从不兼容的void*中分配int*

我有这个代码:

int * generate_code(int *bits, int Fs, int size, int *signal_size, float frameRate)
{
  int sign_prev, i;
  int bit, t, j=0;
  int *x;
  float F0, N, t0, prev_i, F1;
  int temp = 0, temp1, temp2;

  F0 = frameRate * BITS_PER_FRAME;      // Frequency of a train of '0's = 2.4kHz
  F1 = 2*F0;       // Frequency of a train of '1's = 4.8kHz
  N = 2*(float)Fs/F1;   // number of samples in one bit

  sign_prev = -1;
  prev_i = 0;
  x = (int *)malloc(sizeof(int));
  for( i = 0 ; i < size ; i++)
  {

    t0 = (i + 1)*N;
    bit = bits[i];
    if( bit == 1 )
    {
      temp1 = (int)round(t0-N/2)-(int)round(prev_i+1)+1;
      temp2 = (int)round(t0)-(int)round(t0-N/2+1)+1;
      temp =j + temp1 + temp2;
      //printf("%d\n", (int)temp);
      x = realloc(x, sizeof(int)*temp);  // 1

      for(t=(int)round(prev_i+1); t<=(int)round(t0-N/2); t++)
      {
        *(x + j) = -sign_prev;
        j++;
      }
      prev_i = t0-N/2;
      for(t=(int)round(prev_i+1); t <= (int)round(t0); t++)
      {
        *(x + j) = sign_prev;
        j++;
      }
    }
    else
    {
      // '0' has single transition and changes sign
      temp =j + (int)round(t0)-(int)round(prev_i);
      //printf("%d\n",(int)temp);
      x = realloc(x, sizeof(int)*(int)temp);  // 2
      for(t=(int)round(prev_i); t < (int)round(t0); t++)
      {
        *(x + j) = -sign_prev;
        j++;
      }
      sign_prev = -sign_prev;
    }
    prev_i = t0;
  }

  *signal_size = j;
  return x;
}

这两个realloc线,标有//1和//2上一码,给我这个错误信息:

从不兼容的类型void*分配给int*

因为我不希望这段代码表现得格外奇怪或者让我崩溃,显然,我会问:如果我只是int *通过做什么来表达它,我将来会遇到一些问题

x = (int*)realloc(x, sizeof(int)*(int)temp);

谢谢

在C中,类型的值void*(例如返回的值realloc)可以分配给类型的变量int*或任何其他对象指针类型.该值是隐式转换的.

对错误消息最可能的解释是您将代码编译为C++而不是C.确保源文件名以或.c不结束,并确保编译器配置为C编译而不是C++编译..C.cpp

(在C++中转换结果realloc或被malloc认为是不好的样式.在C++中,强制转换是必要的,但你通常不会在C++中使用realloc或malloc首先使用.)