user: 请从给定的汇编指令推导出C源代码:

```asm
<ftoa>:
  endbr64
  push   %r15
  mov    $0x64,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  movss  %xmm0,0xc(%rsp)
  call   <malloc@plt>
  movss  0xc(%rsp),%xmm0
  pxor   %xmm1,%xmm1
  mov    %rax,%r13
  comiss %xmm0,%xmm1
  ja     L34
  mov    %rax,%r14
  mov    $0x1,%r12d
  xor    %r15d,%r15d
L46:
  cvttss2si %xmm0,%ebx
  pxor   %xmm1,%xmm1
  mov    $0x6,%ebp
  cvtsi2ss %ebx,%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,0xc(%rsp)
  nop
L17:
  mov    %ebp,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  test   %eax,%eax
  jg     L35
  sub    $0x1,%ebp
  jae    L17
L59:
  movss  D19(%rip),%xmm1
  movss  0xc(%rsp),%xmm0
  pxor   %xmm2,%xmm2
  movslq %r12d,%rdx
  movb   $0x2e,(%r14)
  mulss  %xmm1,%xmm0
  cvttss2si %xmm0,%eax
  cvtsi2ss %eax,%xmm2
  lea    0x30(%rax),%ecx
  mov    %cl,0x0(%r13,%rdx,1)
  lea    0x1(%r12),%eax
  cltq
  subss  %xmm2,%xmm0
  pxor   %xmm2,%xmm2
  mulss  %xmm1,%xmm0
  cvttss2si %xmm0,%edx
  cvtsi2ss %edx,%xmm2
  lea    0x30(%rdx),%ecx
  mov    %cl,0x0(%r13,%rax,1)
  lea    0x2(%r12),%eax
  cltq
  subss  %xmm2,%xmm0
  pxor   %xmm2,%xmm2
  mulss  %xmm1,%xmm0
  cvttss2si %xmm0,%edx
  cvtsi2ss %edx,%xmm2
  lea    0x30(%rdx),%ecx
  mov    %cl,0x0(%r13,%rax,1)
  lea    0x3(%r12),%eax
  cltq
  subss  %xmm2,%xmm0
  pxor   %xmm2,%xmm2
  mulss  %xmm1,%xmm0
  cvttss2si %xmm0,%edx
  cvtsi2ss %edx,%xmm2
  lea    0x30(%rdx),%ecx
  mov    %cl,0x0(%r13,%rax,1)
  lea    0x4(%r12),%eax
  cltq
  subss  %xmm2,%xmm0
  pxor   %xmm2,%xmm2
  mulss  %xmm1,%xmm0
  cvttss2si %xmm0,%edx
  cvtsi2ss %edx,%xmm2
  lea    0x30(%rdx),%ecx
  mov    %cl,0x0(%r13,%rax,1)
  lea    0x5(%r12),%eax
  add    $0x6,%r12d
  cltq
  movslq %r12d,%r12
  subss  %xmm2,%xmm0
  mulss  %xmm1,%xmm0
  cvttss2si %xmm0,%edx
  add    $0x30,%edx
  mov    %dl,0x0(%r13,%rax,1)
  mov    %r13,%rax
  movb   $0x0,0x0(%r13,%r12,1)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L35:
  movslq %r15d,%r14
  lea    0x1(%rbp),%r12d
  add    %r13,%r14
  nopw   0x0(%rax,%rax,1)
L22:
  sub    $0x1,%r12d
  mov    $0xa,%edi
  add    $0x1,%r14
  mov    %r12d,%esi
  call   <pow@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  add    $0x30,%eax
  mov    %edx,%ebx
  mov    %al,-0x1(%r14)
  test   %r12d,%r12d
  jne    L22
  lea    0x1(%r15,%rbp,1),%r12d
  movslq %r12d,%r14
  add    $0x1,%r12d
  add    %r13,%r14
  jmp    L59
  nopl   0x0(%rax,%rax,1)
L34:
  movb   $0x2d,(%rax)
  lea    0x1(%rax),%r14
  xorps  D10(%rip),%xmm0
  mov    $0x2,%r12d
  mov    $0x1,%r15d
  jmp    L46
```

tool_calls: name: parse_data
arguments: data_tag: D19
datatype: f32
name: parse_data
arguments: data_tag: D10
datatype: byte[16]

tool: D19: 10

tool: {"D10": [0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
char *ftoa(float f) {
  float ff;
  ff = f;
  char *str = (char *)malloc(sizeof(char) * 100);
  int a, b, c, k, l = 0, m, i = 0, j;
  if (f < 0.0) {
    str[i++] = '-';
    f *= -1;
  }
  a = f;
  f -= a;
  k = 6;
  while (k > -1) {
    l = pow(10, k);
    m = a / l;
    if (m > 0) {
      break;
    }
    k--;
  }
  for (l = k + 1; l > 0; l--) {
    b = pow(10, l - 1);
    c = a / b;
    str[i++] = c + 48;
    a %= b;
  }
  str[i++] = '.';
  for (l = 0; l < 6; l++) {
    f *= 10.0;
    b = f;
    str[i++] = b + 48;
    f -= b;
  }
  str[i] = '\0';
  return (str);
}
```