Why are C macros not type-safe?

Question

If have encountered this claim multiple times and can\'t figure out what it is supposed to mean. Since the resulting code is compiled using a regular C compiler it will end up b

Answer 1

Consider the typical "max" macro, versus function:

#define MAX(a,b) a < b ? a : b
int max(int a, int b) {return a < b ? a : b;}

Here's what people mean when they say the macro is not type-safe in the way the function is:

If a caller of the function writes

char *foo = max("abc","def");

the compiler will warn.

Whereas, if a caller of the macro writes:

char *foo = MAX("abc", "def");

the preprocessor will replace that with:

char *foo = "abc" < "def" ? "abc" : "def";

which will compile with no problems, but almost certainly not give the result you wanted.

Additionally of course the side effects are different, consider the function case:

int x = 1, y = 2;
int a = max(x++,y++); 

the max() function will operate on the original values of x and y and the post-increments will take effect after the function returns.

In the macro case:

int x = 1, y = 2;
int b = MAX(x++,y++);

that second line is preprocessed to give:

int b = x++ < y++ ? x++ : y++;

Again, no compiler warnings or errors but will not be the behaviour you expected.