user: 请从给定的汇编指令推导出C源代码：

```asm
<test_dynmem>:
  endbr64
  push   %r14
  xor    %eax,%eax
  push   %r13
  push   %r12
  push   %rbp
  lea    D41(%rip),%rbp
  sub    $0x8,%rsp
  call   <dynmem_avail@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x40,%edi
  call   <dynmem_alloc@plt>
  mov    %rax,%r13
  xor    %eax,%eax
  call   <dynmem_avail@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x40,%edi
  call   <dynmem_alloc@plt>
  mov    %rax,%r12
  xor    %eax,%eax
  call   <dynmem_avail@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %r13,%rdi
  lea    D1(%rip),%rsi
  call   <strcpy@plt>
  lea    D37(%rip),%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %r13,%rdi
  call   <puts@plt>
  mov    %r12,%rdi
  call   <puts@plt>
  mov    %r13,%rdi
  call   <dynmem_free@plt>
  xor    %eax,%eax
  call   <dynmem_avail@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x64,%edi
  call   <dynmem_alloc@plt>
  mov    %rax,%r13
  xor    %eax,%eax
  call   <dynmem_avail@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  lea    D12(%rip),%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <puts@plt>
  mov    %r13,%rdi
  call   <puts@plt>
  mov    %r13,%rdi
  call   <dynmem_free@plt>
  mov    %r12,%rdi
  call   <dynmem_free@plt>
  mov    $0xf8,%edi
  call   <dynmem_alloc@plt>
  test   %rax,%rax
  je     L3
  mov    %rax,%r12
  lea    D38(%rip),%rdi
  call   <puts@plt>
  mov    %r12,%rdi
  call   <dynmem_free@plt>
L3:
  mov    $0x3c,%edi
  call   <dynmem_alloc@plt>
  mov    $0x3c,%edi
  mov    %rax,%r13
  call   <dynmem_alloc@plt>
  mov    $0x3c,%edi
  mov    %rax,%r12
  call   <dynmem_alloc@plt>
  mov    %r12,%rdi
  mov    %rax,%r14
  call   <dynmem_free@plt>
  mov    $0x48,%edi
  call   <dynmem_alloc@plt>
  mov    $0x82,%edi
  mov    %rax,%r12
  call   <dynmem_alloc@plt>
  mov    %rax,%rdi
  call   <dynmem_free@plt>
  mov    %r14,%rdi
  call   <dynmem_free@plt>
  mov    %r13,%rdi
  call   <dynmem_free@plt>
  mov    %r12,%rdi
  call   <dynmem_free@plt>
  xor    %eax,%eax
  call   <dynmem_avail@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  mov    %eax,%esi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    10d0 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D41
data_type: string
name: extract_symbolic_data
arguments: data_tag: D1
data_type: string
name: extract_symbolic_data
arguments: data_tag: D37
data_type: string
name: extract_symbolic_data
arguments: data_tag: D12
data_type: string
name: extract_symbolic_data
arguments: data_tag: D38
data_type: string

tool: D41: [ free mem: %i ]


tool: D1: 11111111

tool: D37: 2222222222222

tool: D12: 333

tool: D38: success

assistant: ```c
void test_dynmem(void) {
  printf("[ free mem: %i ]\n", (int)dynmem_avail());
  char *test1 = dynmem_alloc(64);
  printf("[ free mem: %i ]\n", (int)dynmem_avail());
  char *test2 = dynmem_alloc(64);
  printf("[ free mem: %i ]\n", (int)dynmem_avail());
  strcpy(test1, "11111111");
  strcpy(test2, "2222222222222");
  puts(test1);
  puts(test2);
  dynmem_free(test1);
  printf("[ free mem: %i ]\n", (int)dynmem_avail());
  char *test3 = dynmem_alloc(100);
  printf("[ free mem: %i ]\n", (int)dynmem_avail());
  strcpy(test3, "333");
  puts(test2);
  puts(test3);
  dynmem_free(test3);
  dynmem_free(test2);
  if ((test1 = dynmem_alloc(248))) {
    puts("success");
    dynmem_free(test1);
  }
  test1 = dynmem_alloc(60);
  test2 = dynmem_alloc(60);
  test3 = dynmem_alloc(60);
  dynmem_free(test2);
  test2 = dynmem_alloc(72);
  dynmem_free(dynmem_alloc(130));
  dynmem_free(test3);
  dynmem_free(test1);
  dynmem_free(test2);
  printf("[ free mem: %i ]\n", (int)dynmem_avail());
}
```