fun 1 = 8 bytes fun 2 = 14 bytes fun 3 = 18 bytes fun 4 = 22 bytes two input a n

Question

fun 1 = 8 bytes

fun 2 = 14 bytes

fun 3 = 18 bytes

fun 4 = 22 bytes

two input a ntszt fun uints t e, int32_t F uintg )int3t fun3(uint8 t a, int3 4) ituintst d, int32_t e)t, uint8_t a, uint16_t b f, uintat (1) int32-t fun! (uint8_t g) b? For examp uint8?ta, int32_t uint8 t d, int32 t 13. Write a p One sub sum of t b, int64_t o 4)int32 t fun4(uints_t a, int64 t eris) holds the return value in the follo 14. Write a numbe Which regis )int16 t fun1() (2)int t fun2() )int32 t fun3() (4)int64t fun4() 15. Write 32-bi 16. Mat eler inte (5) int64 t fun5() Why does a recursive assembly function have to preserve the link 7. Give two different assembly instructions to make a subroutinere

Explanation / Answer

The int16_t, int8_t, int32_t, int64_t are all exact width type integer data types, which can guarantee you that your program will use these many bits to store an integer. These are to make more compatible, and increase performance accross different platforms.

But as for your question, The exact register which will be storing these return values can depend on the architecture or on the implementation of the program. But you can probably narrow it down with the size of the register.

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