user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rdi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x3c(%rbp),%rdx
  lea    -0x40(%rbp),%rsi
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  movslq -0x3c(%rbp),%rdx
  mov    %rsp,%rcx
  lea    0x0(,%rdx,4),%rax
  mov    %rax,-0x58(%rbp)
  movslq -0x40(%rbp),%rax
  imul   %rdx,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L75
L44:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L44
L75:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L37
L105:
  lea    D11(%rip),%rdi
  mov    %rsp,%rbx
  call   <puts@plt>
  cmpl   $0x0,-0x40(%rbp)
  jle    L8
  mov    -0x58(%rbp),%rax
  movl   $0x0,-0x44(%rbp)
  xor    %r14d,%r14d
  lea    D42(%rip),%r12
  shr    $0x2,%rax
  mov    %rax,-0x50(%rbp)
  nopw   0x0(%rax,%rax,1)
L71:
  mov    -0x3c(%rbp),%ecx
  lea    (%rbx,%r14,4),%r13
  xor    %r15d,%r15d
  test   %ecx,%ecx
  jle    L4
  xchg   %ax,%ax
L125:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <scanf@plt>
  add    $0x4,%r13
  cmp    %r15d,-0x3c(%rbp)
  jg     L125
L4:
  addl   $0x1,-0x44(%rbp)
  add    -0x50(%rbp),%r14
  mov    -0x44(%rbp),%eax
  cmp    %eax,-0x40(%rbp)
  jg     L71
L8:
  lea    D18(%rip),%rdi
  call   <puts@plt>
  mov    -0x40(%rbp),%edx
  test   %edx,%edx
  jle    L66
  mov    -0x58(%rbp),%r14
  movl   $0x0,-0x44(%rbp)
  lea    D16(%rip),%r13
  shr    $0x2,%r14
  nopl   0x0(%rax)
L83:
  mov    -0x3c(%rbp),%eax
  movslq -0x44(%rbp),%r12
  xor    %r15d,%r15d
  test   %eax,%eax
  jle    L117
  cs nopw 0x0(%rax,%rax,1)
L19:
  movslq %r15d,%rax
  mov    %r13,%rdi
  add    $0x1,%r15d
  imul   %r14,%rax
  add    %r12,%rax
  mov    (%rbx,%rax,4),%esi
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r15d,-0x3c(%rbp)
  jg     L19
L117:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x44(%rbp)
  mov    -0x44(%rbp),%eax
  cmp    %eax,-0x40(%rbp)
  jg     L83
L66:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L124
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L37:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L105
L124:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D3
type_guessed: string
name: unpack_data
arguments: label: D14
type_guessed: string
name: unpack_data
arguments: label: D11
type_guessed: string
name: unpack_data
arguments: label: D42
type_guessed: string
name: unpack_data
arguments: label: D18
type_guessed: string
name: unpack_data
arguments: label: D16
type_guessed: string

tool: D3: Enter the order of the matrix (m*n): 

tool: D14: %d*%d

tool: D11: Enter the elements of the matrix: 

tool: D42: %d

tool: D18: Transpose of the matrix: 

tool: D16: %d 

assistant: ```c
int main() {
  int i, j, m, n;
  printf("Enter the order of the matrix (m*n): ");
  scanf("%d*%d", &m, &n);
  int matrix[m][n];
  printf("Enter the elements of the matrix: \n");
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      scanf("%d", &matrix[i][j]);
    }
  }
  printf("Transpose of the matrix: \n");
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      printf("%d ", matrix[j][i]);
    }
    printf("\n");
  }
}
```