movzbl (%rdi),%ecx
  xor    %r8d,%r8d
  test   %cl,%cl
  je     L43
  movzbl (%rsi),%r9d
  xor    %r8d,%r8d
  nopl   (%rax)
L33:
  test   %r9b,%r9b
  je     L43
  mov    %rsi,%rdx
  mov    %r9d,%eax
  jmp    L50
  nopl   (%rax)
L18:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L43
L50:
  cmp    %al,%cl
  jne    L18
  add    $0x1,%r8
  movzbl (%rdi,%r8,1),%ecx
  test   %cl,%cl
  jne    L33
L43:
  mov    %r8,%rax
  ret
```

assistant: ```c
size_t strspn(const char *s, const char *accept) {
  size_t num = 0;
  const char *start = (const char *)accept;
  if (!accept)
    return strlen(s);
  while (*s) {
    while (*accept) {
      if (*s == *accept) {
        num++;
        break;
      } else
        accept++;
    }
    if (!*accept)
      return num;
    else {
      s++;
      accept = start;
    }
  }
  return num;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strspn>:
  endbr64
  movzbl (%rdi),%ecx
  xor    %r8d,%r8d
  test   %cl,%cl
  je     L0
  movzbl (%rsi),%r9d
  xor    %r8d,%r8d
  nopl   (%rax)
L14:
  test   %r9b,%r9b
  je     L0
  mov    %rsi,%rdx
  mov    %r9d,%eax
  jmp    L40
  nopl   (%rax)
L57:
  movzbl 0x1(%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  je     L0
L40:
  cmp    %al,%cl
  jne    L57
  add    $0x1,%r8
  movzbl (%rdi,%r8,1),%ecx
  test   %cl,%cl
  jne    L14
L0:
  mov    %r8,%rax
  ret
```

assistant: ```c
size_t strspn(const char *s, const char *accept) {
  size_t num = 0;
  const char *start = (const char *)accept;
  if (!accept)
    return strlen(s);
  while (*s) {
    while (*accept) {
      if (*s == *accept) {
        num++;
        break;
      } else
        accept++;
    }
    if (!*accept)
      return num;
    else {
      s++;
      accept = start;
    }
  }
  return num;
}
```
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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L0:
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rdx
  lea    -0x15(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L1
  mov    -0x1c(%rbp),%edx
  lea    -0x15(%rbp),%rax
  mov    %rax,%rsi
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L1:
  mov    $0x0,%eax
  call   <flushinp@plt>
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1c(%rbp),%rdx
  lea    -0x15(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  jne    L34
  mov    -0x1c(%rbp),%edx
  lea    -0x15(%rbp),%rax
  mov    %rax,%rsi
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L11
L34:
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L11:
  mov    $0x0,%eax
  call   <flushinp@plt>
  jmp    L0
```

tool_calls: name: disassemble_data
arguments: label: D34
type_infered: string
name: disassemble_data
arguments: label: D14
type_infered: string
name: disassemble_data
arguments: label: D65
type_infered: string
name: disassemble_data
arguments: label: D52
type_infered: string
name: disassemble_data
arguments: label: D21
type_infered: string
name: disassemble_data
arguments: label: D36
type_infered: string

tool: D34: Please input a string with no more than 12 characters: 

tool: D14: %12[^
]%n

tool: D65: Nice work! And your input is: <%s>(%d characters)


tool: D52: Please input a hexadecimal number with no more than 10 chars: 

tool: D21: %10[A-Fa-f0-9]%n

tool: D36: Are you kidding???

