将参数传递给函数（int传递给size_t）

Question

In function writeDmpFile I'm calling writeFile .

Please see the code and the comments.

My problem is size. In writeDmpFile I see it is 380316. I try to pass it to writeFile.

But while stepping here I'm getting very large number(3832907636190596508).

What am I doing wrong. I'd expect, that 380316 will be passed.

int writeFile(char *name,  unsigned char *buff, size_t *size,const char *dir )
{
  FILE * pFile;
  chdir (dir);
  pFile = fopen ( name, "wb");
  //(gdb) print *size
  //$5 = 3832907636190596508
  fwrite (buff , sizeof(unsigned char), *size, pFile);
  fclose (pFile);

  return 1;
}
int writeDmpFile(GTree *tree, char *filename)
{
  char dmpfilename[32];

  dmpfilename[0] ='\0';
  dmpParams_t params;
  params.buff[0]   ='\0';
  int size =0;
  params.size=&size ;
  g_tree_foreach(tree, (GTraverseFunc)writeDmpFileLine, &params);
  sprintf (dmpfilename, "InstrumentList_FULL.csv_%.*s", 15, filename);
  //here (gdb) print size
  //$1 = 380316
  writeFile(dmpfilename,  ( unsigned char *)params.buff, ( size_t *)&size , dmpdir);//(size_t *)params.size, dmpdir);
}

Answer 1

( size_t *)&size

This is bad as size_t and int are different types with different representations. size_t is an alias for an unsigned integer type, often unsigned long . Here you should declare size as a size_t variable.

Answer 2

There are two fundamental problems here.

1. You declared size to be an int , but really it needs to be a size_t . Change its type to size_t .
2. You should not be passing the address of size to writeFile . You should be passing it as a const value param. You don't need to modify it, nor do you. So make that clear in the signature of the function.

As a general rule, any time you encounter a type mismatch compiler error and are tempted to suppress the error with a cast, you are almost certainly making a mistake. The compiler reported an error because you made a mistake. Sure you can shut the compiler up, but experience tells me that the compiler is usually right, and we humans are very good at making mistakes.

So, don't suppress type mismatch compiler errors with casts. Seek to understand why the types don't match and thus resolve the problem.

Answer 3

Despite the question why the writeFile() function insists on takeing the size via its address, there are two possiblities to solve this issue:

1. Declare an intermediate size_t -typed variable:

    { size_t _s = size; writeFile(dmpfilename, (unsigned char *) params.buff, &_s, dmpdir); } 

2. Or use a (nice) compound statement:

    writeFile(dmpfilename, (unsigned char *) params.buff, &((size_t){size}), dmpdir); 

Answer 4

( size_t *)&size

This is a big mistake which you face while migrating from 32bit system to 64bit system. On 32 bit system this code may work properly as both int and size_t will be of 32 bit. but on 64bit system int will be 32 bit but size_t may be of size 64 bit.

Basically size_t was designed to hold pointer arithmetic. This is the type which is returned from sizeof operator.

For your problems solution better you convert int size =0; to size_t size = 0