The size of a pointer in a 64-bit architecture is 8 bytes. This is because a pointer's size is determined by the addressable memory of the system. In a 64-bit system, the addressable memory is 2^64 bits. Since 1 byte is equal to 8 bits, this means the addressable memory is 2^64 / 8 = 2^61 bytes. Therefore, the pointer needs to be able to address any of these 2^61 possible addresses, which requires 8 bytes.

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The size of a pointer in a 64-bit architecture is 8 bytes. This is because a pointer's size is determined by the addressable memory of the system. In a 64-bit system, the addressable memory is 2^64 bits. Since 1 byte is equal to 8 bits, this means the addressable memory is 2^64 / 8 = 2^61 bytes. Therefore, the pointer needs to be able to address any of these 2^61 possible addresses, which requires 8 bytes.

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The size of a pointer in a 64-bit architecture is 8 bytes. This is because a pointer's size is determined by the addressable memory of the system. In a 64-bit system, the addressable memory is 2^64 bits. Since 1 byte is equal to 8 bits, this means the addressable memory is 2^64 / 8 = 2^61 bytes. Therefore, the pointer needs to be able to address any of these 2^61 possible addresses, which requires 8 bytes.

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What is the size of a pointer to an int (on a 64-bit architecture)

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