Discussion
Home ‣ C Programming ‣ C Preprocessor See What Others Are Saying!
- Question
What will be the output of the program?
#include<stdio.h> #define FUN(arg) do\ {\ if(arg)\ printf("CuriousTab...", "\n");\ }while(--i) int main() { int i=2; FUN(i<3); return 0; }
Options- A. CuriousTab...
CuriousTab...
CuriousTab - B. CuriousTab... CuriousTab...
- C. Error: cannot use control instructions in macro
- D. No output
- Correct Answer
- CuriousTab... CuriousTab...
ExplanationThe macro FUN(arg) prints the statement "CuriousTab..." untill the while condition is satisfied.Step 1: int i=2; The variable i is declared as an integer type and initialized to 2.
Step 2: FUN(i<3); becomes,
do { if(2 < 3) printf("CuriousTab...", "\n"); }while(--2)After the 2 while loops the value of i becomes '0'(zero). Hence the while loop breaks.
Hence the output of the program is "CuriousTab... CuriousTab..."
More questions
- 1. Every function must return a value
Options- A. Yes
- B. No Discuss
Correct Answer: No
Explanation:
No, If a function return type is declared as void it cannot return any value.- 2. What will be the output of the program assuming that the array begins at the location 1002 and size of an integer is 4 bytes?
#include<stdio.h> int main() { int a[3][4] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }; printf("%u, %u, %u\n", a[0]+1, *(a[0]+1), *(*(a+0)+1)); return 0; }
Options- A. 448, 4, 4
- B. 520, 2, 2
- C. 1006, 2, 2
- D. Error Discuss
Correct Answer: 1006, 2, 2
- 3. What will be the output of the program assuming that the array begins at location 1002?
#include<stdio.h> int main() { int a[2][3][4] = { {1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 2}, {2, 1, 4, 7, 6, 7, 8, 9, 0, 0, 0, 0} }; printf("%u, %u, %u, %d\n", a, *a, **a, ***a); return 0; }
Options- A. 1002, 2004, 4008, 2
- B. 2004, 4008, 8016, 1
- C. 1002, 1002, 1002, 1
- D. Error Discuss
Correct Answer: 1002, 1002, 1002, 1
- 4. What will be the output of the program?
#include<stdio.h> int main() { char *p; p="%d\n"; p++; p++; printf(p-2, 23); return 0; }
Options- A. 21
- B. 23
- C. Error
- D. No output Discuss
Correct Answer: 23
- 5. What will be the content of 'file.c' after executing the following program?
#include<stdio.h> int main() { FILE *fp1, *fp2; fp1=fopen("file.c", "w"); fp2=fopen("file.c", "w"); fputc('A', fp1); fputc('B', fp2); fclose(fp1); fclose(fp2); return 0; }
Options- A. B
- B. A
B - C. B
B - D. Error in opening file 'file1.c' Discuss
Correct Answer: B
Explanation:
Here fputc('A', fp1); stores 'A' in the file1.c then fputc('B', fp2); overwrites the contents of the file1.c with value 'B'. Because the fp1 and fp2 opens the file1.c in write mode.Hence the file1.c contents is 'B'.
- 6. A preprocessor directive is a message from compiler to a linker.
Options- A. True
- B. False Discuss
Correct Answer: False
Explanation:
FALSEExample: #define symbol replacement
When the preprocessor encounters #define directive, it replaces any occurrence of symbol in the rest of the code by replacement. This replacement can be an statement or expression or a block or simple text.
- 7. Bitwise can be used to perform addition and subtraction.
Options- A. Yes
- B. No Discuss
Correct Answer: No
- 8. Which of the following statements are correct about an if-else statements in a C-program?
1: Every if-else statement can be replaced by an equivalent statements using ?: operators 2: Nested if-else statements are allowed. 3: Multiple statements in an if block are allowed. 4: Multiple statements in an else block are allowed.
Options- A. 1 and 2
- B. 2 and 3
- C. 1, 2 and 4
- D. 2, 3, 4 Discuss
Correct Answer: 2, 3, 4
- 9. What will be the output of the program?
#include<stdio.h> int addmult(int ii, int jj) { int kk, ll; kk = ii + jj; ll = ii * jj; return (kk, ll); } int main() { int i=3, j=4, k, l; k = addmult(i, j); l = addmult(i, j); printf("%d, %d\n", k, l); return 0; }
Options- A. 12, 12
- B. 7, 7
- C. 7, 12
- D. 12, 7 Discuss
Correct Answer: 12, 12
Explanation:
Step 1: int i=3, j=4, k, l; The variables i, j, k, l are declared as an integer type and variable i, j are initialized to 3, 4 respectively.The function addmult(i, j); accept 2 integer parameters.
Step 2: k = addmult(i, j); becomes k = addmult(3, 4)
In the function addmult(). The variable kk, ll are declared as an integer type int kk, ll;
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'k'.
Step 3: l = addmult(i, j); becomes l = addmult(3, 4)
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'l'.
Step 4: printf("%d, %d\n", k, l); It prints the value of k and l
Hence the output is "12, 12".
- 10. What will be the output of the program?
#include<stdio.h> int main() { int i; char c; for(i=1; i<=5; i++) { scanf("%c", &c); /* given input is 'a' */ printf("%c", c); ungetc(c, stdin); } return 0; }
Options- A. aaaa
- B. aaaaa
- C. Garbage value.
- D. Error in ungetc statement. Discuss
Correct Answer: aaaaa
Explanation:
for(i=1; i<=5; i++) Here the for loop runs 5 times.Loop 1:
scanf("%c", &c); Here we give 'a' as input.
printf("%c", c); prints the character 'a' which is given in the previous "scanf()" statement.
ungetc(c, stdin); "ungetc()" function pushes character 'a' back into input stream.Loop 2:
Here the scanf("%c", &c); get the input from "stdin" because of "ungetc" function.
printf("%c", c); Now variable c = 'a'. So it prints the character 'a'.
ungetc(c, stdin); "ungetc()" function pushes character 'a' back into input stream.This above process will be repeated in Loop 3, Loop 4, Loop 5.
Comments
There are no comments.
More in C Programming:
Programming
Copyright ©CuriousTab. All rights reserved.