problem 3. I think tag is a reference so its size is 4 bytes.
On Tue, Jul 12, 2011 at 1:29 AM, nicks <[email protected]> wrote:
> Guys plz help me in understanding the following output
>
> *PROBLEM 1>.*
> *
> *
> #include<stdio.h>
> main()
> {
> int scanf=78;
> //int printf=45;
> int getchar=6;
> printf("%d",scanf);
> printf("\n%d",getchar);
> }
>
> *OUTPUT-*
> 78
> 6
>
> in this problem my problem is using printf and scanf as variable
> names.....they are functions in the stdio.h then how are they available for
> variable name ??...generally what happens is that whenever we use some name
> for the function and then use that name for some variable then compiler
> gives error then why in this case error is not coming ??
>
> *PROBLEM 2>.*
>
> #include<stdio.h>
> main()
> {
> int i=1;
> printf("\n%d %d",i^=1%2,i<<=1%2); // does it evaluate the final value of i
> before printing ??
> **}
> *OUTPUT-*
> 3 3
>
> In gcc what i have observed is that arguments of printf are evaluated from
> right to left i.e i<<=1%2 is evaluated before i^=1%2...hence i first becomes
> 2 then 3 after XOR with 1....now output according to me should be "3
> 1".....but what actually is happening is that it us evaluating i and then
> printing it 3 3.....can someone explain why this output is coming ?
>
> and the last problem
>
> *PROBLEM 3>.*
> *
> *
> #include<stdio.h>
> main()
> {
> enum {low='a',high='b'}tag;
> char try=low;
> printf("Size=%d",sizeof(tag));
> switch (try)
> {
> case 'a':printf("aaa");break;
> case 'b':printf("bbb");
> case 'c':printf("ccc");
> }
> //system("pause");
> }
>
> in this program size of enum comes out to be 4..help me in understanding
> the size of enum...how it is stored in memory??...does the size of enum
> depend on number of constant in it ?....anyone link regarding that ??
>
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--
Varun Pahwa
B.Tech (IT)
7th Sem.
Indian Institute of Information Technology Allahabad.
Ph : 09793899112
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