[Solved] Why is this output 27 when #define x 5+2

Question

After this code uses #define 5 + 2, why does i = x * x * x make i = 27?
https://code.sololearn.com/cty512lQ58i5/?ref=app

Answer 1

The key is that x is not 7, but "5 + 2". The #define macro is a simple text replacement, meaning whenever the compiler encounters x in the source code, it is replaced by "5 + 2" (not by 7). From there on, it is simple precedence:
i = 5 + 2 * 5 + 2 * 5+ 2
i = 5 + 10 + 10 + 2
i = 27

Answer 2

And this why you shouldn't use macros unless you understand how they work. It tells the compiler to replace every instance of x with 5+2, not 7.

x * x * x is replaced to

5+2*5+2*5+2

Proceed with proper operator precedence.

5+(2*5)+(2*5)+2

5+10+10+2

27

Answer 3

x is 7, x*x is 17, x*x*x is 27, why though?

Answer 4

It was a question in challenges, I verify the answers afterwards if I can't solve them so I can learn from my mistakes. I still don't quite understand why this happens though.. Visually it looks like 7 * 7 * 7, why does it behave this way?

Answer 5

Hatsy Rei And this is why I ask, thanks for the clarification on the challenge question, I now understand :)