user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x4,%edi
  push   %r14
  push   %r13
  lea    D7(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    $0x14,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    0x4(%rbp),%edx
  mov    $0xd,%esi
  lea    D41(%rip),%rdi
  mov    %rax,%r12
  xor    %eax,%eax
  movl   $0xd,0x0(%rbp)
  call   <printf@plt>
  mov    %r13,%rdi
  mov    %rbp,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  mov    %rbp,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D52(%rip),%rdi
  lea    D45(%rip),%r13
  call   <puts@plt>
L1:
  pxor   %xmm0,%xmm0
  mov    %r13,%rdi
  mov    $0x1,%eax
  cvtsi2ss %ebx,%xmm0
  movss  %xmm0,-0x4(%r12,%rbx,4)
  add    $0x1,%rbx
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  cmp    $0x6,%rbx
  jne    L1
  lea    D55(%rip),%rdi
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  xor    %ebx,%ebx
  cvtss2sd 0xc(%r12),%xmm0
  call   <printf@plt>
  mov    %r12,%rdi
  lea    D29(%rip),%r13
  call   <free@plt>
  lea    D4(%rip),%rdi
  call   <puts@plt>
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  mov    %rax,%r12
  mov    %rax,%r14
L44:
  movslq 0x0(%rbp),%rdi
  mov    %rdi,%r15
  shl    $0x2,%rdi
  call   <malloc@plt>
  mov    %rax,(%r12)
  test   %r15d,%r15d
  jle    L6
  xor    %r15d,%r15d
  add    $0x1,%ebx
  jmp    L58
  nop
L50:
  mov    (%r12),%rax
L58:
  lea    (%rbx,%r15,1),%esi
  mov    %r13,%rdi
  mov    %esi,(%rax,%r15,4)
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15d,0x0(%rbp)
  jg     L50
L67:
  mov    $0xa,%edi
  add    $0x8,%r12
  call   <putchar@plt>
  cmp    $0x5,%ebx
  jne    L44
  mov    0x8(%rsp),%rax
  lea    0x28(%rax),%rbx
L30:
  mov    (%r14),%rdi
  add    $0x8,%r14
  call   <free@plt>
  cmp    %r14,%rbx
  jne    L30
  mov    0x8(%rsp),%rbx
  mov    %rbx,%rdi
  call   <free@plt>
  add    $0x18,%rsp
  mov    %rbx,%rdx
  mov    %rbx,%rsi
  lea    D54(%rip),%rdi
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10c0 <printf@plt>
L6:
  add    $0x1,%ebx
  jmp    L67
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
datatype: string
name: extract_symbolic_data
arguments: label: D41
datatype: string
name: extract_symbolic_data
arguments: label: D52
datatype: string
name: extract_symbolic_data
arguments: label: D45
datatype: string
name: extract_symbolic_data
arguments: label: D55
datatype: string
name: extract_symbolic_data
arguments: label: D29
datatype: string
name: extract_symbolic_data
arguments: label: D4
datatype: string
name: extract_symbolic_data
arguments: label: D54
datatype: string

tool: D7: %d 

tool: D41: p = %d, p+1 = %f


tool: D52: 
malloc one-dimensional array:

tool: D45: %4.2f	

tool: D55: pm = %4.1f


tool: D29: %-4d

tool: D4: 
malloc two-dimensional array:

tool: D54: %ld %f

assistant: ```c
void main() {
  int *p = malloc(sizeof(int));
  float *array = (float *)malloc(5 * sizeof(float)), *pm;
  int **arr;
  *p = 13;
  int i, j;
  printf("p = %d, p+1 = %f\n", *p, *(p + 1));
  printf("%d ", p);
  printf("%d ", p);
  printf("\nmalloc one-dimensional array:\n");
  for (i = 0; i < 5; i++) {
    array[i] = i + 1;
    printf("%4.2f\t", array[i]);
  }
  pm = &array[5 - 3];
  printf("pm = %4.1f\n", *(pm + 1));
  free(array);
  printf("\nmalloc two-dimensional array:\n");
  arr = (int **)malloc(5 * sizeof(int *));
  for (i = 0; i < 5; i++) {
    arr[i] = (int *)malloc((*p) * sizeof(int));
    for (j = 0; j < (*p); j++) {
      arr[i][j] = i + j + 1;
      printf("%-4d", arr[i][j]);
    }
    printf("\n");
  }
  for (i = 0; i < 5; i++) {
    free(arr[i]);
  }
  free(arr);
  printf("%ld %f", arr, arr);
}
```