assistant: ```c
int main() {
  char msg[12 + 1];
  while (true) {
    int len;
    printf("Please input a string with no more than "
           "12"
           " characters: ");
    if (scanf("%"
              "12"
              "[^\n]%n",
              msg, &len) == 1)
      printf("Nice work! And your input is: <%s>(%d characters)\n", msg, len);
    flushinp();
    printf("Please input a hexadecimal number with no more than 10 chars: ");
    if (scanf("%10[A-Fa-f0-9]%n", msg, &len) == 1)
      printf("Nice work! And your input is: <%s>(%d characters)\n", msg, len);
    else
      printf("Are you kidding???\n");
    flushinp();
  }
  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
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D10(%rip),%r13
  lea    0x4(%rsp),%rbp
  lea    0xb(%rsp),%rbx
  lea    D37(%rip),%r12
  jmp    L47
L34:
  mov    0x4(%rsp),%edx
  mov    %rbx,%rsi
  lea    D50(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L25
L10:
  mov    0x4(%rsp),%edx
  mov    %rbx,%rsi
  lea    D50(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L24:
  mov    $0x0,%eax
  call   <flushinp@plt>
L47:
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L34
L25:
  mov    $0x0,%eax
  call   <flushinp@plt>
  lea    D53(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L10
  lea    D42(%rip),%rdi
  call   <puts@plt>
  jmp    L24
```

tool_calls: name: parse_data
arguments: data_tag: D10
data_type: string
name: parse_data
arguments: data_tag: D37
data_type: string
name: parse_data
arguments: data_tag: D50
data_type: string
name: parse_data
arguments: data_tag: D53
data_type: string
name: parse_data
arguments: data_tag: D0
data_type: string
name: parse_data
arguments: data_tag: D42
data_type: string

tool: D10: Please input a string with no more than 12 characters: 

tool: D37: %12[^
]%n

tool: D50: Nice work! And your input is: <%s>(%d characters)


tool: D53: Please input a hexadecimal number with no more than 10 chars: 

tool: D0: %10[A-Fa-f0-9]%n

tool: D42: Are you kidding???

assistant: ```c
int main() {
  char msg[12 + 1];
  while (true) {
    int len;
    printf("Please input a string with no more than "
           "12"
           " characters: ");
    if (scanf("%"
              "12"
              "[^\n]%n",
              msg, &len) == 1)
      printf("Nice work! And your input is: <%s>(%d characters)\n", msg, len);
    flushinp();
    printf("Please input a hexadecimal number with no more than 10 chars: ");
    if (scanf("%10[A-Fa-f0-9]%n", msg, &len) == 1)
      printf("Nice work! And your input is: <%s>(%d characters)\n", msg, len);
    else
      printf("Are you kidding???\n");
    flushinp();
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  lea    D39(%rip),%r15
  push   %r14
  lea    D23(%rip),%r14
  push   %r13
  lea    D19(%rip),%r13
  push   %r12
  lea    D41(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  lea    0xb(%rsp),%rbx
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L3:
  mov    0x4(%rsp),%edx
  mov    %rbx,%rsi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L10:
  xor    %eax,%eax
  call   <flushinp@plt>
L34:
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L21
L2:
  xor    %eax,%eax
  call   <flushinp@plt>
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L3
  lea    D25(%rip),%rdi
  call   <puts@plt>
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L21:
  mov    0x4(%rsp),%edx
  mov    %rbx,%rsi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L2
```

tool_calls: name: parse_data
arguments: data_tag: D39
datatype: string
name: parse_data
arguments: data_tag: D23
datatype: string
name: parse_data
arguments: data_tag: D19
datatype: string
name: parse_data
arguments: data_tag: D41
datatype: string
name: parse_data
arguments: data_tag: D14
datatype: string
name: parse_data
arguments: data_tag: D25
datatype: string

tool: D39: Please input a string with no more than 12 characters: 

tool: D23: %12[^
]%n

tool: D19: Please input a hexadecimal number with no more than 10 chars: 

tool: D41: %10[A-Fa-f0-9]%n

tool: D14: Nice work! And your input is: <%s>(%d characters)


tool: D25: Are you kidding???

