C Aptitude Questions and answers with explanation

C Aptitude 21
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The C aptitude questions and answers for those questions along with explanation for the interview related queries.

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Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

Predict the output or error(s) for the following:

C aptitude 21.1

 char *Fun1()
{
char temp[ ] = “example";
return temp;
}
char *Fun2()
{
char temp[ ] = {‘e’, ‘x’,’a’,’m’,’p’,’l’,'e'};
return temp;
}
int main()
{
puts(Fun1());
puts(Fun2());
}


Answer: Garbage values.

Explanation: Both the functions suffer from the problem of dangling pointers. In Fun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function Fun2() also suffers from the same problem but the problem can be easily identified in this case.

C aptitude 21.2

   char *strexp()
{
char *temp = "example string";
return temp;
}
int main()
{
puts(strexp);
}

Answer: example string

Explanation: The character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.

C aptitude 21.3

     main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}

Answer: -1

Explanation: Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is, while(i–!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

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