Re: what is happening in C when increment this way?

From: CBFalconer (cbfalconer_at_yahoo.com)
Date: 01/30/05 

Date: Sun, 30 Jan 2005 15:09:39 GMT

puzzlecracker wrote:
>
> If you run this program, it will give very unexpected results. Can
> anyone explain the nature of this anamaly? (also what is the function
> call [and library to include linux/windows] to execute 'pause');
>
> #include<stdio.h>
>
> //void pause(){ unsigned long i=0; while(i++<10000000); }
>
> int main(){
> float i;
>
> for(i=0;i<100;i+=0.1) // pay attention
> { // pause();
> printf("%d\n",i);
> }
>
> printf("\n");
> return 0;
>
> }

Don't use C99 comments in C90 code. The result:

[1] c:\c\junk>gcc junk.c
junk.c:3: parse error before '/' token
junk.c: In function `main':
junk.c:8: parse error before '/' token
junk.c:10: warning: int format, double arg (arg 2)
junk.c: At top level:
junk.c:12: parse error before string constant
junk.c:12: warning: type defaults to `int' in declaration of
`printf'
junk.c:12: warning: conflicting types for built-in function
`printf'
junk.c:12: ISO C forbids data definition with no type or storage
class

After removing those execrences we get:

[1] c:\c\junk>gcc junk.c
junk.c: In function `main':
junk.c:8: warning: int format, double arg (arg 2)

and lying to the compiler about the type of an argument to a
variadic function is undefined behaviour.

There is no "pause" function in standard C. 

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