int a=0; int b[] = new int[5]; int c=3; b[a]=a=c; //a=b[a]=c; Both "b[a]=a=c;" and "a=b[a]=c;" has the same result . But I want to know the reason . How does it work ?
Anonymous
Ranch Hand
Joined: Nov 22, 2008
Posts: 18944
posted
0
Originally posted by Neal Hanson: int a=0; int b[] = new int[5]; int c=3; b[a]=a=c; //a=b[a]=c; Both "b[a]=a=c;" and "a=b[a]=c;" has the same result . But I want to know the reason . How does it work ?
Ok, step by step; the assignment operator (=) is a right associative operator, i.e. the expressions above are evaluated from right to left. The one little trick that might have caused some confusion is that the subexpression 'b[a]' is evaluated first, i.e. this subexpression denotes the zeroth elemt of array b (b[0]). The first double assignment expression evaluates as follows: b[a]= a= c; / b[0]= 0, a= 0, c= 3; b[a]= (a= c); / b[0]= 0, a= 3, c= 3; (b[a]= 3); / b[0]= 3, a= 3, c= 3; The second double assignment expression evaluates as follows: a= b[a]= c; / a= 0, b[0]= 0, c= 3; a= (b[a]= c); / a= 0, b[0]= 3, c= 3; (a= 3); / a= 3, b[0]= 3, c= 3; And indeed, both assignment expressions have identical results. kind regards
0
