从不兼容的void *中分配int *

Question

I have this code:

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;
}

Both realloc lines, marked with //1 and //2 on the previous code, give me this error message:

assigning to int * from incompatible type void *

Because I don't want this code behaving weirdly or crashing on me, obviously, I ask will: I have some problem in the future if I simply cast it to int * by doing

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

Thanks

Answer 1

In C, a value of type void* (such as the value returned by realloc ) may be assigned to a variable of type int* , or any other object pointer type. The value is implicitly converted.

The most likely explanation for the error message is that you're compiling the code as C++ rather than as C. Make sure the source file name ends in .c , not .C or .cpp , and make sure your compiler is configured to compile as C rather than as C++.

(Casting the result of realloc or malloc is considered poor style in C. In C++, the cast is necessary, but you normally wouldn't use realloc or malloc in C++ in the first place.)

Answer 2

This should work in C. Are you perhaps using a C++ compiler to compile this? For example, some big company based in Redmond refuses to properly support a contemporary C implementation. Their compiler is C++ by default and needs some option to whack it into a C compiler.

You have stdlib.h included? Then you don't need the casts. In fact, it is best practice to not cast the malloc return.

Answer 3

C中的所有alloc style函数都返回具有最严格对齐的内存地址，因此强制转换不能给出不是有效int指针的指针。