assistant: ```c
int main() {
  char msg[12 + 1];
  while (true) {
    int len;
    printf("Please input a string with no more than "
           "12"
           " characters: ");
    if (scanf("%"
              "12"
              "[^\n]%n",
              msg, &len) == 1)
      printf("Nice work! And your input is: <%s>(%d characters)\n", msg, len);
    flushinp();
    printf("Please input a hexadecimal number with no more than 10 chars: ");
    if (scanf("%10[A-Fa-f0-9]%n", msg, &len) == 1)
      printf("Nice work! And your input is: <%s>(%d characters)\n", msg, len);
    else
      printf("Are you kidding???\n");
    flushinp();
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D9(%rip),%r15
  push   %r14
  lea    D22(%rip),%r14
  push   %r13
  lea    D32(%rip),%r13
  push   %r12
  lea    D49(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  lea    0xb(%rsp),%rbx
  jmp    L29
  nopw   0x0(%rax,%rax,1)
L24:
  mov    0x4(%rsp),%edx
  mov    %rbx,%rsi
  lea    D46(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L20:
  xor    %eax,%eax
  call   <flushinp@plt>
L29:
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L21
L14:
  xor    %eax,%eax
  call   <flushinp@plt>
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L24
  lea    D68(%rip),%rdi
  call   <puts@plt>
  jmp    L20
  nopl   0x0(%rax,%rax,1)
L21:
  mov    0x4(%rsp),%edx
  mov    %rbx,%rsi
  lea    D46(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L14
```

tool_calls: name: unpack_data
arguments: tag: D9
data_type: string
name: unpack_data
arguments: tag: D22
data_type: string
name: unpack_data
arguments: tag: D32
data_type: string
name: unpack_data
arguments: tag: D49
data_type: string
name: unpack_data
arguments: tag: D46
data_type: string
name: unpack_data
arguments: tag: D68
data_type: string

tool: D9: Please input a string with no more than 12 characters: 

tool: D22: %12[^
]%n

tool: D32: Please input a hexadecimal number with no more than 10 chars: 

tool: D49: %10[A-Fa-f0-9]%n

tool: D46: Nice work! And your input is: <%s>(%d characters)


tool: D68: Are you kidding???

assistant: ```c
int main() {
  char msg[12 + 1];
  while (true) {
    int len;
    printf("Please input a string with no more than "
           "12"
           " characters: ");
    if (scanf("%"
              "12"
              "[^\n]%n",
              msg, &len) == 1)
      printf("Nice work! And your input is: <%s>(%d characters)\n", msg, len);
    flushinp();
    printf("Please input a hexadecimal number with no more than 10 chars: ");
    if (scanf("%10[A-Fa-f0-9]%n", msg, &len) == 1)
      printf("Nice work! And your input is: <%s>(%d characters)\n", msg, len);
    else
      printf("Are you kidding???\n");
    flushinp();
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_latlong>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbx
  mov    %rsi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    D8(%rip),%rax
  mov    (%rax),%r13d
  add    $0x1,%r13d
  movslq %r13d,%r13
  lea    0xf(%r13,%r13,1),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L47:
  cmp    %rdx,%rsp
  je     L71
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L47
L71:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L57
  orq    $0x0,-0x8(%rsp,%rax,1)
L57:
  mov    %rsp,%r15
  mov    %rsp,%r12
  mov    $0x0,%r14d
L40:
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  movslq %edx,%rcx
  add    %rbx,%rcx
  cmpb   $0x5f,(%rcx)
  je     L3
  sub    $0x2,%eax
  cltq
L7:
  mov    %eax,%edx
  lea    (%rbx,%rax,1),%rcx
  sub    $0x1,%rax
  cmpb   $0x5f,0x1(%rbx,%rax,1)
  jne    L7
L3:
  movb   $0x0,(%rcx)
  add    $0x1,%edx
  movslq %edx,%rdx
