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Re: ANSI C / assignment makes pointer from integer

On Tue, 7 Aug 2001, (Bohdan Vlasyuk) wrote:
> Date: Tue, 7 Aug 2001 15:25:08 +0300
> To: Sebastian Ude <>
> From: (Bohdan Vlasyuk)
> CC:
> Subject: Re: ANSI C / assignment makes pointer from integer
> On Tue, Aug 07, 2001 at 02:25:15PM +0200, Sebastian Ude wrote:
> > Assume this (very common) situation:
> > Looks pretty harmless, doesn't it ?
> > But both gcc 2.95.3 and egcs 1.1.2 give me the following warning
> > when compiling this code with -ansi:
> > warning: assignment makes pointer from integer without a cast
> > But why ?
> You, obviously, hadn't declared your function, and hadn't included
> header file declaring it [<string.h> on my system], so it's supposed
> to be declated as [remember, C remembers no type info in shared object
> files, in contrary to C++]
> int strdup(...)
> but, in given context you use it as char* = int, hence the warning.
> Usually, sizeof(int)==sizeof(char*), thus if a is char*, (char*)(int)a
> == a , but this DOES differ on many platforms other than your.
> You may deal with it by saying:
> char *strdup(const char *s);
> on the very top of your program.

I *did* include string.h.
Guess why I placed the


Yep, to tell you that I left out some lines of code which seemed natural to

Try this complete example:

#include <string.h>

int main(void)
     char *s = strdup("foo");
     return 0;

string.h is included, so the definition of strdup() should be present.
But the warning is still there when compiling with -ansi ...

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