text: in C the evaluation for an expression goes like this:\nEvaluate the right part of expression and then the left one\nand i=5 returns 5\nso:\n1. x-=k*=j/=i%=5;\n - i%=5 -> 2 and i=2\n - j/=i%=5 -> j/=2 -> 1 and j = 1\n - k*=j/=i%=5 -> k*= 1 -> 4\n - x-=k*=j/=i%=5 -> x-=4 -> -3 and x= -3\n2.x*=1 + (i%= 1+(j/=-1+ ++k)\n - ++k - > 3 and k=3 \n - j/=-1+ ++k -> j/=-1 + 3 -> j/= 2 -> 1\n -... i%= 1+(j/=-1+ ++k) -> 0\n - x*=1 + (i%= 1+(j/=-1+ ++k)) -> x*= 1+0\n - x = 5\n3.i%=j=(j=5)%(i=3)\n - i=3 -> 3 and i become 3\n - j=5 -> same\n - (j=5)%(i=3) 0 -> 2\n - j=(j=5)%(i=3) -> j=2 -> 2 and j become 2\n - i%=j=(j=5)%(i=3) -> i%= 2 -> i = 3 % 2 = 1