Question

#define minl (x, y) ( (x < y) ? x : y)
#define min 10
#include <stdio.h>

int min2(int x, int y) {
        if (x < y)   
                return x;  
        else     
                return y;
}

main() {
        int a, b;  
        scanf("%d %d", &a, &b);  
        if (b < min)    
                printf("input out of range\n");  
        else {    
                a = minl(a, b++);    
                printf("a = %d, b %d\n", a, b);    
                a = min2(a, b++);    
                printf("a = %d, b %d\n", a, b);  
        }
}

-------------------------
1  Give the exact C code of the statement "a = min1(a, b + + ) ; " after macro processing.
2. Give the exact C code of the statement "a = min2 (a, b++) ; "after macro processing.
3  Give the exact C code of the statement "if ( b < min) " after macro processing.
4  Assume 60 and -30 are entered as inputs. What is the exact output of the program?   
5  Assume 50 and 30 are entered as inputs. What is the exact output of the program?  

Answer 1

1. a= min1(a,b++);

Macro min1 is:

min1(x,y) ((x<y) x:y)

After macro processing the code will be:

a= ((a<b++) a:b++);

2. a= min2(a,b++)

min2 is a function,

After macro processing the exact C code:

a= { if ( a< b ) return a;

else

return b;

}

b++;

3. Exact C code after macro processing of

if ( b < min) is:

if(b < 10)

4. 60 for a and -30 for b gives output as:

input out of range

This is because, the value of b (-30) is less than value of min(10). Hence the above statement is printed out.

5. 50 for a and 30 for b gives output as:

a = 31, b 32
a = 32, b 33

The first statement is from macro expansion:

Thus b value is incremented twice once while comparing and another while assigning to a. Thus the above values are resulted. The second line is from the function, here the increment is done only once that is done(after returning from function min2).