void main() { int a,*b,**c; a=90; b=&a; c=&b; printf("%u",*(c++)); } can u pl. say the o/p with reasons !!!!!!!!! thank u:undecided
let me explain the whole thing with output.. for that i have changed the code little bit for ur understanding keeping the logic same... Code: # include<stdio.h> # include<conio.h> void main() { int a,*b,**c; a=90; b=&a; printf("%u\n",b); //first printf c=&b; printf("%u\n",c); //second printf printf("%u\n",*(c++)); //third printf printf("%u\n",c); //fourth printf getch(); } explaination:- the value of a is 90 and let itz address is "12345" b is a pointer and expression b=&a assigns the address of a to b...i.e b=12345 let the address of b is "54321" after the expression c=&b, c contains 54321 Now, the first printf gives the output:- 12345 second printf gives the output:- 54321 now let me explain the third printf:- *c++ means first apply * over c then increase the value of c by 1. When printf is used the "*c" value is printed in the console and next time the value of c is changed +1. so the output wud be:- *c i.e value of b i.e 12345 now finally in the fourth printf the output wud be either 54323 or 54325 depending upon whether the size of "int" is 2 or 4 bytes.In turbo C itz 2 and in visual or borland c itz 4.By pointer operation u know that if the pointer c is increased by 1 then c will acquire the new value i.e b's addr increased by int size.
ya ya i got it thank u.... but i have a doubt ..... in ur third printf(); statement , *(c++) is the argument... so the first precedence will go for the brackets only na..... if it is *c++ , it's ok .. but the c++ is given inside the brackets , so how it will take *c first for evaluation?????
