c - MIPS Code Cycle Time -

i need task of determining number of cycles mips code below.

my main issue determining if condition not met, if statement still executed (thus adding number of total cycles)

assuming single cycle implementation scheme i.e., each instruction requires 1 clock cycle execute. 1. number of cycles required execute code when a. s==0? b. s==1?

this have come myself:

a. 9 cycles

b. 8 cycles (it not instruction contained in if statement, , not jump endif statement - (goes function) final endif.

this sample of mips:

main:    # evaluate expression.   # put final result in a0 prepare syscall.  addi    $sp, $sp, -4    # make space on stack. sw  $ra, 0($sp) # save return address.  li  $t0, 1      # put 0 in register  li  $a1, 4      # put 4 in register  li  $a2, 6      # put 6 in register   if: bne   $t0, $zero, else    # (i == 0) ?  add $v0, $a1, $a2   # v0 = a1 + a2  j  endif  else:   jal func  endif: add $a0, $v0, $zero  li  $v0, 1 

so need determine number of cycles if a. s = 1, , b. s = 0 assuming each instruction take 1 cycle.

the c code is

main() { int a; a=4; b=6; s=3; int function(a,b); if (s==0) = a+b; else = function(a,b) return; } # function multiply 2 numbers function(z,y) { int tmp; tmp = z × y; ret 

apologies formatting. first post on forum , still working things out. appreciated.

depending on whether want approximate time (by instruction counting) or determine actual execution time, have count if instruction no matter what.

in modern cpus, flavor of jump instruction includes penalties—maybe (for cpus) when conditional branch not taken!

penalties arise cache misses, pipeline perturbations, , branch prediction.

since don't understand statement means, not dissecting code path further.

tmp = z × y;