Translate this code to MIPS64 assembly language. Addresses: a @ 400 b@ 408 result @ 416 long int a, b, result; // 64-bit if( a
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- Write an assembly language program ( using the IA-32 instructuion set x86 architecture for NASM ) with the following features An array with 10 numbers The program searches for the smallest number in the array The program then prints the number and its position in the array has to be x86 architecture for NASMConsider a MIPS processor and a program that is loaded in the main memory between the addresses 130h and 161Ch (h means hexadecimal). Calculate the number of blocks of the main memory that this program uses if the size of a block is 2^7 words (give the result in decimal).All microprocessors have same number of address lines but different number of data lines. Select one: True False
- Evaluate the C code statements in MIPS assembly codefor(i = 0; i < a; i++) for(j = 0; j < b; j++) D[4*j] = i + j; Assume that i is assigned to $s0, j is assigned to $s1, a is assigned to register $s2, b is assigned to $s3, and the base address of the array D is assigned to $s4Write an assembly language program for the 8085 Microprocessor to subtract three 8-bit numbers 12H, FFH, 03H. Store the result in the memory location 2050H and carry (borrow) in 2051H.In 8051 assembly language: If one of the elements in a list in ROM is more than 8 bits in binary as shown below list: DB 9H,12H,2021H How can I separate 2021 into 2 bytes in RAM? So that one location in RAM holds 20, and the other 21. (Values in RAM should be in BCD format)