Understanding a C Preprocessor Macro's output

Asked Jan 27 '19 at 18:00

Active Jan 27 '19 at 18:00

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The following is a snippet of my C program:

 a = 5; b = 7;
 printf("\ncubef(--a) = %d", cubef(--a));
 a = 5; b = 7;
 printf("\ncubem(--a) = %d", cubem(--a));

Where cubef and cubem are defined as follows:

#define cubem(a) a*a*a
int cubef(int a) {  return a * a * a; }

When executed, I get the following output:

cubef(--a) = 64
cubem(--a) = 18

I understand that cubem "expands" to --a*--a*--a which should result in 4*3*2 which equals 24, but for some reason I am getting 18. Can someone help me understand how I am getting that output?

Here is a link to the full program, if required: https://pastebin.com/7hGYJkTi

asked Jan 27 '19 at 18:00

3

No. What you are invoking is undefined behavior. Sequence point here is not defined. – unalignedmemoryaccess Jan 27 '19 at 18:04
1

Undefined behavior – Some Name Jan 27 '19 at 18:04
1

Gotcha, so to help me understand a bit better behavior with multiple callings of a single variable in a macro can't be predicted? – Jan 27 '19 at 18:05
@rclever2 that's correct. If you use macro, then using `a++` or similar is very risky and should be avoided. – unalignedmemoryaccess Jan 27 '19 at 18:07

Understanding a C Preprocessor Macro's output

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