GATE Overflow Test Series | Programming | Test 1 | Question: 10

Question

Which of the following assignments causes an overflow? (Assume size of int is $4$ bytes and long int is $8$ bytes)

• A. int a = 123456789;
• B. int a = 123456789123456789L;
• C. unsigned a = 123456789;
• D. long int a = 123456789123456789;

Answer 1

When an integer constant overflows the size of integer it automatically gets converted to the higher type like from int to long. So, we do not need "L" suffix unless we want to explicitly do this conversion (for example to ensure the result of an operation using this does not overflow).

A $4$ byte integer has $31$ bits excluding the sign bit. So, this can represent ${31} \times \log 2 \approx 9$ digits. So, Clearly A and C won't overflow. long int being of size 8 bytes can represent at least $2 \times 9 = 18$ digits and so D option also cannot overflow.

Correct Answer: B

Comments

Can you please explain the logic behind how the log2 is multiplied?

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Maximum value with $31$ bits is $31$ 1's which equals $2^{31} - 1.$ In decimal let this value be of $n$ digits. Maximum value with $n$ digits is $n$ $9's$ which is $10^n - 1.$

So, we get $2^{31} - 1 \geq 10^n - 1$

($\geq$ is used because we have to represent decimal numbers accurately with bits and not the other way around)

Taking logarithm, $n \leq 31 \log 2.$

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Thank you so much for the immediate response.I understood now the logic.

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@gatecse sir , @Sachin Mittal 1 Sir , we are using L in option B which means it forces the variable to store value of size of long typr of varibble. So how come it will give overflow?