Why Does Multiplying Integer Literals Result in Overflow Even with a `long long` Variable?

Understanding Long Long Integer Overflow in C Arithmetic

Consider the following code snippet:

long long int n = 2000 * 2000 * 2000 * 2000;    // overflow

Why does this code result in an overflow, despite the declaration of a long long variable to hold the result?

Reason for Overflow

The issue lies in the type of the integer literal constants 2000. By default, integer literals are assigned the smallest type that can hold their value without being smaller than int. Since 2000 can be stored in a 32-bit int, arithmetic operations on these literals are performed as int operations.

Type Promotion

Although long long variables like n can store larger values, arithmetic operators are called with the larger of the types present in the multiplication operation. However, in this case, the larger type remains int, regardless of the destination type.

Implications for Overflow

This means that the result of the multiplication 2000 * 2000 * 2000 * 2000 is calculated as a 32-bit integer, which overflows because the result exceeds the maximum value an int can hold.

Alternative Approaches

To avoid this overflow, you can explicitly cast the integer literals to long long using the LL suffix for literal constants or the explicit cast operator (long long):

long long int n = (long long)2000 * 2000 * 2000 * 2000;  // no overflow

Alternatively, you can use the pow function to calculate the result, ensuring the calculation is performed with long long values:

long long int n = pow(2000, 4);  // no overflow

By explicitly casting or using the pow function, you ensure that the calculations are performed with long long values, preventing overflow.

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