Page 94 - C-Language

#include <stdio.h>

int main(void) {
int array[3] = {1,2,3}; // 4 bytes * 3 allocated
unsigned char *ptr = (unsigned char *) array; // unsigned chars only take 1 byte
/*
* Now any pointer arithmetic on ptr will match
* bytes in memory. ptr can be treated like it
* was declared as: unsigned char ptr[12];
*/

return 0;
}

Macros are simple string replacements

Macros are simple string replacements. (Strictly speaking, they work with preprocessing tokens,
not arbitrary strings.)

#include <stdio.h>

#define SQUARE(x) x*x

int main(void) {
printf("%d\n", SQUARE(1+2));
return 0;
}

You may expect this code to print 9 (3*3), but actually 5 will be printed because the macro will be
expanded to 1+2*1+2.

You should wrap the arguments and the whole macro expression in parentheses to avoid this
problem.

#include <stdio.h>

#define SQUARE(x) ((x)*(x))

int main(void) {
printf("%d\n", SQUARE(1+2));
return 0;
}

Another problem is that the arguments of a macro are not guaranteed to be evaluated once; they
may not be evaluated at all, or may be evaluated multiple times.

#include <stdio.h>

#define MIN(x, y) ((x) <= (y) ? (x) : (y))

int main(void) {
int a = 0;
printf("%d\n", MIN(a++, 10));
printf("a = %d\n", a);
return 0;

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