“warning: type defaults to ‘int’ in type name” when declaring const char*

Question

I'm writing a very simple C program, which begins like this:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int
main( int argc, char **argv ){

  // check that the program has been invoked correctly
  if( argc < 3 ){
    fprintf( stderr, "Usage: find_char <string> <string>.\n" );
    exit( EXIT_FAILURE );
  }

  char const *source = (const) (*++argv); 

For the last line, I get the following warning:

main.c:17:3: warning: type defaults to 'int' in type name [enabled by default] char const *source = (const) (*++argv);

I tried:

char const *source = NULL;
source = (const) (*++argv);

but gives the same varning for the second line. What is going on here?

Answer 1

casting to const amounts to casting to const int . When type is omitted and only qualified is set, the compiler just assumes int .

Just remove the (const) cast. You already did the right thing by declaring the pointed values as const .