Thread: initializing structure variables

  1. #1
    Registered User
    Join Date
    Jan 2008
    Posts
    69

    initializing structure variables

    Given the following code:

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    typedef struct foo {
      int test;
      char b;
    } foo;
    
    int
    main(int argc, char *argv[]) {
    
      foo f = {2, 'c'};
      int i = f->test;
      printf("%d\n", i);
    
      return 0;
    }
    Why do I get an error when compiling?

    test.c: In function 'main':
    test.c:13: error: invalid type argument of '->'

    When I use decimal notation, such as f.test, it works fine.

  2. #2
    Kernel hacker
    Join Date
    Jul 2007
    Location
    Farncombe, Surrey, England
    Posts
    15,677
    a->b is the same as (*a).b - since your f is not a pointer, it is not allowed to use ->, but you should use the "dot" notation (I wouldn't call it "decimal", since that's just one usage of "dot/point").

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  3. #3
    Registered User
    Join Date
    Apr 2008
    Posts
    12

    Struct Definition

    I think you may have too many "foo"s:

    Code:
    typedef struct foo {
      int test;
      char b;
    } foo;
    Could also be written:

    Code:
    typedef struct {
      int test;
      char b;
    } foo;
    I'm pretty sure that first foo is unnecessary.

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