使用strncat进行字符串连接会导致签名错误

Question

update: the point of whether char, signed char, or unsigned was ultimately moot here. it was more appropriate to use memcpy in this situation, since it works indiscriminately on bytes.

Couldn't be a simpler operation, but I seem to be missing a critical step. In the following code, I am attempting to fill bufferdata with buffer for which the compiler warns me of a difference in signedness.

unsigned char  buffer[4096] = {0};
char *bufferdata;

bufferdata = (char*)malloc(4096 * sizeof(bufferdata));

if (! bufferdata)
  return false;

while( ... )
{
    // nextBlock( voidp _buffer, unsigned _length );
 read=nextBlock( buffer, 4096);

 if( read > 0 )
 {

  bufferdata = strncat(bufferdata, buffer, read); // (help)
  // leads to: pointer targets in passing argument 2 of strncat differ in signedness.

  if(read == 4096) {

   // let's go for another chunk
   bufferdata = (char*)realloc(bufferdata, ( strlen(bufferdata) + (4096 * sizeof(bufferdata)) ) );
   if (! bufferdata) {
    printf("failed to realloc\n");
    return false;
   }

  }

 }
 else if( read<0 )
 {
  printf("error.\n");
  break;
 }
 else {
  printf("done.\n");
  break;    
 }
}

Answer 1

显然，在您的编译器中char是带signed char ，因此是警告消息。

Answer 2

char * strncat ( char * destination, char * source, size_t num );

so as your destination buffer is unsigned char so there will be a warning of sign. You can either change your buffer to char array if signed is not necessary else you can ignore warnings using -w option in compiler.