  lea    (%rbx,%rdx,1),%rsi
  movzbl (%rsi),%ecx
  test   %cl,%cl
  je     L14
  lea    0x1(%rbx,%rdx,1),%rdx
  mov    $0x1,%eax
L23:
  mov    %cl,-0x1(%r12,%rax,1)
  movb   $0x0,(%rsi)
  mov    %eax,%edi
  mov    %rdx,%rsi
  movzbl (%rdx),%ecx
  add    $0x1,%rax
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L23
L52:
  movslq %edi,%rdi
  movslq %r14d,%rax
  imul   %r13,%rax
  add    %r15,%rdi
  movb   $0x0,(%rdi,%rax,1)
  add    $0x1,%r14d
  add    %r13,%r12
  cmp    $0x2,%r14d
  jne    L40
  mov    %r15,%rdi
  call   <atof@plt>
  mov    -0x50(%rbp),%rax
  movss  %xmm0,(%rax)
  lea    (%r15,%r13,1),%rdi
  call   <atof@plt>
  mov    -0x48(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L95
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L14:
  mov    $0x0,%edi
  jmp    L52
L95:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void get_latlong(char fname[BUFSIZ + 1], float *lat, float *lon) {
  int i, j, delim;
  char latlong[2][BUFSIZ + 1];
  for (i = 0; i < 2; i++) {
    delim = strlen(fname) - 1;
    while (fname[delim] != '_')
      delim--;
    fname[delim] = '\0';
    delim++;
    j = 0;
    while (fname[delim] != '\0') {
      latlong[i][j] = fname[delim];
      fname[delim] = '\0';
      delim++;
      j++;
    }
    latlong[i][j] = '\0';
  }
  *lon = atof(latlong[0]);
  *lat = atof(latlong[1]);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<get_latlong>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %rsi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rsp,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%r15d
  add    $0x1,%r15d
  movslq %r15d,%r15
  lea    0xf(%r15,%r15,1),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L114
L26:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L26
L114:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L74
L36:
  mov    %rsp,%rbx
  mov    %rsp,%r14
  xor    %r12d,%r12d
L75:
  mov    %r13,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edi
  movslq %edi,%rdx
  lea    0x0(%r13,%rdx,1),%r8
  cmpb   $0x5f,(%r8)
  je     L65
  lea    -0x1(%r13,%rdx,1),%rdx
  nopl   0x0(%rax)
L98:
  mov    %rdx,%r8
  sub    $0x1,%rdx
  mov    %edi,%eax
  sub    $0x1,%edi
  cmpb   $0x5f,0x1(%rdx)
  jne    L98
L65:
  cltq
  movb   $0x0,(%r8)
  lea    0x0(%r13,%rax,1),%rdx
  movzbl (%rdx),%edi
  test   %dil,%dil
  je     L56
  mov    $0x1,%eax
  xchg   %ax,%ax
L67:
  mov    %dil,-0x1(%r14,%rax,1)
  movslq %eax,%r8
  movb   $0x0,-0x1(%rdx,%rax,1)
  add    $0x1,%rax
  movzbl -0x1(%rdx,%rax,1),%edi
  test   %dil,%dil
  jne    L67
L73:
  movslq %r12d,%rax
  add    %rbx,%r8
  add    %r15,%r14
  imul   %r15,%rax
  movb   $0x0,(%r8,%rax,1)
  cmp    $0x1,%r12d
  jne    L90
  mov    %rbx,%rdi
  call   <atof@plt>
  mov    -0x50(%rbp),%rax
  lea    (%rbx,%r15,1),%rdi
  movss  %xmm0,(%rax)
  call   <atof@plt>
  mov    -0x48(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L91
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L90:
  mov    $0x1,%r12d
  jmp    L75
  nopl   0x0(%rax,%rax,1)
L56:
  xor    %r8d,%r8d
  jmp    L73
L74:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L36
L91:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void get_latlong(char fname[BUFSIZ + 1], float *lat, float *lon) {
  int i, j, delim;
  char latlong[2][BUFSIZ + 1];
  for (i = 0; i < 2; i++) {
    delim = strlen(fname) - 1;
    while (fname[delim] != '_')
      delim--;
    fname[delim] = '\0';
    delim++;
    j = 0;
    while (fname[delim] != '\0') {
      latlong[i][j] = fname[delim];
      fname[delim] = '\0';
      delim++;
      j++;
    }
    latlong[i][j] = '\0';
  }
  *lon = atof(latlong[0]);
  *lat = atof(latlong[1]);
}
```