#include <stdio.h>

#define FUDGE(k)		k+3.14159
#define PR(a)			printf(#a " = %d\t",(int)(a))
#define PRINT(a)		PR(a); putchar('\n')
#define PRINT2(a,b)		PR(a); PRINT(b)
#define PRINT3(a,b,c)	PR(a); PRINT2(b,c)
#define MAX(a,b)		(a<b ? b : a)

// These are macro substitution so..

// int x=2; 
// PRINT( x*FUDGE(2) ); becomes PRINT( x*2+3.14159 ); that becomes PRINT(2*2+3.14159)
// that becomes PRINT(4+3.14159) becomes PRINT(7.14159) becomes PR(a); putchar('\n');
// PR(a) becomes PR(7.14159) becomes printf(#7.14159 " = %d\t",(int)(7.14159))
// cast 7.14159 to an int so 7 is output. 

// So PRINT( x*FUDGE(2) ); is replased with
// printf("x*2+3.14159 = %d\t",(int)((x*2+3.14159))); putchar('\n');
// the unwanted interaction between the replacement string and its context in this problem 
// is avoided if FUDGE(k) is defined to be (k+3.14159)
// Parenthesize all macro bodies that contain operators.
/*******		
PRINT2(a,b)		PR(a); PRINT(b)
PR(a)			printf(#a " = %d\t",(int)(a))
PRINT(a)		PR(a); putchar('\n')


int cel;
for( cel=0; cel<=100; cel+=50 )   PRINT2( cel, 9./5*cel+32 ); 
PR(cel); PRINT(9./5*cel+32); 
printf(#cel " = %d\t",(int)(cel)); PR(9./5*cel+32); putchar('\n');

After all the MACRO substitutions... 
int cel;
for( cel=0; cel<=100; cel+=50 )
	printf(#cel " = %d\t",(int)(cel)); this prints the three cel = 
printf(#9./5*cel+32 " = %d\t",(int)(9./5*cel+32)); 
Then with cel == 150  9./5*cel+32 = 302 printed once.
putchar('\n'); this does the new line.
For this problem using commas in place of the semicolons in the body of the PRINT macros 
will fix this one.   *** See FEUER pg.163 remember MACROS expand out to full code. If you have a 
for(a;b;c) Macro3; and calls Macro2 that calls Macro1 and these Macros all end with ';' only Macro3
will be repeted by the for loop.  Then Macro2 and Macro1 will exe once.     
*******/


main()	
{
	{ // nest 1
		int x=2;
		PRINT( x*FUDGE(2) );
		//printf("x*2+3.14159 = %d\t",(int)((x*2+3.14159))); putchar('\n');
	} // end nest 1

	{ // nest 2
		int cel;
		for( cel=0; cel<=100; cel+=50 )		
			PRINT2( cel, 9./5*cel+32 );		
		// for( cel=0; cel<=100; cel+=50 ) printf("cel = %d\t",(int)(cel)); prints cel = three times 
		// now cel == 150     
		// printf("9./5*cel+32 = %d\t",(int)(9./5*cel+32)); putchar('\n');

	} // end nest 2

	{ // nest 3
		int x=1, y=2;
		PRINT3( MAX(x++,y),x,y);
		PRINT3( MAX(x++,y),x,y);
	} // end nest 3

}// end main...