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C++ 8- Suppose the computer on which the program below is executed is 32-bit little endian. uint32_t uint8_t * A = 0x12345678; B = (uint8_t*)
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8- Suppose the computer on which the program below is executed is 32-bit "little endian". uint32_t uint8_t * A = 0x12345678; B = (uint8_t*) (&A); B += 0x2; *B = 0x89; At the end of the execution, the debugger indicates that the address of A is 0x55550000. Provide the values of the expressions below, in hexadecimal: A= B = *B= Step by Step Solution
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