user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<minimum_k1>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  mov    %rdi,%r14
  lea    0x0(,%rdx,4),%edi
  push   %r13
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,0x8(%rsp)
  call   <malloc@plt>
  movslq 0x8(%rsp),%r10
  mov    %rax,%r13
  mov    %r10,%rcx
  test   %r10d,%r10d
  jle    L83
  nopl   0x0(%rax)
L6:
  mov    -0x4(%r14,%r12,4),%edx
  mov    %r13,%rbx
  test   %ebp,%ebp
  je     L111
  xor    %eax,%eax
  jmp    L68
  nopl   0x0(%rax,%rax,1)
L96:
  add    $0x1,%eax
  add    $0x4,%rbx
  cmp    %ebp,%eax
  je     L1
L68:
  cmp    %edx,(%rbx)
  jle    L96
  cmp    %ebp,%eax
  jge    L111
  mov    %ebp,%edx
  mov    %ecx,0x14(%rsp)
  sub    %eax,%edx
  movslq %ebp,%rax
  mov    %r10,0x8(%rsp)
  lea    -0x1(%rdx),%esi
  shl    $0x2,%rax
  mov    %rsi,%rdx
  lea    0x4(,%rsi,4),%r8
  neg    %rdx
  shl    $0x2,%rdx
  lea    -0x4(%rax,%rdx,1),%rsi
  add    %rdx,%rax
  mov    %r8,%rdx
  add    %r13,%rsi
  lea    0x0(%r13,%rax,1),%rdi
  call   <memmove@plt>
  mov    -0x4(%r14,%r12,4),%edx
  mov    0x14(%rsp),%ecx
  mov    0x8(%rsp),%r10
L111:
  mov    %edx,(%rbx)
  cmp    %ebp,%r15d
  jle    L134
  add    $0x1,%ebp
  cmp    %r10,%r12
  je     L15
  add    $0x1,%r12
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L134:
  movslq %ebp,%r15
  shl    $0x2,%r15
  lea    -0x4(%r15),%r9
  lea    0x0(%r13,%r15,1),%r8
  cmp    %r10,%r12
  je     L15
L31:
  mov    (%r14,%r12,4),%edx
  mov    %r13,%rbx
  xor    %eax,%eax
  test   %ebp,%ebp
  jg     L179
  jmp    L42
L77:
  add    $0x1,%eax
  add    $0x4,%rbx
  cmp    %eax,%ebp
  je     L193
L179:
  cmp    %edx,(%rbx)
  jle    L77
  cmp    %eax,%ebp
  jle    L42
  mov    %ebp,%edx
  mov    %r8,0x18(%rsp)
  sub    %eax,%edx
  mov    %ecx,0x14(%rsp)
  sub    $0x1,%edx
  mov    %r9,0x8(%rsp)
  mov    %rdx,%rax
  lea    0x4(,%rdx,4),%rdx
  neg    %rax
  shl    $0x2,%rax
  lea    (%r9,%rax,1),%rsi
  add    %r15,%rax
  add    %r13,%rsi
  lea    0x0(%r13,%rax,1),%rdi
  call   <memmove@plt>
  mov    (%r14,%r12,4),%edx
  mov    0x18(%rsp),%r8
  mov    0x14(%rsp),%ecx
  mov    0x8(%rsp),%r9
L42:
  add    $0x1,%r12
  mov    %edx,(%rbx)
  cmp    %r12d,%ecx
  jg     L31
L15:
  test   %ebp,%ebp
  jle    L83
  lea    -0x1(%rbp),%eax
  mov    %r13,%rbx
  lea    D4(%rip),%rbp
  lea    0x4(%r13,%rax,4),%r12
  nopl   0x0(%rax)
L124:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L124
L83:
  add    $0x28,%rsp
  mov    %r13,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1080 <free@plt>
  nopl   0x0(%rax)
L1:
  movslq %ebp,%rax
  lea    0x0(%r13,%rax,4),%rbx
  jmp    L111
  nopl   (%rax)
L193:
  mov    %r8,%rbx
  jmp    L42
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
data_type: string

tool: D4: %d


assistant: ```c
void minimum_k1(int *arr, int len, int k) {
  int *sorted = (int *)malloc(sizeof(int) * k);
  int i, sorted_bound = 0;
  for (i = 0; i < len; i++) {
    int j = 0;
    while (j < sorted_bound && sorted[j] <= arr[i])
      j++;
    if (sorted[j] > arr[i]) {
      int move = sorted_bound;
      while (j < move) {
        sorted[move] = sorted[move - 1];
        move--;
      }
    }
    sorted[j] = arr[i];
    if (sorted_bound < k)
      sorted_bound++;
  }
  for (i = 0; i < sorted_bound; i++)
    printf("%d\n", sorted[i]);
  free(sorted);
}
```