Backward Analysis of Expression - c

Question

Suppose I have following code segment in C.
i = j = k = 1;
j = (i++) + (++k);
result = i + j + k; //POI
//Expected result = 7
Here, I want to find value of result through backward analysis. When I perform backward analysis, I will go through following expressions in order.
j = (i++) + (++k);
++k;
i++;
i = j = k = 1;
j = k = 1;
k = 1;
During backward analysis I will replace each variable by corresponding expression, whenever it is applicable. But I'm confused how to deal with increment/decrement operations.
My current strategy will produce following result
result = i + j + k
//after j = i++ + ++k
result = (i+(i+(k+1)))+k
//after ++k
result = (i+(i+((k+1)+1)))+(k+1)
//after i++
result = ((i+1)+((i+1)+((k+1)+1)))+(k+1)
//after i = j = k = 1
result = ((1+1)+((1+1)+((k+1)+1)))+(k+1)
//after k = 1
result = (((1+1)+((1+1)+((1+1)+1)))+(1+1))
//Simplifying
result = 9
which is ofcourse not true.
Can anyone help me with this?

Answer 1

I think it should be more like this
result = i + j + k
//after j = X + Y // You didn't analyse ++k and i++ yet
result = i+(X+Y)+k
//after Y = ++k
result = (i+(X+(k+1))+(k+1)
//after X = i++
result = ((i+1)+(i+(k+1))+(k+1)
//after i = j = k = 1
result = ((1+1)+(1+(k+1))+(k+1)
//after k = 1
result = ((1+1)+(1+(1+1))+(1+1)
//Simplifying
result = 7

Answer 2

The thing to notice here is i++ increases after evaluating, not before.
i=j=k=1
++k //happens before expression
//k=2,i=j=1
j = k + i
//k=2,i=1,j=3
i++ //happens after expression
//k=2,i=2,j=3
result = i+ j + k
//result = 7
Your mistake was increasing i before ealuation of j, and then getting j=4, instead of j=3
If you want to analyze it backward, something along the lines of
result = i + j + k
i = i+1 = 1 + 1
//result = (1+1) + j + k
j = i + k
k = k+1 = 1+1
//j = 1 + (1+1)
//result = (1+1) + (1+(1+1)) + (1+1)

Answer 3

I suggest that you first decompose the statements into elementary ones as a compiler would do. For instance, using Frama-C to decompose the statements would give something like:
k = 1;
j = k;
i = j;
tmp = i;
i ++;
k ++;
j = tmp + k;
result = (i + j) + k;
It is then a lot easier to do the analysis.
Consider the sequence stating from the bottom :
// 1 + 1 + 1 + 1 + 1 + 1 + 1
k = 1;
// k + 1 + k + k + 1 + k + 1
j = k;
// j + 1 + j + k + 1 + k + 1
i = j;
// i + 1 + i + k + 1 + k + 1
tmp = i;
// i + 1 + tmp + k + 1 + k + 1
i ++;
// i + tmp + k + 1 + k + 1
k ++;
// i + tmp + k + k
j = tmp + k;
// i + j + k
result = (i + j) + k;
// result ?
Which gives you 7 as expected.