Programming

Question

Struct  exam

{

int a;

char *c;

}

st[]={4,"choice",5,"battle",6,"minute",8,"bookmark",7,"address"}

main()

{

struct exam *p=st;

printf("%s",p++->c);

++p;

printf("%s",++p->c);

printf("%d",p[0].a);

printf("%s",++p++->c);

printf("%s",++p->c);

}

Comments

Here p++->c  means p->c and then do p++           // -> has higher priority than ++

++p->c means ++(p->c)

P is a pointer to an array of structure.

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printf("%s",p++->c);    // Print p->c (Choice)then do p++ , now p is pointing to next element of array. 

++p; // one more increment now p is pointing to 6,"minute"

printf("%s",++p->c);  //access p->c (minute and then do one increment in this string =>inute)

printf("%d",p[0].a); // currently p is pointing to 6,"minute" so its p[0] so p[0].a =6

printf("%s",++p++->c);  //  one increment in string p->c and then p++ so o/p nute

printf("%s",++p->c);     // do one increment in string bookmark so o/p ookmark

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Thank you :)

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X = *p-->c++ and x = (*p-->c)++

How this expression will be evaluated?

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For first expression, *p-->c++

As per precedence, --> has more highest precedence in the expression

So first

p-->c

Then, among * and ++,   ++ has more precedence

So,

(p-->c)++

and finally *

So

*((p-->c)++)

 

For other expression, (*p-->c)++

Since brackets are given so among --> and *, --> has more precendence

So,

(p-->c)

and then

(*p-->c)

and finally

(*p-->c)++

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Better you do it yourself. help from here

http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm

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according to this table prefix has lower precedence that postfix

then

printf("%s",++p++->c);  //  one increment in string p->c and then p++ ,how this is evaluated this way

here prefix is given higher priority than postfix and ->

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I think pre and post are (and should be) compiler dependent...