user: 请从给定的汇编指令推导出C源代码：

```asm
<sortedSquares>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %esi,(%rdx)
  movslq %esi,%r12
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%rsi
  cmpl   $0x0,0x0(%rbp)
  js     L103
  test   %ebx,%ebx
  jle    L6
  mov    %ebx,%ebx
  lea    (%rsi,%rbx,4),%rcx
  mov    %rsi,%rax
  sub    %rsi,%rbp
L30:
  mov    0x0(%rbp,%rax,1),%edx
  imul   %edx,%edx
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L30
  jmp    L6
L103:
  test   %ebx,%ebx
  jle    L6
  lea    0x4(%rbp),%rdx
  mov    $0x0,%eax
L196:
  mov    %eax,%edi
  add    $0x1,%eax
  cmp    %eax,%ebx
  je     L106
  add    $0x4,%rdx
  cmpl   $0x0,-0x4(%rdx)
  js     L196
  test   %edi,%edi
  js     L141
  cmp    %eax,%ebx
  jle    L141
  mov    %rsi,%rcx
  mov    $0x0,%r8d
  jmp    L56
L165:
  add    $0x1,%eax
  mov    %r9d,%edx
L140:
  add    $0x1,%r8d
  imul   %edx,%edx
  mov    %edx,(%rcx)
  add    $0x4,%rcx
  test   %edi,%edi
  js     L185
  cmp    %ebx,%eax
  jge    L185
L56:
  movslq %edi,%rdx
  mov    0x0(%rbp,%rdx,4),%edx
  movslq %eax,%r9
  mov    0x0(%rbp,%r9,4),%r9d
  mov    %edx,%r11d
  add    %r9d,%r11d
  js     L165
  sub    $0x1,%edi
  jmp    L140
L141:
  mov    $0x0,%r8d
L185:
  test   %edi,%edi
  js     L35
  movslq %edi,%r9
  lea    0x0(,%r9,4),%rcx
  lea    0x0(%rbp,%rcx,1),%rdx
  lea    -0x4(%rbp,%rcx,1),%r10
  mov    %edi,%ecx
  shl    $0x2,%rcx
  sub    %rcx,%r10
  movslq %r8d,%rcx
  add    %r9,%rcx
  lea    (%rsi,%rbp,1),%r9
  lea    (%r9,%rcx,4),%r11
L176:
  mov    (%rdx),%ecx
  mov    %r11,%r9
  sub    %rdx,%r9
  imul   %ecx,%ecx
  mov    %ecx,(%r9)
  sub    $0x4,%rdx
  cmp    %r10,%rdx
  jne    L176
  lea    0x1(%rdi,%r8,1),%r8d
L35:
  cmp    %eax,%ebx
  jle    L6
  movslq %r8d,%r8
  lea    (%rsi,%r8,4),%rdx
  sub    $0x1,%ebx
  sub    %eax,%ebx
  add    %r8,%rbx
  lea    0x4(%rsi,%rbx,4),%rdi
  cltq
  shl    $0x2,%rax
  sub    %rsi,%rax
  neg    %r8
  lea    (%rax,%r8,4),%rcx
  add    %rbp,%rcx
L194:
  mov    (%rcx,%rdx,1),%eax
  imul   %eax,%eax
  mov    %eax,(%rdx)
  add    $0x4,%rdx
  cmp    %rdi,%rdx
  jne    L194
  jmp    L6
L106:
  test   %ebx,%ebx
  jle    L6
  shl    $0x2,%r12
  lea    0x0(%rbp,%r12,1),%rax
  lea    -0x4(%rbp,%r12,1),%rdi
  lea    -0x1(%rbx),%edx
  shl    $0x2,%rdx
  sub    %rdx,%rdi
  lea    (%rax,%rsi,1),%r8
L77:
  mov    -0x4(%rax),%edx
  mov    %r8,%rcx
  sub    %rax,%rcx
  imul   %edx,%edx
  mov    %edx,(%rcx)
  sub    $0x4,%rax
  cmp    %rdi,%rax
  jne    L77
L6:
  mov    %rsi,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *sortedSquares(int *A, int ASize, int *returnSize) {
  *returnSize = ASize;
  int *returnA = (int *)malloc(sizeof(int) * ASize);
  int j = 0;
  int type = A[0] < 0 ? 0 : 2;
  if (type == 0) {
    for (int i = 0; i < ASize; i++) {
      if (A[i] >= 0) {
        type = 1;
        j = i;
        break;
      }
    }
  }
  if (type == 0) {
    for (int i = 0; i < ASize; i++) {
      returnA[i] = A[ASize - 1 - i] * A[ASize - 1 - i];
    }
  } else if (type == 2) {
    for (int i = 0; i < ASize; i++) {
      returnA[i] = A[i] * A[i];
    }
  } else {
    int l = j - 1, r = j;
    int i = 0, k = 0;
    while (l >= 0 && r < ASize) {
      if (A[l] + A[r] >= 0) {
        k = A[l--];
      } else {
        k = A[r++];
      }
      returnA[i++] = k * k;
    }
    while (l >= 0) {
      k = A[l--];
      returnA[i++] = k * k;
    }
    while (r < ASize) {
      k = A[r++];
      returnA[i++] = k * k;
    }
  }
  return returnA;
}
```