Main() { int i; clrscr(); printf("%d", &i)+1; scanf("%d", i)-1; }  

A) Runtime error. Access violation
B) Compile error. Illegal syntax
C) None of the above
D) Runtime error
Explanation: printf( ) prints address/garbage of i, scanf() dont have & sign, so scans address for i +1, -1 dont have any effect on code.
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pandi

  • Dec 30th, 2005
 

Ans   : Complier time error

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srinivasan

  • Jan 5th, 2006
 

ans is -12

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Chump

  • Jan 17th, 2006
 

Note that there is & which means it probably prints the address of i...n my guess is that + 1 has no effect on the output...
n regarding scanf - there is no & symbol...so still it will accept a value but not store it in i n my guess is that -1 has no effect on the code....
any comments

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Vijay

  • Jan 23rd, 2006
 

Hey guys look at the code carefully it will give a segmentation fault as scanf expects the address of the variable in which the value will be scanned.

so the code should be like

scanf("%d",&i) -1;

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sreelu

  • Mar 15th, 2006
 

ans is -12because it prints address of i.

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vishal

  • Mar 27th, 2006
 

answer will be A)

As scanf expects address of i.

As soon as user enters any no. for i, compiler will try to access memory location equal to value of i and it will give acess voilation on windows and core dump on Solaris/Linux.

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dev07

  • Jun 11th, 2006
 

answer is A) Runtime error. Access violation.

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