The size of the union will be determined by the largest data type within the union. In this case, we have a pointer to a uint8_t (1 byte), a uint32_t (4 bytes), and a double pointer to a uint8_t.

Given that this is a 32-bit architecture, pointers are 4 bytes long. Therefore, the double pointer to a uint8_t is also 4 bytes long.

So, the size of the union is 4 bytes.

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The size of the union will be determined by the largest data type within the union. In this case, we have a pointer to a uint8_t (1 byte), a uint32_t (4 bytes), and a double pointer to a uint8_t.

Given that this is a 32-bit architecture, pointers are 4 bytes long. Therefore, the double pointer to a uint8_t is also 4 bytes long.

So, the size of the union is 4 bytes.

Knowee AI · Accepted Answer

The size of the union will be determined by the largest data type within the union. In this case, we have a pointer to a uint8_t (1 byte), a uint32_t (4 bytes), and a double pointer to a uint8_t.

Given that this is a 32-bit architecture, pointers are 4 bytes long. Therefore, the double pointer to a uint8_t is also 4 bytes long.

So, the size of the union is 4 bytes.

What is the size of this union definition below given the MSP432 32-bit architecture?union example { uint8_t * ptr; uint32_t data; uint8_t ** dptr; };1 point6 bytes4 bytes8 bytes1 byte

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