Exercise: C Preprocessor

Questions for: C Preprocessor

Point out the error in the program 
#include<stdio.h>
#define SI(p, n, r) float si; si=p*n*r/100;
int main()
{
    float p=2500, r=3.5;
    int n=3;
    SI(p, n, r);
    SI(1500, 2, 2.5);
    return 0;
}
A:
26250.00 7500.00
B:
Nothing will print
C:
Error: Multiple declaration of si
D:
Garbage values
Answer: C

The macro #define SI(p, n, r) float si; si=p*n*r/100; contains the error. To remove this error, we have to modify this macro to

#define SI(p,n,r) p*n*r/100

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What will be the output of the program? 
#include<stdio.h>
#define MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c)

int main()
{
    int x;
    x = MAX(3+2, 2+7, 3+7);
    printf("%d\n", x);
    return 0;
}
A:
5
B:
9
C:
10
D:
3+7
Answer: C

The macro MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c) returns the biggest of given three numbers.

Step 1: int x; The variable x is declared as an integer type.

Step 2: x = MAX(3+2, 2+7, 3+7); becomes,

=> x = (3+2 >2+7 ? 3+2 > 3+7 ? 3+2 : 3+7: 2+7 > 3+7 ? 2+7 : 3+7)

=> x = (5 >9 ? (5 > 10 ? 5 : 10): (9 > 10 ? 9 : 10) )

=> x = (5 >9 ? (10): (10) )

=> x = 10

Step 3: printf("%d\n", x); It prints the value of 'x'.

Hence the output of the program is "10".

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What will be the output of the program? 
#include<stdio.h>
#define PRINT(i) printf("%d,",i)

int main()
{
    int x=2, y=3, z=4;
    PRINT(x);
    PRINT(y);
    PRINT(z);
    return 0;
}
A:
2, 3, 4,
B:
2, 2, 2,
C:
3, 3, 3,
D:
4, 4, 4,
Answer: A

The macro PRINT(i) print("%d,", i); prints the given variable value in an integer format.

Step 1: int x=2, y=3, z=4; The variable x, y, z are declared as an integer type and initialized to 2, 3, 4 respectively.

Step 2: PRINT(x); becomes printf("%d,",x). Hence it prints '2'.

Step 3: PRINT(y); becomes printf("%d,",y). Hence it prints '3'.

Step 4: PRINT(z); becomes printf("%d,",z). Hence it prints '4'.

Hence the output of the program is 2, 3, 4.

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What will be the output of the program? 
#include<stdio.h>
#define MESS junk

int main()
{
    printf("MESS\n");
    return 0;
}
A:
junk
B:
MESS
C:
Error
D:
Nothing will print
Answer: B

printf("MESS\n"); It prints the text "MESS". There is no macro calling inside the printf statement occured.

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What will be the output of the program? 
#include<stdio.h>
#define str(x) #x
#define Xstr(x) str(x)
#define oper multiply

int main()
{
    char *opername = Xstr(oper);
    printf("%s\n", opername);
    return 0;
}
A:
Error: in macro substitution
B:
Error: invalid reference 'x' in macro
C:
print 'multiply'
D:
No output
Answer: C

The macro #define str(x) #x replaces the symbol 'str(x)' with 'x'.

The macro #define Xstr(x) str(x) replaces the symbol 'Xstr(x)' with 'str(x)'.

The macro #define oper multiply replaces the symbol 'oper' with 'multiply'.

Step 1: char *opername = Xstr(oper); The varible *opername is declared as an pointer to a character type.

=> Xstr(oper); becomes,

=> Xstr(multiply);

=> str(multiply)

=> char *opername = multiply

Step 2: printf("%s\n", opername); It prints the value of variable opername.

Hence the output of the program is "multiply"

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