Convert the following C-like code into MIPS assembly. The names of the variables reflect which registers must be used for the MIPS assembly. Do not assume any initial values for the registers. You may use additional registers. $t0 = 10; $t1 = 4; $t2 = ($t0 + $t1) * 3;

Convert the following C statement to MIPS assembly language instructions. Assume all variables in the statement are 32-bit integers. Variables to registers mapping are as follows f maps to $s0, g maps to $s1, b maps to $s2, i maps to $s3, j maps to $s4. f = ((g+h) - (1+j)) *4 v șto, $s1, $s2 v $t1, $s3, $s4 $t3, $ $s0, $

1.) Using unsigned 16-bit numbers, convert the following statement in C++ to Assembly language. (Example of an assembly language that is useful: movf i,w subwf j,w movf i+1, w subwfb j+1,w bc end_if movf j,w addwf i,w movwf m movf j+1,w addwfc i+1,w movwf m+1 end_if)   C++ code: unsigned char i, j, k;   if (i && j) {                         k = j + i;                         } // … rest of code …

1)Write the following code segment in MARIE assembly and test it. int x, y; scanf(“%d”, &x); If ( x > 1) { y = x + x; x = 0; } y = y + 1; printf(“%d”, y); exit(0);