C program(malloc) is not compiling in ubuntu 11.04

Question

I have installed the gcc compiler from this sudo apt-get install build-essential command

and my program code is

 #include<stdio.h>

 main()
   {
      int *b;

      b = (int*)malloc(10*sizeof(int));  

      printf("b=%u\n\n",b);
      printf("b+1=%u\n\n",(b+1));
      printf("b+2=%u\n\n",(b+2));

      b[2]=4;
      printf("*(b+2)=%d\n\n",*(b+2));

  }

when i try to compile this program from cc -c program.c command then i get some error

enter image description here

You should not cast the return value of `malloc`. That's just a good way to hide errors. — Cody Gray - on strike, Jan 15 '12 at 18:28
Your college video tutorial was made for an older version of gcc and should be updated :) Hmm wait a moment? Video tutorials for programming? What happened to reading and comprehending text? You're going to need it anyway to program well... — Torp, Jan 15 '12 at 18:29
It's also strange to print out the value of a pointer. It's a completely meaningless value. Not sure what this demo is showing you, considering that you'll probably never do this again. The compiler is simply warning you that pointers are *not* `unsigned int`s. — Cody Gray - on strike, Jan 15 '12 at 18:29
Also, there is no such thing as `void main()` in C. I'm really getting suspicious of that college you're attending and the study materials that they're providing! — Cody Gray - on strike, Jan 15 '12 at 18:31
@CodyGray Casting the return value of `malloc` is a way to be compatible with C++. But it surely isn't necessary here. — pmr, Jan 15 '12 at 18:32
@pmr: Yes, I'm aware of its necessity in C++, but the question is clearly tagged C here. It's also quite strange that you would ever use `malloc` in C++. — Cody Gray - on strike, Jan 15 '12 at 18:35
@CodyGray Of course. But some people consider it good practice. It really depends on what you want to do with the code and sometimes it's easier to compile C code with C++ code and then this kind of thing helps. — pmr, Jan 15 '12 at 18:37
Why the h*** are you logged in as root user to compile your code? — knittl, Jan 15 '12 at 18:50
knittl sir i haven't use any linux environment this is my first time. in turboc3 i haven't got problem/error like that — , Jan 15 '12 at 19:12

score 6 · Accepted Answer · edited May 23 '17 at 10:32

6

You're missing #include <stdlib.h> (for malloc), and the format strings are wrong. Use %p to print pointers.

Also, you don't need to (and probably shouldn't) cast the return value of malloc (in C).

And the correct signature for main without parameters is:

int main(void)

Corrected code:

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

int main()
{
    int *b;

    b = (int*)malloc(10*sizeof(int));

    printf("b=%p\n\n",  (void*) b);
    printf("b+1=%p\n\n",(void*) (b+1));
    printf("b+2=%p\n\n",(void*) (b+2));

    b[2]=4;
    printf("*(b+2)=%d\n\n",*(b+2));

    return 0;
}

edited May 23 '17 at 10:32

Community

1
1

answered Jan 15 '12 at 18:29

Mat

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1

I don't really believe that, sorry. I did the changes I described above to your code and it compiles without any warning. – Mat Jan 15 '12 at 18:34
Also you should cast the pointer to be printed to `void*` (in the unusual case `void*` and `int*` representations differ). – pmg Jan 15 '12 at 18:37
@pmg: I wonder how much code out there breaks horribly on such an implementation... – Mat Jan 15 '12 at 18:44
@mayank: I posted the full corrected code. It does not issue any warnings with GCC 4.5, 4, 4.7, and neither with clang nor icc. – Mat Jan 15 '12 at 18:45

score 0 · Answer 2 · answered Jan 15 '12 at 18:35

0

I don't know why it worked in the video, it's probably using some strange non-standard compiler.

But your errors are because you are using int instead of unsigned int and you pass pointers to printf when it expects unsigned int.

answered Jan 15 '12 at 18:35

Net4All

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2

C program(malloc) is not compiling in ubuntu 11.04

2 Answers2