CODEWITHSUNDEEP
c
HAnwopn
HAnwopn

Reputation: 3

What is the problem of this simple #define C code?

Here's the example code.

#define A 100
#define B A+1

I've been reading C basic books and the writer said that code could cause fatal problems and there is no explanation about it.

why this code has a problem? I want to understand

Upvotes: 0

Views: 67

Answers (1)

Barmar
Barmar

Reputation: 782344

Suppose you try to write:

int foo = B * 10;

You probably expect this to set B to 1010. But it will actually set it to 110 because it expands to

int foo = 100+1 * 10;

Because of operator precedence, the multiplication is done first, so it's 100 + 10, not 101 * 10.

To prevent problems like this, you should put parentheses around the macro expansion.

#define B (A+1)

Then it will expand to

int foo = (100+1) * 10;

and you'll get the expected result.

Upvotes: 4

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